ENGINEERING ECONOMY (MNG 102) Prepared by : Dr Hosny Abbas Abouzeid

Contents 1. CHAPTER ONE : Foundation Of Engineering Economy 1.1 Introduction to engineering economy(Objectives -terminology- Firm Objectives-Kinds of market structure – Types of economic systems) 1.2 Laws of supply and demands – Balance sheet- Eng. Economy symbols - interest rate types, cash flow diagrams) –- Exercises 1.3 Sheet 1 2 CHAPTER 2 : Single & Uniform Series and Gradient Payments 2.1 Present & future worth for single , uniform series factors ,and discrete payments calculations – Uses of interest tables - Exercises 2.2 Present & future worth and EUAS for uniform gradient payments - Exercises 2.3 Calculation for unknown interest rate and number of years - Exercises 2.4 Solved problems – Sheet 2 3. CHAPTER 3 : Geometric Gradient payments & Single Project Evaluation Methods 3.1 Present /Future Worth calculation for Geometric Gradient payments 3.2 Single Project Evaluation Methods : Net Present Value – Internal Rate Of Return 3.3 Sheet 3 4. CHAPTER 4 : Selection Methods Between Proposed Projects 4.1 Selection methods between different alternatives of equal and different lived finite time periods (present / future worth calculation methods)- Exercises. 4.2 Selection methods between different alternatives of infinite time periods (capitalized cost ,and benefit/ cost ratio calculations methods) - Exercises 4.3 Sheet 4 5. CHAPTER 5 :Inflation Impact On Economic Calculation 5.1 Inflation Impact and Effects 5.2 Present/Future Worth Calculation Adjusted for Inflation- Exercises 5.3 Sheet 5 6. CHAPTER 6 : Basics of a Replacement Analysis 6.1 Introduction(Advantages &Terminologies Used of replacement analysis) 6.2 Different Approaches for comparing Defender and Challenger . 6.3 Sheet 5 7. CHAPTER 7 : Depreciation And Depletion Calculation Models 7.1 Different Depreciation Methods - switching between models- Exercises 7.2 Depletion calculations models for natural resources- Exercises 7.3 Sheet 7 8. CHAPTER 8 :Breakeven & Payback Analysis 8.1 Breakeven Analysis & applications for single and two alternatives – Exercises 8.2 Payback Analysis models & applications - Exercises 8.3 Sheet 8

CHAPTER ONE Foundation Of Engineering Economy 1.1 Introduction: Engineering : Defined as the profession in which knowledge of mathematics and nature science gained by experience and practice is applied to develop ways to utilize materials and nature forces for the benefit of mankind. Economy: Defined as the attainment an of objective at low cost in terms of resources input . Engineering Economy : Defined as a collection of mathematical techniques which simplify economic comparison to support making decision for selecting the best alternative or solution. 1.2 Importance of Engineering Economy 1. It determine which engineering projects is worthwhile . 2. It determine which engineering project should have a higher priority . 3. It estimate ,formulate and evaluate the economic outcomes out of available alternatives 4. It make analysis of present and past situations based on observed data to predict the future . 5. Engineering economics involves the systematic evaluation of the economic benefits of proposed solutions to engineering problems. 6. Engineering Economics is devoted to the problem solving and decision making at the operations level. 7. Engineering Economics integrates economic aspects with engineering practice to assist decision making . 1.3 Basic terminology: The main parameters used in engineering economy are: Alternative – evaluation criterion – time value of money – interest - principle 1.3.1 Alternative: Define as a stand alone solution for a given situation. This means that there are several ways of accomplishing a given task. Engineering economy is used to make a choice between different alternatives which is of a great importance in decision making processes e.g rent or buy house….travel by bus or train …etc . 1.3.2 Evaluation Criterion: Is the rule by which correct choice between different alternatives can be made? It answers the question “how will I know which one is best?” . As an example the Evaluation criterion may be the lowest overall cost or maximum expected benefits… etc. 1.3.3 Time Value of Money (TVM): Is the change in the amount of money over a given time period. It is based on the concept that money received earlier is worth more than the same amount of money received later; because it can be invested to earn interest with time. For example if you invest amount a sum of money in a bank with interest of 10% ,then this sum will become (1.1) of its original value after one year . TVM calculations can be used to compare between different alternatives as will be explained later. 1.3.4 Interest: Is a measure of increase between the original sum borrowed or invested and the final amount owed or accrued as follow: a. For invested money : Interest = total amount accumulated – original investment b. For borrowed money : Interest = Present amount owed – original loan 1.3.5 Interest Rate and Rate of Return (ROR): Defined as the ratio between the interest accrued per unit time and the original amount expressed as percentage . In case of borrowed money , this value is called “Interest Rate” while in case of invested money is called “ Rate of Return(ROR)” . Both values are calculated as follow: Interest rate % = Interest accrued per unit time X 100% / original amount 1.3.6 Principle: Is the original investment or loan. Example 1: A company borrowed $100000 and repaid 110000 after one year , compute the interest and the interest rate ? Solution: Interest =110000-10000 = 100000 $ , Interest rate = 100000x100%/100000 =10 % Example 2: A company invested $100000 for one year at 15% interest. Compute the interest gain and the amount accumulated. Solution: Interest = 0 .15 X 100000 = $ 15000 Total amount accumulated = 100000 + 15000 = $ 115000 1.4 The Firm 1.4.1 Definition : A firm is an organization that employs resources to produce and sell goods and Services using suitable technology . 1.4.2 Firm Objectives • Firms exist to perform useful functions in society by producing and distributing goods and services. • Firms use society's scarce resources, provide employment, and pay taxes. • The firm can work toward long-term sustainability or aim to produce high profit levels in a short period of time. • The firm perform four functions : - Produce product or service . - Market and sell product /service . - Track the accounting and financial transactions . - Perform basic human resources tasks such as hiring and training employees. 1.5 Types of Economic systems Economic Defined as is the social science that analyzes the production, distribution, and consumption of goods and services . The type of economy is determined by the extent of government involvement in economic decision making. Therefore , there are 4 types of systems as follow : 1.5.1 Traditional Economic System • Economic system is based on customs and traditions (handed down from 1 generation to another). • Allocation of scarce resources stems from habit, or customs !system in trade! No money ) مقايضة )Use BARTER • • No two traditional economies are the same, so it is impossible to describe typical economic mechanisms in this type . Examples: parts of Africa, parts of India, the original people of Australian, Eskimos Disadvantages : Discourages new ideas - Lack of progress - Lower standard of living 1.5.2 Command (Communism)Economy • A central authority (government or state) controls the economy by deciding how to use and distribute resources . • Government decides the needs of the people, the best way to produce it and for everyone! • There is very little if any input from the people. • The government regulates prices and wages; it may even determine what sorts of work individuals do. Example: Vietnam, North Korea, Former Soviet Union . Advantages : Basic Needs (Education, public health, other services )cost very little . And very little unemployment . Disadvantages : Doesn’t meet wants – no motivation- Requires a large bureaucracy - New and different ideas are discouraged 1.5.3 Capitalism/Market (or Free Market Economy Capitalism)

Advantages : Individual Freedom for all - Lack of government interference - Consumers have many choices - High degree of consumer satisfaction Disadvantages : Rewards only productive people - Workers and businesses face high Competition - Not enough public goods/services (Education, health, defense) – Unemployment - Must guard against market failure 1.5.4 Mixed Economy • A mix of all of the other three economies . • Basic means of production are owned and managed by government • Many economic decisions are made in the market by individuals. • The government also plays a role in the allocation and distribution of resources • In mixed economy the government and the private sector interact in solving economic problems. • Government controls a significant share of the output through taxation, transfers, provision of public goods . Advantages : High standard of living and economic security - High tax provide free health care and advanced education . 1.6 Kinds of Market structure A market is any arrangement through which buyers & sellers exchange goods & Services . There are four kinds of market organizations (based on their characteristics): perfect competition , monopolistic competition, oligopoly, and monopoly 1.6.1 Perfect competition is a market structure with: - Many firms - Each sells an identical product–Many buyers - No restrictions on entry of new firms to the industry - Both firms and buyers are all well informed of the prices and products of all firms in the industry. 1.6.2 Monopoly is a market structure in which: . One firm produces the entire output of the industry . There are no close substitutes for the product . There are barriers to entry that protect the firm from competition by entering firms 1.6.3 Monopolistic competition is a market structure with: –Many firms –Each firm produces similar but slightly different products—called product differentiation –Each firm possesses an element of market power –No restrictions on entry of new firms to the industry 1.6.4 Oligopoly is a market structure in which : - A small number of firms compete . - The firms might produce almost identical products or differentiated products . - Barriers to limit entry into the market . - Profits are interdependent . Actions by any one firm will affect sales & profits of other firms . Comparison between Different kinds of Market Structure Type of market Number Nature of products Selling price Condition of structure of firms control market entry Perfect competition Large Homogeneous Non- existent Easy Monopolistic Large Are close substitutes Non- Easy competition existent(low) Oligopoly Few May be homogeneous Medium or high Difficult or close substitutes Monopoly One Unique product Very high Impossible 1.7 Demand – Price Relationship 1.7.1 What is Demand? Demand for a commodity refers to the quantity of the commodity that people are willing to purchase at a specific price per unit of time, other factors (such as price of related goods, income, advertising, etc) being constant. 1.7.2 The Law of Demand The law of demand states that, if all other factors remain equal, the higher the price of a good, the lower the quantity demanded from this good. 1.7.3 Demand Curve Demand curve is a graphical representation of price- quantity relationship. Individual demand curve shows the highest price which an individual is willing to pay for different quantities of the commodity .

Demand curve has a negative slope, i.e, it slopes downwards from left to right depicting that with increase in price, quantity demanded falls and vice versa. The factors affecting the downward sloping demand curve are : 1. Income effect- With the fall in price of a commodity, the purchasing power of consumer increases. 2. Substitution effect - the consumers tend to substitute a commodity of high price with other cheaper commodities . 3. Diminishing marginal utility-. When the price of commodity falls, a rational consumer purchases more so as to equate the marginal utility and the price level. 1.8 Supply – Price Relationship 1.8.1 What is Supply? Supply refers to the amount of a certain good producers are willing to supply when receiving a certain price. 1.8.2 The Law of Supply The law of supply states that the higher the price, the higher the quantity supplied. But unlike the law of demand, the supply relationship shows an upward slope. 1.8.3 Supply Curve Supply curve is a graphical representation of price- quantity relationship . The supply curve of a commodity usually slopes upward. In other words, an industry will offer to sell more quantity of a good at a higher price than at a lower one.

A, B and C are points on the supply curve. Each point on the curve reflects a direct correlation between quantities supplied (Q) and price (P). At point B, the quantity supplied will be Q2 and the price will be P2, and so on. Time and Supply Unlike the demand relationship, however, the supply relationship is a factor of time. Time is important to supply because supplier takes time to adjust himself to a change in the demand condition according to the nature of technical conditions of production.

1.9 Balance Sheet 1.9.1 What is a Balance Sheet The balance sheet is a very useful tool that shows one view of the firm's financial condition published at a particular time (e.g monthly ,quarterly or yearly) to display it’s owns and owes, as well as the amount invested by shareholders. 1.9.2 Balance Sheet Structured There are commonly three items (Assets, Liabilities, and shareholder Equity )that are included in the balance sheet equation as follow : Total Assets (Current Assets +Non- Current Assets) = Total Liabilities(Current Liability + Non-Current Liabilities) + Total Shareholders Equity( Share Capital + Retained Earnings) This means that a company has to pay for all the things it owns (assets) by either borrowing money (taking on liabilities) or taking it from investors (issuing .(حقوق المساهمين( shareholders' equity Assets, liabilities and shareholders' equity are each comprised of several smaller accounts that break down the specifics of a company's finances as shown below .

1.9.2.1 Assets It includes all resources owned by or owed to the firm or company and have two main classes of assets : Current assets and fixed assets (sometimes called non-current or long-term assets) A. Current assets: Represent shorter-lived working capital (cash, accounts receivable, etc.), which is more easily converted to cash, usually within 1 year . It include the following: • Cash and cash equivalents: the most liquid assets, these can include assets that have short-term re-payments under three months . )أوراق مالية قابلة للتداول( :Marketable securities • Assets that the company can liquidate on short notice. • Accounts receivable: money which customers owe the company such as deposits • Inventory: goods available for sale, valued at the lower of the cost or market price • Prepaid expenses: representing value that has already been paid for, such as insurance, advertising contracts or rent . B. Non Current Assets (Long-term assets) Assets that the business will own beyond the next year and Include the following: • Long-term investments: securities that will not or cannot be liquidated in the next year • Fixed assets: these include land, machinery, equipment, buildings and other durable, generally capital-intensive assets, that used to produce and deliver goods and/or services, and they are not intended for sale. • Intangible Assets :This line item will include all of the companies intangible fixed assets, such as patents, licenses, and secret formulas. 1.9.2.2 Liabilities Defined as all financial obligations or money that a firm owes to outside parties such as : etc.. . Liabilities have two main classes, (ودائع, سندات)loans, bonds , (فواتير) debts, bills Current Liabilities and non-Current Liabilities. A.Current liabilities are those that are due within one year and are listed in order of their due date. Current liabilities accounts might include: • Current portion of long-term debt or current debt payable )مديونيات البنك)Bank indebtedness • • Interest payable • Rent, tax, utilities • Wages payable • Customer prepayments . and others)أرباح مستحقة الدفع( Dividends payable • B.Non-Current liabilities :or Long-term liabilities are due at any point after one year and include : • Long-term debt: outlines all the companies outstanding debt, the interest expense and the principal repayment for every period. This account includes the sinking fund : سندات مستحقة Bonds Payable • amount of any bonds the company has issued. • Pension fund liability: the money a company is required to pay into its employees' retirement accounts taxes that have been accrued but willإلتزامات ضريبيه مؤجلة :Deferred tax liability • not be paid for another year . رهونات عقاريه Mortgages • 1.9.2.3. Shareholders' Equity Defined as all the financial value of ownership or the money attributable to a business' owners(or shareholders) . It is also known as "net assets or net worth ," since it is equivalent to what’s left of the company’s assets after paying off liabilities. It includes Share capital , stocks issued and retained earning by a firm . A. Share Capital This is the value of funds that shareholders have invested in the company when a company is first formed . B. Retained Earnings Retained earnings are the net earnings a company either reinvests in the business or uses to pay off debt; the rest is distributed to shareholders in the form of .أرباح dividends األسهم المفضلة والشائعة C. Preferred and common stocks Common stock and preferred stock are the two main types of stocks that are sold by companies and traded among investors on the open market. A preferred stock pay an agreed-upon dividend at regular intervals. Common stocks may pay dividends depending on how profitable the company is . Example Of Balance Sheet For a Firm

1.10 Symbols used in engineering economy computation and their Meanings: P = Value or sum of money at the Present time. F = Value or sum of money at the Future time. A = A series of consecutives, end of period payments of money per unit time (month, year...) n = number of interest periods (month, year...) i = interest rate per interest period (1 year, 1 month) t = time , stated in periods ; years, months , days . Pr = Principle I = Interest = Final amount - Principle 1.11 Simple And Compound Interest Rates : 1.11.1 Simple Interest Interest paid (earned) on only the original amount, or principal, borrowed (lent) as follow: Total simple interest = Pni ,AND Final sum after n periods = P (1+ni) Example 3: A company borrow $ 100000 for 3 years at simple interest rate of 6 % per year , how much money will be the owe at the end of 3 years . Solution: Interest = Pni = 100000 * 3 * .06 = $ 18000 Final owe = Principle + Interest = 100000 + 18000 = $ 118000 or calculated as follow: Final owe after 3 years = P(1+ni) = 100000 (1+3*0.06) = $ 118000 1.11.2 Compound Interest Interest paid (earned) on any previous interest earned, as well as on the principal borrowed (lent). Total compound interest = P [(1+i)n -1] Final sum after n periods = P (1+i) n Note: The above mentioned equations are used only when values of „ i„ is constant and cases of single payment occurred at the beginning and end of interest periods only without payments in between Example 4: Repeat the previous problem in case of using compound interest. Solution : Total compound interest = 100000[(1+.06)3 -1] = $ 19102 Final owe after 3 years = Principle + Interest = 100000 + 19102 = $ 119102 OR Final owe after 3 years = 100000(1+0.06)3 = $ 119102 Example 5: If you have 2000 L.E in a saving account and you deposit 200 L.E each year for 5 years. List the values of engineering economy symbols and the final value at the end of interest periods if the interest rates are 10% per year. Solution: P = 2000 L.E ; A = 200 L.E ; i = 10% ; n = 5 years; F ? P1 = sum at end of year 1 = P(1+i) + 200 = 2000(1.1) +200 = 2400 L.E P2 = sum at end of year 2 = P1 (1+ i ) + 200 = 2400(1.1)+200 = 2840 l..E P3 = sum at end of year 3 = P2 (1+ i ) + 200 = 2840(1.1) + 200 = 3324 L.E P4= sum at end of year 4 = P3 (1+ i ) + 200 = 3324 (1.1) + 200 = 3856.4 L.E P5 = sum at end of year 5 = P4 (1+ i ) + 200 = 3856.4(1.1) + 200 = 4444.04 L.E Final sum at the end of the five years = 4444.04 L.E Note: Generally interest rates refer to compound interest unless specified otherwise.

