Topic 4: of (Copyright © 2020 Joseph W. Trefzger)

In this unit, we discuss time value of money ( and value) applications. These applications have many uses in both personal and business financial management. A typical use in personal money management is planning and for retirement. A typical use in business financial management is the capital budgeting decision: determining whether it is appropriate to spend a large amount of money to buy new long-lived equipment (the question addressed in Topic 6, Capital Budgeting Analysis).

The basic factors for computing time value-adjusted dollar amounts are as follows:

What textbooks call “ of $1:” (1 + 푟)푛

1 1 푛 What textbooks call “ of $1:” or, equivalently, ( ) (1 + 푟)푛 1 + 푟

(1 + 푟)푛−1 Future Value of a Level Ordinary Annuity: ( ) 푟

1 푛 1−( ) Present Value of a Level Ordinary Annuity: ( 1 + 푟 ) 푟

In each of these formulas, r represents the or growth that occurs each time period (the specified period is often presented as a full in introductory examples, but in actual financial transactions we are more likely to see payments occurring every half year, quarter of a year, or , and time intervals of days or even more unusual periods are possible). [Note that the rate of return is represented variously in different reference sources as g (indicating a rate of growth in size or value), i ( rate), k (the cost of capital in a business setting), or r (rate of return, the most general designation, which we might favor for its versatility).] The other variable, n, represents the number of those time periods (, , etc.) over which the periodic rate of return is earned or paid.

Before the early 1980s, when calculators became cheap enough to be widely available, financial analysts computed time value-adjusted amounts with the help of printed tables, which provided factors (computed with the formulas above) for a limited number of selected r and n values. Today we have powerful calculators and spreadsheets to facilitate these computations. But while most textbooks attempt to teach time value via financial calculator function keys, the beginning student should be encouraged instead to work directly with the above factors, employing the exponent and logarithm keys on a financial or basic scientific calculator. Using shortcut computational techniques, such as financial calculators’ function keys, when we do not yet fully understand the basic ideas is akin to taking shortcut travel routes when we do not yet know major roads and landmarks in an area: we are apt to get lost, and even if we get to the right destination we may not truly understand what we have done. Once we understand the underlying logic, we can figure out which buttons to hit on a particular financial calculator in a matter of .

The outline that follows presents a “tool box” approach to time value that differs in many ways from presentations typically offered in textbooks. I developed this approach, over many years of teaching, to better address questions students asked, and concepts with which they routinely encountered difficulties, when working with the standard time value of money (often abbreviated as TVM) tools. The tool box is a user-friendly way of organizing traditional time value tools; nothing in our tool box is incompatible with traditional textbook coverage. Indeed, no one who truly understands the time value of money from earlier courses needs to change anything in the approach he or she has become accustomed to (including the use of financial calculator function keys, which are legal, though discouraged, for use on our exams). But the tool box approach presented below has generated much positive feedback from students over the years for simplifying and demystifying time value of money computations and the underlying ideas. Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 1 I. UNDERSTANDING REQUIRED RATES OF RETURN r (sometimes shown as g or i or k)

In every time value of money situation, we work with an expected or required periodic rate of return. (This return is often called an “,” terminology that we frequently employ for convenience, although technically the return earned is “interest” only if a lender provides money to a borrower. The rate would instead be a return on equity if the money provider were an owner rather than a lender.) How do we determine, before the fact, the average annual or other periodic rate of return that would leave us feeling properly compensated for all the risks we encounter as money providers? And why is the expected/required average periodic rate of return 4% in some instances and 19% in others?

Consider the case of a lender. We can think of an interest rate in a building block fashion, embodying the bits and pieces that compensate for the expected cost to the lender of bearing the various risks faced. We represent a lender’s periodic (typically annual) required rate of return in a straightforward additive form as r = r* + IP + DRP + LP + MRP + FP or, combining the first two terms to the right of the equals sign:

r = rRF + DRP + LP + MRP + FP

[A more technically correct, but cumbersome, multiplicative form is represented as

(1 + r) = (1 + r*) (1 + IP) (1 + DRP) (1 + LP) (1 + MRP) (1 + FP)]

The meaning of the symbols is as follows: r = nominal interest rate we actually observe (perhaps the annual rate at which a bank offers to lend); r* = real risk-free interest rate (historically measured in the 2 – 4% per year range); IP = inflation premium to cover the potential average periodic loss in purchasing power; rRF = r* + IP [or the multiplicative (1 + r*)(1 + IP) – 1] is the nominal risk-free interest rate that lenders charge the U.S. federal government when it borrows money on a short- basis; DRP = default risk premium to cover the risk of not being repaid in full and in a timely manner (applicable to private borrowers, and even to city and state governments when they borrow, but not to the U.S. government); LP = liquidity premium to cover the risk of losing money by potentially having to sell an in a thinly-traded, illiquid market; MRP = maturity risk premium to cover the added risks that accompany long-term financial commitments; FP = foreign exchange risk premium to cover risks of uncertainty in the relative values of different countries’ currencies – including concerns over inflation in other regions – that can accompany outside the country in whose currency the normally transacts.

The equation essentially tells us that the average annual rate of return a money provider expects to earn consists of a basic “risk-free” rate that would be anyone’s minimum required average yearly return under any conditions, plus additional amounts (premiums) to cover various risks that apply to the circumstances at hand. If a financial arrangement covers multiple time periods, the value used as expected or required periodic rate of return r in our time value factors should be the average (technically the multiplicative geometric average) of the individual periods’ returns expected over the entire time interval in question.

Example 1: We plan to lend money to the U.S. Government for one year. The annual real risk-free interest rate is 2.5%, and we expect inflation to be 3% over the next year. When we lend to the federal government there is no default or liquidity risk (the borrower is very strong financially, and the market is huge and very active). Here there is no maturity risk (the loan is short-term) or foreign exchange risk premium (both the borrower and lender are domestic parties). What annual interest rate should we charge on this 1-year loan? Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 2 r = approximately 2.5% + 3% = 5.5%, or more technically

(1 + r) = (1.025) (1.03) = 1.05575, so r = .05575, or 5.575%

Thus we would expect to lend to the federal government (buy “T-Bills”) for one year in return for approximately a 5½% interest rate. This 5½% rate is called a nominal rate; it is the one we actually expect to observe in the marketplace. But we do not really feel as though we will be gaining 5½% in wealth, since 3% of that 5½% constitutes compensation for expected inflation (lost purchasing power).

Our expected annual real rate of return (our increased ability to buy actual goods and services) therefore is approximately 2.5%. Of course, if inflation turns out to be less than 3% for the year our real return, measured after the fact, will have been more than 3%. If inflation turns out to be more than 3% our real annual return, measured after the fact, will have been less than 3%. We refer to this relationship – that the nominal “risk-free” interest rate (free of default, liquidity, etc. risks) consists of the real risk-free rate plus an inflation premium – as the Fisher effect, after the late economist Irving Fisher.

Example 2: We plan to lend to a private individual for five years. Again the real risk-free rate is 2.5% per year, and we expect the average annual rate of inflation to be 3% over the next five years. We also assign an annual default risk premium of 2%, an annual liquidity risk premium of 1%, and an annual maturity risk premium of .5%. There is no foreign exchange risk premium (since we are lending to another domestic party). What annual interest rate should we charge on this loan?

r = approximately 2.5% + 3% + 2% + 1% + .5% = 9.0%, or more precisely

(1 + r) = (1.025) (1.03) (1.02) (1.01) (1.005) = 1.093072, so r = .093072, or 9.3072%

When we do time value of money computations, we might think of the above method as the basis for determining the required average annual rate of return. The main point to recognize for now is that an investor who faces more risk should expect to earn a higher average periodic return (and will be unhappy after the fact if the realized return was not high enough to have compensated for the identified risks).

Of course, if there is more risk, then there is more of a chance that the investor’s requirement will not be met, that the actual (after-the-fact) rate of return will turn out to be less than the “required” or “expected” (before-the-fact) return. So in our time value coverage, when a problem states a 4% average annual required rate of return, we should assume that the provider of money in that situation perceives little risk of not being repaid promptly, or of losing purchasing power through inflation. A situation specifying a 19% required average annual rate of return suggests a money provider who fears serious potential adverse effects from inflation, default, liquidity, maturity, foreign exchange, or other problems.

A few points to consider:

▪ One way to think about risk and return is that if we make risky investments, we know it is likely that some of those investments will not turn out well. So we spread our money across a mix of risky investments (a portfolio, as we will call it in a later discussion), hoping that the higher returns earned on the ones that do pay off will compensate for losses on the ones that inevitably will not pay off.

For example, let’s say we have $500,000 to lend. We could lend it all to a financially strong party in return for a 5% agreed-on annual interest rate, and if there is essentially no chance the borrower will default (fail to pay some or all of what is owed when it is due) then our expected annual rate of return is 5%. Or we could make 100 loans of $5,000 each to less financially strong parties, charging each 36% interest per year (“payday” loans could be an example). If experience-based analysis indicates that 40 borrowers will pay as the contract specifies, 30 will pay back principal and half of the agreed- on interest (18% annual rate of return), 16 will repay principal but no interest (0% annual rate of Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 3 return), and 14 will pay no interest or principal (negative 100% annual return), then our expected annual return is .40 (.36) + .30 (.18) + .16 (0) + .14 (–1.00) = .144 + .054 + 0 – .14 = .058, or 5.8%. This expected annual rate of return is slightly higher than the safe investment provides, to compensate for the higher perceived risk (and for any added administrative cost of dealing with many small, riskier loans to the extent that these costs are not covered by loan application fees).

▪ A lender must charge an interest rate to reward himself for giving up the chance to consume today. The amount charged depends on how productive the lender’s opportunities would be if he put the money directly to use in the economy. The periodic percentage measure of this productivity, which economists call the marginal productivity of capital, can be a useful way to think about r*. If we could use our money to buy machines that would produce goods whose real value would increase our wealth by an average of 3% each year, then we would not lend money to any party – even the U.S. federal government, and even if we did not expect any inflation – for less than r* = 3% per year.

▪ In a similar manner, money users that have better opportunities (because they are more efficient, perhaps) are able, and willing, to pay higher interest rates than are less efficient users. For example, if efficient User A can earn a 15% average annual return on its investments, it will pay up to 15% per year to borrow money. Inefficient User B averages only 7% annual investment earnings, and thus can afford to pay only up to 7% per year for borrowed money. User B, offering to pay only 7% interest, will find no willing money providers, and will be driven out of the marketplace by its inefficiency.

▪ Why is a multiplicative approach to constructing an expected annual interest rate more technically correct than the additive approach? Consider a case with r* = .03, IP = .02, and all other premiums are 0 (just for computational simplicity). The additive approach tells us that a lender should charge a one-year interest rate of .03 + .02 = .05, or 5%, in order to get back 3% more than the purchasing power of the principal lent. If we lend an amount that would buy 100 pounds of flour today, we want to get back enough money in one year to buy 103 pounds of flour at that time.

Now assume that there is, in fact, 2% inflation in the subsequent year. Someone who lent $100, which initially would have paid for 100 pounds of flour costing $1 each, gets back $105. Since flour has risen in price by 2% with inflation, to $1.02 per pound, the $105 the lender gets back pays for only $105 ÷ $1.02 = 102.94 pounds. To be able to buy a full 103 pounds at $1.02 each, the lender should have gotten back $105.06 – an amount that he would have received if the interest rate had been quoted as (1 + r*)(1 + IP) – 1 = (1.02)(1.03) – 1 = 1.0506 – 1 = .0506, or 5.06%.

But the additive method provides a close approximation, and makes it easier for people to understand the building block explanation of interest rates. Because the additive approach gives a reasonable approximation, and perhaps especially because input figures like the average annual inflation rate or the appropriate amount to charge for possible default are subject to guess work and error, we typically view the additive approach as an acceptable tool for explaining and examining rates of return.

With the understanding of expected average periodic rates of return developed in this section, we should find the following discussion of time value concepts and computations to be more real-world relevant.

II. PROBLEMS WITHOUT SERIES OF EQUAL (OR RELATED) PAYMENTS

We can view any basic time value of money problem as falling into one of two categories: those that involve series of equal or related payments [annuity], and those that do not [we might coin the term “non-annuity”]. Indeed, the first step in using our tool box to do time value analysis is to identify whether we are dealing with an annuity or non-annuity case. We ask the question: is there a series of repeated

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 4 payments that are equally spaced in time and equal or related in amount? If there is no series, just the comparison of two values at different points in time, then we have a non-annuity situation.

A. What Textbooks Call Future Value of a Single Dollar Amount In the simplest time value application there is no series of payments, and the unknown to solve for is the dollar amount we will have at a future ; textbooks typically call this non-annuity situation “future value of a single dollar amount.” Example: we deposit $100 in the bank today. If our account balance earns a 6% compounded average annual interest rate (“compounded” because we do not spend the interest as we earn it, but rather leave it in the account and then earn interest on the interest – we often say that we compound an initial amount to an ending or future amount), how much money will be in the account after three years? We deposit $100 out of pocket only one time; there is no series of equal or related payments.

“Time value” of money suggests that the passage of time plays an important role in our analysis. While the time lines some textbook authors favor may not be essential tools for understanding time value, it is important for us to recognize that every time value situation plays out over a span of time that begins on the first of year (or other period) 1 and ends on the last day of year (or other period) n. Years over which financial arrangements occur could involve any dates, e.g. May 17 of year 1 to May 16 of year 8, but here we will illustrate with January 1 to December 31 periods because they create less mental clutter.

During year 1 we earn 6% on our $100 deposit (which is $6), so by the end of year 1 we have $100 + $6, or $100 (1.06) = $106. During year 2 we earn 6% on our $106 balance (original deposit plus first year’s interest), so by the end of year 2 we have $106 (1.06) = $112.36. Finally, during year 3 we earn 6% on the $112.36 balance, so by the end of year 3 we have $112.36 (1.06) = $119.10.

1-01-Yr 1 _____ 12-31-Yr 1/1-01-Yr 2 _____ 12-31-Yr 2/1-01-Yr 3 _____ 12-31-Yr 3 Deposit $100 Have $106.00 Have $112.36 Have $119.10

We can compute the answer more directly by simply compounding the original $100 balance at a 6% rate for three periods. [Multiplication is characterized by the associative property: instead of multiplying the $100 deposit by 1.06, and then multiplying that product by 1.06, and then multiplying by 1.06 again, we can simply multiply the $100 deposit by (1.06) (1.06) (1.06), or (1.06)3, which textbooks call the future value of $1 factor.] With our time value tool box we handle the computations for any non-annuity problem (with no series of equal or related payments into or out of an account, such that there are just two single dollar values compared at different points in time) using the following equation:

Beginning Amount (1 + r)n = Ending Amount, or

BAMT (1 + r)n = EAMT $100 (1.06)3 = EAMT $100 (1.191000) = $119.10

What if the expected average annual rate of return were instead 10%? Then the answer would be

$100 (1.10)3 = EAMT $100 (1.331000) = $133.10

Note that if r is higher, the computed ending amount (what textbooks call the future value) in a problem like this one is higher. It should be intuitively clear that if we can earn a higher average periodic rate of return, then we will have accumulated a higher total by the end of a specified number of time periods.

B. What Textbooks Call Present Value of a Single Dollar Amount A slightly more complicated time value application is the non-annuity case in which the unknown to solve for is the amount we must have initially so that we can reach a desired balance by a specified later Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 5 date; textbooks typically call this situation “present value of a single dollar amount.” It may seem more complicated because computing present value has us thinking in reverse, in some sense. But the process is straightforward if we apply our tool box and recognize the situation as a non-annuity problem with the beginning amount as the unknown to solve for. Example: we want our bank balance to have grown to $119.10 after three years. If the account earns a compounded return averaging 6% per year, how much do we need to have on deposit today? (Again we are depositing money out of our pocket only one time.)

It should be evident that this problem is the reverse of the 6% example above, and that the answer should be $100. If depositing $100 today leaves us with $119.10 in three years, then if we want to have $119.10 in three years we need to start with $100 today; as seen earlier:

1-01-Yr 1 _____ 12-31-Yr 1/1-01-Yr 2 _____ 12-31-Yr 2/1-01-Yr 3 _____ 12-31-Yr 3 Deposit $100 Have $106.00 Have $112.36 Have $119.10

Recall that with our tool box we set up any problem involving an initial deposit/investment, an average periodic rate of growth or return, and an ending amount (no repeated payment series) with the equation:

BAMT (1 + r)n = EAMT

With one equation, we can find a unique solution only if there is just one unknown from among beginning amount, r, n, and ending amount. In the earlier example, the $100 beginning amount was known, as were the average periodic rate of return (6% per year) and the number of time periods (three years), so we solved for the ending amount ($119.10). In this example, we know the $119.10 ending amount, periodic rate (6%), and number of time periods (three years), so we solve for the beginning amount:

BAMT (1 + r)n = EAMT so BAMT = EAMT ÷ (1 + r)n

BAMT = $119.10 ÷ (1.06)3 = $100.00

or, in the multiplicative form more traditionally used,

1 3 BAMT = $119.10 ( ) = $100.00 1.06

That is, by tradition we often multiply the ending amount by what textbooks call the present value of $1 factor (the reciprocal of the future value of $1 factor) instead of dividing by the future value of $1 factor, although those operations provide the same direct result (just as dividing something by 2 is the same as multiplying by ½). Multiplying an ending amount by the present value of $1 factor is often referred to as discounting a future amount to an initial value. It is very useful in investment analysis to determine a present value (how much we must deposit or invest today to be able to receive a specified amount of money in the future or, from a different perspective, how much we are willing to pay today – that is, what it is worth to us now – to have the right to realize a particular anticipated value at some future date).

How much must we open an account with today if we hope to have $200 in five years (or, in a slightly different context, what are we willing to pay today for the right to receive $200 in five years) if the expected average annual rate of return is 7%?

BAMT (1 + r)n = EAMT BAMT (1.07)5 = $200 BAMT = $200 ÷ (1.07)5 = $142.60 or 1 5 BAMT = $200 ( ) = $142.60 1.07

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 6 What if the appropriate average annual rate of return were instead 10%? Then the answer would be

BAMT = $200 ÷ (1.10)5 = $124.18 or 1 5 BAMT = $200 ( ) = $124.18 1.10

If r is higher, the computed beginning amount (so-called present value of a single dollar amount) is lower (whereas when computing the ending amount, the so-called future value of a single dollar amount case, a higher r leads to a higher computed value). There are two ways to think about this idea. One is that if we can earn a higher average yearly return, then we can deposit less today and still reach our $200 goal over five years. The other is that if we require a higher average periodic rate of return because we feel that the investment entails higher risk, then we are not willing to hand over as many dollars today, in return for an expected $200 (or other amount) at a specified future date, as we would be if less risk were involved. But by setting up any non-annuity problem with our BAMT (1 + r)n = EAMT equation we can easily compute correctly without worrying about how changing input values should affect the unknown being solved for.

C. Unknown Rate of Return Sometimes the unknown we wish to solve for in a time value problem is not a dollar amount, but rather the average periodic rate of return that a commitment of money generates. If an investment involves not a series of equal or related deposits or withdrawals, but rather only an initial dollar commitment and an ending value (again, what our tool box identifies as a non-annuity situation), we can solve directly for the unknown rate r. If we put $100 in the bank today, and then find three years later that our balance has grown to $119.10, what average annual compounded rate of return has been earned?

BAMT (1 + r)n = EAMT $100 (1 + r)3 = $119.10 (1 + r)3 = $119.10 ÷ $100 = 1.1910

One way to solve for r at this stage would be to use trial and error, trying different r values until the equation’s left-hand side equals about 1.1910. A more efficient method is to eliminate the exponent by taking an appropriate root. Since we have an equation, we can do the same thing to both sides and still have an equation. Here, if we take the cube root of each side, we solve as:

3√(1 + 푟)3 = 3√1.1910 or 1.19101/3 or 1.1910.33333 (1 + r) = 1.06, so r = .06, or 6%

(from our initial examples in parts A and B above, we knew that r had to be 6%). Unlike in annuity situations we will examine later, in which solving for r requires trial and error, in a non-annuity case we can solve directly for an unknown r because that rate appears only once in the factor that connects the beginning dollar figure with the ending amount to which it compounds, and thus we need only take the root corresponding to the number of periods over which the growth occurs. If we bought a painting for $1,350,000 ten years ago, and then sold it today for $3,000,000, what has been the average annual rate of increase in the art work’s value?

BAMT (1 + r)n = EAMT $1,350,000 (1 + r)10 = $3,000,000 (1 + r)10 = $3,000,000 ÷ $1,350,000 = 2.22222 Solve by taking the tenth root of each side:

10√(1 + 푟)10 = 10√2.22222 or 2.222221/10 or 2.22222.10 (1 + r) = 1.083125, so r = .083125, or 8.3125%

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 7 This latter example illustrates three ideas worth noting. First is that a collectible item is a common example of a non-annuity situation. Second is that we have a reason for showing the non-annuity equation in our tool box as Beginning Amount (1 + r)n = Ending Amount rather than textbooks’ typical Present Value (1 + r)n = Future Value. In the example above, $3,000,000 might seem to be the “present” value since it exists today, but that amount is what the art investor gets at the ending date. Indeed, a time value story could all have occurred in the (e.g., someone deposited $10,000 in 1992, added nothing to the account out of pocket over the years, had $26,500 by 2017 – all in the past – and wants to determine the average annual interest rate that was earned). The simple substitution of BAMT for PV and EAMT for FV thus can help prevent considerable confusion. Finally, distinguishing between “present value of a single dollar amount” (solving for the beginning amount) and “future value of a single dollar amount” (solving for the ending amount), as textbooks typically show, is also needlessly confusing since the unknown we solve for makes no difference in how a problem is structured (a point that becomes more evident when we are solving for r or n). The unknown value solved for affects only the last algebra step, not the broader thought process and analysis. With the BAMT (1 + r)n = EAMT equation we can see that all we are doing is solving, with simple algebra, for whatever is unknown in any non-annuity problem.

D. Unknown Number of Time Periods A final unknown we might solve for in a non-annuity case is the number of time periods needed for an initial dollar amount to grow to a specified ending amount. Let’s say we put $4,000 into the bank today, and then add nothing more to the account out of pocket (there is not a series of $4,000 deposits). If we earn a 7% average compounded annual return, how long will it take for our $4,000 to grow to $12,000?

BAMT (1 + r)n = EAMT $4,000 (1.07)n = $12,000 (1.07)n = 3

One way to solve for n would be with trial and error, trying different n values until the equation’s left- hand side equals about 3. A more efficient method is to use logarithms (a logarithm is the exponent to which we raise a base value to reach a specified quantity). We can do the same thing to both sides of the equation – like taking the “natural” logarithm ln (for which financial and scientific calculators have built- in tables) of each side – and still have an equation. Taking the ln of each side yields the result

ln [(1.07)n] = ln 3

Because a logarithm is an exponent, when we take the “log” of something that has an exponent the exponent’s value factors out as a multiplier: n (ln 1.07) = ln 3

From a table of natural logarithms, accessed with a financial or scientific calculator’s LN key, we find that ln 1.07 = .067659, and ln 3 = 1.098612. (We could also use the LOG key for base-10 “common” logarithms, but financial calculators typically offer only tables. The natural log is the power to which we raise the irrational number e, which equals about 2.7182818, to reach a targeted value. e is computed as the limit of 1 푛 (1 + ) 푛 as n approaches ∞. To get 3 we would have to raise 2.7182818 to a power just greater than 1; e1.098612 = 3, so ln 3 = 1.098612. Since an exponent of 0 applied to any value equals 1, to get 1.07 we would have to raise 2.7182818 to a power a little greater than 0; e.067659 = 1.07, so ln 1.07 = .067659.) Thus we solve as

n (ln 1.07) = ln 3 n (.067659) = 1.098612 n = 1.098612 ÷ .067659 = 16.237574 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 8 So it takes slightly more than 16 years for an initial $4,000 deposit to grow to $12,000 if we earn a 7% average annual interest rate. Check: $4,000 (1.07)16.237574 = $12,000. ✓

Simple “non-annuity” recap: TVM analysis is about structure, not about some unknown being solved for. Think of someone who buys a collectible coin for $1,350; its average annual increase in value is 4.6%, and after holding it for 18 years she sells it for $3,033.22. The structure we would use to analyze, no matter which of the four possible unknowns we are solving for (in one equation there can be only one unknown if we are to find a unique solution, so the other three related values would have to be given), is

BAMT (1 + r)n = EAMT $1,350.00 (1.046)18 = $3,033.22

We would multiply to find the $3,033.22 EAMT, divide (or multiply by the factor’s reciprocal) to find the $1,350.00 BAMT, take an 18th root to find the 4.6% average annual r, or take a ratio of logarithms to find the n of 18 years needed, but any of those steps is just applying algebra once the structure is in place.

E. Series of Payments That Are Not Equal (or Related) 1. Payments Corresponding to Large Dollar Amount That Will Not Exist Intact Until a Future Date Recall that we always begin by asking the question: is there a series of repeated payments that are equally spaced in time, and equal or related in amount? If there is no series, just the comparison of two values at different points in time, then we have a simple or pure non-annuity situation. But what if there is a series of payments that are equally spaced in time, but not equal or related in amount? Then we still are dealing with a non-annuity case – but must work several non-annuity problems and sum the individual solutions to compute the correct overall answer. With year-to-year payments (“cash flows”) that are not equal, and not related (e.g., changing by a constant percentage), there is no choice but to use a period-by-period approach; we can not group the payments for computational purposes. Let’s say we plan to deposit $2,000 at the end of year 1; $2,100 at the end of year 2; $2,800 at the end of year 3; and $3,200 at the end of year 4 (so $2,000; $2,100; $2,800; and $3,200 are a series of BAMTs that will occur at different ). [We could analyze from the view of the person making deposits or the bank getting these amounts from the account holder; the numbers look the same from either side of the transaction.] A time line shows:

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Deposit $2,000 Deposit $2,100 Deposit $2,800 Deposit $3,200 With $0 Plan Ends

If we can earn a 10% average annual interest rate on the account’s growing balance, the total we will have by the end of year 4 can be computed with four applications of our BAMT (1 + r)n = EAMT equation:

$2,000 (1.10)3 = $2,000 (1.331000) = $2,662.00 as EAMT deposit #1 will reach by end of year 4 $2,100 (1.10)2 = $2,100 (1.210000) = $2,541.00 as EAMT deposit #2 will reach by end of year 4 $2,800 (1.10)1 = $2,800 (1.100000) = $3,080.00 as EAMT deposit #3 will reach by end of year 4 $3,200 (1.10)0 = $3,200 (1.000000) = $3,200.00 as EAMT deposit #4 will reach by end of year 4 for a combined total ending balance, sometimes called the net future value, of ($2,662.00 + $2,541.00 + $3,080.00 + $3,200.00) = $11,483.00. We could think of this plan as involving four separate accounts, but it could all occur within a single account, the progress of which is shown in the following table:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Accumulated Deposit Balance 1 $0 $0 $0 $2,000.00 $ 2,000.00 2 $2,000.00 $200.00 $2,200.00 $2,100.00 $ 4,300.00 3 $4,300.00 $430.00 $4,730.00 $2,800.00 $ 7,530.00 4 $7,530.00 $753.00 $8,283.00 $3,200.00 $11,483.00 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 9 Now let’s assume we instead will deposit those dollar amounts at the beginning of each respective year:

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Deposit $2,100 Deposit $2,800 Deposit $3,200 Plan Ends Deposit $2,000

Again assuming we can earn a 10% average annual rate on the account’s growing balance, the total we will have by the end of year 4 is again computed with four applications of BAMT (1 + r)n = EAMT:

$2,000 (1.10)4 = $2,000 (1.464100) = $2,928.20 as EAMT deposit #1 will reach by end of year 4 $2,100 (1.10)3 = $2,100 (1.331000) = $2,795.10 as EAMT deposit #2 will reach by end of year 4 $2,800 (1.10)2 = $2,800 (1.210000) = $3,388.00 as EAMT deposit #3 will reach by end of year 4 $3,200 (1.10)1 = $3,200 (1.100000) = $3,520.00 as EAMT deposit #4 will reach by end of year 4 for a combined total value at the future maturity date of ($2,978.20 + $2,795.10 + $3,388.00 + $3,520.00) = $12,631.30. Again we could think of this plan as involving four separate accounts, but it all could occur within a single account, progress for which is shown in the table below. Notice that the final answer of $12,631.30 is merely the $11,483.00 answer found for the same dollar amounts if paid in at the end of each year, multiplied by (1 + r), here (1.10): $11,483.00 (1.10) = $12,631.30. The structures of the two series are identical, except that if payments occur at the beginning of each period interest is applied to the remaining, growing balance one differential number of times (here, one more) over the plan’s life.

Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Accumulated Interest Balance 1 $0 $2,000.00 $ 2,000.00 $ 200.00 $ 2,200.00 2 $2,200.00 $2,100.00 $ 4,300.00 $ 430.00 $ 4,730.00 3 $4,730.00 $2,800.00 $ 7,530.00 $ 753.00 $ 8,283.00 4 $8,283.00 $3,200.00 $11,483.00 $1,148.30 $12,631.30

2. Payments Corresponding to Large Dollar Amount That Exists Intact In the Present Now consider a case in which we find the total amount needed initially to fund a series of withdrawals by adding a series of BAMT figures. (We could be analyzing from the perspective of the individual making withdrawals or the bank that is paying these amounts to the account holder; the numbers look the same from either side of the transaction.) Let’s say we would like to withdraw $2,000 from our bank account at the end of year 1; $2,100 at the end of year 2; $2,800 at the end of year 3; and $3,200 at the end of year 4 (so $2,000; $2,100; $2,800; and $3,200 are a series of EAMTs that will occur at different times). |If interest averaging 10% per year is received by the account holder/paid by the bank on the remaining balance, how much must be on deposit today to fund the plan? A time line perspective would show:

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Withdraw $2,000 Withdraw $2,100 Withdraw $2,800 Withdraw $3,200 With ??? Plan Ends

We can compute the answer (shown as ??? above) with four applications of our BAMT (1 + r)n = EAMT equation [notice that dividing by (1 + r)n is equivalent to multiplying by its reciprocal]: 1 1 $2,000 ÷ (1.10)1 = $2,000 ( ) = $2,000 (.909091) = $1,818.18 as BAMT that funds withdrawal #1 1.10 1 2 $2,100 ÷ (1.10)2 = $2,100 ( ) = $2,100 (.826446) = $1,735.54 as BAMT that funds withdrawal #2 1.10 1 3 $2,800 ÷ (1.10)3 = $2,800 ( ) = $2,800 (.751315) = $2,103.68 as BAMT that funds withdrawal #3 1.10 1 4 $3,200 ÷ (1.10)4 = $3,200 ( ) = $3,200 (.683013) = $2,185.64 as BAMT that funds withdrawal #4 1.10 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 10 for a combined total beginning balance needed of ($1,818.18 + $1,735.54 + $2,103.68 + $2,185.64 ) = $7,843.04. (This amount is sometimes called the , though we will use that term primarily in connection with more complex investment situations, such as capital budgeting analysis in Topic 6.) We could think of this plan as involving four separate accounts, but it could as easily occur within just one account, with year-by-year figures as demonstrated in the following table:

Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $7,843.04 $784.30 $8,627.34 $2,000.00 $6,627.34 2 $6,627.34 $662.73 $7,290.07 $2,100.00 $5,190.07 3 $5,190.07 $519.00 $5,790.07 $2,800.00 $2,909.07 4 $2,909.07 $290.91 $3,200.00 $3,200.00 $0

Now let’s assume we instead are taking those dollar amounts out at the beginning of each respective year:

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Withdraw $2,100 Withdraw $2,800 Withdraw $3,200 Plan Officially With ??? $0 Balance Remains Ends Withdraw $2,000

Once again we can compute the unknown answer (shown as ??? above) with four applications of our BAMT (1 + r)n = EAMT equation [dividing by (1 + r)n is equivalent to multiplying by the reciprocal]:

1 0 $2,000 ÷ (1.10)0 = $2,000 ( ) = $2,000 (1.000000) = $2,000.00 as BAMT that funds withdrawal #1 1.10 1 1 $2,100 ÷ (1.10)1 = $2,100 ( ) = $2,100 (.909091) = $1,909.09 as BAMT that funds withdrawal #2 1.10 1 2 $2,800 ÷ (1.10)2 = $2,800 ( ) = $2,800 (.826446) = $2,314.05 as BAMT that funds withdrawal #3 1.10 1 3 $3,200 ÷ (1.10)3 = $3,200 ( ) = $3,200 (.751315) = $2,404.21 as BAMT that funds withdrawal #4 1.10 for a combined total beginning balance needed of ($2,000.00 + $1,909.09 + $2,314.05 + $2,404.21) = $8,627.35 – a larger answer than in the corresponding case with end-of-year withdrawals; more money is needed up-front to fund a payment stream that starts right away. Again we could envision this plan as involving four separate accounts, but it could as easily occur within just one account, with year-by-year figures as demonstrated in the table below. Just as when a payment stream corresponds to a large dollar amount that will not exist intact until a future date, the $8,627.35 final answer here for this beginning-of- year payments case is merely the $7,843.04 answer found for the same dollar amounts if taken at each year’s end multiplied by (1 + r): $7,843.04 (1.10) = $8,627.35. The two series structures are identical, except that if payments occur at the beginning of each period interest is applied to the remaining, declining balance one differential number of times (here, one less) over the life of the plan.

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $8,627.35 $2,000.00 $6,627.35 $662.74 $7,290.08 2 $7,290.08 $2,100.00 $5,190.08 $519.01 $5,709.09 3 $5,709.09 $2,800.00 $2,909.09 $290.91 $2,300.00 4 $2,300.00 $3,200.00 $0 $0 $0

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 11 III. FUTURE AND PRESENT VALUES OF LEVEL ANNUITIES

A. Introduction to Annuities Each example in parts A – D of the prior section involves the comparison of a single initial deposit or investment with a single dollar value realized after some number of time periods has passed. Of course, many a financial situation instead involves a series of payments into, or withdrawals from, an account that is accompanied by an average periodic percentage return or cost. Part E of the prior section deals with series of unequal, unrelated payments, in which we must use a period-by-period “brute force” approach to computing; there is no alternative to working a group of separate non-annuity problems and summing their answers. But what if there is a series of equally spaced expected payments, made into or taken out of some account or plan, that are equal in amount, or related in that they change by a constant percentage from period to period? We call a situation in which money is paid into or withdrawn from a plan repeatedly, at equally spaced time intervals and with equal or related payments, an annuity. (It is not incorrect to label a series of equally spaced but unrelated payments as a type of annuity, but in our coverage we reserve the term “annuity” for cases in which we can group multiple payments in a single computation. And dollar amounts that are equal or related but not spaced equally in time, such as $200 deposited in an account today, $200 six later, $200 eight months later, and $200 two years later, can not be grouped computationally.) In any annuity example we are dealing with a series of equal or related “cash flows” paid or received, an average expected periodic rate of return or cost r, a number of time periods n, and a lump sum of money that is equivalent, in time value-adjusted terms, to the series of cash flows. An annuity problem turns out to be just the sum of a group of non-annuity problems, as seen in Part E above, that we can treat as a group for computing purposes because of the distributive property.

We identify the situation as present value or future value of an annuity by looking at when the lump sum exists intact. If the lump sum, be it known or unknown, exists intact today (e.g., a loan repayment problem – the regular payments relate to a big amount the lender hands over now), we have a present value of an annuity situation. If the lump sum will not exist intact until some future date (e.g., saving up for retirement – the regular deposits and the interest earned will not have grown to the maximum amount until the day the saver retires), it is a future value of an annuity situation.

As noted, annuities’ expected cash flow streams can be equal in amount (“level” annuities, the most common examples in most textbook applications) or related (changing by a constant percentage from period to period). An ordinary annuity (also called annuity in arrears) is a series of related cash flows, each occurring at the end of the year or other specified time period. An annuity due (also called annuity in advance) is a series of related cash flows, each occurring at the beginning of the year or other specified time period. We can encounter ordinary annuity and annuity due situations in both present value and future value of annuity applications. However, while useful ordinary annuity and annuity due examples exist for both future and present value of annuity scenarios, it sometimes can be awkward to think of the PV of an annuity due (which would involve an initial cash flow on the day the lump sum is paid or received, such as borrowing money and then immediately making the first loan payment) or FV of an ordinary annuity (an example would be opening a retirement account on paper today but then waiting a year to make the first deposit).

Finally, always remember that there are two sides to every transaction or series of transactions, and what one side pays in is what the other side effectively receives or takes out – and the numbers are the same for both parties to the transaction. So do not oversimplify by saying: if someone pays into a plan it is an FV of annuity situation, while withdrawing from a plan is a PV of annuity situation – because the structure of the numbers is what matters. In a retirement savings plan that grows with equal or related deposits over time (an FV of annuity case), it is equally valid to view the series of deposits and the interest applied to the growing balance from the perspective of the saver (who makes the deposits and receives the interest) or that of the bank (which receives the deposits and pays the interest), and the large future total is the same amount regardless of whether we think of it as being property of the saver, or owed to the saver

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 12 by the bank. In a plan for depleting an endowed account with equal or related payments over time (the “rich uncle” problem, a PV of annuity case), the perspective of the account holder (who takes withdrawals from the large initial endowment and earns interest on the declining balance) is no more or less relevant than that of the bank (which, in effect, makes payments to the account holder from the same large initial balance and owes the interest). All dollar amounts are identical for the paying and receiving parties. We identify the type of problem by its structure, not by the side of a transaction we might envision being on.

B. A General Overview of the Annuity Idea 1. Future Value of an Annuity a. Level Payments Think of a level (all payments are equal) ordinary annuity situation, in which we plan to deposit $2,300 at the end of each of years 1 through 4. A very rudimentary time line shows

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Deposit $2,300 Deposit $2,300 Deposit $2,300 Deposit $2,300 With $0 Plan Ends

If we can earn a 10% average annual interest rate on the account’s growing balance, one way to solve for the total we will have by the end of year 4 is with four applications of our BAMT (1 + r)n = EAMT non- annuity equation, just as we did with four unrelated deposits at the end of the prior section:

$2,300 (1.10)3 = $2,300 (1.331000) = $3,061.30 as amount deposit #1 will reach by end of year 4 $2,300 (1.10)2 = $2,300 (1.210000) = $2,783.00 as amount deposit #2 will reach by end of year 4 $2,300 (1.10)1 = $2,300 (1.100000) = $2,530.00 as amount deposit #3 will reach by end of year 4 $2,300 (1.10)0 = $2,300 (1.000000) = $2,300.00 as amount deposit #4 will reach by end of year 4 for a combined total ending balance of ($3,061.30 + $2,783.00 + $2,530.00 + $2,300.00) = $10,674.30. We could think of making annual deposits into four separate accounts, but all could be made into the same account, with a growing balance for this ordinary annuity shown as follows:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Accumulated Deposit Balance 1 $0 $0 $0 $2,300.00 $ 2,300.00 2 $2,300.00 $230.00 $2,530.00 $2,300.00 $ 4,830.00 3 $4,830.00 $483.00 $5,313.00 $2,300.00 $ 7,613.00 4 $7,613.00 $761.30 $8,374.30 $2,300.00 $10,674.30

It is perfectly acceptable to compute the combined ending balance for a series of equal payments just as we would for a series of unrelated payments, in the manner shown above. But there is a more expedient approach to reaching this total when the payments are equal. Note that in the above example we have

[$2,300 (1.10)3] + [$2,300 (1.10)2] + [$2,300 (1.10)1] + [$2,300 (1.10)0] which, by the distributive property (the mathematical basis for all annuity computations), can be restated as $2,300 [(1.10)3 + (1.10)2 + (1.10)1 + (1.10)0]

(1.10)4−1 = $2,300 ( ) = $2,300 (4.641) = $10,674.30 .10

(fortunately, a smart mathematician long ago figured out that

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 13 (1+푟)푛−1 [(1 + r)n – 1 + ⋯ + (1 + r)1 + (1 + r)0] = ( )). 푟

That is, the future value of a level ordinary annuity factor is just the sum of the so-called future value of $1 factors for the same periodic rate and number of time periods. (See section A of the Appendix to this outline for more detail behind the mathematics of the FV of a level annuity factor.) A level annuity therefore is just a series of payments that we can treat for computational purposes as a team, because they are equal in value so the distributive property applies, rather than using the brute force (series of non- annuity problems) approach necessary when the payments are unrelated. (The future value of an annuity is just a net future value, if we wish to use that term, in a case with equal or related payments.) Three interconnected characteristics are seen that help in identifying every FV of an annuity situation:

• The lump sum value that corresponds to the stream of equal or related payments will not be intact until a future date; we start with nothing and end up with a large dollar amount. (The classic example of a FV of an annuity is saving up for retirement, with the large nest egg building up over time and not being complete until the future date when the saver will retire.) • The principal amount to which the interest or other rate of return is applied grows larger (and therefore the amounts of interest do, as well) with each successive year or other time period. • The FV of an annuity factor is larger than the number of payment periods. Here the 4.641 future value of an annuity factor is larger than the 4 payments: if we make four equal deposits into a 10% savings account, the eventual total will be 4.641 times each deposit. We end up with more than the amounts deposited because interest is applied on top of the four deposits made.

What if instead we plan to deposit $2,300 at the beginning of each of years 1 through 4, and expect to earn a 10% average interest rate per year on the growing balance? The payments’ timing would be

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Deposit $2,300 Deposit $2,300 Deposit $2,300 Plan Ends Deposit $2,300

Of course we can solve for the total we will have in this annuity due situation by the end of year 4 with four applications of the BAMT (1 + r)n = EAMT equation our tool box specifies for non-annuity cases:

$2,300 (1.10)4 = $2,300 (1.464100) = $3,367.43 as amount deposit #1 will reach by end of year 4 $2,300 (1.10)3 = $2,300 (1.331000) = $3,061.30 as amount deposit #2 will reach by end of year 4 $2,300 (1.10)2 = $2,300 (1.210000) = $2,783.00 as amount deposit #3 will reach by end of year 4 $2,300 (1.10)1 = $2,300 (1.100000) = $2,530.00 as amount deposit #4 will reach by end of year 4 for a combined total ending balance of ($3,367.43 + $3,061.30 + $2,783.00 + $2,530.00) = $11,741.73. We can follow the year-by-year activity in a single account holding these deposits with the table:

Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Accumulated Interest Balance 1 $0 $2,300.00 $ 2,300.00 $ 230.00 $ 2,530.00 2 $2,530.00 $2,300.00 $ 4,830.00 $ 483.00 $ 5,313.00 3 $5,313.00 $2,300.00 $ 7,613.00 $ 761.30 $ 8,374.30 4 $8,374.30 $2,300.00 $10,674.30 $1,067.43 $11,741.73

Note that in this example we have

[$2,300 (1.10)4] + [$2,300 (1.10)3] + [$2,300 (1.10)2] + [$2,300 (1.10)1] which, by the distributive property, can be restated as Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 14 $2,300 [(1.10)4 + (1.10)3 + (1.10)2 + (1.10)1], which can be factored as

$2,300 [(1.10)3 + (1.10)2 + (1.10)1 + (1.10)0] (1.10)

(1.10)4−1 = $2,300 [( ) (1.10)] = $2,300 (5.1051) = $11,741.73 .10

(If deposits are made at the beginning of each year, then interest is earned during year 1, causing the final balance to be 1.10 times as large as it would be if there were year-end deposits, in which case no interest would be earned in the first year. The exponent in the FV of an ordinary annuity factor is the number of cash flows, but in that factor the number of interest applications is one less than the cash flow count.) The future value of a level annuity factor therefore is

(1 + 푟)푛−1 ( ) for cases with end-of-period cash flows (FV of a level ordinary annuity) 푟

(1 + 푟)푛−1 and [( ) (1 + 푟)] for cases with beginning-of-period cash flows (FV of a level annuity due). 푟 b. Payments Changing by a Constant Periodic Percentage [FIL 240 Students: Just skim, not required] Now consider a case in which the first of four deposits we plan to make at the ends of each of years 1 through 4 is $2,300, and then the amounts are to increase by 2% per year. Therefore the deposits will be $2,300 (1.02)0 = $2,300; $2,300 (1.02)1 = $2,346; $2,300 (1.02)2 = $2,392.92; and $2,300 (1.02)3 = $2,440.78. If a 10% average annual interest rate can be earned on the account’s growing balance, we again can solve for the total we will have by the end of year 4 with four applications of our BAMT (1 + r)n = EAMT non-annuity equation, just as we did with four unrelated deposits at the end of the prior section:

$2,300.00 (1.10)3 = $2,300.00 (1.331000) = $3,061.30 as amount deposit #1 will reach by end of year 4 $2,346.00 (1.10)2 = $2,346.00 (1.210000) = $2,838.66 as amount deposit #2 will reach by end of year 4 $2,392.92 (1.10)1 = $2,392.92 (1.100000) = $2,632.21 as amount deposit #3 will reach by end of year 4 $2,440.78 (1.10)0 = $2,440.78 (1.000000) = $2,440.78 as amount deposit #4 will reach by end of year 4 for a combined total ending balance of ($3,061.30 + $2,838.66 + $2,632.21 + $2,440.78) = $10,972.95. If the four deposits all were made into the same account, the growing balance would show as follows:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Accumulated Deposit Balance 1 $0 $0 $0 $2,300.00 $ 2,300.00 2 $2,300.00 $230.00 $2,530.00 $2,346.00 $ 4,876.00 3 $4,876.00 $487.60 $5,363.60 $2,392.92 $ 7,756.52 4 $7,756.52 $775.65 $8,532.17 $2,440.78 $10,972.95

Recall that when we say we can combine, for computational purposes, a series of payments that are equally spaced apart in time and equal or related in amount, “or related” means changing by a constant percentage from period to period. In this example, with the 2% constant periodic change in cash received or paid out, a more expedient approach is to note that in the above example we have

[$2,300 (1.02)0 (1.10)3] + [$2,300 (1.02)1 (1.10)2] + [$2,300 (1.02)2 (1.10)1] + [$2,300 (1.02)3 (1.10)0] which, by the distributive property, can be restated as $2,300 [(1.02)0 (1.10)3 + (1.02)1 (1.10)2 + (1.02)2 (1.10)1 + (1.02)3 (1.10)0]

(1.10)4 − (1.02)4 = $2,300 ( ) = $2,300 (4.770848) = $10,972.95 .10 − .02 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 15 (see sections B and I of the Appendix to this outline for more detail on the mathematics behind the FV of a changing annuity factor). What if we instead make the deposits described above at the beginning of each of years 1 through 4, still expecting to earn a 10% average interest rate or other rate of return per year on the growing balance? The total expected by the end of year 4 would be

$2,300.00 (1.10)4 = $2,300.00 (1.461000) = $3,367.43 as amount deposit #1 will reach by end of year 4 $2,346.00 (1.10)3 = $2,346.00 (1.331000) = $3,122.53 as amount deposit #2 will reach by end of year 4 $2,392.92 (1.10)2 = $2,392.92 (1.210000) = $2,895.43 as amount deposit #3 will reach by end of year 4 $2,440.78 (1.10)1 = $2,440.78 (1.100000) = $2,684.86 as amount deposit #4 will reach by end of year 4 for a combined total ending balance of ($3,367.43 + $3,122.53 + $2,895.43 + $2,684.86) = $12,070.25. We can follow the year-by-year activity in a single account holding these deposits with the table:

Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Accumulated Interest Balance 1 $0 $2,300.00 $ 2,300.00 $ 230.00 $ 2,530.00 2 $2,530.00 $2,346.00 $ 4,876.00 $ 487.60 $ 5,363.60 3 $5,363.60 $2,392.92 $ 7,756.52 $ 775.65 $ 8,532.17 4 $8,532.17 $2,440.78 $10,972.95 $1,097.30 $12,070.25

Note in this example we have [$2,300 (1.02)0 (1.10)4] + [$2,300 (1.02)1 (1.10)3] + [$2,300 (1.02)2 (1.10)2] + [$2,300 (1.02)3 (1.10)1] which, by the distributive property, can be restated as

$2,300 [(1.02)0 (1.10)4 + (1.02)1 (1.10)3 + (1.02)2 (1.10)2 + (1.02)3 (1.10)1], which can be factored as

$2,300 [(1.02)0 (1.10)3 + (1.02)1 (1.10)2 + (1.02)2 (1.10)1 + (1.02)0 (1.10)0] (1.10)

(1.10)4 − (1.02)4 = $2,300 [( ) (1.10)] = $2,300 (5.247993) = $12,070.25 .10 − .02

2. Present Value of an Annuity a. Level Payments Now consider a different type of level ordinary annuity situation, in which some individual or institution hopes to withdraw or receive $2,300 at the end of each of years 1 through 4.

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Withdraw $2,300 Withdraw $2,300 Withdraw $2,300 Withdraw $2,300 With ??? Plan Ends

If a 10% average annual interest rate can be earned on the account’s remaining balance, one way to solve for the total that must be on hand today to fund the withdrawal stream is with four applications of the BAMT (1 + r)n = EAMT  EAMT ÷ (1 + r)n = BAMT equation our tool box specifies for non-annuity situations (just as was done with four unrelated withdrawals at the end of the prior section):

1 1 $2,300 ÷ (1.10)1 = $2,300 ( ) = $2,300 (.909091) = $2,090.91 as BAMT that funds withdrawal #1 1.10 1 2 $2,300 ÷ (1.10)2 = $2,300 ( ) = $2,300 (.826446) = $1,900.83 as BAMT that funds withdrawal #2 1.10 1 3 $2,300 ÷ (1.10)3 = $2,300 ( ) = $2,300 (.751315) = $1,728.02 as BAMT that funds withdrawal #3 1.10 1 4 $2,300 ÷ (1.10)4 = $2,300 ( ) = $2,300 (.683013) = $1,570.93 as BAMT that funds withdrawal #4 1.10 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 16 for a combined total beginning balance needed of ($2,090.91 + $1,900.83 + $1,728.02 + $1,570.93) = $7,290.69. We could think of this plan as involving four separate accounts, but it could as easily occur within just one account, with progress as demonstrated in the following table:

Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $7,290.69 $729.07 $8,019.76 $2,300.00 $5,719.76 2 $5,719.76 $571.98 $6,291.74 $2,300.00 $3,991.74 3 $3,991.74 $399.17 $4,390.91 $2,300.00 $2,090.91 4 $2,090.91 $209.09 $2,300.00 $2,300.00 $0

Again it is perfectly acceptable to compute the combined beginning balance needed to support a series of equal and equally spaced withdrawals just as we would for a series of unrelated cash flows. But, again, the existence of equal payments allows a more expedient computational approach. In this example we have 1 1 1 2 1 3 1 4 [$2,300 ( ) ] + [$2,300 ( ) ] + [$2,300 ( ) ] + [$2,300 ( ) ] 1.10 1.10 1.10 1.10 which, by the distributive property (which works with multiplication but not division), can be restated as

1 1 1 2 1 3 1 4 $2,300 [( ) + ( ) + ( ) + ( ) ] 1.10 1.10 1.10 1.10

1 4 1−( ) = $2,300 ( 1.10 ) = $2,300 (3.1699) = $7,290.69 .10

(fortunately, that same smart mathematician long ago figured out that

1 푛 1 1 1 2 1 3 1 푛 1−( ) [( ) + ( ) + ( ) + ⋯ + ( ) ] = ( 1 + 푟 )). 1 + 푟 1 + 푟 1 + 푟 1 + 푟 푟

That is, the present value of a level ordinary annuity factor is just the sum of the so-called present value of $1 factors for the same periodic rate and number of time periods. (See section C of the Appendix to this outline for more detail regarding the mathematics of the PV of a level annuity factor.) As noted above, a level annuity is just a cash flow series we can treat as a team, because equal (or also related) payments let us use the distributive property and avoid the brute force computations needed with unrelated payments. (The present value of an annuity is just a net present value, if we wish to use that term here, in a case with equal or related cash flows.) Three interconnected characteristics are at work that help us identify a PV of an annuity situation:

• The lump sum value that corresponds to the stream of equal or related payments is intact in the present; we start with a large dollar amount and end up with nothing. (The classic example of PV of an annuity is borrowing a large sum of money today and then paying back principal and interest down to a zero balance with equal payments over many subsequent periods; a twist on repaying a loan is the “rich uncle” problem involving payments from the bank to a depositor.) • The principal to which interest is applied becomes smaller (and therefore the interest amounts also decline) with each successive year or other time period. • The PV of an annuity factor is smaller than the number of payment periods. Here the 3.1699 PV of an annuity factor is smaller than the 4 payments: if we make four equal withdrawals from an account that earns a 10% annual interest rate, each withdrawal will be about 1 ÷ 3.1699 or 31.5% Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 17 of the original balance. (Or if we make four equal repayments on a loan charging a 10% annual interest rate, each payment will be about 1 ÷ 3.1699 or 31.5% of the principal borrowed.) We take out (pay back) 4 (31.5%) = approximately 126% of what we started with (borrowed), because we will earn (pay) interest on top of having the initial endowment (loan principal).

What if instead we hope to withdraw $2,300 at the beginning of each of years 1 through 4, and expect to earn an average of 10% in interest per year on the account’s declining balance?

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Withdraw $2,300 Withdraw $2,300 Withdraw $2,300 Plan Officially Withdraw $2,300 $0 Balance Remains Ends

We can find the total that must be available today to fund this plan with four applications of our tool box’s BAMT (1 + r)n = EAMT equation:

1 0 $2,300 ÷ (1.10)0 = $2,300 ( ) = $2,300 (1.000000) = $2,300.00 as BAMT that funds withdrawal #1 1.10 1 1 $2,300 ÷ (1.10)1 = $2,300 ( ) = $2,300 (.909091) = $2,090.91 as BAMT that funds withdrawal #2 1.10 1 2 $2,300 ÷ (1.10)2 = $2,300 ( ) = $2,300 (.826446) = $1,900.83 as BAMT that funds withdrawal #3 1.10 1 3 $2,300 ÷ (1.10)3 = $2,300 ( ) = $2,300 (.751315) = $1,728.02 as BAMT that funds withdrawal #4 1.10 for a combined total beginning balance needed of ($2,300.00 + $2,090.91 + $1,900.83 + $1,728.02) = $8,019.76. We could think of providing for our withdrawals with four separate accounts, but all the activity could occur within just one account, with progress illustrated as:

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $8,019.76 $2,300.00 $5,719.76 $571.98 $6,291.74 2 $6,291.74 $2,300.00 $3,991.74 $399.17 $4,309.91 3 $4,309.91 $2,300.00 $2,090.91 $209.09 $2,300.00 4 $2,300.00 $2,300.00 $0 $0 $0

Note that in this example we have

1 0 1 1 1 2 1 3 [$2,300 ( ) ] + [$2,300 ( ) ] + [$2,300 ( ) ] + [$2,300 ( ) ] 1.10 1.10 1.10 1.10 which, by the distributive property, can be restated as

1 0 1 1 1 2 1 3 $2,300 [( ) + ( ) + ( ) + ( ) ], which can be factored as 1.10 1.10 1.10 1.10

1 1 1 2 1 3 1 4 $2,300 [( ) + ( ) + ( ) + ( ) ] (1.10) 1.10 1.10 1.10 1.10

1 4 1−( ) = $2,300 [( 1.10 ) (1.10)] = $2,300 (3.48685) = $8,019.76 .10

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 18 (If withdrawals occur at the beginning of each year, then the original balance does not remain fully intact to earn interest during year 1, requiring that the initial balance be 1.10 times as large as it would have to be if some money were not taken out of the interest-earning pool right away. The exponent in the PV of an annuity due factor is the number of cash flows, but in that factor the number of interest applications is one less than the cash flow count.) The present value of a level annuity factor therefore is

1 푛 1−( ) ( 1 + 푟 ) for cases with end-of-period cash flows (PV of a level ordinary annuity) 푟

1 푛 1−( ) and [( 1 + 푟 ) (1 + 푟)] for cases with beginning-of-period cash flows (PV of a level annuity due). 푟

b. Payments Changing by a Constant Periodic Percentage [FIL 240 Students: Just skim, not required] Now go to a situation involving a plan to take four end-of-year withdrawals, with the first being $2,300 and the subsequent amounts growing by 2% per year, such that the dollar figures will be $2,300 (1.02)0 = $2,300; $2,300 (1.02)1 = $2,346; $2,300 (1.02)2 = $2,392.92; and $2,300 (1.02)3 = $2,440.78. If a 10% average annual interest rate will be earned or paid on the account’s remaining balance, one way to solve for the total that must be available today to fund the changing withdrawal stream is with four applications of the BAMT (1 + r)n = EAMT  EAMT ÷ (1 + r)n = BAMT equation our tool box specifies for non-annuity situations (just as was done with four unrelated withdrawals as discussed in an earlier section):

1 1 $2,300.00 ÷ (1.10)1 = $2,300.00 ( ) = $2,300.00 (.909091) = $2,090.91 as BAMT that funds withdrawal #1 1.10 1 2 $2,346.00 ÷ (1.10)2 = $2,346.00 ( ) = $2,346.00 (.826446) = $1,938.84 as BAMT that funds withdrawal #2 1.10 1 3 $2,392.92 ÷ (1.10)3 = $2,392.92 ( ) = $2,392.92 (.751315) = $1,797.84 as BAMT that funds withdrawal #3 1.10 1 4 $2,440.78 ÷ (1.10)4 = $2,440.78 ( ) = $2,440.78 (.683013) = $1,667.08 as BAMT that funds withdrawal #4 1.10 for a combined total beginning balance needed of ($2,090.91 + $1,938.84 + $1,797.84 + $1,667.08) = $7,494.67. We could think of this plan as involving four separate accounts, but it could as easily occur within just one account:

Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $7,494.67 $749.47 $8,244.14 $2,300.00 $5,944.14 2 $5,944.14 $594.41 $6,538.55 $2,346.00 $4,192.55 3 $4,192.55 $419.26 $4,611.81 $2,392.92 $2,218.89 4 $2,218.89 $221.89 $2,440.78 $2,440.78 $0

But here, again, we can use a less burdensome computational approach. In the above example we have

1 1 1 2 1 3 1 4 [$2,300 (1.02)0 ( ) ] + [$2,300 (1.02)1 ( ) ] + [$2,300 (1.02)2 ( ) ] + [$2,300 (1.02)3 ( ) ] 1.10 1.10 1.10 1.10 which, by the distributive property, can be restated as

1 1 1 2 1 3 1 4 $2,300 [(1.02)0 ( ) + (1.02)1 ( ) + (1.02)2 ( ) + (1.02)3 ( ) ] 1.10 1.10 1.10 1.10 1.02 4 1−( ) = $2,300 ( 1.10 ) = $2,300 (3.258553) = $7,494.67 .10 − .02 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 19 (see sections D and J of the Appendix to this outline for more detail on the mathematics relating to the PV of a changing annuity factor). What if we plan to take four beginning-of-year withdrawals that start at $2,300 and increase by 2% annually, from an account that earns a 10% average annual interest rate on the declining balance? The total that must be available today to fund this plan with four applications of our tool box’s BAMT (1 + r)n = EAMT equation is:

1 0 $2,300.00 ÷ (1.10)0 = $2,300.00 ( ) = $2,300.00 (1.00000) = $2,300.00 as BAMT that funds withdrawal #1 1.10 1 1 $2,346.00 ÷ (1.10)1 = $2,346.00 ( ) = $2,346.00 (.909091) = $2,132.73 as BAMT that funds withdrawal #2 1.10 1 2 $2,392.92 ÷ (1.10)2 = $2,392.92 ( ) = $2,392.92 (.826446) = $1,977.62 as BAMT that funds withdrawal #3 1.10 1 3 $2,440.78 ÷ (1.10)3 = $2,440.78 ( ) = $2,440.78 (.751315) = $1,833.79 as BAMT that funds withdrawal #4 1.10 such that the combined total initial balance needed would be ($2,300.00 + $2,132.73 + $1,977.62 + $1,833.79) = $8,244.14. If everything occurred within just one account, the activity would show as:

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $8,244.14 $2,300.00 $5,944.14 $594.41 $6,538.55 2 $6,538.55 $2,346.00 $4,192.55 $419.26 $4,611.81 3 $4,611.81 $2,392.92 $2,218.89 $221.89 $2,440.78 4 $2,440.78 $2,440.78 $0 $0 $0

But here, again, we can use a less tedious approach to computing. In the above example we have

1 0 1 1 1 2 1 3 [$2,300 (1.02)0 ( ) ] + [$2,300 (1.02)1 ( ) ] + [$2,300 (1.02)2 ( ) ] + [$2,300 (1.02)3 ( ) ] 1.10 1.10 1.10 1.10 which, by the distributive property, can be restated as

1 0 1 1 1 2 1 3 $2,300 [(1.02)0 ( ) + (1.02)1 ( ) + (1.02)2 ( ) + (1.02)3 ( ) ], factored as 1.10 1.10 1.10 1.10

1 1 1 2 1 3 1 4 $2,300 [(1.02)0 ( ) + (1.02)1 ( ) + (1.02)2 ( ) + (1.02)3 ( ) ] (1.10) 1.10 1.10 1.10 1.10 1.02 4 1−( ) = $2,300 [( 1.10 ) (1.10)] = $2,300 (3.584409) = $8,244.14 .10 − .02

(If withdrawals happen at the start of each year, money comes out of the account before any interest can be applied. The initial balance therefore has to be 1.10 times as large as would be needed if the first withdrawal were to come at the end of year 1, after 10% interest had been earned.) c. Overall Review Our time value of money tool box calls for setting up any annuity problem with the equation

Payment x Factor = Total, or PMT x FAC = TOT; the regular cash flow (think of it as a payment PMT made by us or to us each year or month or other period) multiplied by the annuity factor FAC equals the lump-sum total TOT to which the stream of regular cash flows relates. This general structure works with both FV and PV of annuity cases involving equal payments or payments changing by a constant percentage from period to period (in the changing Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 20 annuity case PMT is the first payment in the series). The factor FAC is the PV (FV) of an annuity factor if the large total value TOT exists in full in the present (will not exist intact until some future date).

