Engineering Economics Topics on PE Exams

− Single payments ◦ present worth factor (F/P,i,N) ◦ capital recovery factor (P/F,i,N) − Unequal payment series − Equal payment series ◦ compound amount factor (F/A,i,N) ◦ sinking fund factor (A/F,i,N) ◦ present worth factor (P/A,i,N) ◦ capital recovery factor (A/P,i,N) Engineering economics topics on PE exams

− Annual cost − Breakeven analysis − Cost-benefit analysis − Future worth or value − Present worth − Valuation and depreciation Retirement planning

A 21-year old inherits $100,000 from a distant relative who has deceased. She decides to spend some and invest the rest immediately in order to retire at 65 with a $1,000,000 savings account. At 8% interest compounded annually, how much must be invested? Solution

$1,000,000 = P (1+0.08)44 P = $1,000,000/29.56 = $33,834 $100,000 - $33,834 = $66,166 Multiple payments

How much do you need to deposit today (P) to $25,000 withdraw $25,000 at n

$3,000 $5,000 =1, $3,000 at n = 2, 0

1 2 3 4 and $5,000 at n =4, if your account earns 10% annual interest? P

Set up spreadsheet solution Uneven $25,000 payment series $3,000 $5,000 0 1 2 3 4

$25,000

$3,000 $5,000 0 0 0 ++ 1 2 3 4 1 2 3 4 1 2 3 4

P2 P4 P1

PP2 = $3,000( / F,10%,2) PP4 = $5,000( / F,10%,4) PP1 = $25,000( / F,10%,1) = $22,727 = $2,479 = $3,415

P = P1 + P2 + P3 = $28,622 Check

Beginning Interest Payment Ending balance earned balance n = 0 0 0 +28,622 28,622 n = 1 28,622 2,862 -25,000 6,484 n = 2 6,484 649 -3,000 4,133 n = 3 4,133 413 0 4,546 n = 4 4,546 455 -5,000 1

Rounding error It should be “0.” College fund

Suppose you make an annual contribution of $100 each year to a college education fund for a niece. She is 4 years old now, and you will start next year and make the last deposit when she is 18. The fund is a money market account earning 6.5%/year. What will it be worth immediately after the last deposit?

Set up spreadsheet solution beginning ending Age n balance deposit interest balance 4 0 0.00 0 0.00 0.00 5 1 0.00 100 0.00 100.00 6 2 100.00 100 6.50 206.50 7 3 206.50 100 13.42 319.92 8 4 319.92 100 20.79 440.72 9 5 440.72 100 28.65 569.36 10 6 569.36 100 37.01 706.37 11 7 706.37 100 45.91 852.29 12 8 852.29 100 55.40 1007.69 13 9 1007.69 100 65.50 1173.19 14 10 1173.19 100 76.26 1349.44 15 11 1349.44 100 87.71 1537.16 16 12 1537.16 100 99.92 1737.07 17 13 1737.07 100 112.91 1949.98 18 14 1949.98 100 126.75 2176.73 Equal payment series

012 N AA A

01 2 N

0 N Equal payment series – compound amount factor F

012 N AA A

F 01 2 N

01 2 N

AA A Compound amount factor F

A(1+i)N-2 AA A

A(1+i)N-1

01 2 N 0 12 N

F = A(1 + i)N-1 + A(1 + i)N-2 + ⋅⋅⋅ + A Compound amount factor

F = A(1 + i)N-1 + A(1 + i)N-2 + ⋅⋅⋅ + A

multiply by (1 + i): F(1 + i) = A(1 + i)N + A(1 + i)N-1 + ⋅⋅⋅ + A(1 + i)

subtract:

F(1 + i) – F = A(1 + i)N – A

rearrange: (1 + i)N – 1 F = A Fi = A(1 + i)N – A i Equal payment series compound amount factor (future value of an annuity)

F ()11+−i N 0 1 2 3 FA= N i A = AF(/A,i,N) Example • Given: A = $5,000, N = 5 years, and i = 6% •Find: F • Solution: F = $5,000(F/A,6%,5) = $28,185.46 Validation

beginning ending n balance deposit interest balance 0 0.00 0 0.00 0.00 1 0.00 5000 0.00 5000.00 2 5000.00 5000 300.00 10300.00 3 10300.00 5000 618.00 15918.00 4 15918.00 5000 955.08 21873.08 5 21873.08 5000 1312.38 28185.46 Finding an annuity value (sinking fund factor) F i AF= N 0 1 2 3 ()11+−i N A = ? = FA(/F,i,N)

Example: • Given: F = $5,000, N = 5 years, and i = 7% •Find: A • Solution: A = $5,000(A/F,7%,5) = $869.50 Equal payment series (uniform series) Find the future worth of the following cash flow, assuming interest rate i.