1.11.3 Nominal , Effective Interest , and Effective annual Interest Rates Many financial transactions require that interest be compounded often than once a Year (e.g quarterly ,monthly , daily , etc..) In such situations ,there are three Expressions for the interest rates as follow : 1. The nominal interest rate , r , is expressed on an annual basis . 2. The effective interest rate , ieff , is the rate that corresponding to the actual interest Periods (m) . Value of ieff is obtained dividing the nominal interest rate “r” by the number interest periods per year “m” as follow : ieff = r% / m (per Compounding period m) . 3. The effective annual interest rate , ieffa ,and calculated as follow : m ieffa = (1 + r/m) -1 Example 6 : A bank claims to pay interest to its depositors at rate of 6% per year compounded quarterly . What are the nominal , effective interest rates and the effective annual interest rates ? The nominal interest rates is r = 6% .since there are four interest periods per year (m) then the effective interest rate is : ieff =6%/4 = 1.5 % per quarter or quarterly . The effective annual interest rate , ieffa , obtained as follow : ieffa = (1 + r/m)m -1= (1+.06/4)4 -1 = .06136 = 6.136 % Example 7 : Three different bank loan rates for electric generation equipment are listed below : Determine the effective rate on the basis of the compounding period for each rate. (a) 9% per year , compounded quarterly (b) 9% per year , compounded monthly (c) 4.5 % per 6 months , compounded weekly . solution Apply the equation ieff = r% / m to determine the effective rate per compounding period (cp) for different compounding periods as shown in the following table :

1.12 Cash Flow Diagram : The cash flow defined as the flow of receipts (income) and cash disbursement (costs) which occur over a given time interval. The cash flow diagram is simply a graphical representation of cash flows drawn on a time scale as shown.

General Rules: - The cash flow follows the “end – period conventions “i.e. all cash flows occur at the end of interest period. End of period means one time period from date of transaction. - The direction of arrows is very important as “+” represents income & “– “represents disbursements. - All unknown values for the symbols P, A, F, n, i are represented by a question mark “? - Time 0 represent the present time while the times 1, 2, ….. , 5 are the end of time periods 1, 2,3,4,5. Example 1.8: A person deposited 10000 L.E now into his account which pays 10% per year. He plans to withdraw an equal end of year amount of 2000 L.E for 5 years starting next year and closing the account by with drawing the remaining money at the of sixth year . Construct the cash flow diagram. Find the amount of exit in each of end interest periods and the remaining sum. Solution:

P1 = sum at end of year 1 = P (1+i ) - 2000 = 10000(1.1) - 2000 = 9000 L.E P2 = sum at end of year 2 = P1 (1+ i ) - 2000 = 9000( 1.1 ) – 2000 = 7900 l..E P3 = sum at end of year 3 = P2 (1+ i ) - 2000 = 7900( 1.1 ) - 2000 = 6690 L.E P4= sum at end of year 4 = P3 ( 1+ i ) -2000 = 6690 ( 1.1 ) - 2000 = 5359 L.E P5 = sum at end of year 5 = P4 (1+ i ) - 2000 = 5359( 1.1 ) - 2000 = 3894.9L.E P6 = sum at end of year 5 = P5(1+ i ) – 0.0 = 3894.9( 1.1 ) – 0.0 = 4284.39 L.E The remaining sum = 4284.39 L.E Note: The height of arrows represents the value of money e.g height of arrow at 10000 L.E < height of 2000 L.E. Example 1.9: One want to make a deposit into his account so that he can withdraw an equal amount of 500 L.E per year for the first 3 years starting one year after deposit and a different annual amount of 750 L.E per year for the following two years . Draw the cash flow diagram and find the value of the deposit if i = 10 % per year. A1=500 L.E A2=750 L.E Solution:

0 1 2 3 4 5 i = 10 %

P=? P1= sum at end of year 1= P (1+i ) - 500 = 1.1*P - 500 P2= sum at end of year 2= P1 (1+ i ) - 500 = (1.1*P - 500 )( 1.1 ) – 500 = 1.21*P-1050 P3= sum at end of year 3= P2 (1+ i ) - 500 = (1.21*P-1050 )( 1.1 ) - 500 = 1.331*P-1655 P4 sum at end of year 4 = P3 ( 1+ i ) -750 = (1.331*P-1655 )( 1.1 ) - 750 = 1.4641*P-2570.5 P5=sum at end of year 5= P4(1+ i )- 750 = (1.4641*P-2570.5)( 1.1 )- 750 = 1.61051*P-3577.55 P5= The remaining sum = 0.0 =1.61051*P-3577.55 Then; P = Deposited value = 3577.55/1.61051 = 2221.377 L.E .

Solved Problems 1.1 The ABC Company deposited $100 000 in a bank account on June 15 and withdrew a total of $115 000 exactly one year later. Compute: (a) the interest which the ABC Company received from the $100 000 investment, and (b) the annual interest rate which the ABC Company was paid.

1.3 Compare the interest earned from an investment of $1000 for 15 years at 10% per year simple interest, with the amount of interest that could be earned if these funds were invested for 15 years at 10°/o per year, compounded annually. The simple interest is given by I = (15)(0.10)($1000=) $1500 Compound interest = F-P = P(1+ i)n - P = $1000(1+0.10)15-$1000 = $1000(4.17725)- $1000 = $3177.25 or more than double the amount earned using simple interest. 1.4 How it would take for an investor to double his money at 10% per year compounded annually ?

Actually, since the interest is compounded only at the end of each year, the investor would have to wait 8 years. 1.5 Suppose that a man lends $1000 for four years at 12% per year simple interest. At the end of the four years, he invests the entire amount which he then has for 10 years at 8% interest per year, compounded annually. How much money will he have at the end of the 14-year period? n2 10 F = P(1+ n1 x il)(l+ i2) = $1000[1+ (4)(0.12)](1+ 0.08) = $1000(1.48)(2.15892) = $3195.21 1.6 Suppose that the interest rate is 10% per year, compounded annually. What is the minimum amount of money that would have to be invested for a two-year period in order to earn $300 in interest? I = 300 = F – P = P(1.1)^2 –P P = 300/ (1.1^2 – 1) = 1428.57

Higher Technological Institute Engineering economy MNG102 Tenth Of Ramadan Sheet 1______I. Answer The Following Questions : 1.1 Mention the Objectives and functions of a firm . 1.2 Mention the Types of Economic systems indicating for each one two of the following terms: main characteristics - Examples - Advantage – disadvantages . 1.3 Mention the Kinds of Market structure indicating two main characteristics of each 1.4 Explain in brief the demand curve and its characteristics . 1.5 Explain in brief the supply curve and its characteristics . 1.6 Explain in brief the meaning of balance sheet and it, s structure using a suitable diagram . 1.7 Calculate the principle and the present value of a sum that has been deposited three years ago to become 12000 L.E after one year from now in both cases of simple and compound interest of 12 % per year. Calculate also the interest. 1.8 If you invest 10000 L.E now in a business venture that promises to return 14641 L.E, how many years required to receive this return in order to make interest rat of 10 % per year compounded yearly. On your investment? 1.9 Assume that you have been offered an investment opportunity in which you may invest $1000 at 7%per year simple interest for 3 years or you may invest the same $1000 at 6% per year compound interest for 3 years. Which investment offer you accept? 1.10 Sales revenues for a lift – truck product line are estimated to be 500,000 L.E in the first year, then decrease by 40000 L.E per year up to year 5 at interest rate of 15 % per year. Draw the cash flow diagram, and then calculate the future worth at the end of year five. 1.11 As a result of an old loan for a bank, there remain 5 equal payments each of 10000 L.E with interest rate of 8 % per year. The house just been sold to a new owner, who wishes to renegotiate the loan to reduce the annual payments by increasing it’s number to ten instead of five years. The bank agree but with interest rate of 10 % per year. Calculate the amount of the new annual payments and the total amounts of money received by the bank in both cases. 1.12 The costs of production in a factory is 86120 L.E in the first year , 97100 L.E in the second year and 105630 L.E in the fourth year with interest rate of 14% per year . A tooling investment of 12000 L.E is carried out now to reduce all production costs by 12% per year. Calculate the present worth before and after carrying out the tooling investment. If this investment is delayed for one year from now; Calculate the present worth of the cost of delay. Draw the cash flow diagrams in all cases. 1.13 Calculate the present worth and the future worth of an expenditure of 17000 L.E per year for 6 years starting 3 years from now if the interest rate if 15% per year . Draw the cash flow diagram and list the values of the engineering economy symbols used in this problem. Calculate the equivalent annual expenditure if it starts from first year up to the end of the interest period. 1.14 Suppose that a person invests $3000 at 10% per year, compounded annually, for 8 years. (a) Will this effectively protect the purchasing power of the original principal, given an annual inflation rate of 8%?(b) If so, by how much? [(a) yes; (b) $474.34] . II. Select the correct Answer from the following Questions : 2.1 The market organization structure which consists of many firms and each firm possess an element of market power is called : a. Perfect competition . b. Monopoly . b. Oligopoly . c. Monopolistic competition …. 2.2 The type of industry organization that is characterized by recognized interdependence and non-price competition among firms is called a. monopoly…. b. perfect competition. c. oligopoly. d. monopolistic competition. 2.3 Which of the following is a characteristic of monopolistic competition? a. Few sellers…. b. All of the above are characteristics of monopolistic competition. c. Easy entry into and exit from the industry. d.A differentiated product. 2.4 Which of the following industries is most likely to be monopolistically competitive? a. The automobile industry b. The steel industry …… c. The car repair industry d. The electrical generating industry 2.5 If the price of a good increases while the quantity of the good exchanged on markets decreases, then the most likely explanation is that there has been a. an increase in demand. b. a decrease in demand. c. an increase in supply. d. a decrease in supply. … 2.6 If a market is at equilibrium and if there is a sudden increase in demand, you have a A. surplus B. shortage…. C. Stays the same 2.7 When quantity supplied is greater than quantity demanded, you have a ------A. shortage…. B. surplus… C. deficit D. equilibrium 2.8 Which of the following is a category or element of the balance sheet? A. Expenses B. Gains C. Liabilities 2.9 Which of the following would not be a current asset? A. Accounts Receivable B. Land…. C. Prepaid Insurance D. Supplies 2.10 When an owner draws $5,000 from a sole proprietorship or when a corporation declares and pays a $5,000 dividend, the asset Cash decreases by $5,000. What is the other effect on the balance sheet? A. Owner's/Stockholders' Equity Decreases…. B. None 2.11 On December 1, ABC Co. hired Juanita Perez to begin working on January 2 at a monthly salary of $4,000. ABC's balance sheet of December 31 will show a liability of a. $4,000 b. $48,000 c. No Liability…. 2.12 ABC Co. has current assets of $50,000 and total assets of $150,000. ABC has current liabilities of $30,000 and total liabilities of $80,000. What is the amount of ABC's owner's equity? A. $20,000 B. $30,000 C. $70,000…. D. $120,000 2.13 Which of the following is an economic system in which economic decisions are made according to social roles & culture? a. Market economy b. Command economy c. Traditional economy…… d. Mixed economy 2.14 Which of the following is an economic system in which a central, governmental authority decides how to use a country's scarce resources? a. Market economy b. Command economy……. C. traditional economy d. Mixed economy 2.15 Which of the following is an economic system in which the government and individuals are used to decide how to use scarce resources? a. Market economy b. Command economy c. traditional economy d. Mixed economy……. 2.16 Type of economy that is based on trading and bartering? a. Socialist b. Traditional c. Free Enterprise d. Command 2.17 An economic system in which private property is almost totally restricted is called ... A. A mixed economy B. Competition C. Free enterprise D. A centrally planned economy III. Answer The Following Questions Using Either (√ ) OR ( χ ) Only: 3.1 Market structure refers to the competitive environment in which the buyers and sellers of a product operate… T 3.2 A market is defined as a place where buyers go to purchase units of a commodity …..F 3.3 Oligopoly refers to a type of market organization that is characterized by large number of firms selling a differentiated commodity….F 3.4 Monopolistic competition is a form of market organization that combines elements of perfect competition and monopoly……T 3.5 Monopoly is a market structure in which there is only one buyer of a product for which there are no close substitutes….F 3.6 Oligopoly is a market structure in which there are few sellers of a product and additional sellers cannot easily enter the industry T…... 3.7 The demand curve shows that the demanded quantity increases with increasing the price ….F 3.8 In supply curve , the supplied quantity increases with increasing the price 3.9 The demand curve shows that the demanded quantity increases with increasing the price….F 3.10 A price ceiling imposed above the market equilibrium price will result in a shortage of the product…. F 3.11The law of demand refers to the relationship between consumer income and the quantity of a commodity demanded per time period…F 3.12 The substitution effect holds that an increase in the price of a commodity will cause an individual to search for substitutes….T 3.13 The income effect holds that a decrease in the price of a commodity is, some respects, the same as an increase in income….T 3.14 The balance sheet heading will specify a Period Of Time…..F 3.15 The balance sheet heading will specify a Point Of Time…..T 3.16 Repaid expenses are considered as one of current liabilities in the balance sheet….F 3.17 Fixed assets are considered as one of current assets in the balance sheet . …… F 3.19 Customer prepayments are considered as one of current liabilities in the balance sheet….T 3.20 Pension fund liabilities are considered as one of current liabilities in the balance sheet…..F 3.21 If current assets > current liabilities ,the then company does not has enough cash to run the business…..F Dr HOSNY ABBAS ABOUZEID CHAPTER 2 SINGLE PAYMENT FACTORS 2.1 Introduction The aim of this chapter is to derive a formula for the following engineering economy factors: - Single – Payment Factors ( Compound Amount Factor & Present Worth Factor ) - Uniform Series Compound – Amount Factor. 퐹 2.2 Derivation of Single – Payment formulas Factors: Compound Amount ( ) 푃 푃 &present worth ( ): 퐹 In this case , it is required to get the final value “F” , in terms of given values of a single payment “P” with compound interest rate “ i” , and after time period “ n” , as shown in the cash flow diagram . Since F1= final sum at year 1=P +P*i , F2=F1+F1*i , P i= …%