In summary: when cash flows are expected to be equal or change by a constant percentage each time period, we can use the PMT x FAC = TOT shortcut computational technique, based on the distributive property, instead of the longer brute force approach in either the future or present value of an annuity case. We solve an annuity due problem by multiplying the applicable FV or PV of an ordinary annuity factor by (1 + r), and then using that adjusted factor as FAC in our PMT x FAC = TOT tool box, in solving for any possible unknown (the only difference between the ordinary annuity and annuity due cases is that the number of times interest is applied differs by 1).

With this foundation now in place, we are ready to do computations for various unknown values in a more comprehensive set of annuity situations.

C. Future Value of Level Annuity Applications: Dollar Amounts Unknown Example 1: Lump Sum Future Value is Unknown We plan to deposit $10,000 per year into a bank account in each of the next four years. What will the account balance be at the end of year 4 if we can earn a 10% average annual interest rate on the account’s growing balance? (Note that the lump sum TOT will not exist until a future date, the end of year 4. Note also that if no interest were earned, the balance would grow to exactly $40,000 – although $40,000 plays no direct role in the analysis, because in this situation we will never see $40,000 in one place at one time.) If deposits are to occur at the end of each year, we can solve as

PMT x FAC = TOT (1.10)4−1 $10,000 ( ) = $10,000 (4.641) = $46,410.00 .10

Check to see that our answer is correct by working through a year-by-year progression:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Accumulated Deposit Balance 1 $0 $0 $0 $10,000.00 $10,000.00 2 $10,000.00 $1,000.00 $11,000.00 $10,000.00 $21,000.00 3 $21,000.00 $2,100.00 $23,100.00 $10,000.00 $33,100.00 4 $33,100.00 $3,310.00 $36,410.00 $10,000.00 $46,410.00

Recall that an annuity is just a group of non-annuity problems that we can treat as a team. Here we could think of making $10,000 end-of-year deposits into four separate accounts, with the first to be made on the last day of year 1 and thus earning interest three times by the end of year 4, while the fourth is to be made on the last day of year 4 and thus earning interest 0 times by the end of year 4 (if we assumed beginning- of-year deposits the corresponding numbers of interest applications would be 4 through 1). As above:

(1 + 푟)푛−1 [(1 + 푟)0 + (1 + 푟)1 + ⋯ + (1 + 푟)푛−1] = ( ) 푟 and, as might seem less confusing in the case relating to beginning-of-year deposits:

(1 + 푟)푛−1 [(1 + 푟)1 + (1 + 푟)2 + ⋯ + (1 + 푟)푛] = ( ) (1 + 푟). 푟

So when equally spaced cash flows are expected to be equal or related each period, we can use the expedient PMT x FAC = TOT computational technique, based on the distributive property, instead of the longer brute force technique. If these deposits instead were to occur at the beginning of each period, we would solve as Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 21 (1.10)4−1 $10,000 [( ) (1.10)] = $10,000 [(4.641)(1.10)] = $10,000 (5.1051) = $51,051.00 .10

The answer in the latter case is greater; we will have a higher total if we start immediately than if we waited until year’s end before making the first deposit. Again, work through the account year by year:

Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Accumulated Interest Balance 1 $0 $10,000.00 $10,000.00 $1,000.00 $11,000.00 2 $11,000.00 $10,000.00 $21,000.00 $2,100.00 $23,100.00 3 $23,100.00 $10,000.00 $33,100.00 $3,310.00 $36,410.00 4 $36,410.00 $10,000.00 $46,410.00 $4,641.00 $51,051.00

In this particular case, the annuity due answer is just the ordinary annuity answer multiplied by (1 + r); ($46,410)(1.10) = $51,051.00. But we can not always merely multiply an ordinary annuity problem’s answer by (1 + r) to find the annuity due answer; this approach works only when the unknown to solve for is the lump sum (TOT). If the unknown were the regular payment, or cash flow (PMT), we would have to divide the ordinary annuity answer by (1 + r) to compute the annuity due answer. The situation may seem confusing, but it does not have to be. If we just follow the PMT x FAC = TOT tool box plan, basic algebra causes the correct numbers to fall nicely into place as we solve for whatever the unknown may be.

Example 2: Regular Payment (Cash Flow) is Unknown We hope to have $46,410 in a bank account at the end of year 4. The growing account will earn a 10% average annual interest rate. If we wish to reach this $46,410 future balance with four equal annual deposits plus the accompanying interest buildup, how much money should we put into the account each year? (Again, the lump sum TOT will not exist until a future date, the end of year 4, so we have a future value of an annuity problem.)

We can again solve as PMT x FAC = TOT (so that TOT ÷ FAC = PMT)

If deposits are to occur at the end of each year, each one should be

(1.10)4−1 PMT ( ) = PMT (4.641000) = $46,410.00 .10

so PMT = $46,410 ÷ 4.641 = $10,000.00

Check by working through the plan year by year:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Accumulated Deposit Balance 1 $0 $0 $0 $10,000.00 $10,000.00 2 $10,000.00 $1,000.00 $11,000.00 $10,000.00 $21,000.00 3 $21,000.00 $2,100.00 $23,100.00 $10,000.00 $33,100.00 4 $33,100.00 $3,310.00 $36,410.00 $10,000.00 $46,410.00

If deposits are to be made at the beginning of each year, we can solve as

(1.10)4−1 PMT [( ) (1.10)] = PMT [(4.641)(1.10)] = PMT (5.1051) = $46,410 .10

so PMT = $46,410 ÷ [(4.641)(1.10)] = $46,410 ÷ 5.1051 = $9,090.91

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 22 (Is it clear how this process has the effect of dividing our $10,000 ordinary annuity answer by 1.10?) The answer here is smaller; if the first deposit is made right away, we can put less in each time and still reach the $46,410 future target. Again, check by working through the account’s year-by-year standing:

Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Accumulated Interest Balance 1 $0 $9,090.91 $ 9,090.91 $ 909.09 $10,000.00 2 $10,000.00 $9,090.91 $19,090.91 $1,909.09 $21,000.00 3 $21,000.00 $9,090.91 $30,090.91 $3,009.09 $33,100.00 4 $33,100.00 $9,090.91 $42,190.91 $4,219.09 $46,410.00

D. Present Value of Level Annuity Applications: Dollar Amounts Unknown Example 1: Lump Sum Present Value is Unknown We want to be able to withdraw $10,000 per year from our bank account in each of the next four years. If we can expect to earn a 10% average interest rate per year on the account’s remaining balance, how much must we deposit today to endow the account? (Here we have to solve for an unknown lump sum TOT that must exist intact today, so we have a present value of an annuity problem. Note also that if we did not earn any interest on the account’s declining balance, we would have to start out with $40,000 to fund the plan – but again $40,000 plays no direct role in the analysis, since $40,000 will never exist in one place at one time in this example.) If withdrawals are to occur at the end of each year, we can solve as

PMT x FAC = TOT 1 4 1−( ) $10,000 ( 1.10 ) = $10,000 (3.169865) = $31,698.65 .10

Double-check by working through the account year by year:

Beginning Plus 10% Total Minus End of Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $31,698.65 $3,169.87 $34,868.52 $10,000.00 $24,868.52 2 $24,868.52 $2,486.85 $27,355.37 $10,000.00 $17,355.37 3 $17,355.37 $1,735.54 $19,090.91 $10,000.00 $ 9,090.91 4 $ 9,090.91 $ 909.09 $10,000.00 $10,000.00 $0

As noted before, an annuity is just a group of non-annuity problems that we can treat as a team. Here we could think of making $10,000 year-end withdrawals from four separate accounts, with the first to be taken the last day of year 1 (account #1 earns interest once before the withdrawal) and the fourth to occur on the last day of year 4 (account #4 earns interest four times before year 4’s withdrawal); the exponents therefore are 1 – 4 (if we assumed beginning-of-year withdrawals the corresponding number of interest applications would be 0 through 3). As shown earlier: 푛 푛 1 1 1 1 2 1 3 1 1−( ) [( ) + ( ) + ( ) + ⋯ + ( ) ] = ( 1 + 푟 ) 1 + 푟 1 + 푟 1 + 푟 1 + 푟 푟 and, as might seem a bit more confusing in the case relating to beginning-of-year withdrawals:

1 푛 1 0 1 1 1 2 1 푛−1 1−( ) [( ) + ( ) + ( ) + ⋯ + ( ) ] = [( 1 + 푟 ) (1 + 푟)] 1 + 푟 1 + 푟 1 + 푟 1 + 푟 푟

So if withdrawals were to occur at the beginning of each year, our tool box would tell us to solve as

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 23 1 4 1−( ) $10,000 [( 1.10 ) (1.10)] = $10,000 [(3.169865) (1.10)] = $10,000 (3.486852) = $34,868.52 .10

The answer is greater; we have to start out with a higher bank balance in order to start making $10,000 withdrawals immediately than if we could wait and earn interest for a year before taking the first payment. [In this situation, on the first day of Year 1 we endow the account with $34,868.52 and then immediately take $10,000 out; the part of the account that funds the first withdrawal earns no interest.] Again, check to see that the answer is correct by examining the account year by year:

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $34,868.52 $10,000.00 $24,868.52 $2,486.85 $27,355.37 2 $27,355.37 $10,000.00 $17,355.37 $1,735.54 $19,090,91 3 $19,090.91 $10,000.00 $ 9,090.91 $ 909.09 $10,000.00 4 $10,000.00 $10,000.00 $0 $0 $0

In this case, the annuity due answer is simply the ordinary annuity answer multiplied by (1 + r); ($31,698.65)(1.10) = $34,868.52. But, as in the future value of annuity example discussed earlier, multiplying the ordinary annuity answer by (1 + r) works only when the unknown is the lump sum (TOT). If the unknown were the regular payment, or cash flow (PMT), we would have to divide the ordinary annuity answer by (1 + r) to compute the annuity due answer. Still, as noted above, there is no need for confusion. Just work with the PMT x FAC = TOT equation our tool box specifies for annuities, and the correct numbers fall nicely into place as we solve for any possible unknown.

Example 2: Regular Payment (Cash Flow) is Unknown Today we deposit $31,698.65 into a bank account. The balance that remains in the account from year to year earns a 10% average annual rate of return. To deplete the balance to zero with four equal annual withdrawals, how much should we take out of the account each year? (Or we could ask: if we borrow $31,698.65 and the loan terms call for a 10% annual interest rate and equal year-end payments to be made over four years, how much will each payment be? Either way the TOT of $31,698.65 is known rather than an unknown to solve for, and since that large total exists intact today we have a PV of annuity problem.)

We can solve as PMT x FAC = TOT (so that TOT ÷ FAC = PMT)

If withdrawals (or loan repayments) are to occur at the end of each year, each will be

1 4 1−( ) 1.10 PMT ( ) = PMT (3.169865) = $31,698.65 .10

so PMT = $31,698.65 ÷ 3.169865 = $10,000.00

Check by working through the account’s year-by-year progress:

Beginning Add 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $31,698.65 $3,169.87 $34,868.52 $10,000.00 $24,868.52 2 $24,868.52 $2,486.85 $27,355.37 $10,000.00 $17,355.37 3 $17,355.37 $1,735.54 $19,090.91 $10,000.00 $ 9,090.91 4 $ 9,090.91 $ 909.09 $10,000.00 $10,000.00 $0

If withdrawals (or repayments) were to occur at the beginning of each year, we would solve as

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 24 1 4 1−( ) 1.10 PMT [( ) (1.10)] = PMT [(3.169865)(1.10)] = PMT (3.486852) = $31,698.65 .10

so PMT = $31,698.65 ÷ [(3.169865)(1.10)] = $31,698.65 ÷ 3.486852 = $9,090.91

(Is it clear how this process has the effect of dividing our $10,000 ordinary annuity answer by 1.10?) The answer here is smaller; if the first withdrawal occurs right away, we can not take out as much each time (or a loan could be settled with four smaller beginning-of-year payments). Again, double-check:

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $31,698.65 $9,090.91 $22,607.74 $2,260.77 $24,868.51 2 $24,868.51 $9,090.91 $15,777.60 $1,577.76 $17,355.36 3 $17,355.36 $9,090.91 $ 8,264.45 $ 826.46 $ 9,090.91 4 $ 9,090.91 $9,090.91 $0 $0 $0

E. Applications Involving Rates of Return and Numbers of Periods The distributive property, with the accompanying PMT x FAC = TOT equation from our tool box, is useful in all annuity situations, whether present value or future value (recall that we look at when TOT exists intact to identify the problem correctly as present or future value of annuity), and no matter what value is the unknown to solve for. There can be only one unknown quantity if we are dealing with just one equation. In the above annuity examples our unknowns were either PMT or TOT amounts.

What if we know both the regular cash flow PMT and the lump sum TOT to which the cash flows equate in time value-adjusted terms? Then FAC contains the unknown. But even within FAC there can be only one unknown if the problem is to have a unique solution. So either n is known and r is the unknown to solve for (a rate of return problem), or else r is known and we must solve for n, the number of periods needed for the cash flows PMT to equate, in time value-adjusted terms, to the lump sum TOT.

Example 1: Finding Rate of Return for Future Value of Annuity We plan to deposit $1,000 in the bank at the end of each year, and hope that by the end of year 10 our account balance will have grown to $12,500 (so the $12,500 TOT is a future amount, and this is a future value of an ordinary annuity situation). What average annual interest rate will the bank have to pay if we are to reach the $12,500 target? PMT x FAC = TOT (1 + 푟)10−1 $1,000 ( ) = $12,500 푟

It turns out that solving for r here requires trial and error iterations (the is that we have both r1 and r10 in the same equation; in an annuity case there is no way to get r on one side of the equals sign and only terms that do not include r on the other side of the equals sign). If we tried different numbers (we will not do trial and error examples on our exams), we would find that 4.8669% is the correct answer.

(1.048669)10−1 Check: $1,000 ( ) = $1,000 (12.500000) = $12,500.00 ✓ .048669

Example 2: Finding Rate of Return for Present Value of Annuity We deposit $28,750 in the bank today. What average annual interest rate will we have to earn on the account’s remaining balance if we want to be able to withdraw $3,000 at the beginning of each of the next 12 years? (The $28,750 TOT is an amount that exists intact in the present, so we have a present value of an annuity due case.)

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 25 PMT x FAC = TOT 1 12 1−( ) $3,000 [( 1 + 푟 ) (1 + 푟)] = $28,750 푟

Again, solving for r requires trial and error iterations (both r1 and r12 are in the equation). Trying different rate possibilities (again, not required on our exams), we would find that 4.3807% is the correct answer.

1 12 1−( ) Check: $3,000 [( 1.043807 ) (1.043807)] = $3,000 (9.583333) = $28,750.00 ✓ .043807

Example 3: Finding Number of Time Periods for Future Value of Annuity We deposit $2,000 in a bank account at the start of each year. If the bank pays a 5% average annual interest rate on our growing balance, how many years will it take for the account to grow to $22,000? (The $22,000 TOT is a future amount, so here we are dealing with a FV of annuity due situation.)

PMT x FAC = TOT (1.05)푛−1 $2,000 [( ) (1.05)] = $22,000 .05

(1.05)푛−1 [( ) (1.05) ] = 11 .05

(1.05)푛−1 ( ) = 10.476190 [we divided each side by 1.05] .05

(1.05)n – 1 = .523810 [we multiplied each side by .05] (1.05)n = 1.523810 [we added 1 to each side]

As in the single dollar amount case, we could solve for n with trial and error, trying different n values until the left-hand side equals about 1.523810. But, again, logarithms offer the more direct approach:

ln [(1.05)n] = ln 1.523810 n (ln 1.05) = ln 1.523810

From a table of natural logarithms, accessed through the LN function key on a financial or scientific calculator, we see that ln 1.05 = .048790 and ln 1.523810 = .421213, so we can solve as

n (.048790) = .421213 n = .421213 ÷ .048790 = 8.633164

It takes about 8½ years for the account to grow to $22,000 with $2,000 deposits and 5% annual interest.

(1.05)8.633164−1 Check: $2,000 [( ) (1.05)] = $2,000 (11.0000) = $22,000.00 ✓ .05

Example 4: Finding Number of Time Periods for Present Value of Annuity A retired person wants to withdraw $12,000 at the end of each year from her mutual fund account to help pay living expenses. She has $100,000 today (the lump sum TOT of $100,000 exists intact in the present, so we have a PV of an ordinary annuity situation). If she expects to earn a 9% average annual return on the account’s declining balance, for how many years can she take $12,000 year-end withdrawals?

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 26 PMT x FAC = TOT 1 푛 1−( ) $12,000 ( 1.09 ) = $100,000 .09

1 푛 1−( ) 1 푛 1.09 = 8.33333  1 – ( ) = .75 [we multiplied each side by .09] .09 1.09

1 푛 1 푛 – ( ) = – .25  ( ) = .25 [we subtracted 1 from each side and multiplied by – 1] 1.09 1.09

(.917431)n = .25 [we restated 1 ÷ 1.09 as the decimal .917431]

Again, we could use trial and error, though we can more easily complete the task with logarithms:

ln [(.917431)n] = ln .25 n (ln .917431) = ln .25 n (– .086178) = – 1.386294 n = 1.386294 ÷ .086178 = 16.086463

(since e0 = 1 we would have to take e to a power less than 0 to get a value less than 1, so the logarithm of a fraction, like .25, is negative). Our retiree could continue to take out $12,000 yearly for about 16 years.

1 16.086463 1−( ) Check: $12,000 ( 1.09 ) = $12,000 (8.333333) = $100,000.00 ✓ .09

Annuity recap: TVM analysis is about structure, not about some unknown being solved for. Think of someone who saves $7,500 at the end of each year for 33 years in an account whose growing balance earns a 5.2% average annual return, and ends up with $624,151.50 when he retires. The structure we would use to analyze, no matter which of the four possible unknowns we are solving for (in one equation there can be only one unknown if we are to find a unique solution, so the other three related values would have to be given), is PMT x FAC = TOT (1.052)33−1 $7,500.00 ( ) = $624,151.50 .052

We would multiply to find the $624,151.50 TOT, divide to find the $7,500.00 PMT, take a ratio of logarithms to find the n of 33 years needed, or use trial and error to find the 4.6% average annual r, but any of those steps is just applying algebra once the structure is in place.

Think of a business that repays a $215,000 loan with a 3.9% annual interest rate by paying $15,680.31 at the end of each year for 20 years. The structure we would use to analyze, no matter which of the four possible unknowns we are solving for, is 1 20 1−( ) $15,680.31 ( 1.039 ) = $215,000.00 .039

We would multiply to find the $215,000.00 TOT, divide to find the $15,680.31 PMT, take a ratio of logarithms to find the n of 20 years needed, or use trial and error to find the 3.9% average annual r, but any of those steps is just applying algebra once the structure is in place.

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 27 F. So What Is the Deal With Financial Calculators? Financial calculators are designed to do, quickly, the kinds of computations worked “by hand” above. Students who try to learn the basics by hitting financial calculator buttons frequently fail to grasp the important underlying ideas, but with our tool box framework for setting up the problems we can easily apply time value of money concepts with any type of computing equipment (such as setting up our own cell formulas in a spreadsheet, or using the financial function keys on a financial calculator). Consider a case in which we borrow $25,000 to be repaid over 10 years, with a 7.5% interest rate applied each year to any unpaid principal. To compute the annual payment using a standard financial calculator, punch in 25000 and hit the “PV” key (since $25,000 is our lump-sum present value TOT in this present value of a level ordinary annuity problem), punch in 0 and hit the “FV” key (there is no large future amount, and we do not want the calculator trying to include a future amount remaining in the FV register from an earlier computation), punch in 10 and hit the “N” key (10 is the number of periods), punch in 7.5 and hit the “i” or “I/Y” key (7.5% is the annual interest rate), and then hit “Comp” for compute and the “PMT” key to compute the payment. The screen should show – 3,642.15 (negative because that amount comes out of our pocket each year); $3,642.15 is the annual end-of-year payment. A complication of using a financial calculator is that one of the dollar values must have a negative sign, so if we are solving for an unknown r or n and do not enter a negative sign for one of the dollar amounts a confusing “error” message results.

Or if we save $5,000 at the end of each year and hope to have $1,000,000 when we retire in 45 years, what average annual rate of return must we earn? The beauty of using a financial calculator for a problem like this one is that the calculator is programmed to do trial and error iterations quickly. Punch in 5000, then hit the “+/-” key to make it a negative 5000 (since that amount will come out of our pocket each year), and hit the “PMT” key. Then punch in 1000000 and hit the “FV” key (the TOT amount we hope to have at a future date in this FV of a level ordinary annuity problem), and punch in 0 and hit the “PV” key to make sure no unintended value is hiding in the PV register. Punch in 45 and hit the “N” key, and then hit “Comp” and the “i” or “I/Y” key to compute the rate. The screen goes blank momentarily as the calculator does trial and error; then it should show 5.787172 (the average annual percentage we would have to earn for a series of forty-five $5,000 year-end deposits to grow to $1,000,000). The calculator should have a “BGN” mode for annuity due cases, when cash flows occur at the beginning of each period.

In FIL 240 we use a formula-based “nuts-and-bolts” approach to build understanding of time value logic and computations. But those who will take more course work are encouraged to get, and practice using, financial calculators. The “Handout on Using a Financial Calculator” with Topic 4 material on the course web site shows how several Topic 4, 6, and 10 problems are solved with our tool box equations (“main roads”), and then with the popular Texas Instruments BAII Plus calculator (shortcut approach).

IV. SPECIAL ANNUITY SITUATIONS A. Present Value of a Level Perpetuity What if a series of equal annual cash flows is expected to be received not for 20 years or even 100 years, but forever? This type of perpetual annuity – known as a perpetuity – is easy to analyze. The present value of a level ordinary perpetuity is simply the expected periodic cash flow divided by the required periodic rate of return (the factor FAC for use in our PMT x FAC = TOT breakdown simplifies to 1/r, because the fraction term in the PV of a level annuity factor’s numerator approaches zero as the time periods approach infinity). For example, if a donor wants to give the university enough money to fund a $10,000 scholarship to be awarded at the end of each year indefinitely, and if the university foundation office can earn a 10% average annual return on scholarship endowments, then the amount the benefactor should contribute is PMT x FAC = TOT 1 ∞ 1−( ) 1 − 0 1 $10,000 $10,000 ( 1.10 ) = $10,000 ( ) = $10,000 ( ) = or $10,000 (10) = $100,000 .10 .10 .10 .10

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 28 (far more than the $31,689.65 needed to fund only four $10,000 withdrawals). A time line would show:

1-01-Yr 1 __ 12-31-Yr 1/1-01-Yr 2 … 12-31-Yr 9/1-01-Yr 10 … 12-31-Yr 98/1-01-Yr 99 … 12-31-Yr ∞ Plan Starts Withdraw $10,000 Withdraw $10,000 Withdraw $10,000 Withdraw $10,000 With $100,000 Plan Never Ends

Why is the perpetuity’s value so easy to compute? Consider that if the donor gives $100,000 today, and the university earns a 10% return on the $100,000 every year, then at the end of each year the $10,000 interest earned will be paid out as a scholarship – and the original $100,000 will remain to earn $10,000 again, every year, forever. Now turn the analysis around: if we want to be able to pay out $10,000 at the end of each year forever, and the situation’s risk is such that we would expect those $10,000 payments to represent a 10% annual rate of return, then we need an endowment of $100,000 today to establish the plan. In a true perpetuity we leave the principal intact, collecting only interest each period. If a perpetual stream of withdrawals were to begin at the start of the first period we would – no surprise – multiply the ordinary factor (1/r) by (1 + r) before computing; the product yields a factor of [(1/r) + 1]: if the campus wanted to award the $10,000 scholarship at the start of each year and still keep the needed $100,000 principal intact, the donor’s original endowment would have to be an amount greater by exactly one period’s cash flow. Based on values in the example above, we compute

1 1 $10,000 $10,000 [( ) (1.10)] = $10,000 ( + 1) = ( + $10,000) = $100,000 + $10,000 = $110,000: .10 .10 .10 start with $110,000; pay out $10,000 immediately; and have $100,000 left to fund later years’ awards. Note what happens year by year, indefinitely, with end-of-year and beginning-of-year payments, respectively: Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $100,000.00 $10,000.00 $110,000.00 $10,000.00 $100,000.00 2 $100,000.00 $10,000.00 $110,000.00 $10,000.00 $100,000.00      175 $100,000.00 $10,000.00 $110,000.00 $10,000.00 $100,000.00

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $110,000.00 $10,000.00 $100,000.00 $10,000.00 $110,000.00 2 $110,000.00 $10,000.00 $100,000.00 $10,000.00 $110,000.00      175 $110,000.00 $10,000.00 $100,000.00 $10,000.00 $110,000.00

(The future value of a perpetuity is essentially a meaningless concept, because the ultimate value of any stream that will continue forever must be infinite. Finding an unknown TOT would result in an infinite value, while finding an unknown PMT would require dividing infinity – an undefined operation – by 1/r.)