0 1 2 3 4 N-3 N-2 N-1 N

$A $A $A $A $A $A $A $A Custodial account

Suppose you decide to open a custodial account for your niece, who was born today. The minimum deposit is $100 on opening the account today, and you will put in $100 each year up to and including her 18th birthday. What is the account worth when it is turned over to the child at age 18? You expect to earn 10% interest per year. Custodial account cash flow

0 1 2 …

18 $100 …

0 1 2 … ⎡⎤(1 +−i) N 1 FA= ⎢⎥i ⎣⎦ 18 $100 … ⎡⎤(1.10)18 −1 F ==$100 ⎢⎥$4559.92 ⎣⎦0.10 Custodial account cash flow

0 1 2 … 18

$100 … 0 1 2 … 18

$100 … Custodial account cash flow

0 1 2 … 18

$100 … 0 1 2 … 19

$100 … -1 0 1 … 18

$100 … Sinking fund

You are saving up money to make a 20% down payment on a $100,000 house when you graduate in 4 years. You plan to invest $A at the end of each summer in a money market account earning 6.5%/year. Find A. Sinking fund

4 (1 + i)N – 1 (1 + 0.065) – 1 F = A = A = $100,000 i 0.065

4.41 A = $100,000

A = $100,000/4.41 = $22,690 Annuity factor (capital recovery factor)

You want to obtain a loan of $20,000 to buy a used car. You will pay off the loan in yearly payments over the next 5 years. The salesman quotes a 6% annual interest rate and yearly payments of $4,878. Is $4,878 an accurate payment for this loan? Annuity factor (equal series capital recovery factor)

(1 + i)N – 1 F = A i

i i N A = F = P (1 + i) N (1 + i)N – 1 (1 + i) – 1

i (1 + i)N A = P (1 + i)N – 1

A = P(A/P,i,N) Annuity factor (capital recovery factor)

i (1 + i)N 0.06 (1 + 0.06)5 A = P = $20,000 (1 + i)N – 1 (1 + 0.06)5 -1

0.237 x $20,000 = $4,748 Deferred payments

Suppose you get a student loan for $8,000, and your payments are deferred until after you graduate, 2 years from now. Then, you will make 15 yearly payments (starting 2 years from now). What are your payments? The interest rate is 8%/year. Deferred payments

i (1 + i)N 0.08 (1 + 0.08)15 A = P = $8,000 (1 + i)N – 1 (1 + 0.08)15 -1

0.1168 x $8,000 = $934

Capital recovery factor (annuity factor) Present worth

Your father is about to get downsized out of his position. He has been with the previous company through 3 previous mergers, and is disgusted with that nature of the business. He is considering retiring rather than seeking a new job. What would his retirement savings have to be worth today in order to withdraw $50,000/year for the next 15 years? He expects to invest conservatively, earning 5% per year during his retirement years. Present worth

(1 + i)N -1 (1 + 0.05)15 -1 P = A = $50,000 i (1 + i)N 0.05 (1 + 0.05)15

10.38 x $50,000 = $519,000

Present worth factor Example: early savings plan – 8% interest

Option 1: Early Savings Plan

0 1 2 3 4 5 6 7 8 9 10

$2,000 ?

Option 2: Deferred Savings Plan

0 1 2 3 4 5 6 7 8 9 10 11 12 44

$2,000 Option 1 – early savings plan

F10 = $2,000 (F/A,8%,10) = $28,973 Option 1: Early Savings Plan

F44 = $28,973 (F/P,8%,34) 0 1 2 3 4 5 6 7 8 9 10 = $396,645 44

$2,000

Age 31 65 Option 2: Deferred Savings Plan

F = $2,000 (F/A,8%,10) 44 Option 2: Deferred Savings Plan = $317,233

0 11 12 44

$2,000 At what interest rate would these two options be equivalent?

Option 1:

FF44 = $2,000( / A,i,10)(F/ P,i,34) Option 2:

FF44 = $2,000( / A.i,34) Option 1 = Option 2 $2,000(FA/ ,i,10)(F/ P,i,34) = $2,000(FA/ .i,34) Solve for i AB CD E F 1 2 Year Option 1 Option 2 3 0 4 1 $ (2,000) 5 2 $ (2,000) Interest rate 0.08 6 3 $ (2,000) 7 4 $ (2,000) FV of Option 1 $ 396, 645.95 8 5 $ (2,000) 9 6 $ (2,000) FV of Option 2 $ 317, 253.34 10 7 $ (2,000) 11 8 $ (2,000) Target cell $ 79, 392.61 12 9 $ (2,000) 13 10 $ (2,000) 14 11 $ (2,000) 15 12 $ (2,000) 16 13 $ (2,000) 17 14 $ (2,000) 18 15 $ (2,000) 19 16 $ (2,000) 20 17 $ (2,000) 21 18 $ (2,000) 22 19 $ (2,000) 40 37 $ (2,000) 41 38 $ (2,000) 42 39 $ (2,000) 43 40 $ (2,000) 44 41 $ (2,000) 45 42 $ (2,000) 46 43 $ (2,000) 47 44 $ (2,000) Using excel’s goal seek function Result