ퟐ Then F2= Final sum at end of year 2 = P*(ퟏ + 풊) , 0 1 2 n=… F3= Final sum at end of year 3 =F2+F2*i F? ퟑ 풏 = F2(1+i)= P*(ퟏ + 풊) , So, Fn =Final sum at end of year n =P*(ퟏ + 풊) 푭? = (ퟏ + 풊)풏 called single payment compound amount factor 푷 푷? ퟏ = called single payment present worth factor 푭 (ퟏ+풊)풏 푷 2.3 Derivation of Uniform – Series Present Worth Factor ( ): 푨 In this case, it is required to get the present worth value P in terms of known values Of A, i, n as shown in the cash flow diagram. P? i= …. % P=P1+P2+P3+…..= Pn = Sum of all the present worth values 0 1 2 3 n-1 n 푨 푨 푨 = + +……… A = …… L. E ퟏ+풊 (ퟏ+풊)ퟐ (ퟏ+풊)풏

ퟏ Multiply both sides by we get : ퟏ+풊 푷 ퟏ ퟏ ퟏ ퟏ = A[ + + ⋯ + + ] ퟏ+풊 (ퟏ+풊)ퟐ (ퟏ+풊)ퟑ (ퟏ+풊)풏 (ퟏ+풊)풏+ퟏ

Then, 푷 ퟏ ퟏ ퟏ ퟏ ퟏ ퟏ -P=A[ + + ⋯ + + − { + + ퟏ+풊 (ퟏ+풊)ퟐ (ퟏ+풊)ퟑ (ퟏ+풊)풏 (ퟏ+풊)풏+ퟏ ퟏ+풊 (ퟏ+풊)ퟐ ퟏ ퟏ + ⋯ + }] (ퟏ+풊)ퟑ (ퟏ+풊)풏 풏 −풊 ퟏ ퟏ 푨 ퟏ−(ퟏ+풊) P [ ] = A ( ) [ - 1] = [ ] ퟏ+풊 ퟏ+풊 (ퟏ+풊)풏 ퟏ+풊 (ퟏ+풊)풏

푷 (ퟏ+풊)풏−ퟏ  = , Called Series Present Worth Factor 푨 풊(ퟏ+풊)풏

푨 풊(ퟏ+풊)풏  = , which is called the Capital Recovery Factor “CRF” which 푷 (ퟏ+풊)풏−ퟏ gives The value of A in terms of p , i , n. Note that P must always be located one period prior to the first A.

2.4 Derivation of the uniform- series compound (given A, F is unknown) : Amount Factor “F/A”; (USCAF); And the sinking Fund Factor “A/F”; (SFF): (given F, A is unknown)

푭 푭 푷 (ퟏ+풊)풏−ퟏ (ퟏ+풊)풏−ퟏ = ∗ = (ퟏ + 풊)풏 ∗ = = 푼푺푪푨푭 푨 푷 푨 풊(ퟏ+풊)풏 풊

푨 풊 ∴ = = SFF which gives the value of A in terms of F, i, n. 푭 (ퟏ+풊)풏−ퟏ In these cases A starts at year 1 and ends at year n where F is existed as shown.

0 1 2 3 …………. n-1 nF

2.5 Standard Factor Notation and Use of Interest Tables: The various factors will be represented by a standard notation of the form (X|Y, i%, n) where: X = what you want to find (unknown) Y = what is given (known) i = interest rate in percent n = number of periods involved. e.g (F/P, 6%, 20) means it required to get value of the factor 푭 at i= 6% ; n=20. To get value of 푷 F; multiply this factor by P i.e F = P*푭 or F = P (F/P, 6%, 20) 푷 The following table summarize the standard factors notation used in the interest tables .

Notes: - In all above formulas one parameter only is unknown, the remainders must be given. - Annual payment start in year 1 while gradient starts in year 2. - “i “ must be constant . The interest tables are prepared the values of all the above mentioned factors starting from i=0.5% to 50% and times from 1 to 100 years as follow: Table A- 1 Discrete Cash Flow 0.5% Discrete Compound interest factors Single Payment Uniform Series Payments Compound Present Sinking Compound Capital Present N amount worth Fund amount Recovery Worth F/P P/F A/F F/A A/P P/A

1 1 1.005 0.995 1.0 1.0 1.005 0.995 2 2 1.0151 0.9851 0.33167 3.015 0.33667 2.9702 3 3

4 .

5 .

. .

. .

. .

. .

. .

100 100

Examples: • (F/A, 0.5%, 3) = 3.015 from table A-1 at i=0.5%; n=3 (ퟏ+풊)풏−ퟏ (ퟏ+ퟎ.ퟎퟎퟓ)ퟑ−ퟏ Or (F/A, 0.5%, 3) = = = ퟑ. ퟎퟏퟓퟎퟐퟓ 풊 ퟎ.ퟎퟎퟓ • (P/A, 5%, 10) = 7.7217 from table A-7 at i=5%; n=10 (ퟏ+풊)풏−ퟏ (ퟏ.ퟎퟓ)ퟏퟎ−ퟏ Or (P/A, 5%, 10) = = = 7.7217 풊(ퟏ+풊)풏 ퟎ.ퟎퟓ(ퟏ.ퟎퟓ)ퟏퟎ • (P/F, 25%, 35) = 0.0004 from table A-25 풊 풊 Or (P/F, 25%, 35) = = = 0.0004 (ퟏ+풊)풏 (ퟏ.ퟐퟓ)ퟑퟓ

2.6 Interpolation in Interest Tables: For factors which corresponds to values of either i or n that is not exist in the interest table, we locate the values of these parameters between two known values in the tables i1, i2 or n1, n2 then the values of these factors are obtained from the equation: 풊−풊 ퟏ where 푭 corresponds to F/P or P/F or P/A, A/p, ….. etc 푭풊 = 푭풊ퟏ + (푭풊ퟐ − 푭풊ퟏ), 풊 풊ퟐ−풊ퟏ Example: Determine the value of the equivalent uniform annual series for an investment of 100,000 L.E within ten years from now with interest rate of 7.3% per year. Solution: A= 100,000 (A/P, 7.3%, 10) From Tables we get i N A/P

풊ퟏ = ퟕ% 10 0.14238 = 푭풊ퟏ i=? 10 F=? 풊 =8% 10 ퟐ 0.14903=푭풊ퟐ 푨 ퟕ. ퟑ − ퟕ ∴ | = ퟎ. ퟏퟒퟐퟑퟖ + (ퟎ. ퟏퟒퟗퟎퟑ − ퟎ. ퟏퟒퟐퟑퟖ) = ퟎ. ퟏퟒퟒퟑퟕퟓ 푷 ퟕ.ퟑ ퟖ − ퟕ ∴ 푨 = 풆풒풖풊풗풂풍풆풏풕 풖풏풊풇풐풓풎 풂풏풏풖풂풍 풔풆풓풊풆풔 푨 =푷 ∗ | = ퟏퟎퟎ, ퟎퟎퟎ ∗ ퟎ. ퟏퟒퟒퟑퟕퟓ = ퟏퟒퟒퟑퟕퟓ 푳. 푬 푷 ퟕ.ퟑ 2.7 Definition and Derivation of Gradient Formulas: A uniform gradient is a cash-flow series which either increases or decreases uniformly i.e changed by the same amount. In this case the cash flow is different with time (i.e. not constant like the uniform series). The cash flow will be as follow: 0 1 2 3 …………. n-1 n

400 Ab 425 450

Ab Ab 400+ (n-2)25 +G +2G 400+ (n-1)25 Ab+ (n-2) G Ab+ (n-1) G Cash Flow Diagram For A Uniform - Gradient With a gradient G G =25 in this case the value of G is obtained as follow : 풇풊풏풂풍 풑풂풚풎풆풏풕풔 풐풇 풖풏풊풇풐풓풎 품풓풂풅풊풆풏풕 − 푩풂풔풆 풑풂풚풎풆풏풕 푮 = 풏풖풎풃풆풓 풐풇 풚풆풂풓 풇풐풓 풕풉풆 풖풏풊풇풐풓풎 품풓풂풅풊풆풏풕 풑풆풓풊풐풅 − ퟏ 푨풇 − 푨풃 ∴ 푮 = 푵 − ퟏ If we ignore the base payment, the generalized uniformly increasing-gradient cash-flow diagram will be as follow:

This gradient is called a “Conventional gradient” when it starts at year 2 from the present time or when the base payment starts at year 1. In all other cases the gradient is called “unconventional gradient”.

2.8 Derivation of Uniform Gradient Factors Formulas: 2.8.1 Uniform-gradient present- worth factor 푷 (UGPWF) 푮 for conventional gradient : P = G(P/F, i%, 2) + 2G(P/F, i%, 3) +3G(P/F, i%, 4) +……+ (n-2)G(P/F, i%, n-1) + (n-1)G(P/F, i%, n) ퟏ ퟐ 풏−ퟐ 풏−ퟏ P = G[ + + ⋯ + + ] ………….. (A) (ퟏ+풊)ퟐ (ퟏ+풊)ퟑ (ퟏ+풊)풏−ퟏ (ퟏ+풊)풏 Multiply both sides by (1+i) we get: ퟏ ퟐ ퟑ 퐧−ퟐ 풏−ퟏ P(1+i) = G[ + + + ⋯ + + ]…. (B) ퟏ+퐢 (ퟏ+퐢)ퟐ (ퟏ+퐢)ퟑ (ퟏ+퐢)퐧−ퟐ (ퟏ+풊)풏−ퟏ (B) – (A) will give: ퟏ ퟐ−ퟏ ퟑ−ퟐ (풏−ퟏ)−(풏−ퟐ) 풏−ퟏ P[ (1+i) -1] = G [ + + + ⋯ + − ] ퟏ+풊 (ퟏ+풊)ퟐ (ퟏ+풊)ퟑ (ퟏ+풊)풏−ퟏ (ퟏ+풊)풏 ퟏ ퟏ ퟏ ퟏ 풏−ퟏ Pi = G {[ + + + ⋯ + ] − } ퟏ+풊 (ퟏ+풊)ퟐ (ퟏ+풊)ퟑ (ퟏ+풊)풏−ퟏ (ퟏ+풊)풏 푷 ퟏ 풏−ퟏ P= 푮 [ ( - ) - ] 풊 푨 (ퟏ+풊)풏 (ퟏ+풊)풏 i=….% 푮 (ퟏ+풊)풏−ퟏ 풏 0 1 2 3 …………. n-1 n = [ - ] 풊 풊(ퟏ+풊)풏 (ퟏ+풊)풏

푷 ퟏ (ퟏ+풊)풏−ퟏ 풏 ∴ = [ - ] G=…… 푮 풊 풊(ퟏ+풊)풏 (ퟏ+풊)풏 P?

2.8.2 Uniform gradient annual series factor 푨 (UGASF): 푮 This factor is used to convert from uniform gradient to an equivalent annual series as shown below: i=….% i=….% 0 1 2 …………. n-1 n 0 1 2 3 …………. n-1 n G Ab A It’s value is obtained as follow: 푨 푨 푷 풊(ퟏ+풊)풏 ퟏ (ퟏ+풊)풏−ퟏ 풏 ퟏ 풏 = ∗ = ∗ [ - ] = - 푮 푷 푮 (ퟏ+풊)풏−ퟏ 풊 풊(ퟏ+풊)풏 (ퟏ+풊)풏 풊 (ퟏ+풊)풏−ퟏ These two gradient factors are obtained from tables P/G; A/G as follows in next examples: Example: find the value of (P/G,4%,6) Solutions: ퟏ (ퟏ+ퟎ.ퟎퟒ)ퟔ−ퟏ ퟔ (P/G, 4%, 6) = [ - ] = 12.50624 ퟎ.ퟎퟒ ퟎ.ퟎퟒ(ퟏ+ퟎ.ퟎퟒ)ퟔ (ퟏ+ퟎ.ퟎퟒ)ퟔ Or from the table: Present Worth Gradient Factors N (P/G)1% 2% 3% 4% 5% 6% N

2 2 3 ...... 6 14.321 13.68 13.076 12.506 11.968 11.459 6 (P/G, 4%, 6) = 12.506 2.9 Present Worth, Future Worth, Equivalent Uniform annual Series; and Conventional Gradients: Steps of Calculations are: 1. Draw the cash flow diagram for the problem 2. Select the suitable engineering economy factors based on which is given and which is required. 3. Specify correctly the conditions for which the formulas apply e.g uniform- series factors could not be used if payments or receipts occurred every other (more than one) year. 4. For conventional gradients (i.e. gradients begins in year 2); the present worth and the equivalent uniform annual series are obtained from the following equations:

PT = Present worth due to base payment (or receipts) “PA” + Present worth due to

gradients value “PG” =

Ab(P/A, i%, n) + G(P/G, i, 10) & AT = Ab + AG = Ab + G(A/G; i% , 10)

Or AT = PT (A/O, i%, n)

Examples: 1. An investor deposited 60000 L.E now; 30000 two years from now; and 40000 L.E five years in order to be able to buy machine after ten years from now. Calculate the price of this machine at same time if the interest rate is 5% per year. Solution : F = price of machine after ten years F? = 60000(F/P, 5%, 10) + 30000(F/P, 5%, 8) + 40000(F/P, 5%, 5) = 60000(1.6289) + 30000(1.4775) + 40000(1.2763) 0 1 2 3 4 5 = 193111 L.E i= 5% n=10 30000 40000 60000 Or to avoid mistakes in compute n: 0 F = PT (F/P, 5%, 10) = [60000 + 30000(P/F, 5%, 2) + 40000 (P/F,5%, 5)] * (F/P, 5%, 10) = 193110.6 L.E 2. A company buys a machine in the forms installments starting by 50000 L.E at year 1 and then increasing by 10000 L.E per year up to year 10. Calculate the present worth and the equivalent uniform annual series of this machine if the interest rate is 5% per year Solution:

PT = 50000(P/A, 5%, 10) + 10000(P/G, 5%, 10) = 50000(7.7217) + 10000(31.652) 0 1 2 3 …………. n = 10 = 702605 L.E. 50000 EUAS = Ab + G(A/G, 5%, 10) = 50000 + 10000 (4.099) = 90990 G=10000 L.E Or i=5% EUAS = 702605 (A/P, 5%, 10) = 90990

2.10 Calculation of Unknown Interest Rates “i” & Unknown years “n”: Examples 1. Calculate the rate of interest for the case of initial deposit of 60000 L.E and final receipt of 100000 L.E after 5 years F = 100000 ퟔퟎퟎퟎퟎ 0 1 2 3 4 n=5 (P/F, i%, 5) = = 0.6 ퟏퟎퟎퟎퟎퟎ i?