One practical use of the perpetuity idea, with its simple attendant computations, is to estimate the value of a long, but not truly perpetual, annuity. Let’s say we have to quickly understand the value of the right to collect an expected $100 per year at the end of each of the next 75 years, if our required average annual rate of return, based on applicable risks, is 8%. This problem involves a series of level payments equally spaced in time, so we can start with the annuity equation from our tool box structure:

PMT x FAC = TOT

If we treat this long-term annuity as though it were a perpetuity, we quickly come up with an approximate answer of 1 $100 $100 ( ) = = $100 (12.500000) = $1,250.00 .08 .08 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 29 This quick step turns out to produce a fairly good estimate; the truly correct answer is just a bit smaller:

1 75 1−( ) $100 ( 1.08 ) = $100 (12.461084) = $1,246.11 .08

The perpetuity model sometimes is used as a practical approximation in estimating the values of long- term investments, such as common stock or real estate. Why does this estimation technique work so well? When we use the perpetuity approximation, we are including the present values of cash flows that, if they were actually to be received, would not come to us until far into the future. Because the present value of anything to be received in the far distant future is essentially zero, we can include the far distant future cash flows (use a perpetuity approximation for a long-term but finite annuity) or exclude them (compute the actual value) without affecting the total computed present value in a substantial way.

Note also that in a perpetuity case we could instead have been solving algebraically for PMT (the amount we could withdraw periodically forever from a given initial balance) or r (the rate of return that a given payment stream represents relative to a given initial balance). There would be no reason to solve for the number of time periods n, of course; in a perpetuity, the number of time periods is always infinite.

Finally, we can use the distributive property to compute the present value of a perpetuity whose payments are expected to change by a constant percentage from period to period, if we expect the rate of growth or change per period to be less than the expected periodic rate of return (a necessary condition over a long time period, since growth is a sub-component of the rate of return). The amount needed today to fund an infinite stream of withdrawals that start at $10,000 and grow by 2% each year forever is computed as

PMT x FAC = TOT 1.02 ∞ 1−( ) 1 − 0 1 $10,000 $10,000 $10,000 ( 1.10 ) = $10,000 ( ) = $10,000 ( ) = ( ) = = $125,000 .10 − .02 .10 − .02 .10 − .02 .10 − .02 .08 for payments taking place at the end of each year; for beginning-of-year payments the initial total would have to be 1.02 ∞ 1−( ) 1 $10,000 $10,000 [( 1.10 ) (1.10)] = $10,000 [( ) (1.10)] = [( ) (1.10)] = $137,500 .10 − .02 .10 − .02 .08

We will use this changing perpetuity computation with traditional common stock models in Topic 13. (See sections D and J of the Appendix to this outline for more detail on the mathematics of the changing perpetuity situation.)

B. Present Value of a Deferred Level Annuity Example 1: Finite Annuities. In each of the earlier annuity examples, we have assumed that the cash flows (to which the lump sum equates in time value-adjusted terms) would begin at the start or end of the first period (a situation for which we sometimes use the term “immediate” annuity). But there could be a case in which we expect to begin collecting money in equal or related annual (or other periodic) amounts, but not until some number of periods has passed. For example, assume that we are offered the right to collect $10,000 per year for four years, but that the first $10,000 will not be available until the end of year 4 (so we will receive money at the ends of years 4, 5, 6, and 7 rather than the ends of years 1, 2, 3, and 4).

1-01-Yr 1 __ 12-31-Yr 1/1-01-Yr 2 … 12-31-Yr 4/1-01-Yr 5 … 12-31-Yr 6/1-01-Yr 7 ___ 12-31-Yr 7 Plan Starts Withdraw $0 Withdraw $10,000 Withdraw $10,000 Withdraw $10,000 With ??? Plan Ends

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 30 We call this situation a “deferred level annuity.” What would we pay for the right to collect these amounts if our required average annual rate of return is 10%?

• Method 1: Brute Force Just sum the present values of all the expected cash inflows:

1 1 $0 ÷ (1.10)1 = $ 0 ( ) = $ 0 (.909091) = $ 0 to fund the year 1 withdrawal 1.10 1 2 $0 ÷ (1.10)2 = $ 0 ( ) = $ 0 (.826446) = $ 0 to fund the year 2 withdrawal 1.10 1 3 $0 ÷ (1.10)3 = $ 0 ( ) = $ 0 (.751315) = $ 0 to fund the year 3 withdrawal 1.10 1 4 $10,000 ÷ (1.10)4 = $10,000 ( ) = $10,000 (.683013) = $6,830.13 to fund the year 4 withdrawal 1.10 1 5 $10,000 ÷ (1.10)5 = $10,000 ( ) = $10,000 (.620921) = $6,209.21 to fund the year 5 withdrawal 1.10 1 6 $10,000 ÷ (1.10)6 = $10,000 ( ) = $10,000 (.564474) = $5,644.74 to fund the year 6 withdrawal 1.10 1 7 $10,000 ÷ (1.10)7 = $10,000 ( ) = $10,000 (.513158) = $5,131.58 to fund the year 7 withdrawal 1.10 or a combined ($6,830.13 + $6,209.21 + $5,644.74 + $5,131.58) = $23,815.66 in total. Note that the $23,815.66 total is less than the $31,698.65 present value of an immediate level ordinary annuity of $10,000 per year for four years beginning at the end of year 1. We would pay more to collect a series of four $10,000 annual payments if we could get the first $10,000 at the end of the current year than if we had to wait until three years had passed before reaching the first year with a $10,000 year-end inflow.

• Method 2: Compute a Factor The brute force method actually is not too cumbersome to use for a short stream of cash flows. But if we expected to collect $10,000 per year for fifty years after some deferral period, we would seek a more efficient computing technique. If we can find the right factor, then we can use the distributive property: multiply $10,000 by the correct factor. This factor is the sum of the relevant individual PV of $1 factors: (.683013 + .620921 + .564474 + .513158) = 2.381566, such that $10,000 (2.381566) = $23,815.66. So our task becomes finding a quick way to compute the 2.381566 deferred level ordinary annuity factor.

Recall that the present value of an annuity factor is the sum of the present value of $1 factors for the included number of time periods. For example, the 7-period present value of a level ordinary annuity factor at a 10% periodic discount rate is

1 7 1−( ) ( 1.10 ) = 4.868419 , which equals the sum of .10

1 1 1 2 1 3 1 4 1 5 1 6 1 7 ( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) 1.10 1.10 1.10 1.10 1.10 1.10 1.10

= .909091 + .826446 + .751315 + .683013 + .620921 + .564474 + .513158 = 4.868419

Since the present value of a level ordinary annuity factor for seven years is the sum of the PV of $1 factors for years 1 through 7, we could find the sum of the year 4 through 7 present value of $1 factors by subtracting the year 1, 2, and 3 PV of $1 factors from the 7-year PV of a level ordinary annuity factor.

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 31 In fact, since the sum of the year 1, 2, and 3 present value of $1 factors is the present value of a 3-year level ordinary annuity factor, we can compute the deferred level ordinary annuity factor as:

Present value of a level ordinary annuity factor for number of periods until the last cash flow is received, Minus present value of a level ordinary annuity factor for number of periods before cash flows begin.

In this case, with the recipient getting payments until the end of year 7 after getting no payments in years 1 through 3, we have the 7-year present value of a level ordinary annuity factor minus the 3-year present value of a level ordinary annuity factor, or

1 7 1 3 1−( ) 1−( ) [( 1.10 ) – ( 1.10 )] = 4.868419 – 2.486852 = 2.381567 .10 .10

We started with the sum of all seven of these … 1 1 1 2 1 3 1 4 1 5 1 6 1 7 ( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) 1.10 1.10 1.10 1.10 1.10 1.10 1.10 ______then got rid of these three … leaving this difference.

We multiply this computed factor FAC by $10,000 in our PMT x FAC = TOT breakdown to get the $23,815.67 value (1¢ rounding difference) we computed using the brute force method above:

PMT x FAC = TOT $10,000 (2.381567) = $23,815.67

• Method 3: Discount the Annuity We could consider, alternatively, that the deferred annuity is still a 4-year annuity, albeit one that does not start until three time periods have passed. So we start by computing the present value of a 4-year level ordinary annuity with our tool box: PMT x FAC = TOT 1 4 1−( ) $10,000 ( 1.10 ) = $10,000 (3.169865) = $31,698.65 .10

That is what it will be worth at the start of year 4 (= end of year 3) to have the right to collect year-end $10,000 payments in the subsequent four years. But what is that series worth today? Like anything else we will not receive until three years have passed, we find its value by discounting for three periods:

1 3 $31,698.65 ÷ (1.10)3 = $31,698.65 ( ) = $31,698.65 (.751315) = $23,815.67 1.10

Note that the value of the right to collect $10,000 at the end of each of years 12 through 31 (a 20-year annuity deferred until after 11 years have passed), if the required rate of return averages 7% per year, is

1 31 1 11 1−( ) 1−( ) $10,000 [( 1.07 ) – ( 1.07 )] = $50,331.40 or .07 .07 1 20 1−( ) 1 11 $10,000 [( 1.07 ) ( ) ] = $50,331.40 .07 1.07

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 32 Which shortcut method is better? The two are equally valid to use, so it seems most sensible to go with whichever is easier to understand. Many students prefer the “compute a factor” method because of its direct link to the straightforward brute force logic. But the “discount the annuity” method (which some prefer because of its focus on the period over which cash flows actually occur) tends to be more versatile; it can be more easily adapted for use in applications more advanced than we usually see in our coverage.

It should not be surprising that, for a deferred level annuity due (with beginning-of-period cash flows), we multiply the factor appropriate to end-of-period cash flows by (1 + r). For example, the value of the right to collect $10,000 at the beginning of each of years 4 – 7, if the required average annual return is 10%, is

1 7 1 3 1−( ) 1−( ) $10,000 [(( 1.10 ) – ( 1.10 )) (1.10)] = $26,197.24 or .10 .10

1 4 1−( ) 1 3 $10,000 [( 1.10 ) (1.10) ( ) ] = $26,197.24 .10 1.10

A variation on the deferred annuity situation is the multi-stage, or sequential, annuity. The value of the right to make year-end collections of $6,000 per year for three years, followed by $10,000 per year for four years, and then followed by $12,000 per year for nine years at a 10% average annual discount rate is computed as the value of a non-deferred or immediate annuity, plus the values of two deferred annuities (deferred for 3 years and for 3 + 4 = 7 years, respectively):

1 3 1 4 1 9 1−( ) 1−( ) 1 3 1−( ) 1 7 $6,000 ( 1.10 ) + $10,000 [( 1.10 ) ( ) ] + $12,000 [( 1.10 ) ( ) ] .10 .10 1.10 .10 1.10

= $6,000 (2.486852) + $10,000 [(3.169865)(.751315)] + $12,000 [(5.759024)(.513158)] = $14,921.11 + $23,815.67 + $35,463.48 = $74,200.26

We would have to deposit just in excess of $74,200 today to be able to withdraw $166,000 over time. Note that the $10,000/year to be received in years 4 – 7 is the same deferred annuity illustrated above.

Example 2: Perpetuities. A perpetuity can be deferred just like a finite annuity can. Perhaps a donor wants to establish a $10,000 annual scholarship that will first be paid out in year 9 (the 25th anniversary of the donor’s graduation), and then will continue indefinitely. How much should the donor give today if the university foundation can expect to earn a 10% average annual rate of return on the endowment? Again we can “Compute a Factor” (think of it as an annuity lasting forever minus one lasting 8 years):

1 ∞ 1 8 1 8 1−( ) 1−( ) 1 1−( ) $10,000 [( 1.10 ) – ( 1.10 )] = $10,000 [( ) – ( 1.10 )] .10 .10 .10 .10

= $10,000 (10 – 5.334926) = $10,000 (4.665074) = $46,650.74 or use the “Discount the Annuity” method, which perhaps is less complicated in this situation (bringing the eventual infinite cash flow series back 8 periods to a present value): 1 1 8 $10,000 [( ) ( ) ] = $10,000 [(10)(.466507)] = $10,000 (4.665074) = $46,650.74 .10 1.10

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 33 if the first $10,000 were scheduled to be paid at the end of year 9, with cash flows timed as follows:

1-01-Yr 1 __ 12-31-Yr 1/1-01-Yr 2 … 12-31-Yr 8/1-01-Yr 9 ___ 12-31-Yr 9/1-01-Yr 10 … 12-31-Yr ∞ Plan Starts Withdraw $0 Withdraw $0 Withdraw $10,000 Withdraw $10,000 With $46,651 Plan Never Ends

If the first $10,000 instead were scheduled to be made at the beginning of year 9, with cash flows occurring as:

1-01-Yr 1 __ 12-31-Yr 1/1-01-Yr 2 … 12-31-Yr 7/1-01-Yr 8 ___ 12-31-Yr 8/1-01-Yr 9 … 1-01-Yr ∞ Plan Starts Withdraw $0 Withdraw $0 Withdraw $10,000 Withdraw $10,000 With $51,316 Plan Never Ends

1 ∞ 1 8 1−( ) 1−( ) then it would be $10,000 [(( 1.10 ) – ( 1.10 )) (1.10)] = $10,000 [(4.665074) (1.10)] .10 .10

= $10,000 (5.131581) = $51,315.81

1 1 8 or $10,000 [( ) (1.10) ( ) ] = $10,000 [(11)(.466507)] = $10,000 (5.131581) = $51,315.81 .10 1.10

It should not be surprising that these two computed deferred level perpetuity values are less than the $100,000 and $110,000 that characterized the immediate level perpetuity cases seen earlier; the generous grad would have to give more to fund an unending $10,000 periodic payment stream beginning this year than for a similar stream that will not begin until eight years have passed. Viewed slightly differently the donor could give just $46,650.74 today and let it earn a 10% average annual rate of return for eight years so it would grow to $46,650.74 (1.10)8 = $100,000 by the end of year 8/start of year 9, after which the university could withdraw $10,000 in interest to provide a scholarship at the end of every year thereafter. Note also that in this example the difference in the deferred ordinary perpetuity and deferred perpetuity due answers is not the $10,000 regular payment, but rather the regular payment multiplied by [1/(1 + r)n], seen here as $10,000 [1/(1.10)8] = $4,665.07 (the difference between $46,650.74 and $51,315.81).

Finally, what if an expected perpetual stream of cash flows that change from period to period by a constant percentage is not expected to begin until some number of periods has passed? The amount needed today to fund a series of scholarships that begin at $10,000 in year 15 (after 14 years have passed) and grows by 2% each successive year forever is computed as

PMT x FAC = TOT 1.02 ∞ 1−( ) 1 14 1 1 14 $10,000 1 14 $10,000 [( 1.10 ) ( ) ] = $10,000 [( ) ( ) ] = [( ) ( ) ] = $32,916.41 .10 − .02 1.10 .10 − .02 1.10 .10 − .02 1.10 if payments occur at the end of each year; for start-of-year scholarships the endowment would have to be a larger 1 1 14 $10,000 1 14 $10,000 [( ) ( ) (1.10)] = [( ) ( ) (1.10)] = $36,208.05 .10 − .02 1.10 .10 − .02 1.10

These amounts are less than the $125,000 or $137,500 needed to fund a series of payments that begin at $10,000 in the current period and grow by 2% per year forever; here many years of interest will build up in the account, moving the balance to $125,000 or $137,500 by the end of year 14. We will use this changing deferred perpetuity computation with traditional common models in Topic 13.

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 34 See sections D and J of the Appendix to this outline for more detail on the mathematics of the changing perpetuity situation.

C. Future Value of a “Truncated” Level Annuity The future value of a deferred annuity is largely a meaningless idea; if we simply wait several periods before making deposits, the account balance is just $0 until we begin. But what about the opposite situation: we want to know the accumulated value a series of deposits will reach by some future date, but the deposits (which will begin in the current period) will terminate before the target date? For example, how much will be in an account by the end of year 7 if we deposit $10,000 at the ends of each of years 1 through 4, and then just let the accumulated balance grow for three more years, with all accumulations earning a 10% average annual rate of return? A time line would show a progression something like:

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 ___ 12-31-Yr 2/1-01-Yr 3 … 12-31-Yr 4/1-01-Yr 5 … 12-31-Yr 7 Plan Starts First $10,000 Deposit Second $10,000 Deposit Fourth $10,000 Deposit Plan Ends

As with deferred annuities, there are three ways to approach this problem, which we might call a “truncated” (cut short) annuity.

• Method 1: Brute Force Just sum the future values of all the expected cash flows:

$10,000 (1.10)6 = $10,000 (1.771561) = $17,715.61 for the year 1 deposit $10,000 (1.10)5 = $10,000 (1.610510) = $16,105.10 for the year 2 deposit $10,000 (1.10)4 = $10,000 (1.464100) = $14,641.00 for the year 3 deposit $10,000 (1.10)3 = $10,000 (1.331000) = $13,310.00 for the year 4 deposit $ 0 (1.10)2 = $ 0 (1.210000) = $ 0 for the year 5 deposit $ 0 (1.10)1 = $ 0 (1.100000) = $ 0 for the year 6 deposit $ 0 (1.10)0 = $ 0 (1.000000) = $ 0 for the year 7 deposit or ($17,715.61 + $16,105.10 + $14,641.00 + $13,310.00) = $61,771.71 in total. This $61,771.71 total is much greater than the $46,410 future value of an ordinary annuity of $10,000 per year for four years terminating at the end of year 4. The reason is that we are letting the money remain in the account to earn a 10% average annual rate of return for three additional periods after the deposits have stopped.

• Method 2: Compute a Factor Brute force is not too cumbersome to use for a payment series of a mere four years. But if we expected to leave the accumulated total from a long series of deposits to earn additional returns, we would look for a more efficient computational technique. Note that if we can find the right factor, we can use the distributive property: multiply $10,000 by the correct factor. This factor is the sum of the relevant individual future value of $1 factors: (1.771561 + 1.610510 + 1.464100 + 1.331000) = 6.177171, such that $10,000 (6.177171) = $61,771.71. Our task becomes finding an efficient way to compute the 6.177171 “truncated” annuity factor.

Recall that the future value of a level ordinary annuity factor is the sum of the future value of $1 factors for the included number of time periods (with the last cash flow expected on the last day of the final period, such that it would earn interest zero times). For example, the future value of a 7-period level ordinary annuity factor at a 10% periodic compounding rate is

(1.10)7−1 ( ) = 9.487171, which equals .10

(1.10)0 + (1.10)1 + (1.10)2 + (1.10)3 + (1.10)4 + (1.10)5 + (1.10)6

= 1.000000 + 1.100000 + 1.210000 + 1.331000 + 1.464100 + 1.610510 + 1.771561 = 9.487171 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 35 Since the 7-year future value of a level ordinary annuity factor is the sum of the future value of $1 factors for years 1 through 7 (with year-end cash flows, such that the exponents range from 0 to 6), we could find the sum of the year 4 through 7 FV of $1 factors by subtracting the year 1 – 3 future value of $1 factors from the 7-year FV of a level ordinary annuity factor. In fact, since the sum of the year 1 – 3 future value of $1 factors is the 3-year FV of a level ordinary annuity factor, we can compute the needed factor as:

Future value of a level ordinary annuity factor for number of periods through which returns are earned, Minus future value of a level ordinary annuity factor for number of periods when no deposits are made.

In this case, with returns earned until the end of year 7 after deposits are made only through year 4, we have the 7-year FV of a level ordinary annuity factor minus the 3-year FV of a level ordinary annuity factor, or (1.10)7−1 (1.10)3−1 [( ) − ( )] = 9.487171 – 3.31000000 = 6.177171 .10 .10

We started with the sum of all seven of these … (1.10)0 + (1.10)1 + (1.10)2 + (1.10)3 + (1.10)4 + (1.10)5 + (1.10)6 then got rid of these three … leaving this difference.

We multiply this computed factor FAC by $10,000 in our PMT x FAC = TOT tool box breakdown to get the $61,771.71 value we computed using the brute force method above:

PMT x FAC = TOT $10,000 (6.177171) = $61,771.71

• Method 3: Compound the Annuity We could consider, alternatively, that the series of deposits constitutes only a 4-year annuity, albeit one whose accumulated total will continue to earn a return for three extra periods. So we start by computing the future value of a 4-year level ordinary annuity:

PMT x FAC = TOT (1.10)4−1 $10,000 ( ) = $10,000 (4.641000) = $46,410.00 .10 That is what the account total will be at the end of year 4, after $10,000 end-of-year deposits have been made for four years. Then to find what the total will be after a 10% average annual return has been earned for three additional years, we compound for three periods:

$46,410.00 (1.10)3 = $46,410.00 (1.331000) = $61,771.71

We often hear about the benefits of saving early for retirement. For example, if someone deposited $2,500 into a Roth IRA at the end of each year when she is ages 25 – 34, and can earn a 7% average compounded annual return on the account’s growing balance, then 30 years later (just before turning 65, which is 40 years after she started) her $2,500 (10) = $25,000 deposited should lead to an account balance of (1.07)40−1 (1.07)30−1 $2,500 [( ) − ( )] = $262,935.81 or .07 .07 (1.07)10−1 $2,500 [( ) (1.07)30] = $262,935.81 .07

(the two methods are equally valid to use, though the first one seems harder for many to understand; confusion may arise over the fact that the FV of a 10-period level ordinary annuity factor is the sum of the FV of $1 factors with exponents 0 to 9 rather than 1 to 10). As always in our time value computations, if deposits are made at the start of each period we simply multiply the factor appropriate to end-of-period cash flows by (1 + r). For example, if the Roth IRA deposits described above were made at the beginning of each of years 1 – 10, the accumulated total at the end of year 40 would be Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 36 (1.07)40−1 (1.07)30−1 $2,500 [(( ) − ( )) (1.07)] = $281,341.32 or .07 .07

(1.07)10−1 $2,500 [( ) (1.07)(1.07)30] = $281,341.32 .07

Where we might use this “truncated” annuity computation a lot is in multi-stage, or sequential, annuities. A 24 year-old saver plans to deposit $3,000 in her IRA account every year for 18 years; then $5,000 every year for the next 13 years; and then $8,000 every year for the 9 remaining years until retirement. If she can earn an 8.5% compounded average annual return on her accumulating balance, how much will she have in 40 years when she retires? Here we have an 18-year level ordinary annuity compounded for 40 – 18 = 22 additional years, followed by a 13-year level ordinary annuity compounded for 40 – 18 – 13 = 9 additional years, followed by a 9-year level ordinary annuity. Making end-of-year deposits, she should retire with:

(1.085)18−1 (1.085)13−1 (1.085)9−1 $3,000 [( ) (1.085)22] + $5,000 [( ) (1.085)9] + $8,000 ( ) .085 .085 .085

= $3,000 [(39.322995)(6.018028)] + $5,000 [(22.210936)(2.083856)] + $8,000 (12.751244)

= $709,940.72 + $231,421.93 + $102,009.95 = $1,043,372.60

Making beginning-of-year deposits instead, she should retire with a larger

(1.085)18−1 (1.085)13−1 $3,000 [( ) (1.085)(1.085)22] + $5,000 [( ) (1.085)(1.085)9] .085 .085 (1.085)9−1 + $8,000 [( ) (1.085)] = $3,000 (256.761894) + $5,000 (50.218559) + $8,000 (13.835099) .085

= $770,285.68 + $251,092.79 + $110,608.79 = $1,132,059.26

(1.085)9−1 The final term in each of the two equations above could also be shown as $8,000 ( ) (1.085)0 .085 (1.085)9−1 and $8,000 [( ) (1.085)(1.085)0]. Multiplying by (1.085)0 means multiplying by 1, of .085 course, so including this factor does not change the value, and thus some people see its inclusion as just adding clutter. But others like the visually similar structure of showing that the total accumulated with the series of $3,000 deposits keeps earning interest for 22 added years until the retirement date and the total accumulated with the $5,000 deposit series earns nine extra years of interest until the retirement date, while the $8,000 deposit series ends in the year when retirement occurs and thus its accumulated total earns additional interest for zero periods before the grand total of the three separate accounts (actually it could all be done within one single account) is intact.

D. Combining Non-Annuity and Annuity Applications in the Same Problem Non-annuity and annuity computations can be encountered in the same analysis. Think of a bank that owes someone the $24,000 that is in his account today; how much will it owe him in four years if he deposits another $10,000 at the end of each year for the next four years and it expects to pay savers a 10% average annual interest rate? The easiest way to solve in this situation is to think of two separate accounts and combine the results of two separate problems. Here the two problems are a non-annuity with the ending amount unknown, and an FV of annuity (the big amount corresponding to a stream of payments will not exist intact until a future date) with the total unknown. By the end of year 4 the existing $24,000 balance should grow to

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 37 BAMT (1 + r)n = EAMT $24,000 (1.10)4 = $24,000 (1.4641) = $35,138.40

Four end-of-year $10,000 deposits are expected to grow by the end of year 4 to

PMT x FAC = TOT (1.10)4−1 $10,000 ( ) = $10,000 (4.641) = $46,410.00 .10

So the total amount the bank should owe in four years is $35,138.40 + $46,410.00 = $81,548.40 (greater than the $46,410.00 found earlier as the amount it would owe the saver at the end of year 4 in an account containing just the four $10,000 deposits and accumulated interest). The $81,548.40 could be spread over two separate accounts containing the individual amounts, or the full total could be in one account.

What large amount of money does someone need to have today if she wants to withdraw $10,000 at the start of each year for the next four years and still have $18,000 remaining to give to charity at the end of year 4, and she expects to earn a 10% average annual return on any money invested? Again the easiest way to solve, in this situation with two goals to meet in the same plan, is to combine the results of two separate problems: now a non-annuity with the beginning amount unknown and a PV of annuity (the big amount relating to a stream of payments exists intact in the present) with the total unknown. The amount needed today to fund a single $18,000 withdrawal in four years is

BAMT (1 + r)n = EAMT BAMT (1.10)4 = $18,000 $18,000 ÷ (1.10)4 = BAMT or 1 4 $18,000 ( ) = $18,000 (.683013) = BAMT = $12,294.24 1.10

The amount needed today to fund the four-year series of $10,000 beginning-of-year withdrawals is

PMT x FAC = TOT 1 4 1−( ) $10,000 [( 1.10 ) (1.10)] = $10,000 (3.486852) = $34,868.52 .10

So the total needed today is $12,294.24 + $34,868.52 = $47,162.76 (more than the $34,868.52 computed earlier as the amount that would fund four $10,000 withdrawals with nothing left over). This total could consist of two separate accounts containing the individual amounts or everything could be in one account.

The unknown could be something other than the large total; if the individual above has a slightly larger $50,000 today how much can she withdraw at the beginning of each year and still have $18,000 to give to charity after four years? In the first part of the two-part problem we determined that $12,294.24 is needed today to fund the $18,000 charitable donation, so the remainder of $50,000 – $12,294.24 = $37,705.76 is the big present total available to fund the four beginning-of-year withdrawals. Each withdrawal can be

PMT x FAC = TOT 1 4 1−( ) PMT [( 1.10 ) (1.10)] = PMT (3.486852) = $37,705.76 .10 $37,705.76 ÷ 3.486852 = $10,813.70

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 38 E. Combining FV of Annuity and PV of Annuity in the Same Problem The common situation of managing a savings and withdrawal plan from start to finish (e.g., saving up for retirement and then working the balance down during retirement years) combines present and future value of annuity situations. Think of someone who decides today that she will put $7,500 into her retirement savings account at the end of each year for 35 years, and then expects to take equal amounts out to have an income stream at the start of each year for the subsequent 27 years while living in retirement. If any balance she holds can be expected to earn a 4.5% average annual rate of return, how much can she expect the account to provide her with each year? First, the account is built up such that the large total amount will not be intact until 35 years after the payment (deposit) stream starts, an FV of annuity application:

PMT x FAC = TOT (1.045)35 −1 $7,500 ( ) = $7,500 (81.496618) = $611,224.64 .045

Thus on the day she retires she has $611,224.64 as the big present amount to draw from over the next 27 years, so the amount she can take out at the start of each year is a PV of annuity computation:

1 27 1−( ) 1.045 PMT [( ) (1.045)] = PMT (16.146611) = $611,224.64 .045

$611,224.64 ÷ 16.146611 = PMT = $37,854.67

Let’s say that our friend is disappointed with that $37,854.67 result; she had thought the $7,500 yearly deposits made during her working/saving years would allow her a considerably better standard of living during retirement. What larger amount would she instead have to deposit at the end of each of the 35 working years to achieve the goal of withdrawing $50,000 at the start of each of the 27 retirement years? Now we reverse the order of the PV of annuity and FV of annuity computations; on the day she retires she will need to have on hand, through having saved it up, the large present (at that time) amount of:

PMT x FAC = TOT 1 27 1−( ) $50,000 [( 1.045 ) (1.045)] = $50,000 (16.146611) = $807,330.57 .045

She can accumulate that amount by making deposits at the ends of each of the 35 saving years of:

(1.045)35 −1 PMT ( ) = PMT (81.496618) = $807,330.57 .045

$807,330.57 ÷ 81.496618 = PMT = $9,906.31

If she deposits $7,500 each saving year she can pay herself $37,854.67 in each retirement year; if she saves a bigger $9,906.31 each saving year she can enjoy a bigger $50,000 income each retirement year.

Which of the two computations is done first? Easy: start with the phase we know the payment for. If we know she wants to pay in $7,500 per year we start with the buildup phase (FV of annuity), and then solve for the amount that can be paid out to her each year accordingly (PV of annuity). If we know she wants payments of $50,000 made to her each year we first compute for the paying out phase (PV of annuity), and then solve for how much she must pay in each year to get there (FV of annuity).