From tables P=600000 L.E i N P/F ퟎ. ퟔퟐ − ퟎ. ퟐퟎퟗ ∴ 풊 = ퟏퟎ + (ퟏퟏ − ퟏퟎ) ퟏퟎ 5 0.6209 ퟎ. ퟓퟗퟑퟓ − ퟎ. ퟔퟐퟎퟗ ? 0.6 = ퟏퟎ. ퟕퟔ% ퟏퟏ 5 0.5935 푷⁄ − 푷⁄ 푭 푭ퟏ ∴ 풊 = 풊ퟏ + (풊ퟐ − 풊ퟏ) 푷⁄ − 푷⁄ 푭ퟐ 푭ퟏ ퟏ Using the equation: (P/F, i%, 5) = 0.6 = (ퟏ+풊)ퟓ ퟏ ∴ 풊 = − ퟏ = ퟏ. ퟏퟎퟕퟔ − ퟏ = ퟎ. ퟏퟎퟕퟔ = ퟏퟎ. ퟕퟔ% ퟎ.ퟔퟎ.ퟐ

2. An investor borrows 100000 L.E now to buy a machine and repay this loan on the form of annual installments with interest rate of 15%. Calculate the number of these installments if it starts one year from now by 10000 L.E and then increased gradually by 1000 L.E per year after that. P= 100000 i= 15% 100000 = 10000(P/A, 15%, n) + 1000(P/G, 15%, n) 0 1 2 n = PT 10000 G=1000 L.E

n P/A P/G PT Notes ퟏퟎퟎퟎퟎퟎ−ퟗ.ퟗퟐퟎퟐ 5 3.3533 5.775 39297 <<10000 ∴ 풏 = ퟐퟐ + (24 -22) 15 5.8474 26.693 85167 ퟏퟎퟏퟔퟑퟖ−ퟗퟗퟐퟎퟐ 20 6.2543 33.582 96175 ∴ 풏 = ퟐퟐ. ퟔퟔ 풚풆풂풓 22 6.3587 35.615 99202 <100000 24 6.4338 37.302 101638 >100000

Solved Problems Containing Single ,Uniform-Series ,Gradient - Series Factors 1. Suppose that a person deposits $500 in a savings account at the end of each year, starting now, for the next 12 years. If the bank pays 8% per year, compounded annually, how much money will accumulate by the end of the 12-year period? F = $500 x (FIA, 8%, 12) = $500(18.9771) = $9488.55 2.A man has deposited $50 000 in a retirement income plan with a local bank. This bank pays 8.75% per year, compounded annually, on such deposits. What is the maximum amount the man can withdraw at the end of each year and still have the funds last for 12 years?

3.A father wants to set aside money for his 5-year-old son's future college education. Money can be deposited in a bank account that pays 8% per year, compounded annually. What equal deposits should be made by the father, on his son's 6th through 17th birthdays, in order to provide $5000 on the son's 18th, 19th, 20th, and 21st birthdays? On the son's 17th birthday, the deposits must have accumulated to P = $5000(P/A, 8%, 4) = $5000(3.312) = $16 560.68 Thus, the deposit size, A, must satisfy $16 560.68 = A (FIA, 8%, 12) $16 560.68 = A (18.9771) therefore : A = $872.67 4. The ABD Company is building a new plant, whose equipment maintenance costs are expected to be $500 the first year, $150 the second year, $200 the third year, $250 the fourth year, etc., increasing by $50 per year through the 10th year. The plant is expected to have a 10-year life. Assuming the interest rate is 8% , compounded annually, how much should the company plan to set aside now in order to pay for the maintenance?

First we compute the present worth at the end of year 1 , p’ P' = $500 + $150(P/A, 8%, 9) + $50(P/G, 8%, 9) = $500 + $I50(6.249) + $50(21.808) = $2527.42 The present worth at the end of year 0 is thus P = P' (P/F, 8%, 1) = $2527.42(1.0800)-' = $2340.20 5. Mr. Jones is planning a 20-year retirement; he wants to withdraw $6000 at the end of the first year, and then to increase the withdrawals by $800 each year to offset inflation. How much money should he have in his savings account at the start of his retirement, if the bank pays 9% per year, compounded annually, on his savings? P = A o (P/A, 9%, 20) + G (P/G, 9%, 20) = $6000(9.17) + $800(61.777) = $104 189.14

6. Mr. Holzman estimates that the maintenance cost of a new car will be $75 the first year, and will increase by $50 each subsequent year. He plans to keep the car for 6 years. He wants to know how much money to deposit in a bank account at the time he purchases the car, in order to cover these maintenance costs. His bank pays 5.5% per year, compounded annually, on savings deposits

The P/A and P/G factors must be evaluated directly from :

ퟏ (ퟏ+ퟎ.ퟎퟓퟓ)ퟔ−ퟏ ퟔ (P/G, 5.5%, 6) = [ - ] = 11.71 ퟎ.ퟎퟓퟓ ퟎ.ퟎퟓퟓ(ퟏ+ퟎ.ퟎퟓퟓ)ퟔ (ퟏ+ퟎ.ퟎퟓퟓ)ퟔ P= 75 (4.9955) + 50 ( 11.71) = $ 960.17

7.Find the eight – year equivalent uniform annual series of uniform annual disbursement of 16000 L.E starting from third year with interest rate of 6% per year . First Solution : EUAS = PT (A/P , 6%, 8) = P'( P/F , 6% , 2) (A/P , 6%, 8) = A( P/A, 6% , 6) ( P/F , 6% , 2) (A/P , 6%, 8) = 16000 (4.9173)(0.890)(0.16104) = 11276.4 L.E

0 1 2 3 4 5 6 7 8 Second Solution : EUAS = F(A/F, 6%, 8) = A(F/A, 6%, 6) (A/F, 6%, 8) = 16000(6.975)(0.10104) = 11274 L.E EUAS 8. An investor makes a long – term investment in a company such that he will pay 20 payments each of 40000 L.E per year beginning 3 years from now ; 20,000 L.E six year from now and 30000 L.E sixteen years from now . Calculate the accumulated amount of money that the investor could obtain at the end of the interest period . Calculate also the present worth value for this investment if i = 6% per year .

F = A(F/A, 6% , 20) +20000(F/P , 6% , 16) + 30000( F/P , 6% , 6 ) = 40000(36.786) + 200000 ( 2.5404) + 30000 (1.4185) = 1564802 L.E PT = F( P/F , 6% , 22) = 1564802 ( 0.277505 ) = 434236.4 L.E 9. Calculate the present worth and the EUAS for annual payments that starts by 2500 L.E at year 1 up to year 3 then increased gradually and uniformly until reach 7500 L.E at year 8 with interest rate of 6% per year . 0 1 2 3 4 5 6 PI i=6% 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8

EUAS PT

7500 PT = PI + PII = 2500(P/A , 6% ,2) + [ 2500(P/A , 6% ,6) +G(P/G, 6% ,6)] (P/F , 6% ,2) G = (7500 – 2500)/(6-1) = 5000/5 = 1000 L.E PT = 2500(1.8334) + [2500(4.9173) + 1000 (11.459)](0.89) = 25723.01 L.E EUAS = PT(A/P , 6% , 8) = 25723.01* 0.16104 = 4142.485 L.E

10. Calculate the present worth and EUAS for five deposits started by 4000 L.E at year 1 then increased uniformly gradually to reach 8000 L.E at year 5 followed by withdrawals started by 20000 L.E at year 6 then decreased uniformly gradually to reach a constant value of 2000 L.E at year 10 ,11 and 12 , if the interest rate is 7% . 10000 G2

2000

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

4000 G1 8000 G1 = (8000 – 4000)/(5-1) = 1000 L.E ; G2 = (2000 -10000)/(5-1) = - 2000 L.E PG1 =present worth of increasing gradient or deposits = 4000(P/A , 7% , 5) + 1000 (P/A , 7%, 5) = 4000(4.1002) +1000(7.646) = 24046.8 L.E P'T = present worth of all withdrawals = PG2(P/F,7%,5)+ P'A (P/F,7%,10) = [10000(P/A, 7%,5)- 2000(P/G ,7% ,5)] (P/F,7%,5) +2000(P/A,7%,2) (P/F,7%,10) = [ 10000(4.1002) – 2000(7.646)](0.713) + 2000(1.808)(0.5084) = 20169.8 L.E Pnet = Net or resultant present worth = P'T- PG1 = 20169.6 – 24046.8 = -3877.2 L.E Which means annual deposits EUAS = Pnet(A/P,7%,12) = -3877.2(0.1259) = -488.14 L.E

11. An engineer wishes to purchase an $80 000 home by making a down payment of $20 000 and borrowing the remaining $60 000, which he will repay on a monthly basis over the next 30 years. If the bank charges interest at the rate of 9.5% per year, compounded monthly, how much money must the engineer repay each month? A = P x (AIP, ieff, mn) , ieff= i/m = annual interest rate/ No of interest periods = .095/12 , n=No of years =30 A=

It is interesting to note that the total amount of money which will be repaid to the bank is $504.51 x 360 = $181 623.60 or three times the amount of the original loan.

12. An engineer deposits $1000 in a savings account at the end of each year. If the bank pays interest at the rate of 6% per year, compounded quarterly, how much money will have accumulated in the account after 5 years? im= interest rate per 3months = .06/4 = .015 =1.5% F= 1000(A/F,.1.5%,4)(F/A,1.5%,20)= 1000[0.015/((1.015^4) -1)][(1.015)^20-1]/.015 = $5652.5 Another solution: 1 + i = (1+im/4)^4 , i = (1.015)^4 -1= .06136 = 6.136% per year =effective interest rate. F= 1000(F/A,6.136%,5) = 1000((1.06136)^5-1)/.06136] = $5652

Higher Technological Institute Engineering economy Tenth Of Ramadan Sheet 2______

1. The fixed cost of a machine maintenance will be 3500 L.E per year ; while the remaining cost will be 200 L.E starting two year from now and increase gradually and uniformly by 200 L.E per year . Calculate the expected present worth ; future worth ; and the equivalent uniform annual amounts for the total cost during the next ten years if the interest rate will be 12.4 % yearly .

2. Determine the present worth of a loan which will be repaid in annual payments starting one year from now and increase gradually by 1000 L.E per year starting from year two up to year ten where the final payments will be 10000 L.E with interest rate of 15 % per year compounded quarterly .

3. Calculate the present worth and annual payments for deposits that will be started at year 2002 and ends at year 2036 with final accumulated value of 275000 L.E and interest rate of 10% per year.

4. A factory is planned to reduce it’s operating cost for the next four year. If this reduction will follow a uniformly decreasing gradient of 12000 L.E per year such that the equivalent uniform annual amount will be 115000 L.E. Calculate the values of reduction in year 1, 2 ,3 ,and 4 if i = 12 % per year .

5. An investor deposits 6000 L.E now then his annual deposits increase uniformly and gradually by 300 L.E up to year ten. Where his total investment will reach 115000 L.E. Calculate the rate of interest.

6. A loan of 4000 L.E has to be repaid in the form of monthly payments each of 400 L.E starting next year with interest rate of 1.25 per month. How many months does it takes; and what is the amount of final payments?

7. An Oil Company is considering the purchase of a new machine that will last 5 years and cost $50 000; maintenance will cost $6000 the first year, decreasing by $1000 each year to $2000 the fifth year. If the interest rate is 8% per year, compounded annually, how much money should the company set aside for this machine?

8. An investor deposits 20000 L.E now, 5000 L.E three years from now, and 10000 L.E five years from now. How many years will it take from now for his total investment to amount 100000 L.E if the interest rate is 6%?

Dr HOSNY ABBAS ABOUZEID CHAPTER 3 Geometric Gradient Factors Formulas 3.1 Derivation of Geometric Gradient to Present Worth In this case a uniform gradient is a cash-flow series which either increases or decreases in a form of geometric series as shown below .

OR

Proof :

n-1 An = A1 / (1+g) , Where : g =uniform rate of cash flow increase/decrease ( + / - )from period to period, that is, the geometric gradient

A1 = Base amount or value of cash flow at Year 1 and g is the growth rate.

An =value of cash flow at any Year n P n = the present worth of any cash flow An at interest rate i n n-1 n = An / (1+i) = A1 ((1+g) / (1+i) In the general case, where i not equal to g ,then : The present worth of the entire gradient series of cash flows , P, may be obtained by

Subtract Equation 4-28 from Equation 4-27 to get :

Replacing the original values for a and b , we get :

Where i ≠ g

Example 1 : The first-year maintenance cost for a new automobile is estimated to be $100, and it increases at a uniform rate of 10%per year. Using an 8% interest rate, calculate the present worth of cost of the first 5 years of maintenance. Step by step solution : Maintenance PW of Year n Cost (P/F,8%,n ) Maintenance 1 100.0 = 100.0 x 0.9259 = $ 92.59 2 100.0 + 10%(100.0) = 110.0 x 0.8573 = 94.30 3 110.0 + 10%(110.0) = 121.0 x 0.7938 = 96.05 4 121.0 + 10%(121.0) = 133.0 x 0.7350 = 97.83 5 133.0 + 10% (133.0) = 146.41 x 0.6806 = 99.65 $ 480.42 Solution using geometric series present worth factor ퟏ−(ퟏ+품)풏(ퟏ+풊)−풏 P = A1 [ ] , where i not equal g 풊−품 ퟏ−(ퟏ.ퟏ)ퟓ(ퟏ.ퟎퟖ)−ퟓ = 100.0 [ ] = $480.42 −ퟎ.ퟎퟐ the present worth of cost of the first 5 years of maintenance is $480.42 . (1) g < 0 :  use formula (4-30) (4) g = I :  use formula (4-31) Example 2 : Engineers at a specific company need to make some modifications to an existing machine . The modification costs only $8,000 and is expected to last 6 years with a $1,300 salvage value . The maintenance cost is expected to be high a $1,700 the first year, increasing by 11% per year thereafter . Determine the equivalent present worth of the modification and maintenance cost. The interest rate is 8% per year

ƒ The present worth value is comprised of three components ƒThe present modification cost = $8,000 ƒThe present value of the future salvage value ƒThe present value of all the maintenance values throughout the 6 years and these are represented by the geometric gradient series

3.2 Derivation of Geometric Gradient to Future Worth To compute a future worth from a geometric gradient series use: 풏 −풏 n ퟏ−(ퟏ+품) (ퟏ+풊) n F=P(1+i) = A1 [ ](ퟏ + 퐢) 풊−품 n n F = A1[((1 + i) - (1 + g) )/(i - g)] use only if i does not equal g. The term [(1-(1 + g)n(1 + i)-n)/(i - g)] is called the geometric-gradient-series future worth factor. n-1 F = nA1(1 + i) , use if i = g. Example 3 : Suppose that your retirement benefits during your first year of retirement are $50.000. Assume that this amount is just enough to meet your cost of living during the first year. However, your cost of living is expected to increase at an annual rate of 5%, due to inflation. Suppose you do not expect to receive any cost-of-living adjustment in your retirement pension. Then. some of your future cost of living has to come from your savings other than retirement pension, If your savings account earns 7% interest a year, how much should you set aside in order to meet this future increase in cost of living over 25 years?

Given: Al = $50.000, g = 5%. i = 7%, and N = 25 years, as shown in above Figure Find: P. Find the equivalent amount of total benefits paid over 25 years:

P1 = 50000 ( P/A,7%,25) = 50000 x 11.65358 = $582679.159

Find the equivalent amount of total cost of living with inflation ,P2

P2= 50000 (P/A,5%,7%,25) =50000 x 18.80334 = $940167.2185 The required additional savings to meet the future increase in cost of living will be P,

P = P2 – P1 = 940167.2185 - $582679.159 = $357488.06

Single Project Evaluation Methods 1. NET PRESENT VALUE The net present worth (NPV) of a given series of cash flows is the equivalent value of the cash flows at the end of year 0 (i.e., at the beginning of year 1). For the case of annual compounding ,NPV is calculated as follow: NPV(i) = Net present Value calculated at interest rate a minimum attractive rate of return (MARR) i

An = net cash flow at the end of period n, i = MARR , and n = service life of the project.

An will be positive if the corresponding period has a net cash inflow and negative if the period has a net cash outflow.

We will first summarize the basic procedure for applying the net-present-worth criterion to a typical investment project, as well as for comparing alternative projects. Evaluation Steps for a Single Project : Step 1: Determine the interest rate that the firm wishes to earn on its investments. This interest rate is often referred to as either a required rate of return or a minimum attractive rate of return (MARR). Usually this selection is a policy decision made by top management. Step 2: Estimate the service life of the project. Step 3: Estimate the cash inflow for each period over the service life. Step 4: Estimate the cash outflow for each period over the service life. Step 5: Determine the net cash flows for each period (net cash flow = cash inflow – cash outflow). Step 6: Find the present worth of each net cash flow at the MARR. Add up these present- worth figures; their sum is defined as the project's NPV Step 7: In this context, a positive NPV means that the equivalent worth of the inflows is greater than the equivalent worth of the outflows, so the project makes a profit. Therefore , if the NPV(i) is positive for a single project, the project should be accepted; if it is negative, the project should be rejected. The process of applying the NPV measure is implemented with the following decision rule: If NPV(i) > 0, accept the investment. If NPV(i) = 0, remain indifferent. If NPV(i) < 0, reject the investment. Example 4 : The cash flows associated with a milling machine are Ao= -$50000. Aj = $15000 ( j = 1,. . . ,5 ) . Determine the economic acceptability of this machine at interest rates of (a) 10°/o, (b) 15%, and (c) 20% per year, all compounded annually .

The machine is seen to be an economically acceptable investment when the interest rate is 1O0/0, and (barely) when the interest rate is 15%. It is not economically justifiable to buy the machine if the interest rate is 20%. Example 5 Tiger Machine Tool Company is considering the acquisition of a new metal cutting machine. The required initial investment of $75,000 and the projected cash benefits over a three-year project life are as follows :

You have been asked by the president of the company to evaluate the economic merit of the acquisition. The firm's MARR is known to be 15%. If we bring each flow to its equivalent at time zero as shown in Figure below, we find that NPV = -$75000 + $24000(P/F,15%,1) +$27340(P/F,15%,2) + $55760(p/f,15%,3) = $3553 . Since the project results in a surplus of $3,553, the project is acceptable. It is returning a profit greater than 15%.

2.Internal Rate Of Return ( IRR ) The internal rate of return (IRR) for a series of cash flows is that particular value, i*, of the interest rate for which the NPV vanishes. Thus, if we plot the NPV as a function of i, using 0ne of the following equations , the curve will cross the i-axis at i*. Alternatively, we could find, by trial and error, i-values for which the NPV is slightly positive and slightly negative, and interpolate linearly between them for i*. If a more accurate approximation for i* is required, a numerical technique can be used to solve these equations for i, with the left side replaced by zero. IRR could be calculated one of the following equations : • PW of benefits - PW of costs = 0 or , • Net present worth(NPV) = 0 or , • PW of benefits / PW of costs =1 or, • PW of costs = PW of benefits or, • EUAB - EUAC = 0 EUAB AND EUAC are the equivalent uniform annual series of benefits and costs respectively .

Example 6 : Find the IRR (internal rate of return ) for the machine of Example 4. We can use NPV = 0 By linear interpolation between the results of Example l (b) and (c):

The IRR criterion (i*) could applied to take a decision for a typical simple investment project according to the following rule : If IRR > MARR. accept the project. If IRR = MARR. remain indifferent. If IRR < MARR, reject the project.

Example 7: An investment resulted in the following cash flow .Compute the internal rate of return

We can use EUAB - EUAC = 0 ,then 100+ 75(A/G, i, 4) -700(A/ P, i, 4) =0 by trial and error. Try i =5% first: At i = 5%, EUAB - EUAC=100+ 75(1.439) -700(0.2820) = 208 -197 = +11 Then try i =8%: At i =8%, EUAB - EUAC = 100+ 75(1.404) - 700(0.3019) = 205 - 211 = -6 We see that the true rate of return is between 5% and 8%. Try i =7%: At i = 7%, EUAB - EUAC = 100+ 75(AfG, 7%, 4) -700(Af P, 7%, 4) = 100+ 75(1.416) - 700(0.2952) = 206- 206= 0 The IRR is 7%.

Example 8 For the series of cash flows

determine the NPV at annual interest rates 0% , 5%, lo%, 20%, 30%, SO%, and 70%. From a graph of the results, find the rate(s) of return. For the given flows, NPV = -$3000 + $6000(P/A,i%, 2)(P/F,i %, 1)- $10 000(P/F,i %, 5) and evaluation at the specified interest rates gives the points

which are plotted in the following figure . It is seen that there are two rates of return in this case, i* - 7% and i* = 54%.

If multiple i*-values exist, it is usually better to abandon the IRR method and instead to investigate the sign of the NPV for various assumed values of the interest rate.

Higher Technological Institute Engineering economy Tenth Of Ramadan Sheet 3______1. What is the amount of 10 equal annual deposits that can provide five annual withdrawals, when a first withdrawal of $1,000 is made at the end of year 11, and subsequent withdrawals increase at the rate of 6% per year over the previous year's if the interest rate is 8%, compounded annually? 2. Facility has aging cooling system which currently runs 70% of the time the plant is open – Pump will only last 5 more years. As it deteriorates, the pump run time is expected to increase 7% per year . New cooling system would only run 50% of the time What is the value of replacing the pump , if : • Either pump uses 250 kWh, Electricity cost $0.05/KWh • Plant runs 250 days per year, 24 hours per day • Firm’s discount rate is 12% 3. Tuition costs are expected to inflate at the rate of 8% per year. The first year’s tuition is due one year from now and will be $10,000. To cover tuition cost for 4 years, a fund is to be set up today in an account that will earn interest at the rate of 5% per year, compounded annually. How much must be deposited into the fund today in order to pay the 4 years of tuition expenses ?

4.

5. A new plant to produce steel tubing requires an initial investment of $10 million. It is expected that after three years of operation an additional investment of $5 million will be required; and after six years of operation, another investment of $3 million. Annual operating costs will be $3 million and annual revenues will be $8 million. The life of the plant is 10 years. If the interest rate is 15% per year, compounded annually, what is the NPV of this plant? 6. Compute the IRR for the following cash flows:

Dr . HOSNY ABBAS ABOUZEID

Chapter 4 Comparison Of Alternatives For Proposed Investments Different methods are used: ◼ Present /Future worth ◼ Capitalized cost ◼ Annual equivalent cash flow ◼ Internal rate of return ◼ Benefit/cost ratios

1. Present / future worth Analysis : This analysis is used for alternatives of finite periods. When comparing Alternatives from cost or disbursements point of view, the lowest present / future alternative should be selected. But when comparing alternatives where income is greater than costs, then the highest present/future worth should be selected. 1.1 Equal lived alternatives : Example 1 It is required to select one of the two following machines for a project. Machine A has a first cost of 100000 L.E, annual operating cost of 36000L.E and salvage of 8000 L.E . Machine B has a first cost of 140000 L.E, annual operating cost of 28000 L.E. and salvage value of 14000 L.E . Determine which of these two machines should selected based on the present worth analysis if the interest rate was 10% per year and both machines has a life time of five years . 14000 L.E Solution: 8000 L.E 0 1 2 3 4 5 0 1 2 3 4 5

28000 L.E 36000 L.E 140000 100000 0 PWA = PA + A(P/A,10%,5) – S.V(P/F,10%,5) = 100000 + 36000(3.7908) – 8000(.6209) = 231501.6 L.E PWB = 140000 + 28000(3.7908) – 14000(.6209) = 237449.8 L.E PWA < PWB , Then machine A should be selected .

1.2 Different lived alternatives: In this case the alternatives must be compared over the same number of years by calculating the least common multiple of the two alternatives (for example the least common multiple of 8 and 12 is 24) as explained in the following example.

Example 2: Determine which of the following two projects should be selected on the basis of a present worth analysis using an interest rate of 15% per year . First cost Annual operating cost Salvage value Life ,years Project A 110000L.E 35000 L.E 10000 L.E 6 Project B 180000L.E 31000L.E 20000L.E 9 Solution : 10000 10000 10000 i=15% 20000

0 1 2 3 4 5 6 12 18 0 6 9 12 18

A=35000L.E A =31000L.E 110000 110000 110000 110000 110000

PWA = 110000+110000(P/F,15%,6)-10000(P/F,15%,6)+110000(P/F,15%,6) -10000(P/F,15%,12) --10000(P/F,15%,18)+35000(P/A,15%,18) = 11000(1+.4323+0.1869)-10000(.4323+.1869)+0.0808)+35000(6.128) = 385592 L.E PWB = 180000+180000(P/F,15%,9)+31000(P/A,15%,18) -20000 (P/F,15%,9) -20000(P/F,15%,18)=180000(1+.2843)+31000(6.128)-20000(.2843+.0808) = 413840 L.E PWA < PWB Then project A should be accepted

2.Capitalized Cost Calculations : Capitalized cost refers to the present – worth value of a projects that assumed to last forever e.g Dames, Bridges , Roads …..etc. Capitalized Cost = EUAC( or EUAB) / i Where , EUAC &EUAB are the equivalent uniform annual series of costs & Benefits respectively and is the interest rate . Example 3 Calculate the capitalized cost of a project that has an initial cost of 300000L.E And additional investment cost of 100000L.E after ten years . The annual operating cost will be 10000 L.E for the first four years and 16000 L.E after that. It is expected that the recurring major rework costs of 30000 L.E every 13 years with interest rate of 5%per year. Solution: Draw the cash flow diagram .

Present worth of non-recurring costs = PNR = 300000+10000(P/A,5%,4)+ +100000(P/F,5%,10)= 300000+10000(3.546)+100000(0.6139) =396830L.E Capitalized cost of recurring costs=[16000(P/F,5%,4) + 30000(A/F,5%,13)]/0.05 =[16000(.8227)+30000(.05646)]/0.05=297140 L.E Total Capitalized cost = 396850+2971.40 = 693990 L.E

Example 4: Two sites are currently under consideration to construct either a suspension bridge or a truss bridge to cross a river . In the north site , the suspension bridge would have to stretch from one hill to another to span the widest part of the river ,railroad tracks ,and local highways below . Its first cost will be 30millon L.E with annual maintenance and inspection costs of 15000 L.E . In addition the concrete deck would have be resurfaced every ten years at cost of 50000 L.E. The cost of purchasing right –of - way is expected to cost 800000 L.E. In the south site the truss bridge will be constructed but would require a new roads construction. The first cost of this bridge will be 10 million L.E while the cost of the new roads will be 80000 L.E . This bridge have to painted every 3 years at cost of 100000 L.E. In Addition , this bridge would have a major reworks every ten years at cost of 450000 L.E. The cost of purchasing right –of - way is expected to be 11million L.E. Determine which of these two sites should be selected if the interest rate is 6% per year . Solution : North site (suspension bridge)

Present worth (or cap. Cost )for non-recurring cost = 30x106 +0.8x106 = 30.8x106 L.E Cap. Cost for recurring cost = (15000/0.06) + [50000(A/F,6%,10)/0.06] = 313223.3 L.E Total Tap. Cost = 30.8x106 +.3132233x106 = 31.113223x106 For South site(or truss bridge)

Present worth (or cap. Cost )for non-recurring cost =(10+3)x106 =23x106L.E Cap. Cost for recurring cost =(80000/0.06) +100000[A/F,6%,3)/0.06] + 450000(A/F,6%,10)/0.06 = 2.425859x 106 L.E Total Tap. Cost = (23+2.425859)x 106 L.E South site (or truss bridge) should be selected. Example 5:

Present Cost of A

Present Cost of B

Example 6 : Three alternatives

Select the best alternative based on : future / present worth Future worth of alternative C

Results of future worth analysis :

Accept alternative C 3. Benefit cost ratio Evaluation Introduction The benefit /cost ratio is one of the methods used to compare two alternatives and would agree with the PW and the EUAC . This method must carried out when the proposal with higher PW (or EUAC) yields a higher benefit , for example : PWA =100000 L.E ; PWB =80000 L.E Benefit A=20000 L.E ; Benefit B = 16000 L.E But it is not necessary to use this method when lower PW (or EUAC) gives lower benefits . Benefits , Disbenefits and Cost Calculation Of A Single Project : The Benefit /Cost ratio “B /C’’ is calculated as follow : B/C =( Benefits – Disbenefits) /Costs If B/C > 1.0 Accept the project economically . If B/C < 1.0 Reject the project economically . Where ; Benefits : Are the advantage , expressed in terms of money units which happened to the owner ; e.g annual income ……. Disbenefit : Are the disadvantage , expressed in terms of money units which happened to the Owner ; e.g loss due to accidents , …… Costs : Are the expected expenditures for construction , operation , maintenance , …… Another method called “B-C” method could be used as follow : If B- C > 0 Accept the project economically . If B- C < 0 Reject the project economically . Example : Two routes N & S are under consideration for a new interstate highway that last 30 years with no salvage value . The north road has an initial cost of 10 millions L.E , annual operation and maintenance cost of 35000 L.E and expected loss due to construction in agriculture land will be 400000 L.E per year .The south route has initial cost of 15 millions L.E ,annual operation and maintenance cost of 55000 L.E. The expected annual income is 1 million L.E for north route and 1.1 million L.E for south route with an interest rate of 5% per year . Determine which route should be selected on the bases of B /C ratio analysis . Solution : 6 EUACN = 10 x 10 ( A/P,5% , 30) + 35000 = 685500 L.E 6 EUACS = 15 x 10 ( A/P,5% , 30) + 55000 = 1030750 L.E 6 6 Then ; B /CN = [10 – 0.4x 10 ] /685500 = 0.875 < 1.0 6 B /CN = 1.1x 10 / 1030750 = 1.067 > 1.0 Accept the south route . Example : An educational research organization is thinking to make an investment of 1.5 millions L.E in Grants to develop new ways for teaching . The grants would extend over a 10 ten years period and would create an estimated saving of 500000 L.E/year in professor’s ; salaries student , ..etc . An estimated 200000L.E per year have to be released from other programs to support the educational research . If the rate of return is 6% ; determine the advisability of the program over 10 year period using : a) B/C method b) B- C analysis . Solution ; EAUC = 1.5 x 106 (A/P,6%,10) =203805 L.E/year Benefit = 500000 L.E /year ; Disbenefit = 200000 L.E a) B/C==( Benefits – Disbenefits) /Costs = (500000 – 200000) /203805 = 1.47 > 1 This project is accepted b) B – C = ( Benefits – Disbenefits) – Costs = (500000 – 200000) – 203805 = 96195 > 0 Also ,this project is accepted 4. Selection From Mutually Exclusive Alternatives Using Incremental B/C Ratio Analysis : This method could summarized as follow ; • Identify all alternatives • Arrange the alternatives from the smallest to large initial investment cost (or EUAC) • Evaluate first two alternatives ,then eliminate the alternative that has B/C <1.0 • Repeat evaluation until all alternatives have been evaluated as shown next example. Example : It is required to construct a dam on a river with the following construction cost and average Annual operating coat and annual income as follow : Cost (L.E) A B C D E F Construction cost 6x106 8 x106 3 x106 10 x106 5x106 11x106 Annual income 350x103 420 x103 125 x103 700x103 350x103 700x103 AOC 11000 12000 10000 13000 11000 13000 Assume AOC is disbenefit . Select the best location using the B /C ratio method if the minimum rate of return is 6% . Solution : Since dam life is infinity ,so EUAC = Px i then the alternatives are first ranked from smallest to largest initial investment cost (or EUAC) according to the following table : Parameters C E A B D F Remarks EUAC x10-3 180 300 360 480 600 660 =Px i (1) Annual 125 350 350 420 700 700 (2) Benefit x10-3 Annual dis- 10 11 11 12 13 13 (3) benefit x10-3 Ben – Disb. 115 339 339 408 687 687 [(2)-(3)] (4) Site C to none E to none A to E B to E D to E F to D (5) comparison Incremental 180 300 60 180 300 60 (6) (or ∆)EAUC (180-0) (300-0) (360-300) (480-300) (600-300) (660-600) Incremental 115 339 0 69 348 0 (7) (or∆)Benefit (115-0) (339-0) (339-339) (408-339) (687-339) (687-687) Incremental 0.639 1.13 0 0.383 1.16 0 (7)/(6) (8) (or∆)B/C Site None E E E D D Selected Location D is selected . Higher Technological Institute Engineering economy I Tenth of Ramadan City Sheet 4______1.It is required to construct a large shopping center in a site within 3 years from now . During this period , the will exposed to 12 heavy thundershowers at rate of one every 3 months . The losses of these thundershowers are expected to be 20000 L.E per thundershower in the project site . Determine if it would be economically feasible to install rainwater drain pipe to protect the site OR not . The cost of drain pipe would be 130 L.E per meter with total length of 2000 meters and expected salvage value of 60000 L.E after 3 years , with nominal interest rate of 5% per year .