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 39 V. TIME VALUE OF MONEY SITUATIONS INVOLVING PERIODS OTHER THAN YEARS

The full-year periods that have characterized our examples up to this point reduce clutter in the numbers as we become familiar with basic time value ideas. But money tends to change hands in financial transactions more frequently than annually; a year is a long time to wait to make or receive a payment. Loan payments are almost always made monthly, U.S. companies usually pay their bondholders interest semiannually, stockholders usually get dividends quarterly, and a bank might pay interest to savers with compounding occurring semiannually, quarterly, monthly, daily, or over other time periods. Two issues that arise are 1) the dollar figures that are suggested by a non-annual arrangement and 2) the impact of compounding that occurs within the year when interest is applied more frequently than once per year.

A. Annual Percentage Rates and Effective Annual Rates By tradition we talk about an interest rate (or other rate of return) in annualized terms, even if such a rate is broken down into payments more frequent than once per year (and even if the investment period is not some number of full years). If a rate is stated in annual nominal terms, but there is compounding more than once per year, then we divide the nominal rate (also called the stated rate or, more formally, the annual percentage rate, or APR) by the number of compounding periods in a year to get the periodic rate (e.g., divide the stated rate by 2 to get a semiannual rate to work with as r in a time value factor). But if compounding takes place more than once per year, then by the end of the year the periodic returns would compound, such that the saver’s true gain in wealth for the year would be greater than the APR. This higher, compounded annual rate measure is the effective annual rate (EAR) of return, computed as

EAR = (1 + periodic rate)number of periods within a year – 1

[Note that the simpler, related APR = periodic rate x number of periods within a year]

If the APR is 4.88% and compounding occurs quarterly, then the quarterly periodic rate r that we work with is .0488 ÷ 4 = .0122, and the EAR is (1.0122)4 – 1 = .0497, or 4.97%. (The APR that lenders must quote to comply with federal consumer laws can be more complicated than the APR as explained above; for example, it can reflect the impact of fees the borrower must pay to obtain the loan.)

Assume that our bank quotes a 6% nominal annual interest rate (an APR) on savings accounts. If the full 6% is credited at the end of the year, then the EAR we earn is also 6%, computed formally as

.06 1 (1 + ) – 1 = (1.06)1 – 1 = .06, or 6% 1

[When compounding is annual the APR and EAR are equal.] But if the bank credits interest semiannually (we get 6% ÷ 2 = 3% after six months and then another 3% six months later, for 2 compounding periods during the year), we compute the EAR as

.06 2 (1 + ) – 1 = (1.03)2 – 1 = .0609, or 6.09% 2

If the bank credits interest quarterly (we get 6% ÷ 4 = 1.5% after three months and then another 1.5% three months later, etc., for 4 compounding periods during the year), we compute the EAR as

.06 4 (1 + ) – 1 = (1.015)4 – 1 = .06136, or 6.136% 4

For monthly compounding, the saver gets 6% ÷ 12 = .5% each month, for an EAR of

.06 12 (1 + ) – 1 = (1.005)12 – 1 = .06168, or 6.168% 12

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 40 Daily compounding’s EAR is

.06 365 (1 + ) – 1 = (1.000164)365 – 1 = .061831, or 6.1831% 365

For continuous compounding – largely a theoretical concept, involving an infinite number of compounding periods during the year – the EAR is (recall that e, the base for natural logarithms, which is essentially [1 + 1/∞]∞, is an irrational number like ; but with a value of approximately 2.718281828):

er – 1 = e.06 – 1 = 2.718281828.06 – 1 = .061836, or 6.1836%

For any given APR, the EAR is higher as the number of intra-year compounding periods increases (though the increase as we go from 365 to  periods per year is quite small).

Think of two investments: A provides a 10.25% annual percentage rate (APR) of return with annual compounding; B offers a lower 10% stated APR, but with quarterly compounding. If we assume that the two alternatives are accompanied by similar risks, which is the better selection? For A (with annual compounding), APR = EAR = 10.25%. B’s APR is 10%, but the EAR, with quarterly compounding, is

.10 4 (1 + ) – 1 = .1038, or 10.38%; B actually provides the higher addition to wealth. 4

In this example, note that if we deposit $1,000 into A at the beginning of the year, we will receive 10.25% in interest at the end of the year, so our account will have grown to $1,000 (1.1025) = $1,102.50. But if we deposit $1,000 into B at the beginning of the year, we will receive 10% ÷ 4 = 2.5% every 3 months, and the account will have grown to $1,000 (1.025) = $1,025 after 3 months; $1,025 (1.025) = $1,050.63 after 6 months; $1,050.63 (1.025) = $1,076.89 after 9 months; and $1,076.89 (1.025) = $1,103.81 after a year. If we leave our money invested in B for an entire year we will end up with 10.38% more than we started with. Getting 10% with quarterly compounding is, in effect, just as good as getting 10.38% in a single year-end payment; and thus 10.38% is B’s effective annual return. When evaluating competing investments we want to compare their EARs, not their APRs.

In time value computations we must work with a rate that corresponds to the timing of the payments and compounding periods. But rates are discussed/quoted in annual terms, so we must convert the annual measure (APR or EAR) to a periodic r before completing time value computations. Consider these examples, which illustrate the relationship between APR, EAR, and the “working” periodic rate r: In a situation with quarterly cash flows and compounding, the given periodic return r is 1.3%. Therefore, APR = .013 (4) = .052, or 5.2% EAR = (1.013)4 – 1 = .053023, or 5.3023%

In a situation with semiannual (twice/year) cash flows and compounding, the given APR is 6.8%. Thus, Semiannual periodic rate of return r is .068 ÷ 2 = .034, or 3.4% EAR = (1.034)2 – 1 = .069156, or 6.9156%

In a situation with monthly cash flows and compounding, the given EAR is 7.3143%. Therefore, Monthly periodic rate of return r is 12√1.073143 – 1 = .0059, or .59% APR = .0059 (12) = .0708, or 7.08%

While it can be comforting to understand the ideas sufficiently that we can toggle back and forth among APR, EAR, and the working r, our task going forward usually will be to convert the annual rate that we talk about – either a simpler APR or a more complex, compounded EAR – into a periodic rate r that we will work with in a time value of money analysis. Converting the simpler APR to a periodic r is easy; we just divide the APR (often by 2 or 4 or 12). In converting the more complicated EAR to a periodic rate, Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 41 we must remember to take the appropriate root of (1 + EAR), not just the root of the EAR, and then to subtract 1 from the answer. Think of the 1 as representing 100% of the original principal balance in a savings plan that accrues interest multiple times during a year. Interest in a later quarter or month is earned not just on the interest earned in earlier periods, but also on all of the original principal. So when we compound we take (1 + periodic rate) to an appropriate power to find (1 + EAR); then we subtract the 1 so that the remainder is the EAR rate. And when we “un-compound” we take the appropriate root of (1 + EAR) to find (1 + periodic rate); then we subtract the 1 so that the remainder is the periodic rate r.

B. Computing the Dollar Amounts When time value of money computations involve compounding/discounting over periods more frequent than yearly, the BAMT (1 + r)n = EAMT and PMT x FAC = TOT structures from our tool box remain fully operative. But we want n, r, and (in annuity cases) payments to be the appropriate monthly, quarterly, etc. values. So for use in the basic time value factors, we must 1) find the total number of periods n by multiplying the number of years discussed by the number of intra-year periods, and 2) find the working periodic rate r either by dividing the stated APR by the number of intra-year periods, or by “un- compounding” the compounded EAR with the applicable root.

Example 1: A saver’s retirement account earns a 7.6% annual percentage rate (APR) of return, with semiannual (twice per year) compounding. If she deposits $1,800 at the start of every six-month period, how much will she have in 13 years? Here we have a future value of a level annuity due problem (the large amount will not exist until a future date), but must note that payments are made semiannually, and thus n must be the number of half-years (13 years x 2 halves/year = 26), and r must be the semiannual periodic rate (the annual measure provided is the simpler, non-compounded APR, so just divide it by 2: .076 ÷ 2 = .038). [Instead of knowing the 7.6% APR we could have been told that the EAR is 7.744%; if so we would have found the working semiannual r as 2√1.07744 – 1 = .038, or 3.8%.] Here we solve as

(1 + 푟)푛−1 PMT [( ) (1 + 푟)] = TOT 푟

(1.038)26−1 $1,800 [( ) (1.038)] = $80,494.99 .038

Example 2: An individual borrows $100,000 to buy a house. Equal payments are to be made at the end of each month for 30 years, and the lender charges a 9.3807% effective annual interest rate (EAR). What will the monthly payment be? This loan situation is a present value of a level ordinary annuity problem (the lump sum TOT borrowed exists intact today). Payments are made monthly, so n must be the total number of months over the life of the plan (30 years x 12 months/year = 360). The annual interest rate measure given in the discussion phase is the more complicated, compounded EAR, so to find the working monthly r we “un-compound” by taking the 12th root of (1 + EAR) and subtracting 1: 12√1.093807 – 1 = .0075. [Instead of being given the 9.3807% EAR, we could have been told that the simpler APR is 9%; if so we would have found the working monthly r as .09 ÷ 12 = .0075, or .75%.] The solution is:

1 푛 1−( ) 1 + 푟 PMT ( ) = TOT 푟

1 360 1−( ) 1.0075 PMT ( ) = $100,000 .0075

PMT (124.281866) = $100,000 PMT = $100,000 ÷ 124.281866 = $804.62

Therefore $804.62 will be the borrower’s payment at the end of each month for 30 years. Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 42 Example 3: What would we pay today for the right to receive $340 at the start of each quarter forever, if the risk of the plan led us to expect a 5.8% EAR as our average annual rate of return? Here we have quarterly payments so r must be the quarterly periodic rate; we un-compound the EAR by taking the 4th root of 1.058 and subtracting 1: 4√1.058 – 1 = .014195. [We could have been given the simpler 5.678% APR as the annual interest rate measure, and would have found the quarterly working r as .05678 ÷ 4 = .014195.] The present value of this level perpetuity due would be

1 1 PMT [( ) (1 + 푟)] = $340 [( ) (1.014195)] = $340 (71.447899) = $24,292.29 푟 .014195

Example 4: Finally, let’s say we deposit $3,000 at the end of each year into retirement savings (obviously a future value of a level ordinary annuity problem). The annual interest rate our account is expected to earn is represented as an 8.64% APR. But while we will make annual deposits, interest is to be paid quarterly. How much money do we expect to have when we retire in 40 years? Our working periodic r should correspond to the timing of the payments – which, here, is annual. So in this unusual instance we need to convert the APR to an EAR before doing the annuity computation.

.0864 4 First, compute EAR = (1 + ) – 1 = (1.0216)4 – 1 = .089240, or 8.9240% 4

Then compute: PMT x FAC = TOT

(1.089240)40−1 $3,000 ( ) = TOT .089240

$3,000 (331.075977) = $993,227.93

A married couple both following this plan could retire with almost $2 million! See how important it is to start saving early for retirement?

Note what has happened in these examples. We like to talk about interest rates (or other rates of return) in annual terms. But in dollar-based computations we have to work with rates that correspond to the frequency of the cash flows and compounding. So if the payment is made every month and interest is compounded monthly we have to use a monthly periodic rate of return as our compounding/discounting rate r. We typically have to find that rate by converting an annual rate and, unfortunately, there are two different ways we routinely talk about those annual rates: APRs (simple because they are not adjusted for the impact of intra-year compounding) and EARs (accurate because they reflect intra-year compounding).

If the annual rate we talk about is in APR terms, just divide that APR by the number of compounding periods in a year to get the working periodic rate r. If the annual rate we talk about is an EAR, we must un-compound it to get the working periodic rate r, by taking a root. If cash flows and compounding occur annually, the APR and EAR are the same. In the first example above, the payments and compounding occur twice annually, so we work with a semiannual periodic r. In the fourth example above, the compounding occurs quarterly, but the payments are annual, so we must use as r the periodic rate that corresponds to the frequency of the payments – which, in that situation, is the EAR.

C. More Complicated Cases The simplest non-annual computational case involves compounding at the same intervals when cash flows occur, such as making quarterly deposits into an account that pays interest quarterly. If deposits are made annually and the account pays interest quarterly, it is still fairly simple to compute the amount we will have by the end of a discrete number of future full years: just convert the quarterly periodic rate to an effective annual rate, compute the appropriate factor corresponding to that EAR, and apply that factor to the annual cash flow; see the fourth example in section B above. Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 43 Example 1: But what if we instead deposit $1,000 each quarter into an account that compounds interest monthly? For example, at the start of every quarter we put $1,000 into an account that pays a stated APR of 7.2%, but with monthly compounding. First, note that the monthly periodic rate is .072 ÷ 12 = .006. Then, note that we are now thinking in terms of quarters instead of years, so instead of an effective annual rate, we want to find what we might call an effective quarterly rate. With three months in a quarter, that effective quarterly rate would be (1.006)3 – 1 = .018108, or 1.8108%. (Above, we converted a quarterly periodic rate into an effective annual rate; here we convert a monthly periodic rate into an effective quarterly rate.)

So how much would we have after following this plan for five years? We would be depositing $1,000 in each of twenty quarters, earning an effective quarterly rate of .018108:

(1.018108)20−1 $1,000 [( ) (1.018108)] = $24,276.70 .018108

Example 2: What if we deposit that same $1,000 at the start of each quarter, but the account compounds interest daily (the stated annual rate, or APR, is 7.2%, but with daily compounding)? Let’s use the simplest case (a bank could use various techniques in relating days to quarters): treat a year as having four 90-day quarters, for a 360-day total. The daily periodic rate is .072 ÷ 360 = .0002, or .02%. We are still thinking in terms of quarters instead of years, and still want to find an effective quarterly rate, but now must do so based on a daily periodic rate and the assumed 90 days in each quarter. That effective quarterly rate would be (1.0002)90 – 1 = .018161.

We would again be depositing $1,000 in each of twenty quarters, but now earning a slightly higher effective quarterly rate of 1.8161% with the daily compounding, and after five years we would have

(1.018161)20−1 $1,000 [( ) (1.018161)] = $24,290.67 .018161

Example 3: One nice thing about this technique is that its use need not be restricted to a whole number of years; if, for example, deposits are made quarterly and interest is compounded quarterly or over shorter time intervals, then we can compute a meaningful answer based on being in the plan for 5 years, 6¼ years, 7½ years, 8¾ years, etc. If we decided that, in the above plan with daily compounding, we would like to make deposits for 5½ years (22 quarters), our ending balance would be computed as

(1.018161)22−1 $1,000 [( ) (1.018161)] = $27,235.79 .018161

Example 4: Finally, what if we deposit $1,000 at the start of each quarter into an account that earns interest just once annually, at 8%? If we want to make quarterly deposits then choosing this account would make little sense; each year we would earn interest on the first $1,000 deposit, but would earn nothing on the other three $1,000’s until a new year had begun. (Each deposit made in quarter 2, 3, or 4 would essentially be sitting in a zero-interest account under the bank’s safe-keeping until the subsequent year began.) By the end of the first year we would have

$1,000 (1.08)1 + $3,000 = $1,080 + $3,000 = $4,080.00

By the end of the second year, we would have

($4,080 + $1,000) (1.08)1 + $3,000 = $5,486.40 + $3,000 = $8,486.40 and so on continuing into the future. •

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 44 VI. APPENDIX

This appendix presents material that may be of interest to those who enjoy greater detail and seeing the “why” behind the time value of money factors and other concepts addressed in earlier sections of this outline. But the appendix material is not essential to a solid working understanding of the TVM applications that are important to the learning and grading process in FIL 240 and 404.

A. Deriving the Future Value of a Level Ordinary Annuity Factor The future value of a level annuity factor is the sum of the future value of $1 factors over the same number of periods. For example, if we add the future value of $1 factors at a 7% periodic discount rate for periods 1 through 3, their sum is (1.07)3 + (1.07)2 + (1.07)1 = 1.225043 + 1.1449 + 1.07 = 3.439943, (1.07)3−1 the same as future value of a level annuity due factor [( ) (1.07)] = 3.439943. The future .07 value of an ordinary annuity factor is the sum of the future value of $1 factors over the same number of periods, with one adjustment: because an ordinary annuity’s cash flows occur at the end of each period, the first deposit would earn the 7% return for only 2 periods by the end of year 3 and the third deposit would earn no interest: (1.07)2 + (1.07)1 + (1.07)0 = 1.1449 + 1.07 + 1.00 = 3.2149, the same as future (1.07)3−1 value of a level ordinary annuity factor ( ) = 3.2149. So we can see that it works. .07

But how can we prove why this relationship works, in an easily understood way? Let’s say we plan to deposit PMT dollars into an account at the end of each of the next 3 years, and that accumulations will earn an annual interest rate of r%. Then by the end of year 3 we should have a total TOT of

TOT = PMT (1 + r)2 + PMT (1 + r)1 + PMT (1 + r)0 TOT = PMT [(1 + r)2 + (1 + r)1 + (1 + r)0] [Equation 1]

(Because we put money in at the end of each year, the first deposit earns interest two times, the second one time, and the third one zero times.) Recall that we can perform any operation we choose on both sides of an equation, and the equivalency remains intact. If we multiply each side by (1 + r) (a geometric transformation that provides two different but related equations), we have

TOT (1 + r) = PMT (1 + r) [(1 + r)2 + (1 + r)1 + (1 + r)0], or TOT + TOT (r) = PMT [(1 + r)3 + (1 + r)2 + (1 + r)1] [Equation 2]

Now subtract Equation 1 from Equation 2 (subtracting equal amounts from both sides of Equation 2; notice that most terms cancel out in the subtracting):

TOT + TOT (r) – TOT = PMT [(1 + r)3 + (1 + r)2 + (1 + r)1] – PMT [(1 + r)2 + (1 + r)1 + (1 + r)0], or TOT (r) = PMT [(1 + r)3 + (1 + r)2 – (1 + r)2 + (1 + r)1 – (1 + r)1 – (1 + r)0] , or TOT (r) = PMT [(1 + r)3 – (1 + r)0], or TOT (r) = PMT (1 + r)3 – 1 [anything to the zero power is 1], (1 + 푟)3−1 so TOT = PMT ( ), 푟

(1 + 푟)3−1 with ( ) representing FAC for FV of a level ordinary annuity in our PMT x FAC = TOT structure. 푟 Here we use three time periods because an exponent of 3 allows for all steps to be shown. But this approach works for any number n of periods, because in the subtracting all terms for the numerator will cancel out except for (1 + r)n and (1 + r)0.

The FV of a level annuity due factor is simply the factor computed above multiplied by (1 + r), since with beginning of period payments interest would be paid or earned one extra time over the account’s life.

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 45 B. Deriving the Future Value of a Changing Ordinary Annuity Factor (see examples in part I below) Let’s say we plan to make a deposit at the end of each of the next 3 years. The first deposit will be in the amount of PMT dollars, and then each subsequent deposit will be higher than its predecessor by a constant percentage amount ց. Therefore our deposits will be PMT [which can also be stated, for consistency, as PMT (1 + ց)0], PMT (1 + ց)1, and PMT (1 + ց)2. Then by the end of year 3 we should have a total TOT of

TOT = [PMT (1 + ց)0 ] (1 + r)2 + [PMT (1 + ց)1] (1 + r)1 + [PMT (1 + ց)2] (1 + r)0] TOT = PMT [(1 + r)2 + (1 + ց)1 (1 + r)1 + (1 + ց)2 (1 + r)0] [Equation 1]

(Because we put money in at the end of each year, the first deposit earns interest 2 times, the second 1 time, and the third zero times.) We can perform any operation we choose on both sides of an equation, and the equivalency remains intact. If we do a geometric transformation, multiplying each side of Equation 1 by (1 + r), we have

TOT (1 + r) = PMT (1 + r) [(1 + ց)0 (1 + r)2 + (1 + ց)1 (1 + r)1 + (1 + ց)2 (1 + r)0], or TOT + TOT (r) = PMT [(1 + r)3 + (1 + ց)1 (1 + r)2 + (1 + ց)2 (1 + r)1] [Equation 2]

Now let’s do another geometric transformation, multiplying each side of Equation 1 by (1 + g); we get

TOT (1 + ց) = PMT (1 + ց) [(1 + ց)0 (1 + r)2 + (1 + ց)1 (1 + r)1 + (1 + ց)2 (1 + r)0], or TOT + TOT (ց) = PMT [(1 + ց)1 (1 + r)2 + (1 + ց)2 (1 + r)1 + (1 + ց)3 (1 + r)0] [Equation 3]

Finally we subtract Equation 3 from related Equation 2 (both are transformations of Equation 1; notice that many terms cancel out in the subtracting):

[TOT + TOT (r)] – [TOT + TOT (ց)] = PMT [(1 + ց)0 (1 + r)3 + (1 + ց)1 (1 + r)2 + (1 + ց)2 (1 + r)1] – PMT [(1 + ց)1 (1 + r)2 + (1 + ց)2 (1 + r)1 + (1 + ց)3 (1 + r)0], or

TOT (r – ց) = PMT [(1 + r)3 – (1 + ց)3]

(1 + 푟)3 − (1 + 푔)3 TOT = PMT ( ), 푟 − 푔 (1 + 푟)3 − (1 + 푔)3 with ( ) representing FAC for FV of a changing ordinary annuity in PMT x FAC = TOT. 푟 − 푔

Notice that PMT, here the first of the multiple payments, is an unchanging piece of every payment in the changing series that lets us use the distributive property, on which all annuity computations are based. The FV of a changing annuity due factor is just the factor computed above multiplied by (1 + r), as start of period payments would be accompanied by one additional interest application over the life of the plan. (See section I of this Appendix for examples that show computations involving various combinations of r and ց values, including negative expected rates of growth ց.)

C. Deriving the Present Value of a Level Ordinary Annuity Factor: Finite or Perpetual The present value of a level annuity factor is the sum of the present value of $1 factors over the same number of periods. For example, if we add the present value of $1 factors at 7% for years 1 through 3, 1 1 1 2 1 3 their sum is ( ) + ( ) + ( ) = .934579 + .873439 + .816298 = 2.624316, the same as 1.07 1.07 1.07 1 3 1−( ) present value of a level ordinary annuity factor ( 1.07 ) = 2.624316. So we can see that it works. .07

But how can we prove why this relationship works, in an easily understood way? Let’s say we can afford to apply PMT dollars toward paying back a loan at the end of each of the next 3 years, and that we must Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 46 pay an annual interest rate of r% on the loan’s declining principal balance. Then we can afford to borrow a total of 1 1 1 2 1 3 TOT = PMT ( ) + PMT ( ) + PMT ( ) 1 + 푟 1 + 푟 1 + 푟

1 1 1 2 1 3 TOT = PMT [( ) + ( ) + ( ) ] [Equation 1] 1 + 푟 1 + 푟 1 + 푟

If we do a geometric transformation, multiplying each side by (1 + r), we get the different but related equation 1 1 1 2 1 3 TOT (1 + r) = PMT (1 + r) [( ) + ( ) + ( ) ], or 1 + 푟 1 + 푟 1 + 푟

1 0 1 1 1 2 TOT + TOT (r) = PMT [( ) + ( ) + ( ) ] [Equation 2] 1 + 푟 1 + 푟 1 + 푟

Now subtract Equation 1 from Equation 2 (subtracting equal amounts from both sides of Equation 2; notice that many terms cancel out in the subtracting):

1 0 1 1 1 2 1 1 1 2 1 3 TOT + TOT (r) – TOT = PMT [( ) + ( ) + ( ) ] – PMT [( ) + ( ) + ( ) ], or 1 + 푟 1 + 푟 1 + 푟 1 + 푟 1 + 푟 1 + 푟

1 0 1 1 1 1 1 2 1 2 1 3 TOT (r) = PMT [( ) + ( ) − ( ) + ( ) − ( ) − ( ) ], or 1 + 푟 1 + 푟 1 + 푟 1 + 푟 1 + 푟 1 + 푟

1 0 1 3 1 3 TOT (r) = PMT [( ) − ( ) ], or TOT (r) = PMT (1 − ( ) ) 1 + 푟 1 + 푟 1 + 푟 1 3 1−( ) 1 + 푟 so TOT = PMT ( ), 푟

1 3 1−( ) 1 + 푟 with ( ) representing FAC for PV of a level ordinary annuity in our PMT x FAC = TOT structure. 푟

(Here we illustrate with 3 periods, but the same process works for any number of periods n.) This PV of annuity factor structure works both for finite (as shown above) and infinite numbers of time periods. An annuity whose cash flows are projected to continue forever, or perpetually, is called a perpetuity. (Perpetuity is a PV of an annuity concept: the amount needed today to provide for an indefinite stream of payments. The perpetuity idea is not meaningful in an FV of annuity case, because if the same amount were deposited into an account every year or other period forever the total would be infinite, and there would not be much more to analyze.) Think of the PV of a level ordinary annuity factor structure as determined above. If r is a percentage value greater than 0 (as a rate of return almost always would be), then as the exponent on the upper-right term approaches infinity that term’s magnitude approaches 0, and the factor becomes 1/r: 1 ∞ 1−( ) 1−0 1 ( 1 + 푟 ) = ( ) = ( ) 푟 푟 푟

The PV of a level ordinary annuity factor simplifies to 1/r in the perpetuity case, such that we multiply the regular payment by (1/r) or divide it by r, the expected periodic rate of return, to find the present value. The perpetuity idea is quite simple: we leave the principal intact and just spend or take out the interest (or whatever the return dollars are characterized as being) every period. The value of the right to collect $4 at the end of every year forever, if the expected annual rate of return r is 4% or .04, is $4 ÷ .04 = $4 (1/.04) = $100, which follows our PMT x FAC = TOT breakdown with FAC = 1/r. If we have $100 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 47 today and can earn 4% in interest each year the account will grow from $100 to $104 each year, and if we take out $4 we are left with the $100 and can earn and withdraw another $4 every year forever (as long as 4% yearly interest is earned).

Finally, the PV of a level annuity due factor, in either the finite periods or perpetuity case, is simply the appropriate ordinary annuity factor computed above multiplied by (1 + r); if payments occur at the beginning of each period interest will be applied one differential number of times over the life of the plan. [(The number of interest applications is, interestingly, one less when there are beginning-of-period payments in a PV of annuity case, because the first payment occurs immediately, before any interest is earned or owed – but we still get the correct result by multiplying the ordinary annuity factor by (1 + r).]

D. Deriving PV of Changing Ordinary Annuity Factor: Finite or Perpetual (see examples in part J below) Think of a plan in which we will make withdrawals at the end of each of the next 3 years. First we will take out PMT dollars, and then each subsequent withdrawal will exceed the preceding period’s figure by constant percentage ց. Thus our withdrawals will be PMT [which can also be stated, for consistency, as PMT (1 + ց)0], PMT (1 + ց)1, and PMT (1 + ց)2. To be able to make this series of withdrawals, the amount TOT we must have on hand today is

1 1 1 2 1 3 TOT = [PMT (1 + ց)0] ( ) + [PMT (1 + ց)1] ( ) + [PMT (1 + ց)2] ( ) ] 1 + 푟 1 + 푟 1 + 푟

1 1 1 2 1 3 TOT = PMT [(1 + 푔)0 ( ) + (1 + 푔)1 ( ) + (1 + 푔)2 ( ) ] [Equation 1] 1 + 푟 1 + 푟 1 + 푟

We can perform any operation on both sides of an equation, and the equivalency remains intact. If we do a geometric transformation, multiplying each side of Equation 1 by (1 + r), we get the related

1 1 1 2 1 3 TOT (1 + 푟) = PMT (1 + 푟) [(1 + 푔)0 ( ) + (1 + 푔)1 ( ) + (1 + 푔)2 ( ) ], or 1 + 푟 1 + 푟 1 + 푟 1 1 1 2 TOT + TOT (r) = PMT [1 + (1 + 푔)1 ( ) + (1 + 푔)2 ( ) ] [Equation 2] 1 + 푟 1 + 푟

If we multiply each side of Equation 1 by (1 + ց), the result is the related

1 1 1 2 1 3 TOT (1 + ց) = PMT (1 + 푔) [(1 + 푔)0 ( ) + (1 + 푔)1 ( ) + (1 + 푔)2 ( ) ], or 1 + 푟 1 + 푟 1 + 푟

1 1 1 2 1 3 TOT + TOT (ց) = PMT [(1 + 푔)1 ( ) + (1 + 푔)2 ( ) + (1 + 푔)3 ( ) ] [Equation 3] 1 + 푟 1 + 푟 1 + 푟

Subtracting Equation 3 from Equation 2 (the two are related because both are geometric transformations of Equation 1), we find:

1 1 1 2 [TOT + TOT (r)] – [TOT + TOT (ց)] = PMT [1 + (1 + 푔)1 ( ) + (1 + 푔)2 ( ) ] 1 + 푟 1 + 푟 1 1 1 2 1 3 – PMT [(1 + 푔)1 ( ) + (1 + 푔)2 ( ) + (1 + 푔)3 ( ) ] 1 + 푟 1 + 푟 1 + 푟 1 + 푔 3 TOT (푟 − 푔) = PMT [1 − ( ) ] 1 + 푟

1 + 푔 3 1−( ) 1 + 푟 TOT = PMT ( ), 푟 − 푔

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 48 1 + 푔 3 1−( ) 1 + 푟 with ( ) representing FAC for PV of a changing ordinary annuity in PMT x FAC = TOT. 푟 − 푔 Notice that PMT, here the first of the multiple payments, is an unchanging piece of every payment in the changing series that lets us use the distributive property, on which all annuity computations are based.