2.A company received two proposals to update the production lines . The first proposal “A” has an initial cost of 60000 L.E and an AOC of 8000 L.E /year for the next four years , then it becomes 10800 L.E / year after that . Proposal “B” has an initial cost of 112000 L.E and AOC of 4800 L.E /year for the first 3 years , then it will increase by 480 L.E /year . The Salvage value of machines in this proposal will be 8000 L.E after 20 years . Which of the two proposals should be accepted on the basis of a present worth analysis if they have same lifetime and rate of interest of 10% per year .

3.Select one of the two machines below on the basis of their present worth , using an Interest Rate Of 18% per year . Machine P Machine Q First cost $290000 $370000 Salvage Value $40000 $50000 Annual maintenance cost $30000 $35000 Overhaul every 2 years $37000 $20000 Life , years 3 5

4.A park has a first cost of 350000 L.E with AOC of 120000 L.E for the first year ; then it Increased by 20000 L.E per year up to year 5 , and after that it remain constant at 200000 L.E /year . The expected costs for improving the park will be 6000 L.E / year for the first five years . The expected profits will be 110000 L.E in the first year , then it increased by 30000 L.E /year up to 8 years , after which the net profit will remain the same . Calculate the capitalized cost and the capitalized profits of the park if the interest rate is 6% per year

5.Compare between the following two machines on the bases of their capitalized cost with Interest rate of 14 % per year . Machine First Cost AOC Salvage ValueOver haul Lift ,years after 6 years X 50000 L.E 62000 L.E 10000 L.E ------7 Y 200000 L.E 24000 L.E 0 5000 L.E ∞

6.There are three sites for flood – control dams ( designated as sites A , B ,and C) . The construction costs are $10 millions , $ 12 millions , and $20 millions , and maintenance costs are expected to be $15000 , $ 20000 , and $23000 respectively , for sites A , B , and C. In addition , a $75000 expenditure will be required every 10 years at each site. The present Cost of flood damage is $2 million per year . If only the dam at site A is constructed , the Flood damage will be reduced to $1.6 million per year . If only the dam at site B is constructed , the flood damage will be $1.2 million per year . If only the dam at site c is built , the flood damage will be reduced to $0.77 million per year . Since the dams would be built on different branch of a large river , either one or all the dams could be constructed and the decrease in flood damages would be additive . If the interest rate is 5% per year , determine which ones , if any , should be built on the bases of their B /C ratios . Assume that the dams will be permanent .

7.A city trying to attract professional strong teams is considering construction a large public Sporting area or a conventional stadium . The public area would cost $90 million to Construct and would have a useful life of 50 years . The maintenance and operation costs would be $300000 the first year , with costs increasing by $10000 per year . Every ten years ,an expenditure of $800000 would be required to renewing the interior . The conventional Stadium would cost $50 million to construct and would also have a useful life of 50 years . The cost of maintenance would be $75000the first year , increasing by $8000 per year . Periodic costs for repairing , resurfacing , etc .., would be $100000 every 4 years . Revenue from the public area is expected to be greater than that from the conventional stadium by $500000 the first year , with amounts increasing by $200000 per year through year 15 . Thereafter , the extra revenue from the public area would remain the same at $3.3 million per year . Assuming that both structures would have a salvage value of $5 million , use an interest rate of 8% and a B / C analysis to determine which structure should be built .

8.Select the best mutually exclusive alternative using the B /C ratio method from the proposals Shown below if the MARR is 10% per year and the project have a useful life of 15 years . Assume that the cost of the land will be recovered when the project is terminated . Treat maintenance costs as disbenefits .

Proposal No 1 2 3 4 5 6 7 Cost Land Cost $50000 $40000 $70000 $80000 $49000 $65000 $75000 Construction Cost $200000 $150000 $170000 $185000 $165000 $175000 $190000 Annual $15000 $16000 $14000 $17000 $18000 $13000 $12000 Maintenance Annual $52000 $49000 $68000 $50000 $81000 $77000 $45000 Income

CHAPTER 5 Inflation Impact On Economic Calculation 5.1 Inflation Impact Inflation is an increase in the amount of money necessary to obtain the same amount of goods or services before the inflated price was present. Inflation effects : 1. Inflation cause prices to increase and decreases the purchasing ability of money. 2. It declines the standard of living as savings and investments are eroded . 3. It declines the economy with resulting increase in greater unemployment . 5.2 Present Worth Calculations Adjusted for Inflation

if = i +f + i x f where i is the real interest rate , f is the inflation rate , is a measure of the rate of change in the value of the currency. if is the inflation-adjusted or market interest rate , is the interest rate that has been adjusted to take inflation into account. It is a combination of the real interest rate i and the inflation rate f. It is also known as the inflated interest rate. Example 1 : A self-employed chemical engineer is on contract with Dow Chemical, currently working in a relatively high-inflation country in Central America. She wishes to calculate a project‟s PW with estimated costs of $35,000 now and $7000 per year for 5 years beginning 1 year from now with increases of 12% per year thereafter for the next 8 years. Use a real interest rate of 15% per year to make the calculations (a) without an adjustment for inflation and (b) considering inflation at a rate of 11% per year.

(a) The PW without an adjustment for inflation is found using i = 15% and g = 12% is: PW = 35,000 - 7000(P/A,15%,12%,4) – 7000[(1-(1+(1.12/1.15)9 )/( 0.15- 0.12)] = 35,000 - 19,985- 28,247 = $83,232 (b) To adjust for inflation, calculate the inflated interest rate as follow : if = 0.15 + 0.11 + (0.15)(0.11) = 0.2765 PW =35,000 - 7000(P/A,27.65%,4) – 7000[(1-(1+(1.12/1.2765)9 )/( 0.15- 0.2765)] = 35,000 - 7000(2.2545) - 30,945(0.3766) = $-62,436 Example 2 : Abbott Mining Systems wants to determine whether it should upgrade a piece of equipment used in deep mining operations in one of its international operations now or later. If the company selects plan A, the upgrade will be purchased now for $200,000. However, if the company selects plan I, the purchase will be delayed for 3 years when the cost is expected to rise to $300,000. Abbott is ambitious; it expects a real MARR of 12% per year. The inflation rate in the country has averaged 3% per year. From only an economic perspective, determine whether the company should purchase now or later (a) when inflation is not considered and (b) when inflation is considered. (a) Inflation not considered: The real rate is i = 12% per year. The cost of plan I is $300,000 after three years. FWA = - 200,000(F/P,12%,3) = $-280,986 , FWI = $−300,000(after three years) Then Select plan A (purchase now). (b) Inflation considered: the real rate is (12%), and inflation of 3% must be accounted for. if = 0.12 + 0.03 + 0.12(0.03) = 0.1536 FWA = - 200,000(F/P,15.36%,3) = $-307,040 , FWI = $−300,000 Purchase later (plan I) is now selected, because it requires fewer equivalent future dollars. Example 3 : What annual deposit is required for 5 years to accumulate an amount of money with the same purchasing power as $680.58 today, if the market interest rate is 10% per year and inflation is 8% per year? Solution The actual number of inflated dollars required 5 years in the future that is equivalent to $680.58 today is calculated as follow:. F = (present purchasing power)(1+ f )5 = 680.58(1.08)5 - $1000 The actual amount of the annual deposit is calculated using the market interest rate of 10%. A is calculated for a given F as follow: A = 1000(A/F,10%,5) = $163.80 . Higher Technological Institute Engineering economy Tenth of Ramadan City Sheet 5 1. It is required to select one of the following three planes which maximize the buying power of the money received : In Plan A , the money received are 60000 L.E now , in Plan B , the money received are 16000 L.E/year for 12 years beginning one year From now , and In Plan C, the money received are 50000 L.E three years from now and 80000 L.E five years from now . The interest rate is 12% per year and the inflation rate is 11% per year .

2. A company trying to decide whether it selects plan E or plan F for upgrading its production facilities . In plan E , the necessary equipment would be purchased . Now for 20000 L.E . In plan F these equipment would be delayed for 3 years where the cost would be 34000 L.E. The minimum attractive rate of return is 18% per year and the inflation rate is expected to be 12% per year . Determine which of the two plans should be selected when the inflation is not considered or is considered .

3. Calculate the present worth and future worth of a uniform series of payments of 10000L.E per year for 5 years if i = 10% and f = 8% .

4. New vehicle has first cost of 80000 L.E and expected to last 6 years with 13000 L.E. L.E salvage value, and annual operating cost 17000 L.E in the first year , then Increasing by 11% year after that . Calculate the equivalent present cost of the vehicle if interest rate 8% per year with and without inflation rate of 10% .

5. A women deposited $1300 per year for 6 years starting one year from now in a saving account . In years 7 to 12 , she deposited $2000 . How much money did She have in the account after last deposit in terms of today, s buying power if The interest rate she received was 12% per year and the inflation rate was 9% per year .

CHAPTER 6 Basics of a Replacement Analysis 6.1 Introduction Replacement Analysis is the economic analysis that carried out by the business firms and government organization as well as individuals to take a decision in which an existing asset should be retired from use or continued in service or replaced with a new asset . Replacement of an existing asset should be considered in case of following reasons: 1. Obsolescence occurs when the technology of an asset is exceeded by newer and/ or different technologies. 2. Depletion : the gradual decrease of market value of an asset as it is being consumed or exhausted 3. Deterioration due to ageing: the general condition of loss in value of some asset due to the ageing process which associated with additional maintenance and operating expenses. 4. Inadequacy: The equipment does not have sufficient capacity to meet the present demands. 5. Rapid Technological Changes: Recognition and handling of replacement problems have paid off quite well for many companies. 6.2 Advantages of replacement analysis 1. Reduction in maintenance cost 2. Keep the company in competitive position with reduction in production cost 3. Reduction in losses, rework, scraps 4. Introduce modernization which helps to take off productivity and returns 5. Reduce delays off down time costs 6. Increase enthusiasm and morale of workers resulting increased human efficiency and better human relations . 6.3 Terminologies Used in Replacement Analysis: 6.3.1 Defender and challenger : a replacement study compares these two alternatives : - Defender : Is the currently installed or owned item . - Challenger: The new item which possibly replace the defender . 6.3.2 Defender first cost: Defender first cost is the initial investment amount ‘P’ used for the defender. The current market value (MV) is the current estimate to use for ‘P’ for the defender in a replacement study. 6.3.3 Challenger first cost: It is the amount of cost that must be recovered when replacing a defender with a challenger. This amount is almost always equal to ‘P’, the first cost of the challenger. 6.3.4 Market price ( Market value or trade-in value) The highest estimated price that a buyer would pay and a seller would receive for an item in an open and competitive market . 6.2.5 Salvage value : Is the estimated value at the end of the expected life. In replacement analysis, the salvage value at the end of one year is used as the market value at the beginning of the next year. 6.2.6 Sunk cost : Is a prior expenditure or loss of capital (money) that cannot be recovered by a decision about the future . Sunk cost = present book value – present realizable value. 6.2.7 Present book value : Is the remaining investment after the total amount of depreciation has been charged to date . 6.2.8 Marginal cost: These are the year by year estimates of the costs to own and operate an asset for that year. It includes loss in value of the asset by retaining it for one or more year, cost and expenses directly related to the project or asset (Insurance, operating and maintenance etc.) 6.2.8 Economic service life (ESL): ESL for an alternative is the number of years at which the lowest EUAC or Annual Worth (AW) occurs .

Total AW = capital recovery – AW of annual operating costs = CR - AW of AOC

6.4 Approach for comparing Defender and Challenger 6.4.1 Cash Flow Approach In this case , the challenger first cost is obtained by subtracting it,s initial cost from the defender present market (or trade-in value) . But the defender first cost is set to be equal zero . Example 1: An old asset may be replaced with either two new assets of data as given below . determine the most economic decision if interest rate =10% . Defender Challenger 1 Challenger 2 First Cost $30,000 (2 years ago) $35,000 $40,000 Trade-in (Def) ------$20,000 $23,000 AOC $3,500 $1,500 $1,000 Salvage value $1,000 $2,000 $5,000 Estimated Life 10 years 10 years 10 years Solution EUACD = $3500 - $1000(A/F,10,10) = $3500 - $1000(0.06275) = $3,437 Note that the first cost was omitted from the EAC of the defender. EUACC1 = $1500 + (35000 - 20000)(A/P,10,10) - $2000 (A/F,10,10) = $1500 + (35000 - 20000)(0.16275) - $2000 (0.06275) = $3,816 Note that the first cost of this challenger was reduced by the defender trade-in cost. EUACC2 = $1000 + (40000 - 23000)(A/P,10,10) - $5000(A/F,10,10) = $1000 + (40000 - 23000)(0.16275) - $5000(0.06275) = $3,453 Again the trade-in cost was subtracted from the Challenger first cost. Keep the Defender. Less cost. 6.4.2 Conventional Approach In this case the , the defender current trade in-value is used as first cost for the Defender . If there are more than one challenger with each offering a different trade - in value for the defender , this approach will give different EUAC values for the Defender when compared with each challenger . Example 2 : • A Machine bought 3 years ago for $100,000. • 8 years life remaining. • Annual operating cost = $23,000/year. • Salvage value = $10,000 (In 8 years) • Sell existing machine for $75,000 • Buy more efficient machine for $150,000. • New machine operating costs = $10,000/year • New machine life = 8 years (with no salvage). Keep old machine or replace with new? MARR = 10%

Defender Challenger P $75,000 $150,000 AOC $23,000 $10,000 SV $10,000 $0 N (years) 8 8 Note that defender first cost is the current price obtained by selling it. Defender EACD = $23,000 + $75,000(A/P,10,8) -$10,000(A/F,10,8) = $23,000 + $75,000(0.18744) -$10,000(0.08744) = $36,184 Challenger EACC = $10,000 + $150,000(A/P,10,8) = $10,000 + $150,000(0.18744) = $38,116 The Defender is less cost. Keep the old machine.