In the case of a perpetual series of payments that grows by ց% per year, with ց smaller than r (periodic ց must average less than periodic r, because over the long run growth is a component of the rate of return, the rate of return is not a component of growth), the PV of a changing ordinary perpetuity factor similarly simplifies to 1/(r – ց): 1 + 푔 ∞ 1−( ) 1−0 1 ( 1 + 푟 ) = ( ) = ( ) 푟 − 푔 푟 − 푔 푟 − 푔

As we have seen in all other annuity factor situations, the PV of a changing annuity due factor, in the case of finite payments or that of payments expected or modeled to occur perpetually, is nothing more than the related ordinary annuity factor, as computed above, multiplied by (1 + r). If payments occur at the beginning of each period interest will be applied one differential number of times over the life of the plan. That interest is applied to the growing stream of payments, and the expected growth in payments is accounted for by the incorporation of growth rate ց in the factor. (See section J of this Appendix for examples that show how different combinations of r and ց values, including negative ց rates, play out.)

E. Difference Between Two Perpetuities: An Alternate Approach to Deriving PV of Annuity Factors Another way to think about a present value of an annuity is as the present value of a perpetuity that we will sell after a stated number of periods. Note that the value of the right to collect $100 per year for five years, if the required average annual rate of return is 10%, is

1 5 1−( ) 1.10 TOT = $100 ( ) = $100 (3.7908) = $379.08 .10

But this five-year annuity also can be thought of as a perpetuity we will keep for five years and then sell. The value of a $100 per year level ordinary perpetuity, if the required annual rate of return is 10%, is

1 $100 Value = $100 ( ) = ( ) = $1,000 .10 .10

The value of this right will also be $1,000 in five years ($1,000 will be the present value for a buyer, at that time, of the right to collect $100 per year forever after that date – the number of remaining payment periods is infinite, whether we look at things today or in five years – so we can plan to sell the claim in five years for $1,000). But what is it worth today to know we can get $1,000 for something in five years? It is the PV of a deferred level perpetuity, as discussed in section IV. A of the main part of the outline: $100 1 5 1 5 Value = ( ) ( ) = $1,000 ( ) = $620.92 .10 1.10 1.10

The value of the right to collect $100 per year for five years therefore is what we would pay for the perpetuity today minus what we could expect to sell it for in five years, or

$1,000 – $620.92 = $379.08, as also seen above.

$100 $100 1 5 1 1 1 5 Note the algebra: Value = ( ) – ( ) ( ) = $100 [( ) – ( ) ( ) ] .10 .10 1.10 .10 .10 1.10

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 49 1 5 1 5 1 ( ) 1−( ) = $100 [( ) – 1.10 ] = $100 ( 1.10 ) = $379.08 .10 .10 .10

(just like 7 - 2 = 7−2). Similar logic can be used in viewing a changing annuity as a changing perpetuity 9 9 9 that we buy today and plan to sell later. Consider a five-year stream of cash flows beginning with $100 and growing by 3% per year (such that the most recent flow in the stream, if there were earlier cash flows, would have been $97.09). If it were a changing perpetuity rather than a changing five-year annuity, and the required average annual rate of return were r = 10%, its value today would be

1 $100 ( ) = $1,428.57 .10 − .03

Then in five years, the changing perpetuity’s value to a potential buyer will be

1 1 ($100)(1.03)5 ( ) = $100 ( ) (1.03)5 = $1,656.11 .10 − .03 .10 − .03

But the value today of the right to receive that expected selling price in five years (PV of a deferred changing perpetuity, as discussed in section IV. B of the main part of the outline) is

1 1 5 1 (1.03)5 [$100 ( ) (1.03)5] ( ) = [$100 ( ) ( )] = $1,028.31 .10 − .03 1.10 .10 − .03 (1.10)5

So the value of holding this growing cash flow stream for five years is what we would have to pay for the changing perpetuity today, minus what it is worth today to expect to be able to sell the changing perpetuity in five years, or $1,428.57 – $1,028.31 = $400.26

Note that we could combine some terms to compute this value as

1 1 (1.03)5 1 (1.03)5 [$100 ( )] – [$100 ( ) ( )] = $100 ( ) [1 − ( )] .10 − .03 .10 − .03 (1.10)5 .10 − .03 (1.10)5

1.03 5 1−( ) = $100 ( 1.10 ) = $100 (4.00260) = $400.26 .10 − .03 1 + 푔 푛 1−( ) Thus we see how the present value of a changing annuity factor can be represented as ( 1 + 푟 ). 푟 − 푔

F. Future Value of PV of Annuity is FV of Annuity; Present Value of FV of Annuity is PV of Annuity Note the algebra in these equations:

1 푛 1 + 푔 푛 1−( ) (1 + 푟)푛−1 1−( ) (1 + 푟)푛−(1 + 푔)푛 [( 1 + 푟 ) (1 + 푟)푛] = ( ) [( 1 + 푟 ) (1 + 푟)푛] = ( ) 푟 푟 푟 − 푔 푟 − 푔

1 푛 1 + 푔 푛 (1 + 푟)푛−1 1 푛 1−( ) (1 + 푟)푛−(1 + 푔)푛 1 푛 1−( ) [( ) ( ) ] = ( 1 + 푟 ) [( ) ( ) ] = ( 1 + 푟 ) 푟 1 + 푟 푟 푟 − 푔 1 + 푟 푟 − 푔

The future value of a present value of an annuity factor is the future value of an annuity factor for the same periodic rate and number of periods, and the present value of a future value of an annuity factor is Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 50 the present value of an annuity factor for the same periodic rate and number of periods. This relationship is helpful to us in solving problems that involve corresponding sets of cash flows with differences in their timing. (It applies to level and changing annuities alike, as shown above, but here we will illustrate primarily with level ordinary annuities.) Consider this series of examples.

Example 1: We just lent someone $10,000 at a 7% annual interest rate, and will receive equal payments at the end of each year for five years. If we then reinvest each of those payments to earn a 7% annual rate of return, how much will we have in our investment account at the end of year 5?

Computing the loan payments is a simple present value of a level ordinary annuity example:

PMT x FAC = TOT 1 5 1−( ) 1.07 PMT ( ) = $10,000 .07 PMT (4.100197) = $10,000 PMT = $10,000 ÷ 4.100197 = $2,438.91

Then reinvesting them is a simple future value of a level ordinary annuity problem:

PMT x FAC = TOT (1.07)5−1 $2,438.91 ( ) = TOT .07 $2,438.91 (5.750739) = $14,025.52

But if all we want to know is the reinvested total, and not the value of each payment, we can find the answer more expediently by recalling that the FV of an annuity is the future value of a corresponding PV of an annuity. Here the $10,000 lent is, by definition, the PV of the payment stream annuity. If we just take the future value of the $10,000 PV of the annuity, we get:

BAMT (1 + r)n = EAMT $10,000 (1.07)5 = EAMT $10,000 (1.402552) = $14,025.52

So the fact that the future value of the present value of an annuity is the future value of an annuity, and the present value of the future value of an annuity is the present value of an annuity, allows us to do some quick computing in some cases. But more importantly, it helps us put cash flow streams with different timing onto an equal footing (see examples 3 – 6 below).

Example 2: How much must we set aside at the start of each of years 1 – 5 to achieve these two goals: • Accumulate $10,000 by end of year 5 in a savings account that provides a 4% average annual interest rate, and • Contribute $500 to charity at the start of each of years 1 through 5?

Here there is no difference in the timing of the two cash flow streams, so we can simply add an extra $500 to the answer we would otherwise obtain in reaching the $10,000 desired savings account total:

PMT x FAC = TOT (1.04)5−1 PMT [( ) (1.04)] = $10,000 .04 PMT (5.632975) = $10,000 PMT = $10,000 ÷ 5.632975 = $1,775.26

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 51 A series of five $1,775.26 beginning-of-year deposits would, by itself, grow with 4% annual interest to $10,000 by the end of year 5, so if we want another $500 to give to charity at the start of each year we should set aside $1,775.26 + $500 = $2,275.26 at the start of each year.

Example 3: How much must we set aside at the start of each of years 1 – 5 to achieve these two goals: • Accumulate $10,000 by end of year 5 in a savings account that provides a 4% average annual interest rate, and • Contribute $500 to charity at the end of each of years 1 through 5?

Now we have an end-of-year cash flow stream and a beginning-of-year stream, so it is helpful to know that the future value of a PV of an annuity is the FV of an annuity. First look at the amount we would need to have in a 4% savings account today to allow withdrawing $500 at the end of each year for five years: PMT x FAC = TOT 1 5 1−( ) 1.04 $500 ( ) = TOT .04 $500 (4.451822) = $2,225.91

Then take the future value, at the end of year 5, of this PV of a 5-year annuity:

BAMT (1 + r)n = EAMT $2,225.91 (1.04)5 = EAMT $2,225.91 (1.216653) = $2,708.16

This $2,708.16 is the FV of the PV of an annuity, so it is the FV of an annuity of the five $500 year-end charitable contributions. We already know that $10,000 is the FV of an annuity of the five beginning-of- year savings deposits. So we simply combine these two FV of an annuity amounts, $2,708.16 + $10,000 = $12,708.16, and find the five beginning-of-year deposits that would grow to this total:

PMT x FAC = TOT (1.04)5−1 PMT [( ) (1.04)] = $12,708.16 .04 PMT (5.632975) = $12,708.16 PMT = $12,708.16 ÷ 5.632975 = $2,256.03

This answer is lower than the $2,275.26 found in Example 2; the deposits can be smaller if interest is earned each year before the $500 is given to charity. Look at the cash flows year by year:

Beginning Plus Beg.-of- Total Plus 4% Minus-End-of- Ending Year Balance Year Deposit Accumulated Interest Year Withdrawal Balance 1 $ 0 $2,256.03 $ 2,256.03 $ 90.24 $500.00 $ 1,846.27 2 $1,846.27 $2,256.03 $ 4,102.30 $164.09 $500.00 $ 3,766.39 3 $3,766.39 $2,256.03 $ 6,022.42 $240.90 $500.00 $ 5,763.32 4 $5,763.32 $2,256.03 $ 8,019.35 $320.77 $500.00 $ 7,840.12 5 $7,840.12 $2,256.03 $10,096.15 $403.85 $500.00 $10,000.00

Example 4: How much must we set aside at the start of each of years 1 – 5 to achieve these two goals: • Accumulate $10,000 by end of year 5 in savings account that provides 4% in annual interest, and • Contribute $500 to charity at the end of each of years 1 through 3?

Again we have an end-of-year cash flow stream and a beginning-of-year stream, but now the year-end stream lasts for only three years while the beginning-of-year stream lasts for five years. So again it is Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 52 helpful to know that the future value of a PV of an annuity is the FV of an annuity. Start with the shorter cash flow stream. Consider the amount we would need to have in a 4% savings account today to allow us to withdraw $500 at the end of each year for three years:

PMT x FAC = TOT 1 3 1−( ) 1.04 $500 ( ) = TOT .04 $500 (2.775091) = $1,387.55

Then create a common time reference point by taking the future value, at the end of year 5, of this PV of a 3-year level ordinary annuity: BAMT (1 + r)n = EAMT $1,387.55 (1.04)5 = EAMT $1,387.55 (1.216653) = $1,688.17

This $1,688.17 is the year-5 FV of the series of three $500 year-end contributions (i.e., it is the total we would reach if we deposited $500 at the end of each year for three years in a 4% savings account, and then let the money remain in the account to earn interest for two more years). We don’t actually plan to do that with the money, but this computational step gives the two cash flow streams a common time frame. We already know that $10,000 is the FV of an annuity of the five beginning-of-year savings deposits. Now we simply combine these two FV of an annuity amounts, $1,688.17 + $10,000 = $11,688.17, and find the five beginning-of-year deposits that would grow to this total:

PMT x FAC = TOT (1.04)5−1 PMT [( ) (1.04)] = $11,688.17 .04 PMT (5.632975) = $11,688.17 PMT = $11,688.17 ÷ 5.632975 = $2,074.95

This answer is lower than the $2,256.03 found in Example 3; deposits can be smaller if only three years (rather than five) of $500 charitable contributions are to be made. Look at the year-by-year cash flows:

Beginning Plus Beg.-of- Total Plus 4% Minus End-of- Ending Year Balance Year Deposit Accumulated Interest Year Withdrawal Balance 1 $0 $2,074.95 $2,074.95 $ 83.00 $500.00 $ 1,657.95 2 $1,657.95 $2,074.95 $3,732.90 $149.32 $500.00 $ 3,382.22 3 $3,382.22 $2,074.95 $5,457.17 $218.29 $500.00 $ 5,175.46 4 $5,175.46 $2,074.95 $7,250.41 $290.02 $ 0 $ 7,540.43 5 $7,540.43 $2,074.95 $9,615.38 $384.62 $ 0 $10,000.00

Example 5: We need money at the start of each of years 1 through 5. The bank will provide a line of credit under these terms: 8% annual interest is charged on any principal remaining unpaid, and all principal plus applicable interest must be repaid by the end of year 5. We expect to collect $10,000 at the end of each of years 3 through 5 as returns on a previous investment we made. If we use those three $10,000 receipts to repay the bank, how much can we borrow each year?

Here, as in Example 4, there are two types of timing differences: five years vs. three, and beginning-of- year vs. end-of-year cash flows. One way to approach this example is to recall that the present value of the FV of an annuity is the PV of an annuity. Again, start with the shorter cash flow stream; look at the amount to which the three $10,000 receipts would grow by the end of year 5 if we could reinvest them to earn 8% annually:

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 53 PMT x FAC = TOT (1.08)3−1 $10,000 ( ) = TOT .08 $10,000 (3.246400) = $32,464.00

Now discount for 5 years to find the present value of this future value of a level ordinary annuity:

BAMT (1 + r)n = EAMT BAMT (1.08)5 = $32,464 BAMT (1.469328) = $32,464 BAMT = $32,464 ÷ 1.469328 = $22,094.45

This $22,094.45 is the amount we could borrow today if we planned to make a single $32,464 repayment in five years. Because we found it by taking the PV of a future value of an annuity, it is the present value of an annuity. Now find the 5-year, beginning-of-year cash flow stream that corresponds to this total:

PMT x FAC = TOT 1 5 1−( ) 1.08 PMT [( ) (1.08)] = $22,094.45 .08 PMT (4.312127) = $22,094.45 PMT = $22,094.45 ÷ 4.312127 = $5,123.79

We can borrow $5,123.79 at the start of each year for 5 years [a total of 5 ($5,123.79) = $25,618.95, but that number is not relevant to the analysis because it is not time-value adjusted], and then repay $10,000 at the end of each of years 3 through 5. Again let’s look at the cash flows year by year (there is a slight 3¢ rounding difference because intermediate figures are rounded to whole cents):

Beginning Plus Beg.-of- Total Plus 8% Minus End-of- Ending Year Balance Year Borrowing Accumulated Interest Year Payment Balance 1 $ 0 $5,123.79 $ 5,123.79 $ 409.90 $ 0 $ 5,333.69 2 $ 5,333.69 $5,123.79 $10,657.48 $ 852.60 $ 0 $11,510.08 3 $11,510.08 $5,123.79 $16,633.87 $1,330.71 $10,000 $ 7,964.58 4 $ 7,964.58 $5,123.79 $13,088.37 $1,047.07 $10,000 $ 4,135.44 5 $ 4,135.44 $5,123.79 $ 9,259.23 $ 740.74 $10,000 $ 0

Examples with both shorter and longer time periods may seem especially tricky. A helpful step is to deal with the shorter annuity first: discount it to a PV, or compound to an FV; the solution is the total TOT in a subsequent PV or FV of annuity computation involving the number of time periods in the longer annuity.

Example 6: To add a level of complexity, re-do Example 5 with these changes: our year 3 – 5 year-end receipts are expected to start at $10,000 and then decrease by 5% per year, while we want the amount we borrow each year to be 2% higher than in the prior year, in line with our prediction of inflation. Again, start with the shorter cash flow stream; look at the amount to which the three decreasing receipts would grow by the end of year 5 if we could reinvest them to earn 8% annually:

PMT x FAC = TOT (1.08)3 − (1−.05)3 $10,000 ( ) = TOT .08 − (−.05) $10,000 (3.094900) = $30,949.00

Now take the present value of this future value of a changing ordinary annuity:

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 54 BAMT (1 + r)n = EAMT BAMT (1.08)5 = $30,949 BAMT (1.469328) = $30,949  BAMT = $30,949 ÷ 1.469328 = $21,063.37

This $21,063.37 is the amount we could borrow today if we planned to make a single $30,949 repayment in five years. The first cash flow in a 5-year, beginning-of-year cash flow stream that increases by 2% annually for five years, and corresponds to this present value of an annuity, is:

PMT x FAC = TOT 1.02 5 1−( ) 1.08 PMT [( ) (1.08)] = $21,063.37 .08 − .02 PMT (4.474461) = $21,063.37 PMT = $21,063.37 ÷ 4.474461 = $4,707.47

The amount we borrow in the first year will be $4,707.47; in year 5 we will borrow $4,707.47 (1.02)4 = $5,095.52. (Even the highest year’s loan advance here is lower than the $5,123.79 each year with level cash flows in example 5; here we have declining amounts coming in to service the loan, so we must borrow less). A year-by-year look at the cash flows shows (with a 4¢ rounding difference):

Beginning Plus Beg.-of- Total Plus 8% Minus End-of- Ending Year Balance Year Borrowing Accumulated Interest Year Payment Balance 1 $ 0 $4,707.47 $ 4,707.47 $ 376.60 $ 0 $ 5,084.07 2 $ 5,084.07 $4,801.62 $ 9,885.69 $ 790.86 $ 0 $10,676.55 3 $10,676.55 $4,897.65 $15,574.20 $1,245.94 $10,000 $ 6,820.14 4 $ 6,820.14 $4,995.60 $11,815.74 $ 945.26 $ 9,500 $ 3,361.00 5 $ 3,361.00 $5,095.52 $ 8,356.52 $ 668.52 $ 9,025 $ 0

While cases like Example 6 may not occur every day in our own lives, it should be comforting to see that once the basic time value of money ideas are understood we can use them to solve time value problems based on such a wide range of situations and assumptions.

G. More on Relationship Between Future Value and Present Value of Level Ordinary Annuity Factors Both of our level ordinary annuity factors come from the geometric progression formula

푠·푑푛 − 푠 ( ), 푑 − 1 a formula in which s is the smallest proportional value in a series of progressing values, d is the degree by which the progression changes, and n is the number of times the series continues. In the future value of a level ordinary annuity case, the progression’s smallest proportional value s is 1 (i.e., 1 times the regular cash flow), and the degree of growth is (1 + r), such that we have

1·(1 + 푟)푛−1 (1 + 푟)푛−1 ( ) = ( ) (1+푟)−1 푟

In the present value of a level ordinary annuity case, the progression’s smallest proportional value s is [1/(1 + r)n], and with the degree of growth again (1 + r) we have

1 푛 1 푛 1 푛 ( ) (1 + 푟)푛− ( ) 1−( ) 1 + 푟 1 + 푟 = ( 1 + 푟 ) (1 + 푟)−1 푟

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 55 So obviously the two factors must be closely related. In fact, each one can be shown as the difference between two perpetuities. Consider, as shown in section E of this Appendix above, that the present value of a level ordinary annuity factor can be restated algebraically as

1 푛 1−( ) 1 1 1 푛 ( 1 + 푟 ) = ( ) – ( ) ( ) , 푟 푟 푟 1 + 푟 an outcome that tells us the PV of a level ordinary annuity is the difference between the value of a level ordinary perpetuity that starts in the current period and the value of a level ordinary perpetuity that would not begin until n periods had passed. (We can view a 7-year annuity as a stream of cash flows running from the end of the current year until infinity, minus a stream running from the end of year 7 through infinity.) In a similar manner, the future value of a level ordinary annuity factor can be restated as

(1 + 푟)푛−1 1 1 ( ) = ( ) (1 + 푟)푛 – ( ), 푟 푟 푟 a result (perhaps less intuitively useful that the one above) telling us that the FV of a level ordinary annuity is simply the difference between the value of a level ordinary perpetuity compounded for n periods and the value of the same level ordinary perpetuity without the compounding. Alternatively, it can be shown that the present value of a level ordinary annuity factor is the present value of the future value of a level ordinary annuity factor, and the future value of a level ordinary annuity factor is the future value of the present value of a level ordinary annuity factor. As shown in the previous section:

1 푛 1 푛 1−( ) (1 + 푟)푛−1 (1 + 푟)푛−1 1 푛 1−( ) ( 1 + 푟 ) (1 + 푟)푛 = ( ) and ( ) ( ) = ( 1 + 푟 ) 푟 푟 푟 1 + 푟 푟

With this observation we can restate the two factors on a more similar basis as

(1 + 푟)푛−1 (1 + 푟)푛−1 FV level ordinary annuity factor = ( ) and PV level ordinary annuity factor = ( ) 푟 푟 (1 + 푟)푛

But perhaps the most elegant way to show the relationship between the two level ordinary annuity factors has been offered by ISU accountancy professor emeritus Dr. Thomas Craig, who shows that both the future value and present value of level ordinary annuity factors can be represented as (note the absolute value nature of the factor’s numerator) |(1 + 푟)±푛−1| ( ) 푟

H. How and Why the Works The “Rule of 72” provides a quick, and often surprisingly good, estimate of how many periods (usually we think in terms of years) it takes for a given amount of money to double if a specified periodic rate of return is earned, or what periodic rate of return must be earned if a given amount of money is to double in a specified number of time periods (again, usually years). The rule is

72 ÷ Specified periodic rate of return = Number of periods it takes for money to double in value, or 72 ÷ Number of periods specified = Periodic rate of return we must earn for money to double.

Why does the rule work? Any time we observe how a single initial sum of money grows, we are working with the non-annuity equation

Beginning Amount (1 + r)n = Ending Amount or BAMT (1 + r)n = EAMT Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 56 If that initial sum doubles, we can restate the equation as

BAMT (1 + r)n = 2 (BAMT) or (1 + r)n = 2

We can clean up the troublesome variable exponent by taking logarithms of both sides of the equation. [In financial applications we usually use the natural logarithm, based on the irrational number called e, 1 푛 which is the limit of (1 + ) as n approaches infinity, a value that turns out to be about 2.7182818.] 푛 (1 + r)n = 2 ln [(1 + r)n] = ln 2 n [ln (1 + r)] = ln 2

So if we divide ln 2 [which is .693147] by ln (1 + any chosen r), our quotient is n. And if we divide .693147 by any chosen n, our quotient is ln (1 + the applicable r). It turns out that when n x ln (1 + r) = .693147, the product n x r typically will equal about .72.

Finding n For a Given r Let’s say we know our expected r and want to find the corresponding n by computing ln 2 ÷ ln (1 + r). For example, perhaps we expect to earn a 9% annual rate of return. Then (1 + r) = 1.09 and ln (1 + r) = ln 1.09 = .086178, and ln 2 ÷ ln (1 + r) = .693147 ÷ .086178 = 8.04. The actual time needed for doubling is 8.04 years; the Rule of 72 (with its estimate of 72 ÷ 9% = 8 years) works well.

If we expect to earn a 4% annual rate of return, then (1 + r) = 1.04 and ln (1 + r) = ln 1.04 = .039221. We can take ln 2 ÷ ln (1 + r) = .693147 ÷ .039221 = 17.67, or about 17⅔. The actual time needed for doubling is 17⅔ years; the Rule of 72 (with its estimate of 72 ÷ 4% = 18 years) again works pretty well. If we expect to earn a 36% annual rate of return, then (1 + r) = 1.36 and ln (1 + r) = ln 1.36 = .307485. We can take ln 2 ÷ ln (1 + r) = .693147 ÷ .307485 = 2.25425, or about 2¼. The time needed for doubling is 2¼ years; even here the Rule of 72 (with its estimate of 72 ÷ 36% = 2 years) would work somewhat.

Finding r For a Given n Let’s say we have a desired n and want to find the corresponding ln (1 + r) by computing ln 2 ÷ n. Of course, we need a means of undoing the logarithmic form so we will end up with r rather than a logarithmic function of r. To undo the log form, we use the quotient of ln 2 ÷ n as the exponent x in ex. For example, if we want a sum to double in ten years, then n = 10 and we compute .693147 ÷ 10 = .069315. If ln (1 + r) = .069315, then (1 + r) = e.069315 = 1.071773. The actual rate needed for doubling is 7.1773%; the Rule of 72 (with its estimate of 72 ÷ 10 years = 7.2%) would work quite well.

If our goal is for money to double in six years, then n = 6 and we find .693147 ÷ 6 = .115525. If ln (1 + r) = .115525, then (1 + r) = e.115525 = 1.122462. The actual rate needed for doubling is 12.2462%; the Rule of 72 (with its 72 ÷ 6 years = 12% estimate) again would work pretty well.

If we want our money to double in just two years, then n = 2 and we find .693147 ÷ 2 = .346574. If ln (1 + r) = .346574, then (1 + r) = e.346574 = 1.414214. The actual rate needed for doubling is 41.4214%; the Rule of 72 (with its estimate of 72 ÷ 2 years = 36%) would not work particularly well.

The Rule of 72 works best for “middle of the road” time periods or rates of return. It is obvious that for money to double in one year we would need a 100% annual rate of return, but the Rule of 72 would indicate that a 72 ÷ 1 year = 72% annual return would suffice.

Why Does This Rule of Thumb Work? A slightly more technical explanation of why the rule works is as follows. If money is to double, then we have the equation Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 57 (1 + r)n = 2 and ultimately find that n = ln 2 ÷ ln (1 + r) ln (1 + r) = ln 2 ÷ n

But to understand the Rule of 72 we want to isolate the value n·r, not simply n or some function of r. How can we get n·r on the left-hand side of the equals sign in the above equation? We can always multiply something by 1 (or by anything equal to 1) without changing its value. Here we will multiply n by r·1/r (which equals 1) in the original equation:

(1 + r)n = 2 (1 + r)n·r·1/r = 2 ln [(1 + r)n·r·1/r] = ln 2 n·r (ln [(1 + r)1/r]) = ln 2 n·r [ln푟√(1 + 푟)] = ln 2

Therefore, n·r = ln 2 ÷ ln푟√(1 + 푟) n·r = .693147 ÷ ln푟√(1 + 푟)

This last equation might not seem earth-shattering, but notice: for “middle of the road” rates of return r, ln푟√(1 + 푟) turns out to be a value just a little less than 1. So we divide .693147 by a value slightly less than 1, and get a quotient slightly greater than .693147 – somewhere close to .72. In other words, for reasonable r values, n·r = approximately .72 (treating r as a percentage figure gives n·r = about 72). To see this result, consider r = 8%. We find ln.08√(1.08)= ln [(1.08)1/.08] = ln [(1.08)12.5] = ln 2.616959 = .962013. Here ln 2 ÷ ln푟√(1 + 푟) = .693147 ÷ .962013 = .720517. The true relationship between n and r is .7205; pretending it should be .72 (as the Rule of 72 tells us to do) yields a pretty good approximation.

For r = 10%, we find ln.10√(1.10) = ln [(1.10)1/.10] = ln [(1.10)10] = ln 2.593742 = .953102. Here ln 2 ÷ ln푟√(1 + 푟) = .693147 ÷ .953102 = .727254. The true relationship between n and r is .727254; pretending it should be .72 (as the Rule of 72 tells us to do) is not too bad an approximation.

For r = 24%, we find ln.24√(1.24) = ln [(1.24)1/.24] = ln [(1.24)4.1667] = ln 2.450531 = .896305. Here ln 2 ÷ ln푟√(1 + 푟) = .693147 ÷ .896305 = .773339. The true relationship between n and r is .773339; pretending it should be .72 (as the Rule of 72 specifies) is not that good (albeit still not horrible) an approximation.

A Couple of Points to Note: • For very small r values, a rule based on 69 can be superior to one based on 72. When r is small and n is large, we approach continuous compounding, based on the formula en·r. If en·r = 2, then n·r = ln 2 ÷ ln e = ln 2 ÷ 1 = .693147. An implication is that if there are many compounding periods in a year, we might want to estimate the time or rate needed for doubling by dividing into 69 rather than 72. For example, with a 12% APR return and annual compounding, it takes ln 2 ÷ ln 1.12 = 6.116 years for a value to double, so 72 ÷ 12% = 6 years is a good estimate. But with monthly compounding the monthly rate is 1%, for an effective annual rate (EAR) of (1.01)12 – 1 = 12.6825%; doubling takes only ln 2 ÷ ln 1.01 = 69.661 months or years 5.805 years, and 69 ÷ 12% = 5.75 is a better estimate than 72 ÷ 12% = 6.

• We can compute other “Rules” for multiples other than doubling. For tripling, we have

n·r = ln 3 ÷ ln푟√(1 + 푟) = 1.098612 ÷ ln푟√(1 + 푟)

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 58 Dividing 1.098612 by a value slightly less than 1 gives a quotient slightly greater than 1.098612 – a figure close to 1.14. A Rule of 114 for tripling gives fairly good estimates for a wide range of rates and time periods. For quadrupling we have

n·r = ln 4 ÷ ln푟√(1 + 푟) = 1.386294 ÷ ln 푟√(1 + 푟)

Dividing 1.386294 by a value slightly less than 1 gives a quotient slightly greater than 1.386294 – a figure close to 1.44. A Rule of 144 (twice the Rule of 72) for quadrupling gives fairly good estimates for a wide range of rates and time periods. This result is not surprising, since quadrupling involves doubling what has already doubled, and .72 doubled is 1.44. So does a Rule of 288 provide estimates for octupling? No; octupling (increasing by a factor of 8, which is 23) is estimated by a Rule of 216 (which is 72 x 3); a Rule of 288 (which is 72 x 4) estimates growth by a factor of 16 (which is 24).

Other possible “Rules” (double-check them for correctness): Rule of 238 for growth by a factor of 10, and Rule of 478 for growth by a factor of 100. For example: about how many years will it take for $100,000 given to a newborn grandchild to grow to $1 million? If the annual rate of return is 6%, an approximate answer is 238 ÷ 6% = 39.67 years (actual answer is ln 10 ÷ ln 1.06 = 39.52 years). About how long would it take for $10,000 to grow to $1,000,000? If the annual rate of return is 9%, an approximate answer is 478 ÷ 9% = 53.11 years (actual answer is ln 100 ÷ ln 1.09 = 53.44 years).

I. FV of Annuity Changing by a Constant Percentage Amount: Different Growth Rates and Other Issues Consider how various constant rates of change in expected periodic cash flows play out when those cash flows equate, in time value-adjusted terms, to a future large lump sum. (The more complicated case of an FV of annuity’s periodic cash flows changing by a constant dollar amount is presented in part L of this Appendix.) Start with a four-year series of end-of-period deposits beginning at $10,000 and growing by 5% per period (with the periods here being years).

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Deposit $10,000 Deposit $10,500 Deposit $11,025 Deposit $11,576 Plan Ends

One method for computing the total to which this stream would grow is brute force, summing the values of all compounded cash flows. Expected deposits would be $10,000 (1.05)0 = $10,000 in year 1; $10,000 (1.05)1 = $10,500 in year 2; $10,000 (1.05)2 = $11,025 in year 3; and $10,000 (1.05)3 = $11,576.25 in year 4. By the end of year 4 these deposits would compound to

$10,000.00 (1.10)3 = $10,000.00 (1.331000) = $13,310.00 for the year 1 deposit $10,500.00 (1.10)2 = $10,500.00 (1.210000) = $12,705.00 for the year 2 deposit $11,025.00 (1.10)1 = $11,025.00 (1.100000) = $12,127.50 for the year 3 deposit $11,576.25 (1.10)0 = $11,576.25 (1.000000) = $11,576.25 for the year 4 deposit or ($13,310.00 + $12,705.00 + $12,127.50 + $11,576.25) = $49,718.75 in total. The future value of this stream of deposits is greater than the $46,410.00 future value of an unchanging stream of four year-end $10,000 deposits compounded at 10%. We would accumulate more with a growing series of deposits than with a level stream or, alternatively, to reach a higher total we would need to consider a plan through which our deposits would increase over time. Consider the account’s buildup year by year:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Available Deposit Balance 1 $0 $0 $0 $10,000.00 $10,000.00 2 $10,000.00 $1,000.00 $11,000.00 $10,500.00 $21,500.00

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 59 3 $21,500.00 $2,150.00 $23,650.00 $11,025.00 $34,675.00 4 $34,675.00 $3,467.50 $38,142.50 $11,576.25 $49,718.75

Because the payments are expected to grow at a constant rate from the $10,000 base, we can combine the payments with the distributive property as

$10,000 (1.05)0 (1.10)3 + $10,000 (1.05)1 (1.10)2 + $10,000 (1.05)2 (1.10)1 + $10,000 (1.05)3 (1.10)0 = TOT $10,000 [(1.05)0 (1.10)3 + (1.05)1 (1.10)2 + (1.05)2 (1.10)1 + (1.05)3 (1.10)0] = TOT,

And as shown in earlier Section C of this Appendix, we can use PMT x FAC = TOT from our main time value of money discussion, with the factor FAC equal to

(1 + 푟)푛 − (1 + 푔)푛 ( ) 푟 − 푔

The amount an investor will accumulate with a 4-year stream of year-end deposits beginning at $10,000 and growing by ց = 5% per year for the three additional years, if the expected average annual return is 10%, therefore can be computed more expediently as

(1.10)4 − (1.05)4 $10,000 ( ) = $10,000 (4.971875) = $49,718.75 .10 − .05

(as found with brute force above). Note that if we set ց = 0 the changing annuity factor simplifies to the future value of a level ordinary annuity factor; the latter is a special case of the former.

Other similarities carry through from our earlier illustrations. For example, the PMT x FAC = TOT format lets us compute any unknown (PMT, TOT, r, or n – or ց) in the changing annuity case just as we would in the case of equal periodic cash flows; we might know that we want to have $49,718.75 (= TOT) in four years and wish to determine how much we must deposit each year if we can earn a 10% average annual return on the growing balance and we want the amount deposited to grow by 5% per year (answer: TOT ÷ FAC = PMT = $10,000 in the initial period).

Another similarity noted earlier is that the factor for beginning-of-period cash flows is simply the factor for end-of-period cash flows multiplied by (1 + r). How much will we have after four years if we deposit a stream of four payments beginning with $10,000 at the start of year 1 and growing at 5% per year for three added years if r averages 10% per year? A time line would show

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Deposit $10,500 Deposit $11,025 Deposit $11,576 Plan Ends Deposit $10,000 and the answer would be computed, based on a distributive property application, as

$10,000 [(1.05)0 (1.10)4 + (1.05)1 (1.10)3 + (1.05)2 (1.10)2 + (1.05)3 (1.10)1] = TOT $10,000 [(1.05)0 (1.10)3 + (1.05)1 (1.10)2 + (1.05)2 (1.10)1 + (1.05)3 (1.10)0] (1.10) = TOT

(1.10)4 − (1.05)4 $10,000 [( ) (1.10)] = $10,000 (5.469063) = $54,690.63 .10 − .05

(a larger total will be achieved if the first deposit is to occur immediately). Double-check:

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 60 Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Available Interest Balance 1 $0 $10,000.00 $10,000.00 $1,000.00 $11,000.00 2 $11,000.00 $10,500.00 $21,500.00 $2,150.00 $23,650.00 3 $23,650.00 $11,025.00 $34,675.00 $3,467.50 $38,142.50 4 $38,142.50 $11,576.25 $49,718.75 $4,971.88 $54,690.63

The rate of change can be negative instead of positive; just be careful to keep the negative signs straight (as we must when the rate of change ց is expected to exceed average rate of return r). The compounded value of a 4-year stream of year-end deposits beginning at $10,000 and declining by 5% per year for the three additional years, if the required annual rate of return averages 10%, therefore can be computed as

(1.10)4 − (1−.05)4 (1.10)4 − (.95)4 $10,000 ( ) = $10,000 ( ) = $10,000 (4.330625) = $43,306.25 .10 − (−.05) .15 with the following year-by-year progression:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Available Deposit Balance 1 $0 $0 $0 $10,000.00 $10,000.00 2 $10,000.00 $1,000.00 $11,000.00 $9,500.00 $20,500.00 3 $20,500.00 $2,050.00 $22,550.00 $9,025.00 $31,575.00 4 $31,575.00 $3,157.50 $34,732.50 $8,573.75 $43,306.25

What if ց is expected to be greater than r? The compounded value of a 4-year stream of year-end deposits beginning at $10,000 and increasing by 15% per year for the three additional years, if the required rate of return averages 10% annually, is computed as (again, the negative signs)

(1.10)4 − (1.15)4 −.284906 $10,000 ( ) = $10,000 ( ) = $10,000 (5.698125) = $56,981.25 .10 − .15 − .05 with the following year-by-year results:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Available Deposit Balance 1 $0 $0 $0 $10,000.00 $10,000.00 2 $10,000.00 $1,000.00 $11,000.00 $11,500.00 $22,500.00 3 $22,500.00 $2,250.00 $24,750.00 $13,225.00 $37,975.00 4 $37,975.00 $3,797.50 $41,772.50 $15,208.75 $56,981.25

An especially interesting case occurs when ց = r (e.g., if deposits were to increase by ց = 10% per year, with the account’s growing balance expected to earn r = 10% per year also). With r = ց, our factor

(1 + 푟)푛 − (1 + 푔)푛 ( ) 푟 − 푔 has both numerator and denominator equal to zero, so we can not work with the FV of a changing ordinary annuity factor directly in this form. But note that a series of four year-end deposits that begin with $10,000 and increase by ց = r = 10% with each passing year will be $10,000 (1.10)0; $10,000 (1.10)1; $10,000 (1.10)2; and $10,000 (1.10)3. Then when we compound them to maturity at a rate of r = ց = 10% per year, we compute a total through brute force of

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 61 $10,000 (1.10)3 + [$10,000 (1.10)1] (1.10)2 + [$10,000 (1.10)2] (1.10)1 + [$10,000 (1.10)3] (1.10)0 = $10,000 (1.10)3 + $10,000 (1.10)3 + $10,000 (1.10)3 + $10,000 (1.10)3 = $10,000 [(1.10)3 + (1.10)3 + (1.10)3 + (1.10)3] = $10,000 [(4)(1.10)3] = $10,000 (5.324000) = $53,240.00

More generally, the FV of a changing ordinary annuity factor when ց = r is n (1 + r)n – 1. So here, as shown, our factor is 4 (1.10)3 = 5.324000, and

PMT x FAC = TOT $10,000 [4 (1.10)3] = $10,000 x 5.324000 = $53,240.00 with the following year-by-year cash flows observed:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Available Deposit Balance 1 $0 $0 $0 $10,000.00 $10,000.00 2 $10,000.00 $1,000.00 $11,000.00 $11,000.00 $22,000.00 3 $22,000.00 $2,200.00 $24,200.00 $12,100.00 $36,300.00 4 $36,300.00 $3,630.00 $39,930.00 $13,310.00 $53,240.00

Another way to approach this situation would be to fool the equation into disclosing an almost-correct answer by setting ց at a value almost, but not exactly, equal to r, thereby replacing 0 in the computations with a very small amount that we can actually compute with. For example, if told that r = 10% and ց = 10%, we can pretend that ց is instead 9.9999%, and compute the almost identical

(1.10)4 − (1.099999)4 .000005323993 $10,000 ( ) = $10,000 ( ) = $10,000 (5.323993) = $53,239.93 .10 − .099999 .000001

And, of course, if cash flows were to occur at the start of each period we would simply multiply the factor n (1 + r)n – 1 by (1 + r), which would be n (1 + r)n: $10,000 (4) (1.10)4 = $10,000 (5.8564) = $58,564.00: Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Available Interest Balance 1 $0 $10,000.00 $10,000.00 $1,000.00 $11,000.00 2 $11,000.00 $11,000.00 $22,000.00 $2,200.00 $24,200.00 3 $24,200.00 $12,100.00 $36,300.00 $3,630.00 $39,930.00 4 $39,930.00 $13,310.00 $53,240.00 $5,324.00 $58,564.00

Finally, what about a changing finite annuity whose accumulated balance will keep earning the periodic rate of return after the cash flows truncate? In computing the FV of a changing truncated annuity (let’s call it that even if no one else does), we compound the changing annuity factor for the number of periods when investment returns will be earned after cash flows have ended. If $10,000 is deposited at the end of year 1, then year-end deposits grow by 5% per year through year 4, and then the accumulated total is left to earn a 10% annual rate of return through the end of year 7, that accumulated total can be computed as

PMT x FAC = TOT (1.10)4 − (1.05)4 $10,000 [( ) (1.10)3] = $10,000 (6.617566) = $66,175.66 .10 − .05

For beginning-of-period deposits, we simply multiply the factor shown above by (1 + r):

(1.10)4 − (1.05)4 $10,000[( ) (1.10)(1.10)3] = $10,000 (7.279322) = $72,793.22 .10 − .05

Double-check by working through the process year by year; for end-of-year payments we find: Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 62 Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Accumulated Deposit Balance 1 $0 $0 $0 $10,000.00 $10,000.00 2 $10,000.00 $1,000.00 $11,000.00 $10,500.00 $21,500.00 3 $21,500.00 $2,150.00 $23,650.00 $11,025.00 $34,675.00 4 $34,675.00 $3,467.50 $38,142.50 $11,576.25 $49,718.75 5 $49,718.75 $4,971.88 $54,690.63 $0 $54,690.63 6 $54,690.63 $5,469.06 $60,159.69 $0 $60,159.69 7 $60,159.69 $6,015.97 $66,175.66 $0 $66,175.66

The beginning-of-year cash flow case unfolds as:

Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Accumulated Interest Balance 1 $0 $10,000.00 $10,000.00 $1,000.00 $11,000.00 2 $11,000.00 $10,500.00 $21,500.00 $2,150.00 $23,650.00 3 $23,650.00 $11,025.00 $34,675.00 $3,467.50 $38,142.50 4 $38,142.50 $11,576.25 $49,718.75 $4,971.88 $54,690.63 5 $54,690.63 $0 $54,690.63 $5,469.06 $60,159.69 6 $60,159.69 $0 $60,159.69 $6,015.97 $66,175.66 7 $66,175.66 $0 $66,175.66 $6,617.57 $72,793.23

J. PV of Annuity Changing by a Constant Percentage Amount: Different Growth Rates and Other Issues Now consider how various constant rates of change in expected periodic payments made or received play out when those cash flows relate, in time value-adjusted terms, to a large lump sum that is intact in the present. (The more complicated case of a PV of annuity’s cash flows changing by a constant dollar amount from period to period is presented in part M of this Appendix.) Start with a four-year series of end-of-period withdrawals, the first of which is $10,000 followed by subsequent growth of 5% per year. Example 1: Finite cash flow streams. Consider a four-year series of end-of-year withdrawals, beginning at $10,000 and then growing by ց = 5% per year.

1-01-Yr 1 ___ 12-31-Yr 1/1-01-Yr 2 __ 12-31-Yr 2/1-01-Yr 3 __ 12-31-Yr 3/1-01-Yr 4 ___ 12-31-Yr 4 Plan Starts Withdraw $10,000 Withdraw $10,500 Withdraw $11,025 Withdraw $11,576 With ??? Plan Ends

One method for computing the present value of this stream would be brute force, summing the present values of all expected cash flows, as seen in many earlier examples. [The expected withdrawals would be $10,000 (1.05)0 = $10,000 in year 1; $10,000 (1.05)1 = $10,500 in year 2; $10,000 (1.05)2 = $11,025 in year 3; and $10,000 (1.05)3 = $11,576.25 in year 4.]

1 1 $10,000.00 ÷ (1.10)1 = $10,000.00 ( ) = $10,000.00 (.909091) = $9,090.91 for year 1 withdrawal 1.10 1 2 $10,500.00 ÷ (1.10)2 = $10,500.00 ( ) = $10,500.00 (.826446) = $8,677.69 for year 2 withdrawal 1.10 1 3 $11,025.00 ÷ (1.10)3 = $11,025.00 ( ) = $11,025.00 (.751315) = $8,283.25 for year 3 withdrawal 1.10 1 4 $11,576.25 ÷ (1.10)4 = $11,576.25 ( ) = $11,576.25 (.683013) = $7,906.73 for year 4 withdrawal 1.10 or ($9,091.91 + $8,677.69 + $8,283.25 + $7,906.73) = $33,958.58 in total. The present value of this stream of year-end withdrawals is greater than the $31,698.65 present value of an unchanging stream of four $10,000 yearly withdrawals discounted at 10% per year. We would willingly pay more for a

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 63 growing series of cash receipts than for a level stream or, alternatively, would have to deposit more today to fund a growing stream of withdrawals than to fund a level stream. Note the year-by-year result:

Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $33,958.58 $3,395.86 $37,354.44 $10,000.00 $27,354.44 2 $27,354.44 $2,735.44 $30,089.88 $10,500.00 $19,589.88 3 $19,589.88 $1,958.99 $21,548.87 $11,025.00 $10,523.87 4 $10,523.87 $1,052.38 $11,576.25 $11,576.25 $0

Because the payments are expected to grow at a constant periodic rate from the initial $10,000, we can combine the payments with the distributive property as

1 1 1 2 1 3 1 4 $10,000 (1.05)0 ( ) + $10,000 (1.05)1 ( ) + $10,000 (1.05)2 ( ) + $10,000 (1.05)3 ( ) = TOT 1.10 1.10 1.10 1.10

1 1 1 2 1 3 1 4 $10,000 [(1.05)0 ( ) + (1.05)1 ( ) + (1.05)2 ( ) + (1.05)3 ( ) ] = TOT 1.10 1.10 1.10 1.10

And as shown in earlier Section D of this Appendix, we can use PMT x FAC = TOT from our time value of money, with the factor FAC equaling 1 + 푔 푛 1−( ) ( 1 + 푟 ) 푟 − 푔

The value of the right to collect a 4-year stream of cash flows beginning at $10,000 and growing by ց = 5% per year for the three added years, if the expected average annual rate of return is 10%, therefore can be computed more quickly and easily as

1.05 4 1−( ) $10,000 ( 1.10 ) = $10,000 (3.395858) = $33,958.58 .10 − .05

(as found more tediously above). Note that if we set ց = 0 this changing annuity factor simplifies to the present value of a level ordinary annuity factor; the latter is a special case of the former.

Other similarities carry through from our earlier illustrations. For example, the PMT x FAC = TOT format allows us to compute any unknown (PMT, TOT, r, or n – or ց) in the changing annuity case as we would in the case of equal periodic cash flows; we might know that we have $33,958.58 (= TOT) available today and wish to determine how much we could withdraw at the end of each year if we can earn a 10% average yearly return on the declining balance and we want the amount withdrawn to grow by 5% per year (answer: TOT ÷ FAC = PMT = $10,000 in the first period). Another similarity is that the factor for beginning-of-period cash flows is simply the factor for end-of-period cash flows multiplied by (1 + r). How much must our initial balance be if our expected average annual rate of return is 10% and we want to make four withdrawals that begin with $10,000 today and grow at 5% per year for three added years?

PMT x FAC = TOT 1 1 1 2 1 3 1 4 $10,000 [(1.05)0 ( ) + (1.05)1 ( ) + (1.05)2 ( ) + (1.05)3 ( ) ] = TOT 1.10 1.10 1.10 1.10 1 0 1 1 1 2 1 3 $10,000 [(1.05)0 ( ) + (1.05)1 ( ) + (1.05)2 ( ) + (1.05)3 ( ) ] (1.10) = TOT 1.10 1.10 1.10 1.10

1.05 4 1−( ) $10,000 [( 1.10 ) (1.10)] = $10,000 (3.735443) = $37,354.43 .10 − .05

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 64 (a larger initial balance is needed to fund the plan if the first withdrawal is to occur immediately). Double-check: Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $37,354.43 $10,000.00 $27,354.43 $2,735.44 $30,089.87 2 $30,089.87 $10,500.00 $19,589.87 $1,958.99 $21,548.86 3 $21,548.86 $11,025.00 $10,523.86 $1,052.39 $11,576.25 4 $11,576.25 $11,576.25 $0 $0 $0

Note also that, as in applications discussed earlier, the rate of change can be negative instead of positive. The value of the right to collect a stream of four year-end receipts beginning at $10,000 and declining by 5% (so the change is ց = – 5% per year) for the three additional years, if the required rate of return averages 10% annually, therefore can be computed as

1 + (−.05) 4 .95 4 1−( ) 1−( ) $10,000 ( 1.10 ) = $10,000 ( 1.10 ) = $10,000 (2.957875) = $29,578.75 .10 − (−.05) .15 with the following year-by-year breakdown:

Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $29,578.75 $2,957.88 $32,536.63 $10,000.00 $22,536.63 2 $22,536.63 $2,253.66 $24,790.29 $ 9,500.00 $15,290.29 3 $15,290.29 $1,529.03 $16,819.32 $ 9,025.00 $ 7,794.32 4 $ 7,794.32 $ 779.43 $ 8,573.75 $ 8,573.75 $0

The value of the cash flow stream is lower if the amounts we can withdraw will decline over time or, alternatively, we can begin with less in the account if our withdrawals will get smaller over time.

If we have a finite annuity, the periodic rate of change ց could logically exceed the expected periodic rate of return r; we would merely use caution in our computations to treat the negative signs correctly. The value of the right to collect a 4-year stream of receipts beginning at $10,000 and growing by 12% per year for the following three years, if the required average annual rate of return is 10%, can be computed as

1.12 4 1−( ) $10,000 ( 1.10 ) = $10,000 (3.736745) = $37,367.45 .10 − .12 with the following year-by-year breakdown:

Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $37,367.45 $3,736.75 $41,104.20 $10,000.00 $31,104.20 2 $31,104.20 $3,110.42 $34,214.62 $11,200.00 $23,014.62 3 $23,014.62 $2,301.46 $25,316.08 $12,544.00 $12,772.08 4 $12,772.08 $1,277.21 $14,049.29 $14,049.28 $0

The value of the cash flow stream is higher if the amounts we can withdraw will increase more over time or, alternatively, we must begin with more in the account if our withdrawals will rise over time by 12% per year than if they rose by 5% per year (recall the earlier 5% growth case with its $33,958.58 value).

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 65 An especially interesting case occurs when ց is equal to r (e.g., if withdrawals were to increase by ց = 10% per year, while the account’s declining balance were expected to earn an average r = 10% per year also). With ց = r, the factor 1 + 푔 푛 1−( ) ( 1 + 푟 ) 푟 − 푔 has both numerator and denominator equal to zero, so we can not compute the answer directly with the factor. But note that a series of four year-end withdrawals that begin with $10,000 and grow by ց = r = 10% with each passing year will be $10,000 (1.10)0; $10,000 (1.10)1; $10,000 (1.10)2; and $10,000 (1.10)3. Then when we discount them to present values at an annual rate of ց = r = 10%, we compute a total with brute force of

1 1 1 2 1 3 1 4 $10,000 ( ) + [$10,000 (1.10)1] ( ) + [$10,000 (1.10)2] ( ) + [$10,000 (1.10)3] ( ) 1.10 1.10 1.10 1.10

1 1 1 1 1 1 1 1 1 1 = $10,000 ( ) + $10,000 ( ) + $10,000 ( ) + $10,000 ( ) = $10,000 [(4) ( ) ] 1.10 1.10 1.10 1.10 1.10

= $10,000 (3.636364) = $36,363.64. More generally, the present value of a changing ordinary annuity 1 1 factor when ց = r is n ( ). So here, as shown, our factor is [(4) ( )] = 3.636364, and 1 + 푟 1.10

PMT x FAC = TOT $10,000 (3.636364) = $36,363.64 with the following year-by-year breakdown: Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $36,363.64 $3,636.36 $40,000.00 $10,000.00 $30,000.00 2 $30,000.00 $3,000.00 $33,000.00 $11,000.00 $22,000.00 3 $22,000.00 $2,200.00 $24,200.00 $12,100.00 $12,100.00 4 $12,100.00 $1,210.00 $13,310.00 $13,310.00 $0

As in the earlier future value of a changing annuity case with r = ց we could directly compute an almost- correct solution by setting ց’s magnitude a tiny bit below r’s so that 0 is replaced with very small values we are able to compute with. Again, if we are told that r = 10% and ց = 10%, we can treat ց instead as 9.9999%, and arrive at a dollar figure only a few cents from the theoretically correct value:

1.099999 4 1−( ) .000003636359 $10,000 ( 1.10 ) = $10,000 ( ) = $10,000 (3.636359) = $36,363.59 .10 − .099999 .000001

And of course if cash flows were to occur at the start of each period we would simply multiply the factor 1 n ( ) by (1 + r), which would be, strange though it might seem, just n. Notice how the cash flows 1+푟 would proceed in this case, with n = 4 such that the stream’s PV would have to be 4 ($10,000) = $40,000:

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $40,000.00 $10,000.00 $30,000.00 $3,000.00 $33,000.00 2 $33,000.00 $11,000.00 $22,000.00 $2,200.00 $24,200.00 3 $24,200.00 $12,100.00 $12,100.00 $1,210.00 $13,310.00 4 $13,310.00 $13,310.00 $0 $0 $0 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 66 Example 2: Perpetuity. As shown in earlier Section D of this Appendix, the present value of a changing ordinary perpetuity is computed as the first expected cash flow in the changing series, divided by the required periodic rate of return minus the expected periodic rate of change. In our PMT x FAC = TOT tool box equation for annuities, PMT becomes the first expected cash flow (rather than an unchanging cash flow), and FAC is 1/(r – ց), such that 1 PMT PMT ( ) = ( ) = TOT 푟 − 푔 푟 − 푔

That first expected cash flow may be a given value, or we may have to compute it as the most recent cash flow multiplied by (1 + ց). Think of a scholarship, an example often used to illustrate perpetuities. If a donor wants to endow a fund that will pay $10,000 at the end of year 1 and then rise by a compounded 4% per year indefinitely to adjust for rising school costs [such that the year 2 award will be $10,000 (1.04)1 = $10,400; the year 3 payout will be $10,000 (1.04)2 = $10,816; and if the plan had existed in the previous year the winning student would in theory have received $10,000 ÷ 1.04 = $9,615.38]:

1-01-Yr 1 __ 12-31-Yr 1/1-01-Yr 2 … 12-31-Yr 9/1-01-Yr 10 … 12-31-Yr 98/1-01-Yr 99 … 12-31-Yr ∞ Plan Starts Withdraw $10,000 Withdraw $14,233 Withdraw $466,947 Huge Withdrawal With ??? Plan Never Ends

If 10% can be earned yearly on the fund, the donation today to endow this growing perpetuity must be

PMT x FAC = TOT 1 1 $10,000 $10,000 ( ) = $10,000 ( ) = = $10,000 (16.666667) = $166,666.67 .10 − .04 .06 .06

[or ($10,000/.06) (1.10) = $183,333.33 if the changing stream’s first award were scheduled for the start of year 1]. Note that even with a small percentage of expected growth the needed endowment increases considerably (from $100,000 for a level $10,000 perpetuity to $166,666.67 in this 4% growth case). As was discussed with our common stock coverage, an assumption that there will be growth in the cash flows can have a substantial upward impact on value estimates. (Viewed differently, someone might try to justify charging a higher price for a given asset by estimating future cash flows using an unrealistically high growth rate.) Awards to be paid out at the start of each year would require a higher initial balance:

1 1 $10,000 [( ) (1.10)] = $10,000 [( ) (1.10)] = $10,000 (18.333333) = $183,333.33 .10 − .04 .06

(a $10,000 payout would leave the plan with exactly the amount needed to fund end-of-period payments starting at $10,000 (1.04) and growing by 4% per period forever: $10,400 [1/(.10 – .04)] = $173,333.33). A changing perpetuity also could involve a negative average constant rate of change (expected decline in cash flows over time); just remember to treat the negative rate of change correctly in the formula. If our benefactor wants the scholarship to be $10,000 at the end of year 1 and then drop by 3% each subsequent year to reflect falling expected education costs in the on-line [such that the year 2 amount will be $10,000 (1 – .03) = $9,700; the third year’s will be $10,000 (1 – .03)2 = $9,409; and if the plan had existed in the prior year the amount would in theory have been $10,000 ÷ .97 = $10,309.28], then with a 10% expected annual return on the fund the donor’s initial contribution could be a smaller

1 1 $10,000 $10,000 ( ) = $10,000 ( ) = = $10,000 (7.692308) = $76,923.08 .10 − (−.03) .13 .13

[or ($10,000/.13) (1.10) = $84,615.38 if the first award in the changing perpetual stream were scheduled for the start of year 1]. The $76,923.08 cost of endowing an unending stream of decreasing scholarships starting out at $10,000 yearly is less than the $100,000 outlay for funding the infinite $10,000 level annual stream, which in turn is less than the $166,667 donation needed to fund the unending series of

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 67 4% increasing awards that starts at $10,000. The cash flows in a declining percentage perpetuity never actually reach zero, at least in theory, but they do eventually become so small – tiny fractions of a cent – that paying anything to the beneficiary would be unworkable. Thus perpetuities with cash flows that are expected to decline in percentage terms do raise practicality questions.

Unlike with finite annuities, a perpetuity’s rate of change g can not exceed (or even equal) the expected rate of return r. For example, if the university could earn a 10% average annual return on scholarship endowments, but the contributor wanted the award to be $10,000 in year 1 and then increase by 12% per year, the scholarship program can not continue forever [the endowment’s balance will eventually reach zero). Note that we get a nonsensical answer of

1 1 $10,000 $10,000 ( ) = $10,000 ( ) = = $10,000 (– 50.000000) = – $50,000.00 .10 − .12 − .02 −.02 or ($10,000/– .02) (1.10) = – $55,000 if the first of the changing awards were to be paid immediately]; a negative initial balance (with the school owing the donor) surely can not fund a series of increasing awards over time. Actually, an infinite initial TOT amount would be needed to fund an infinite series of withdrawals that grow by a periodic rate equal to or greater than the periodic rate of return (the factor [1/(r – ց)] for PV of a changing perpetuity with ց = r would be ). Conceptually, over the long term, the periodic growth rate ց should be a portion of expected average periodic return rate r.