Example 3 : Solve Example 1 using the conventional approach . Solution Comparing Defender with first Challenger EUACD1 = 20000(A/P,10%,10) +3500 – 1000(A/F,10%,10) =20000(0.16275)+3500 - 1000(0.06275) = 6692.25 EUACC1 = $1500 + 35000(A/P,10,10) - $2000 (A/F,10,10) = 1500 + 35000(0.16275) -2000(0.06275) = 7070.75 Choose Defender Comparing Defender with second Challenger EUACD2 = 23000(A/P,10%,10) +3500 – 1000(A/F,10%,10) =23000(0.16275)+3500 - 1000(0.06275) = 7180.5 EUACC2 = $1000 + 40000(A/P,10,10) - $5000 (A/F,10,10) = 1000 + 40000(0.16275) -5000(0.06275) = 7196.25 Choose defender or The final decision is to keep the Defender

6.4.3 Economic Service Life (ESL) Approach : In this case , the respective economic service lives of the defender and the challenger will be used when conducting a replacement analysis. The objective is to find the number of the year (n*) that minimizes the equivalent uniform annual worth of costs is minimum and at which the replacement analysis should conducted. The total annual equivalent costs (Annual Worth “AW”) are calculated as follow : Total AW = capital recovery (CR)– AW of annual operating costs = CR - AW of AOC Where : Capital recovery cost (CR) at year k = -P(A/P,i%,k) + SV(A/F,i%,k) , The salvage value SV , which usually decreases with time, is the estimated market value (MV) in that year. AW OF Annual Operating Cost (OC) = ∑ AOCk (P/F, i%, k). (A/P, i%, k) , k = 1 to n Therefore, the complete equation for total AW of costs over k years (k = 1, 2, 3, . . . ) is :

where P = initial investment or current market value Sk salvage value or market value after k years AOCj = annual operating cost for year j ( j = 1 to k )

Example 4: A 3-year-old backup power system is being considered for early replacement. Its current market value is $20,000. Estimated future market values and annual operating costs for the next 5 years are given in columns 2 and 3 of the following table:

What is the economic service life of this defender if the interest rate is 10% per year? Solution Equation [11.3] is used to calculate total AWk for k _ 1, 2, . . . , 5. In above Table , column 4, shows the capital recovery for the $20,000 current market value ( j = 0) plus 10% return. Column 5 gives the equivalent AW of AOC for k years. As an illustration, the computation of total AW for k = 3 from Equation [11.3] is Total AW3 = P(A/P,i,3) - MV3(A/F,i,3) - [PW of AOC1,AOC2, and AOC3](A/P,i,3) = - 20,000(A/P,10%,3) + 6000(A/F,10%,3) - [5000(P/F,10%,1) + 6500(P/F,10%,2) + 8000(P/F,10%,3)](A/P,10%,3) = - 6230 - 6405 = $-12,635 A similar computation is performed for each year 1 through 5. The lowest equivalent cost ( numerically largest AW value) occurs at k = 3. Therefore, the defender ESL is n = 3 years, and the AW value is $ -12,635. In the replacement study, this AW will be compared with the best challenger AW determined by a similar ESL analysis.

Example 5: Two years ago, Toshiba Electronics made a $15 million investment in new assembly line Machinery . Due to the new standards, coupled with rapidly changing technology, a new system is challenging the retention of these 2-year-old machines. The chief engineer at Toshiba USA realizes that the economics must be considered, so he has asked that a replacement study be performed this year and each year in the future, if need be. The i is 10% and the estimates are below. Challenger: First cost: $50,000 Future market values: decreasing by 20% per year Estimated retention period: no more than 10 years AOC estimates: $5000 in year 1 with increases of $2000 per year thereafter Defender: Current international market value: $15,000 Future market values: decreasing by 20% per year Estimated retention period: no more than 3 more years AOC estimates: $20000 next year, $8000 at year 2 and 12000 at year 3 (a) Perform the replacement study now by determine the AW values and economic service lives necessary to perform the replacement study. (b) After 1 year, it is time to perform the follow-up analysis. The expected market value for the defender is still $12,000 this year, but it is expected to drop to $2000 next year on the worldwide market and zero after that. Also, the estimated AOC next year has been increased from $8000 to $12,000 and to $16,000 two years out. Perform the follow-up replacement study analysis. Solution (a) The results of the ESL analysis, shown in Figure below, For the challenger, note that P= $50,000 is also the MV in year 0. The total AW of costs is for each year, should the challenger be placed into service for that number of years. As an example, the year k = 4 amount of $-19,123 is determined using Equation [11.3], where the A/G factor accommodates the arithmetic gradient series in the AOC. Total AW4 = -50,000(A/P,10%,4) - 20,480(A/F,10%,4) - [5000 - 2000(A/G,10%,4)] = $-19,123

The defender costs are analyzed in the same way up to the maximum retention period of 3 years. The lowest AW cost (numerically largest) values for the replacement study are as follows: Challenger: AWC = $-19,123 for nC = 4 years Defender : AWD = $ - 17,307 for nD = 3 years Select the defender because it has the better AW of costs ($ - 17,307), and expect to retain it for 3 more years The challenger total AW of cost curve in above Figure is classically shaped and relatively flat between years 3 and 6; there is virtually no difference in the total AW for years 4 and 5. For the defender, note that the estimated AOC values change substantially over 3 years, and they do not constantly increase or decrease. (b) After 1 year of defender retention, the challenger estimates are still reasonable, but the defender market value and AOC estimates are substantially different . New AW values are calculated using Equation [11.3] as shown below .

The AW and n values for the new replacement study are as follows: Challenger: unchanged at AWC = $_19,123 for nC = 4 years Defender: new AWD = $-20,819 for nD = 2 more years Now select the challenger based on its favorable AW value. Therefore, replace the defender now, not 2 years from now. Expect to keep the challenger for 4 years, or until a better challenger appears on the scene.

Higher technological Institute Sheet 6 Engineering Economy MNG 102 Dr Hosny Abouzeid 1. An 5 years old machine may be replaced by a new one .Determine the most economic decision for following data if i=12% using both cash flow and conventional approaches . First cost SV now SV after 3 years AOC service life(years) For old machine - 10000 2500 8000 3 For new machine 15000 - 5500 6000 3

2. Three years ago Chicago’s O’Hare Airport purchased a new fi re truck. Because of flight increases, new fire-fighting capacity is needed once again. An additional truck of the same capacity can be purchased now, or a double-capacity truck can replace the current fi re truck. Estimates are presented below. Compare the options at 12% per year using (a) a 12-year study period and (b) a 9-year study period.

3. Apiece of machinery costs $7500and has no salvage value after it is installed .The manufacturer's warranty will pay the first year's maintenance and repair costs. In the second year, maintenance costs will be $900, and this item will increase on a $900 arithmetic gradient in subsequent years, Also, operating expenses for the machinery will be $500 'the first year and will increase on a $400 arithmetic gradient in the following years. If interest is 8%, compute the useful life of the machinery that results in a minimum EUAC .That is, find its minimum cost life.

4. A 5-years old machine , whose Current market value is $5000, is being analyzed to determine its economic life in a replacement analysis .Compute its lowest EUAC using a 10% interest rate Salvage value and maintenance cost are given the following : Remaining life 1 2 3 4 5 6 7 8 9 10 11 n ,years MV at year n 4000 3500 3000 2500 2000 2000 2000 2000 2000 2000 2000 AOC 0 100 200 300 400 500 600 700 800 900 1000

CHAPTER 7 Depreciation And Depletion Calculation Models 7.1 Depreciation Terminology 7.1.1 Depreciation : is the decreasing in market value of an asset(or property) with time with time due to one or more of the following reasons : - Physical depreciation such as wear or tear of assets parts which reduce its ability to perform the intended function . - Functional depreciation which occurs when demands from assets exceeds its production capacity . - Technological depreciation or obsolescence due to existence of newly developed means for performing functions more economically and quality. - Sudden failure or damage due accident or damage . 7.1.2 Book value BVt : represents the remaining, undepreciated capital investment on the books after the total amount of depreciation charges to date has been subtracted from the basis. The book value is determined at the end of each year t ( t = 1, 2, . . . , n ) . 7.1.3 Market value MV : is the estimated amount realizable if the asset were sold on the open market. Because of the structure of depreciation laws, the book value and market value may be substantially different. 7.1.4 Salvage(Scrap)value SV : is the estimated trade-in or market value realized at the end of the asset’s useful life. The salvage value, expressed as an estimated dollar amount or as a percentage of the first cost, may be positive, zero,or negative due to dismantling (disassemble)and carry-away costs. 7.1.5 Recovery period n : is the depreciable life of the asset in years. 7.1.6 Depreciation rate or recovery rate dt : is the fraction of the first cost removed by depreciation each year t. This rate may be the same each year, which is called the straight line rate d , or different for each year of the recovery period. Notes : 1. The rules for depreciation are linked to the classification of business Property(assets)can which can be Classified as follow : a. Tangible Property : physical assets that can been seen touched or felt and classified as : - Personal assets : such as machines , equipment’s , vehicles, ……etc - Real assets : which are anything which erected ,constructed or growing on the land such as building ,but the land itself is not depreciable . b. Intangibles Property : includes all property that has value to the owner but cannot be directly seen or touched. Examples include patents, copyrights, trademarks, trade names ….etc . 2. Conditions for depreciable assets : a. It must be used in business or be hold for income production . b. It must have determinable life time more than one year . c. It must be decays, wears out and become obsolete due to one of reasons above. 7.2 Depreciation Calculation 7.2.1 Introduction The book value of an asset at any time ,t, over the asset's life could be obtained from the following equation : Book value , BVt = Asset initial cost - Depreciation charges made to date Or

where BVt = book value of the depreciated asset at the end of time t , and Cost basis = P = first cost that is being depreciated; this includes the asset's purchase price as well as any other costs necessary to make the asset "ready for use". dj = depreciation rate (or annual depreciation charge) at year j

sum of annual depreciation charges taken from time 1(at year j=1) to any time t ( at year j or may be end of useful Life) .

The value of this summation depends on the depreciation calculation methods . NOTE : generally BVt = Book value after time t= BVt-1 – dt

7.2.2 Depreciation Calculation Methods There four commonly used depreciation methods as follow : 1. The straight-line(SL), 2. The sum-of-the - years' -digits(SYD), and 3. The declining balance(DB) and double declining balance(DDB)methods 4. The modified accelerated cost recovery system(MACRS) . The following figure compares between these methods .

7.2.2.1 Straight-Line Method It represents the simplest depreciation method .To calculate the constant annual depreciation charge, the total amount to be depreciated, P- S, is divided by the depreciable life, in years, N as follow : Annual depreciation charge = dt = (P-SV)/ N SV = Salvage value after depreciable life,N. ,P=First cost BVt = Book value after time t = P – t . dt ……………… (Equation of straight line)

The rate of depreciation ,d, is the fraction by which the depreciable amount (P- SV) is decreased each year . In this method d = 1/n (i.e. constant). N is used for the depreciation period because it may be shorter than n(useful life) . The straight-line (SL) method is often used for intangible property. Example 1: Consider the following: Cost of the asset, P = $900 & Depreciable life, in years, N= 5 & Salvage value, S = $70 Compute the annual depreciation allowances and the resulting book values ,using the straight-line depreciation method . Solution Annual depreciation charge = dt = (P-SV)/n = (900-70)/5 = $166 (constant value with time) BVt = Book value at any time t = P -t . dt (Equation of straight line ) BV1 = 900 – 1x166 = 734 , BV2 = 900 – 2x166 = 568 , BV1 = 900 – 3x166 = 402 BV4= 900 – 4x166 = 236 , BV5 = 900 – 5x166 = 70 = salvage value . These reults could be summarized in the following table .

These data are plotted in the following Figure .

Example 2 If an asset has a first cost of $50,000 with a $10,000 estimated salvage value after 5 years, ( a ) calculate the annual depreciation and ( b ) calculate and plot the book value of the asset after each year, using straight line depreciation. Solution (a) The depreciation each year for 5 years can be found by Equation : Annual depreciation charge = dt = (P-SV)/N = (50000- 10000)/5 = $8000 The BV t values are plotted in the following Figure For years 1 and 5, for example, BV 1 = 50,000 - 1(8000) = $42,000 BV 5 = 50,000 - 5(8000) = $10,000 = S

7.2.2.2 Sum-of-Years'-Digits Depreciation(SOYD) In this case the depreciation charges are very high in the first few years but decease rapidly in later years of the asset life as in figure shown in item 7.2.2 . Annual depreciation charge at any time t , dt , obtained as follow : dt = [(N- t + 1)/SOYD](P- SV) ,t = 1,2,3,….,N BVt = Book value after time t = P – [t(N- 0.5.t + 0.5)/SOYD](P-SV) Where : SOYD = Sum-of-Years'-Digits From 1 TO N = N(N+1)/2 N- t + 1 = remaining depreciable life Or ;

Example 2: Solve Example 1 using the SOYD method . Solution SOYD = 5X6/2 = 15 BV1 =900 – [1(5-0.5x1+0.50)/15](900-70) = 623.3 BV2 =900 – [2(5-0.5x2+0.50)/15](900-70) = 402 BV3=900 – [3(5-0.5x3+0.50)/15](900-70) = 236 BV1 =900 – [4(5-0.5x4+0.50)/15](900-70) = 125.3 BV1 =900 – [5(5-0.5x5+0.50)/15](900-70) = 70 Annual depreciation charges : d1 = [(5 – 1 +1)/15](900-70) = 277 , d2 = [(5 – 2 +1)/15](900-70) = 221 d3 = [(5 – 3 +1)/15](900-70) = 166 , d4 = [(5 – 4 +1)/15](900-70) = 111 d5 = [(5 – 5 +1)/15](900-70) = 55

These data are plotted in the following Figure .

Time , Depreciable Life = N (years)

NOTE : generally BVt = Book value after time t= BVt-1 – dt 7.2.2.3 Declining-Balance(DB) And Double- Declining-Balance(DBB) depreciation Methods Declining balance method of depreciation is a technique of accelerated depreciation in which more depreciation is charged during the beginning of the life time and less is charged during the end. This because assets are usually more productive when they are new and their productivity declines gradually in later years of their life . The calculation for these two methods are carried out as follow :

The depreciation charge for year t , Dt = ( d )BVt-1 t-1 BVt-1 = P(1-d) , or ;

d is the maximum annual depreciation rate and obtained as follow: For DB depreciation: d = 1 - (SV/P)1/n SV >=0 For DBB depreciation : d = 2/n Example 3: Solve Example 1 using the DDB depreciation method Solution d = 2/5 = 0.4 1 2 BV1 = P = 900 BV2 = 900(1-0.4) = 540 , BV3 = 900(0.6) =324 3 4 5 BV4 = 900(0.6) =194.4 , BV4 = 900(0.6) =116.64 , BV5= 900(0.6) =69.98 For annual depreciation charges : D1 =0.4x900=360 , D2 =0.4x540=216 , D3 =0.4x324=129.6 D4 =0.4x194.4=77.76 , D5 =0.4x116.64= 46.65 OR ; Year Depreciation charge Sum of Depreciation Book Value at for year t charges upto year t end of year t

These data are plotted in the following Figure

Example 4: Freeport-McMoRan Copper and Gold has purchased a new ore grading unit for $80,000. The unit has an anticipated life of 10 years and a salvage value of $10,000. Use the DB and DDB methods to compare the schedule of depreciation and book values for each year . Solution An implied DB depreciation rate is determined by Equation

2 BV1 =80000(1-0.1877)=64984 BV2 =80000(1-0.1877) = 52786.5 3 4 BV3 =80000(1-0.1877) = 42878.48 BV4=80000(1-0.1877) = 34830.19 5 10 BV5=80000(1-0.1877) = 28292.56 …….. BV10=80000(1-0.1877) = 10006.86 For annual depreciation charges : D1 = d P =.1877x80000 = 15016 , D2 = dxBV1 = .1877X64984 = 12197.497 D3 = dxBV2 = .1877X52786.5 = 9908.026 , D4= dxBV3 = .1877X42878.48 = 8048.29 ……………. , D10 = dxBV9 = .1877X12317.94= 2312.08 An implied DDB depreciation rate is determined by Equation : d = 2/10 = 0.2 1 2 BV1 = 80000X(1- 0.2) = 64000 , BV2 = 80000X(0.8) = 51200 3 10 BV3 = 80000X(0.8) = 40960 ………….. , BV10 = 80000X(0.8) = 8589.93 For annual depreciation charges : D1 = Dxp = .2x80000 = 16000 , D2 = dxBV1 = 0.2x64000 = 12800 D3= dxBV2 = 0.2x51200 = 10240 …… D10 = dxBV9 = 0.2 x10737.418= 2147.48 Summary of Dt and BVt Values for DB and DDB Depreciation

7.3 Depletion Calculation 7.3.1 Introduction Depletion is similar to depreciaton but it is applied to natural resouces which when removed cannot be replaceable or repurchased . Depltion is defined the decreasing value of a natural resource as it is recovered, removed from a site to produce products or services such as removal of oil , minerals , forests , wells ,quarries , geothermal deposits…etc. 7.3.2 Depletion Calculation Methods 7.3.2.1 Cost (or Factor)Depletion This method is based on the level of activity or usage, not time, as in depreciation. It may be applied to most types of natural resources and must be applied to wood (or timber) production. The cost depletion factor for year t , denoted by CDt , is the ratio of the first cost of the resource to the estimated number of units recoverable as follow :. CDt = initial investiment / resource capacity The annual depletion charge = CDt x year’s activity usage or volume . Note : The total cost depletion cannot exceed the fi rst cost of the resource. Example 5: Company purchased forest of 175 million boared meter of wood with $350000 . Determine the depletion charge if 15 million and 22 million boared meter are removed in first and second year . Solution : CDt = 350000/175 = $2000 per million boared meter depletion charge in the first year = 2000 x 15 = $30000 depletion charge in the second year = 2000 x22= $44000 7.3.2.2 Percentage Depletion In this case, it is assumed that a constant stated percentage of the resource’s gross income may be depleted each year provided it does not exceed 50% of the company’s taxable income .The following table shows examples for the annual percentage depletion rates for some common natural deposits .