Example 3: Deferred Changing Annuity. Finally, what about a changing finite or infinite annuity whose cash flows will not begin until some number of periods has passed? In computing the present value of a deferred changing finite annuity, we discount the changing annuity factor for the number of deferral periods. For example, the right to collect a 4-year stream of year-end cash flows beginning at $10,000 in year 4 and growing by 5% per year for the three additional years (through year 7), if the required average annual rate of return is 10%, can be computed as

PMT x FAC = TOT 1.05 4 1−( ) 1 3 $10,000 [( 1.10 ) ( ) ] = $10,000 (2.551358) = $25,513.58 .10 − .05 1.10

This variation on the “discount the annuity” method for dealing with deferred level annuities should make intuitive sense (there is no easy way to adjust the “compute a factor” method for use with deferred changing annuities). And of course the factor for beginning-of-period cash flows is the factor shown above multiplied by (1 + r). The right to collect a 4-year stream of beginning-of-year cash flows starting at $10,000 in year 4 and growing by 5% per year for the three additional years (through year 7), if the required average annual rate of return is 10%, has a value of

1.05 4 1−( ) 1 3 $10,000 [( 1.10 ) (1.10) ( ) ] = $10,000 (2.806494) = $28,064.94 .10 − .05 1.10

Double-check by working through the plans year by year; the end-of-year case plays out as:

Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $25,513.58 $2,551.36 $28,064.94 $0 $28,064.94 2 $28,064.94 $2,806.49 $30,871.43 $0 $30,871.43 3 $30,871.43 $3,087.14 $33,958.57 $0 $33,958.57 4 $33,958.57 $3,395.86 $37,354.43 $10,000.00 $27,354.43 5 $27,354.43 $2,735.44 $30,089.87 $10,500.00 $19,589.87 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 68 6 $19,589.87 $1,958.99 $21,548.86 $11,025.00 $10,523.86 7 $10,523.86 $1,052.39 $11,576.25 $11,576.25 $0

For the beginning-of-year case we find:

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Available Interest Balance 1 $28,064.94 $0 $28,064.94 $2,806.49 $30,871.43 2 $30,871.43 $0 $30,871.43 $3,087.14 $33,958.57 3 $33,958.57 $0 $33,958.57 $3,395.86 $37,354.43 4 $37,354.43 $10,000.00 $27,354.43 $2,735.44 $30,089.87 5 $30,089.87 $10,500.00 $19,589.87 $1,958.99 $21,548.86 6 $21,548.86 $11,025.00 $10,523.86 $1,052.39 $11,576.25 7 $11,576.25 $11,576.25 $0 $0 $0

If a perpetual stream of changing cash flows were not expected to begin until some number of periods had elapsed we would have a deferred changing perpetuity. Imagine our scholarship donor wanted year-end awards to grow by 2% per year after an initial $10,000 is announced at his 25th class reunion in year 9:

1-01-Yr 1 __ 12-31-Yr 1/1-01-Yr 2 … 12-31-Yr 9/1-01-Yr 10 … 12-31-Yr 98/1-01-Yr 99 … 12-31-Yr ∞ Plan Starts Withdraw $0 Withdraw $10,000 Withdraw $58,266 Huge Withdrawal With ??? Plan Never Ends

If the average annual return the school can earn is 10%, the grad can endow the plan today with a gift of

PMT x FAC = TOT 1 1 8 1 1 8 $10,000 [( ) ( ) ] = $10,000 [( ) ( ) ] = $10,000 (5.831342) = $58,313.42 .10 − .02 1.10 .08 1.10

(less than the $125,000 value of the similar but immediate changing ordinary perpetuity, because eight years of interest would be earned before any withdrawals were made to pay awards, but more than the $46,650.74 needed to fund a deferred stream of unchanging $10,000 annual scholarships). And as always, our tool box tells us to multiply the deferred ordinary changing perpetuity factor by (1 + r) in working with the corresponding annuity due (start-of-period payments) application. (The donation required would be 10% higher, at $64,144.76, if the initial $10,000 were to be paid at the start, rather than the end, of year 9.)

K. Using Financial Calculator to Solve for the Future or Present Value of a Changing Annuity If we want to use the basic financial function keys on the Texas Instruments Business Analyst II Plus calculator to solve for the FV or PV of a changing annuity, the rate of return we punch in as I/Y has to be [(1 + r)/(1 + g)] – 1, with r representing the periodic rate of return earned on the plan’s changing balance and g the periodic rate by which cash flows grow. Why? Start with the FV of a changing ordinary annuity computation as we know it, with PMT representing the first payment in the changing series:

(1 + 푟)푛 − (1 + 푔)푛 PMT [( )] = TOT 푟 − 푔

[(1 + 푟)푛 − (1 + 푔)푛](1 + 푔) (1 + 푔) Multiply each side by (1 + 푔)/(1 + 푔)푛, yielding PMT [( )] = TOT [( )] (푟 − 푔) (1 + 푔)푛 (1 + 푔)푛

Since multiplying the bracketed term in the left-hand side’s numerator by (1 + g) is equivalent to dividing it by 1/(1 + g), this result can be rearranged as

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 69 (1 + 푟)푛− (1 + 푔)푛 (1 + 푔)푛 (1 + 푔) PMT ( (푟 − 푔) ) = TOT [( )], further rearranged as (1 + 푔)푛 (1 + 푔)

(1 + 푟)푛 [ ] − 1 (1 + 푔)푛 (1 + 푔) PMT ( (푟 − 푔) ) = TOT [( )], (1 + 푔)푛 (1 + 푔) and since adding 1 and subtracting 1 in the top portion of the denominator does not change the value:

(1 + 푟)푛 [ ] − 1 (1 + 푔)푛 (1 + 푔) PMT ((1 + 푟) − (1 + 푔) ) = TOT [( )] (1 + 푔)푛 (1 + 푔)

(1 + 푟) 푛 [ ] − 1 (1 + 푔) (1 + 푔) PMT ( (1 + 푟) ) = TOT [( )] − 1 (1 + 푔)푛 (1 + 푔)

And again since adding 1 and subtracting 1 (now to the left-hand side of the numerator) does not change the value: (1 + 푟) 푛 (1 + [ − 1]) − 1 (1 + 푔) (1 + 푔) PMT ( (1 + 푟) ) = TOT [( )] − 1 (1 + 푔)푛 (1 + 푔)

Finally, because (1 + 푔)/(1 + 푔)푛 = 1/(1 + 푔)푛−1, we get

(1 + 푟) 푛 (1 + [ − 1]) − 1 (1 + 푔) 1 PMT ( (1 + 푟) ) = TOT [( 푛−1)] − 1 (1 + 푔) (1 + 푔)

(1 + 푟) 푛 (1 + [ − 1]) − 1 푛−1 (1 + 푔) PMT (1 + 푔) ( (1 + 푟) ) = TOT [Equation 1] − 1 (1 + 푔)

Notice Equation 1’s similarity to the structure of the FV of a level ordinary annuity factor, which the calculator’s basic financial function keys are programmed to work with:

(1 + 푟)푛 − 1 PMT ( ) = TOT [Equation 2] 푟

(1 + 푟) The only differences: quantity [ – 1] holds the place in Equation 1 that r holds in standard FV of (1 + 푔) annuity Equation 2, and PMT (1 + ց)n – 1 holds the place in Equation 1 that PMT holds in Equation 2. The steps above show why we can solve for FV of a changing ordinary annuity with a financial calculator’s (1 + 푟) regular function keys by setting the I/Y rate on the TI BA II Plus as [ – 1] and setting as PMT the (1 + 푔) first payment in the changing series times (1 + ց)n – 1. (A somewhat different version of the algebra shown above is presented in “Discounting at the Spread and Growing Annuities: A Note” by Johnston, Hatem, and Woods in the Summer 2016 Journal of Economics and Finance Education.)

Example: The bank you manage expects to collect a series of 29 yearly deposits that start at $6,000 and grow by 2.3% per year, in keeping with the saver’s estimate of the average annual inflation rate in coming years. If you pay a 5.8% average annual interest rate on the account’s growing balance, how much will you owe the saver by the end of year 29 if she makes the changing deposits at the end of each year? At the beginning of each year? With the basic future value of changing annuity factors we compute

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 70 PMT x FAC = TOT (1.058)29−(1.023)29 $6,000 ( ) = TOT .058 − .023 $6,000 (91.311175) = $547,867.05 for the ordinary annuity end-of-period payments case, and

(1.058)29−(1.023)29 $6,000 [( ) (1.058)] = TOT .058 − .023 $6,000 (96.607223) = $579,643.34 for the annuity due beginning-of-period payments case.

Calculator solution: For the end-of-year deposits case the amount we want to enter as PMT for computing the FV of this changing series is the final expected payment in the series. Here it would be $6,000 (1.023)28 = $11,341.46 ($6,000 is the cash flow for year 1, so the 29th year’s deposit would be the $6,000 compounded n – 1 = 28 times). For I/Y instead of using the 5.8% periodic rate of return we substitute the quantity [(1 + r)/(1 + ց)] – 1, or (1.058/1.023) – 1 = .034213, which in full percentage terms for calculator entry would be 3.4213%. So to solve we enter the values 29 N; 1.058 ÷ 1.023 = – 1 = x 100 = (should show 3.421310) I/Y; 1.023 yx 28 = x 6000 = +/– (should show – 11,341.46) PMT; 0 PV; CPT FV  547,867.05. Then for beginning-of-year cash flows we could just multiply the end-of-year payment answer by (1 + r), but if we want to use the calculator’s automated function keys we must enter BGN mode (2nd BGN 2nd SET 2nd QUIT) and also make another change that may not seem intuitive; change PMT to a phantom final payment that would occur in a 30th year: $6,000 (1.023)29 = $11,602.31. Once in BGN mode enter 1.023 yx 29 = x 6000 = +/– (should show – 11,602.31) PMT; CPT FV  579,643.34.

Because the PV of a changing ordinary annuity’s factor is just the FV of a changing ordinary annuity’s factor divided by (1 + r)n (which also describes the relationship between the PV and FV of level ordinary (1 + 푟) annuity factors), once we show that [ – 1] is the I/Y rate that conforms the FV of a changing (1 + 푔) annuity factor to the logical structure embedded in the financial calculator’s basic function keys, we know that using this quantity in place of rate r also conforms the PV of a changing annuity factor to the financial calculator function keys’ embedded logic.

Example: A financial planner has a client who is retiring today at 65. The retiree expects to live until age 100, another 35 years, and he wants to be able to spend amounts that start at $4,500 in the first month and then rise by .32% per month to cover expected inflation and higher medical costs as he ages. If the planner expects the average annual rate of return earned on the client’s declining account balance to be a 6.6% annual percentage rate (APR), how large a nest egg does he need to have today if the growing withdrawals are to occur at the end, vs. at the start, of each month? Note with monthly payments over 35 years n is 35 x 12 = 420 months, and with the expected average yearly return expressed as a 6.6% APR, the monthly r is .066 ÷ 12 = .0055. Using the basic present value of changing annuity factors we compute

PMT x FAC = TOT 1.0032 420 1−( ) $4,500 ( 1.0055 ) = TOT .0055 − .0032 $4,500 (268.609572) = $1,208,743.07 for end-of-month withdrawals, and

1.0032 420 1−( ) $4,500 [( 1.0055 ) (1.0055)] = TOT .0055 − .0032 $4,500 (270.086925) = $1,215,391.16 for beginning-of-month withdrawals.

Calculator solution: for the end-of-period payments case the amount we want to enter as PMT for computing the PV of this changing series is the phantom payment that would have occurred before the first payment; here it would be $4,500 ÷ (1.0032) = $4,485.65, while for I/Y instead of using the .55% monthly periodic rate of return we substitute the quantity [(1 + r)/(1 + ց)] – 1, or (1.0055/1.0032) – 1 = Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 71 .002293, which in full percentage terms is .229266%. So to solve we enter 420 N; 1.0055 ÷ 1.0032 = – 1 = x 100 = (should show .229266) I/Y; 4500 ÷ 1.0032 = +/– (should show – 4,485.65) PMT; 0 FV; CPT PV  1,208,743.07. Then for beginning-of-period payments we could just multiply the answer computed for end-of-period payments by (1 + r), but if we want to use the calculator’s automated function keys we must make another change that may not seem intuitive; in addition to setting the calculator to BGN mode we change PMT to the initial period’s actual $4,500 payment. Once in BGN mode enter 4500 +/– PMT; CPT PV  1,215,391.16.

L. Future Value of Annuity Changing by a Constant Dollar Amount Consider an annuity example in which the cash flows (to which the lump sum equates in time value- adjusted terms) begin at the end of the first period, but with these cash flows changing by a given dollar amount, rather than a constant percentage, with each successive period. The compounded value of a 4-year stream of cash deposits beginning at $10,000 and increasing by  = $500 per year for the three additional years, with a 10% average annual required rate of return, might most easily be computed as the sum of the future values of the individual cash flows. (The constant dollar change could be negative instead of positive; simply apply correct algebraic treatment to the negative  amount.) But we can use a variation of the PMT x FAC = TOT breakdown (and probably would if the cash flow stream were lengthy and we did not have access to a spreadsheet); it is the somewhat more complex

FAC − 푛 (PMT x FAC) +  ( ) = TOT 푟

(1.10)4−1 4 ( )−4 (1.10) −1 .10 $10,000 ( ) + $500 [ ] = TOT .10 .10

$10,000 (4.6410) + $500 (6.410) = TOT $46,410 + $3,205 = $49,615.00

Check to see that our answer is correct by working through the arrangement year by year:

Beginning Plus 10% Total Plus End-of-Yr. Ending Year Balance Interest Accumulated Deposit Balance 1 $0 $0 $0 $10,000.00 $10,000.00 2 $10,000.00 $1,000.00 $11,000.00 $10,500.00 $21,500.00 3 $21,500.00 $2,150.00 $23,650.00 $11,000.00 $34,650.00 4 $34,650.00 $3,465.00 $38,115.00 $11,500.00 $49,615.00

The foregoing analysis applies to end-of-period cash flows. For start-of-period CFs, we must multiply FAC − 푛 both FAC and ( ) by (1 + r), thereby making TOT greater than in the end-of-period case by (1 + r). 푟

FAC − 푛 [(PMT x FAC)(1 + r)] +  [( ) (1 + 푟)] = TOT 푟 (1.10)4−1 4 ( )−4 (1.10) −1 .10 $10,000 [( ) (1.10)] + $500 [( ) (1.10)] = TOT .10 .10

$10,000 (5.1051) + $500 (7.0510) = TOT $51,051 + $3,525.50 = $54,576.50

Again check by working through the plan year by year:

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 72 Beginning Plus Beg.-of-Yr. Total Plus 10% Ending Year Balance Deposit Accumulated Interest Balance 1 $0 $10,000.00 $10,000.00 $1,000.00 $11,000.00 2 $11,000.00 $10,500.00 $21,500.00 $2,150.00 $23,650.00 3 $23,650.00 $11,000.00 $34,650.00 $3,465.00 $38,115.00 4 $38,115.00 $11,500.00 $49,615.00 $4,961.50 $54,576.50

M. Present Value of Annuity Changing by a Constant Dollar Amount Like the future value situation discussed in the preceding section, the present value of a short series of cash flows changing by a constant dollar amount might be computed most easily as the sum of the individual present values. But again, there is a similar formula that we can use (though it is even a little more complex than its future value counterpart) in a variation of the PMT x FAC = TOT breakdown:

1 푛 FAC − 푛 ( ) 1 + 푟 (PMT x FAC) +  ( ) = TOT 푟

The value of the right to collect a stream of four year-end cash flows beginning at $10,000 and increasing by  = $500 per year for the three additional years, with a 10% required average annual return, can be found as 1 4 1−( ) 4 1.10 1 1 4 ( )−4( ) 1−( ) .10 1.10 1.10 $10,000 ( ) + $500 = TOT .10 .10

[ ]

$10,000 (3.169865) + $500 (4.378116) = TOT $31,698.65 + $2,189.06 = $33,887.71

Check to see that our answer is correct by examining the account year by year: Beginning Plus 10% Total Minus End-of-Yr. Ending Year Balance Interest Available Withdrawal Balance 1 $33,887.71 $3,388.77 $37,276.48 $10,000.00 $27,276.48 2 $27,276.48 $2,727.65 $30,004.13 $10,500.00 $19,504.13 3 $19,504.13 $1,950.41 $21,454.54 $11,000.00 $10,454.54 4 $10,455.54 $1,045.46 $11,500.00 $11,500.00 $0

Again, as in earlier discussions, the constant dollar change could be negative instead of positive; simply apply correct algebraic treatment to the negative  amount (but be careful, because over a long enough time period the cash flow figure will ultimately reach zero). And again, for beginning-of-period cash

1 푛 FAC − 푛( ) 1 + 푟 flows, we must multiply both FAC and ( ) by (1 + r), thereby making TOT greater than in 푟 the end-of-year case by (1 + r): 1 푛 FAC − 푛( ) 1 + 푟 [(PMT x FAC) (1 + r)] +  [( ) (1 + 푟)] = TOT 푟

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 73 1 4 1−( ) 4 1.10 1 1 4 ( )−4( ) 1−( ) .10 1.10 1.10 $10,000 [( ) (1.10)] + $500 (1.10) = TOT .10 .10

[( ) ]

$10,000 (3.486852) + $500 (4.815928) = TOT $34,868.52 + $2,407.96 = $37,276.48

Again check with a year-by-year analysis:

Beginning Minus Beg.-of-Yr. Total Plus 10% Ending Year Balance Withdrawal Accumulated Interest Balance 1 $37,276.48 $10,000.00 $27,276.48 $2,727.65 $30,004.13 2 $30,004.13 $10,500.00 $19,504.13 $1,950.41 $21,454.54 3 $21,454.54 $11,000.00 $10,454.54 $1,045.46 $11,500.00 4 $11,500.00 $11,500.00 $0 $0 $0

It should not be surprising that we can also compute the present value of a perpetual cash flow stream that grows by a constant dollar amount  with a variation of our PMT x FAC = TOT breakdown. In this situation FAC equals 1/r, and we find (PMT x FAC) + ( x FAC2) = TOT 1 1 2 PMT ( ) +  ( ) = TOT 푟 푟

If an initial end-of-year withdrawal is expected to be $100, each subsequent annual withdrawal is expected to grow by $30, and the required annual rate of return is 10%, the value is computed as

1 1 2 $100 ( ) + $30 ( ) = TOT .10 .10 $1,000 + $3,000 = $4,000.00

The cash flows in this type of plan are interesting; the account balance keeps getting larger and larger so that the difference between the amount of interest earned and the amount withdrawn remains stable at /r (i.e., $30/.10 = $300 in the example above). Any increase greater than  would cause the account balance eventually to go negative (in the example above, if each withdrawal exceeded the prior one by $30.01 instead of $30, the account balance would go negative in year 111 and no further withdrawals could be taken). Note also that when there is a constant dollar change in the perpetuity case, our answer makes practical sense only if  is positive. Otherwise, the cash flows would eventually become negative, which makes no sense because the investor would simply abandon the project (unless it were illegal to do so) when the cash flows become negative rather than to eventually start paying money in.

N. Solving for r With Logarithms In earlier discussions we solved for a missing rate of return or growth in a straightforward “non-annuity” case by taking the appropriate root. For example, if a $200 initial investment triples in value to $600 over a 17-year period, we can solve for the average annual rate of growth or return r as

$200 (1 + r)17 = $600 (1 + r)17 = $600 ÷ $200 = 3.00 17√(1 + 푟)17 = 17√3.00 or 3.001/17 or 3.00.058824 (1 + r) = 1.066758, so r = .066758, or 6.6758%

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 74 But another way to solve would be to take the logarithm of both sides of the equation:

$200 (1 + r)17 = $600 (1 + r)17 = 3.00 ln [(1 + r)17] = ln 3.00 17 ln [(1 + r)] = 1.098612 ln [(1 + r)] = .064624 eln [(1 + r)] = e.064624 (1 + r) = 1.066758, so r = .066758, or 6.6758%

Recall that the natural logarithm ln is the exponent to which we raise the value e = 2.7182818 … to get a targeted value, such that when we take the ln of something with an exponent the exponent factors out as a multiplier. Now turn it around; taking e to the power of the ln of x leaves us with x [here x is (1 + r)]. This means of solving for r seems more complicated than taking the applicable root, but some might like the idea of logarithms as a common approach to solving for either r or n in the non-annuity case.

O. Annuities Due and Time Value Tables 1. The future value (FV) of a level ordinary annuity (with end-of-year payments) of $4,000 per year for eight years, if the account’s growing balance earns a 6% annual rate of return, is computed as

$4,000 [(1.06)7 + (1.06)6 + … + (1.06)1 + (1.06)0]

(1.06)8−1 = $4,000 ( ) = $4,000 (9.897468) = $39,589.87 .06

A year-by-year breakdown appears as follows:

Beginning Plus Total Plus Ending Year Balance Interest Available Deposit Balance 1 $0.00 $0.00 $0.00 $4,000.00 $4,000.00 2 $4,000.00 $240.00 $4,240.00 $4,000.00 $8,240.00 3 $8,240.00 $494.40 $8,734.40 $4,000.00 $12,734.40 4 $12,734.40 $764.06 $13,498.46 $4,000.00 $17,498.46 5 $17,498.46 $1,049.91 $18,548.37 $4,000.00 $22,548.37 6 $22,548.37 $1,352.90 $23,901.27 $4,000.00 $27,901.27 7 $27,901.27 $1,674.08 $29,575.35 $4,000.00 $33,575.35 8 $33,575.35 $2,014.52 $35,589.87 $4,000.00 $39,589.87

Notice that the 8-year FV of a level ordinary annuity factor is consistent with eight cash flows but only seven applications of interest (no interest is earned or paid in year 1). If payments instead are to occur at the start of each year (annuity due), the only difference is that interest occurs one additional time over the plan’s life; there are eight cash flows and eight applications of interest, as shown in the breakdown below:

Beginning Plus Total Plus Ending Year Balance Deposit Available Interest Balance 1 $0.00 $4,000.00 $4,000.00 $240.00 $4,240.00 2 $4,240.00 $4,000.00 $8,240.00 $494.40 $8,734.40 3 $8,734.40 $4,000.00 $12,734.40 $764.06 $13,498.46 4 $13,498.46 $4,000.00 $17,498.46 $1,049.91 $18,548.37 5 $18,548.37 $4,000.00 $22,548.37 $1,352.90 $23,901.27 6 $23,901.27 $4,000.00 $27,901.27 $1,674.08 $29,575.35

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 75 7 $29,575.35 $4,000.00 $33,575.35 $2,014.52 $35,589.87 8 $35,589.87 $4,000.00 $39,589.87 $2,375.39 $41,965.26

The most direct and easily understood way to compute the annuity due’s value is to multiply the ordinary annuity factor by (1 + r), and then let algebra guide the steps that follow:

$4,000 [(1.06)8 + (1.06)7 + … + (1.06)2 + (1.06)1]

= $4,000 [(1.06)7 + (1.06)6 + … + (1.06)1 + (1.06)0] (1.06)

(1.06)8−1 = $4,000 [( ) (1.06)] = $4,000 (10.491316) = $41,965.26 .06

But instead of multiplying the level ordinary factor by (1 + r), students using a standard FV of a level ordinary annuity table are told to subtract 1.0 from the factor for 6% and nine periods. Recall that the 8-year FV of a level ordinary annuity factor is consistent with eight cash flows and seven applications of interest, while the 9-period ordinary annuity factor relates to eight applications of interest but also to nine cash flows, the last of which would occur at the end of year 9. Subtracting 1.0, representing that final phantom CF that would accrue no interest, leaves eight CFs and eight applications of interest.

Future Value of a Level Ordinary Annuity Factor @ 6% 6 Years 7 Years 8 Years 9 Years 10 Years 11 Years 6.975319 8.393838 9.897468 11.491316 13.180795 14.971643

So here we get 11.491316 – 1.0 = 10.491316, the same value computed above as the level ordinary factor multiplied by 1.06. This convoluted process of subtracting 1.0 from the ordinary annuity factor for the next higher number of periods seems more difficult than merely multiplying the ordinary annuity factor by (1 + r). If you are going to use a table of FV of ordinary annuity factors it may be less confusing to just multiply the given 8-period factor by (1 + r): 9.897468 (1.06) = 10.491316 as well.

2. The present value (PV) of a level ordinary annuity (with end-of-year payments) of $3,000 per year for ten years, if the account’s declining balance earns a 5% annual rate of return, is computed as

1 1 1 2 1 9 1 10 $3,000 [( ) + ( ) + ⋯ + ( ) + ( ) ] 1.05 1.05 1.05 1.05 1 10 1−( ) = $3,000 ( 1.05 ) = $3,000 (7.721735) = $23,165.20 .05

A year-by-year breakdown appears as follows:

Beginning Plus Total Minus Ending Year Balance Interest Available Withdrawal Balance 1 $23,165.20 $1,158.26 $24,323.47 $3,000.00 $21,323.47 2 $21,323.47 $1,066.17 $22,389.64 $3,000.00 $19,389.64 3 $19,389.64 $969.48 $20,359.12 $3,000.00 $17,359.12 4 $17,359.12 $867.96 $18,227.08 $3,000.00 $15,227.08 5 $15,227.08 $761.35 $15,988.43 $3,000.00 $12,988.43 6 $12,988.43 $649.42 $13,637.85 $3,000.00 $10,637.85 7 $10,637.85 $531.89 $11,169.74 $3,000.00 $8,169.74 8 $8,169.74 $408.49 $8,578.23 $3,000.00 $5,578.23 9 $5,578.23 $278.91 $5,857.14 $3,000.00 $2,857.14 10 $2,857.14 $142.86 $3,000.00 $3,000.00 $0.00 Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 76 Notice that the 10-year PV of a level ordinary annuity factor is consistent with ten cash flows and ten applications of interest. If payments instead are to occur at the start of each year (annuity due), the only difference is that interest occurs one less time over the plan’s life; there are ten cash flows and nine applications of interest, as shown in the breakdown below:

Beginning Minus Total Plus Ending Year Balance Withdrawal Available Interest Balance 1 $24,323.47 $3,000.00 $21,323.47 $1,066.17 $22,389.64 2 $22,389.64 $3,000.00 $19,389.64 $969.48 $20,359.12 3 $20,359.12 $3,000.00 $17,359.12 $867.96 $18,227.08 4 $18,227.08 $3,000.00 $15,227.08 $761.35 $15,988.43 5 $15,988.43 $3,000.00 $12,988.43 $649.42 $13,637.85 6 $13,637.85 $3,000.00 $10,637.85 $531.89 $11,169.74 7 $11,169.74 $3,000.00 $8,169.74 $408.49 $8,578.23 8 $8,578.23 $3,000.00 $5,578.23 $278.91 $5,857.14 9 $5,857.14 $3,000.00 $2,857.14 $142.86 $3,000.00 10 $3,000.00 $3,000.00 $0.00 $0.00 $0.00

The most direct and easily understood way to compute the annuity due’s value is to multiply the ordinary annuity factor by (1 + r) and then let algebra guide the steps that follow:

1 0 1 1 1 8 1 9 $3,000 [( ) + ( ) + ⋯ + ( ) + ( ) ] 1.05 1.05 1.05 1.05

1 1 1 2 1 9 1 10 = $3,000 [( ) + ( ) + ⋯ + ( ) + ( ) ] (1.05) 1.05 1.05 1.05 1.05

1 10 1−( ) = $3,000 [( 1.05 ) (1.05)] = $3,000 (8.107822) = $24,323.47 .05

But instead of multiplying the level ordinary factor by (1 + r), students using a standard PV of a level ordinary annuity table are told to add 1 to the factor for 5% and nine periods. Recall that the 10-year PV of a level ordinary annuity factor is consistent with ten cash flows and ten interest applications, while the 10-year annuity due factor should have nine applications of interest and ten cash flows, the first occurring at the start of year 1, with no interest being applied before it occurs. Adding 1.0, representing the new CF that would occur before any interest could accrue, to the 9-year level ordinary annuity factor leaves ten CFs and nine applications of interest.

Present Value of a Level Ordinary Annuity Factor @ 5% 7 Years 8 Years 9 Years 10 Years 11 Years 12 Years 5.786373 6.463213 7.107822 7.721735 8.306414 8.863252

So here we get 7.107822 + 1.0 = 8.107822, the same value computed above as the level ordinary factor multiplied by 1.05. Again this awkward computation of the annuity due factor as an adjustment to the ordinary annuity factor for an adjacent number of periods is more difficult to understand than merely multiplying the ordinary annuity factor by (1 + r). If you are going to use a table of PV of ordinary annuity factors it may be more intuitively appealing to multiply the given 10-period factor by (1 + r): 7.721735 (1.05) = 8.107822 as well. •

Trefzger/FIL 240 & 404 Topic 4 Outline: Time Value of Money 77