Depletion amount for year t , = percentage of gross income depletion(PD) x anticipated gross income Example 6 : A gold mine was purchased for $10 million. It has an anticipated gross income of $5.0 million per year for years 1 to 5 and $3.0 million per year after year 5. Assume that depletion charges do not exceed 50% of taxable income. Compute annual depletion amounts for the mine. How long will it take to recover the initial investment at i = 0%? Solution The rate for gold is PD = 0.15( from above table) . For years 1 to 5: Depletion amounts(charges)= 0.15(5.0 million) = $750,000/year Years thereafter: Depletion amounts(charges)= 0.15(3.0 million) = $450,000/year Total of is written off in 5 years = 5X750000=$3.75 million The remaining is written off at $450,000 per year =$10million - $3.75million = $6.25 million. The total number of years = 5 +$6.25million/$450,000 = 5+ 13.9 = 18.9 years This means that in 19 years, the initial investment could be fully depleted. EXAMPLE 16.6

Higher technological Institute Sheet 7 Engineering Economy MNG 102 Dr Hosny Abouzeid 1. The ABC company paid $52000 for an asset in 1977 and installed it at a cost of $3000 . The asset was expected to remain in service for 10 years and be sold of 10% of the original purchased price . If the asset was sold in 1982 fir $8700 , write values used in depreciation analysis for the following :first cost , anticipated life , salvage value , actual life, and the market value in 1982 ,book value in 1982 if 75% of the first cost had been written off in depreciation .

2. A new machine costs $160 000, has a useful life of 10 years, and can be sold for $15000 at the end of its useful life. It is expected that $5000 will be spent to dismantle and remove the machine at the end of its useful life. Determine the depreciation schedule for this machine using: A. straight-line method . B. the sum-of-years'-digits method .

3. Underwater electroacoustic transducers were purchased for use in SONAR applications. The equipment will be DDB depreciated over an expected life of 12 years. There is a first cost of $25,000 and an estimated salvage of $2500. ( a ) Calculate the depreciation and book value for years 1 and 4. ( b ) Calculate the implied salvage value after 12 years.

5. Declining – balance depreciation at rate of 1.5 times the straight line rate is to be used for for Automated process-control equipment with P=$175000 , N=12 , and expected salvage value =$32000 . Compute the depreciation rate , the book values for years 1&12 , and compare the expected salvage value and the salvage value used by the DB method .

5. A land of timber was purchased for $500000 . Appraisal data placed a value of $420000 On 2 million board feet (Mbf) of standing timber . The land value was $80000 .What is The depletion allowance for the first year if 0.4Mbf of timber is removed from the land.

6. A coal mining company has owned a mine for the past 5 years . During this time the Following tonnage of raw material has been removed each year : 40000 , 52000 , 58000,60000, and 56000 tons . The mine is estimated to contain a total of 2million tons of coal and the mine had an initial cost of $3.5 million . If the company had a gross income for this coal of $15 per ton for the first two years and $18 per ton for the last three years . A. Compute the depletion charges each year using the larger values for two accepted depletion methods . B. Compute the percent of the initial cost that has been written off in these 5 years .

CHAPTER 8 Breakeven and Payback Analysis 8.1 Breakeven Analysis 8.1.1 Introduction • Applications Of Breakeven analysis Break-even Analysis is a technique widely used to determine the value of a variable that makes two elements equal. In economic terms: A breakeven analysis involves the determination of the value of a specified variable that will make the revenues exactly equal to the costs . • The basic technique involves identifying the variable of interest and then finding the magnitude of that variable that will lead to breakeven . • This technique can be used for one project or for determining the best of two or more alternatives.

- Is commonly used as comparison tool for making a decisions . For example : make-or-buy , buy or lease(rent) decisions when a decision is needed about the source for components, services, etc. - Used as a decision- making aid to determine whether a particular volume of sales will result in losses or profits . - Determining the level of sales volume, sales value or production at which the business makes neither a profit nor a loss (the "break-even point"). - Selecting the best of two or more alternatives. 8.1.2 Breakeven Analysis for a Single Project In this case a break even point is determined for the project by determining the unit volume ,quantity QBE, that balances total costs with total gains or revenues . Simple break even analysis finds QBE by analyzing relationships between just three variables: fixed costs, variable costs, and cash inflows using The simple break even formula as explained below . The analysis must consider additional factors, however, when semi-variable costs or variable pricing are present. The simple break even formula QBE is obtained by equating the total revenue R and total cost TC as follow : R = TC , but R = p x QBE and TC= FC+VC = FC + v * QBE ,where: QBE = Breakeven quantity , FC = Total fixed cost , VC= Total variable cost v = Variable cost per unit , p = Selling price(or cash inflow)per unit QBE = FC/(p – v) or Break Even Point Quantity= fixed costs / ( selling price per unit – variable costs per unit) This means that the break even quantity depends on at least three variables: Fixed cost, variable cost per unit, and revenues per unit as explained below . Fixed Cost (FC) – that cost which does not vary based on production volume. Includes building, insurance, fixed overhead (e.g. Engineering staff), equipment recovery cost, information systems, etc. Variable Cost (VC) – that cost which varies as production volume varies. Includes direct labor, materials, warranty, utilities (power consumption), marketing, etc. Revenue(or Cash inflow ) per unit : This usually means selling price per unit. As a result, analysts sometimes label this variable as revenues. The total cash flow , therefore, is a function of unit volume. The Break-Even Chart In its simplest form, the break-even chart is a graphical representation of costs at various levels of activity(quatities) . The point at which neither profit nor loss is made is known as the "break-even point" and is represented on the chart below by the intersection of the two lines: OA represents the variation of income and OB represents the total fixed costs .

If Q >QBE , there is a predictable profit. For linear models of R and VC, the greater the quantity, the larger the profi t. Profi t is calculated as : Profit = revenue - total cost = R – TC = R - (FC + VC) But if Q < QBE , there is a loss = TC – R = (FC-VC) – R Example 1 : Suppose for instance, a firm produces and sells a product with these values: F = Total Fixed costs = $1,200 , v = Variable cost per unit = $15 , P = Selling price per unit = $40 . Calculate the breakeven point and the total revenue and total costs . Solution QBE = FC/(p – v) = 1200/(40-15) = 48 units Total revenue ( or Break-even Point in Sales prices)=R = 48x 40 = $1920 , Total cost =FC+VC= 1200 +48X15 = $1920=R Example 2 : From the following data of a factory , calculate: (i) Break-even point in terms of sales value and in units. (ii) Number of units that must be sold to earn a profit of $ 90,000. Fixed Factory overhead cost = $60000 Fixed selling overhead cost = $12000 Variable Manufacturing cost per unit = $ 12 Variable selling cost per unit = $ 3 Selling price per unit = $24 Solution: Total fixed cost = 60000 +12000 = $72000 Total variable cost per unit = 12 + 3 = $15 i) QBE in terms of units = FC/(p – v) =72000/(24-15) = 8000 units . QBE in terms of sales values = 8000x24 = $192000 ii) Profit = total revenue – (FC+VC) = 90000 = Qx24 –(72000 – 15XQ) Q = Number of units that must be sold to earn a profit of $ 90,000 = (90000+72000)/(24-15) = 18000 units

Example 3 : Indira Industries is a major producer of diverter dampers used in the gas turbine power industry to divert gas exhausts from the turbine to a side stack, thus reducing the noise to acceptable levels for human environments. Normal production level is 60 diverter systems per month, but due to significantly improved economic conditions in Asia, production is at 72 per month. The following information is available. Fixed costs FC = $2.4 million per month ,Variable cost per unit v = $35,000 , Revenue per unit r = $75,000 (a) How does the increased production level of 72 units per month compare with the current breakeven point? (b) What is the current profit level per month for the facility? (c) What is the difference between the revenue and variable cost per damper that is Necessary to break even at a significantly reduced monthly production level of 45 units, if fixed costs remain constant? Solution (a) Breakeven quantity = QBE = FC/(p – v) = 2.4x10^6/(75000-35000) = 60units per month This means that the increased production level of 72 units is above the breakeven value (b) The current Profit = revenue - total cost = R - (FC + VC) = (r-v)Q -FC =(75000-35000)x72 -2.4 x10^6 = $480000 per month (c) It is required to determine the difference ( p- v) when QBE becom 45 and FC=2.4 x 10^6 Therefore, p - v = FC/ QBE =2.4x 10^6 /45 = 53333.3 per unit

Example 4 The cost of a machine for producing a certain part is $40,000. The machine is expected to have a maintenance cost of $14,000 and an $8,000 salvage value after its 5-year economic life. If the variable cost for producing the part is $1.50 per unit and the part can be sold for $4.00 per unit, how many parts per year must the company sell in order to breakeven at an interest rate of 12% per year? Solution Let x represent the number of parts per year required for breakeven. The annual worth equation is: 0 = -40,000 (A/P, 12%, 5) - 14,000 + 8000 (A/F, 12%, 5) - 1.50x + 4.00x 0 = -40,000 (0.27741) - 14,000 +8000 (0.15741) + 2.50x 2.50x = 23,837 Then , x = 9,534 parts/yr Thus, if the company expects to sell more than 9,534 parts per year, it should produce the part. At any sales level below 9,534 parts per year, the company would lose money and, therefore, should not invest in the machine.

8.1.2 Breakeven Analysis Between Two Alternatives Often breakeven analysis involves revenue or cost variables common to both alternatives. The following Figure illustrates the breakeven concept for two alternatives with linear cost relations.

The fixed cost of alternative 2 is greater than that of alternative 1. However, alternative 2 has a smaller variable cost,as indicated by its lower slope. The intersection of the total cost lines locates the breakeven point, and the variable cost establishes the slope. Thus, if the number of units of the common variable is greater than the breakeven amount, alternative 2 is selected, since the total cost will be lower. Conversely, an anticipated level of operation below the breakeven point favors alternative 1. Instead of plotting the total costs of each alternative and estimating the breakeven point graphically, it may be easier to calculate the breakeven point numerically using engineering economy expressions for the PW or AW at the MARR. The AW is preferred when the variable units are expressed on a yearly basis, and AW calculations are simpler for alternatives with unequal lives. The following steps determine the breakeven point of the common variable and the slope of a linear total cost relation. 1. Define the common variable and its dimensional units. 2. Develop the PW or AW relation for each alternative as a function of the common variable. 3. Equate the two relations and solve for the breakeven value of the variable. Selection between alternatives is based on this guideline: If the anticipated level of the common variable is below the breakeven value, select the alternative with the higher variable cost (larger slope). If the level is above the breakeven point, select the alternative with the lower variable cost. (refer to the above figure.

Example 5 Guardian is a national manufacturing company of home health care appliances. It is faced with a make-or-buy decision. A newly engineered lift can be installed in a car trunk to raise and lower a wheelchair. The steel arm of the lift can be purchased internationally for $3.50 per unit or made in-house. If manufactured on site, two machines will be required. Machine A is estimated to cost $18,000, have a life of 6 years, and have a $2000 salvage value; machine B will cost $12,000, have a life of 4 years, and have a $_500 salvage value (carry-away cost). Machine A will require an overhaul after 3 years costing $3000. The annual operating cost for machine A is expected to be $6000 per year and for machine B is $5000 per year. A total of four operators will be required for the two machines at a rate of $12.50 per hour per operator. In a normal 8-hour period, the operators and two machines can produce parts suffi cient to manufacture 1000 units. Use a MARR of 15% per year to determine the following. (a) Number of units to manufacture each year to justify the in-house (make) option. (b) The maximum capital expense justifiable to purchase machine A, assuming all other Estimates for machines A and B are as stated. The company expects to produce breakeven quantity10,000 units per year. Solution (a) Use steps 1 to 3 stated previously to determine the breakeven point. 1. Define x as the number of lifts produced per year. 2. There are variable costs for the operators and fixed costs for the two machines for the make option. Annual VC = (cost per unit)(units per year) = (4 operators/1000 units)($12.50/hour)(8 hours)x = 0.4x The annual fi xed costs for machines A and B are the AW amounts. Annual VC = (cost per unit)(units per year) The annual fixed costs for machines A and B are the AW amounts. AWA =-18,000(A/P,15%,6) + 2000(A/F,15%,6) – 6000 - 3000(P/F,15%,3)(A/P,15%,6) AWB =-12,000(A/P,15%,4) - 500(A/F,15%,4) - 5000 Total cost is the sum of AWA, AWB, and VC. 3. Equating the annual costs of the buy option (3.50x) and the make option yields -3.50x= AWA + AWB – VC = = -18,000(A/P,15%,6) + 2000(A/F,15%,6) – 6000 - 3000(P/F,15%,3)(A/P,15%,6) -12,000(A/P,15%,4) - 500(A/F,15%,4) – 5000 – 0.4x -3.1x= -18000(0.2642) +2000(0.1142) – 6000 -3000(0.2642)(0.6575) - 12000(0.3503)-500(0.2) – 5000 =-203351.95 Then x= 6565 units per year A minimum of 6565 lifts must be produced each year to justify the make option, which has the lower variable cost of 0.4x. (b) Substitute 10,000 for x and PA for the to-be-determined first cost of machine A (currently$18,000) in above Equation. -3.1x10000= - PA (0.2642) +2000(0.1142) – 6000 -3000(0.2642)(0.6575) - 12000(0.3503)-500(0.2) – 5000 Solution yields PA = $58,295. This is approximately three times the estimated first cost of $18,000, because the breakeven production of 10,000 per year is considerably larger than the breakeven amount of 6565.

Example 6 A construction company has two alternatives to purchase an excavator which is to be employed at a construction site for excavation of earth. The cash flow details of the two alternatives are presented as follows; Alternative-1: - Initial purchase cost = $4865000 , Salvage value = $1250000 , and Useful life = 12 years - The operating cost for excavating 1m3 of earth is $11.0. The excavator can excavate 52 m3 of earth in one hour. Alternative-2: - Initial purchase cost = $5350000 , Salvage value = Rs.1410000 ,Useful life = 12 years - The operating cost for excavating 1m3 of earth is Rs.8.0. The excavator can excavate 60 m3 of earth in one hour. The company's minimum attractive rate of return (MARR) is 10.5% per year. How many hours the excavators have to operate per year at the breakeven point of the two alternatives . Solution: Let ‘y' is the number of operating hours per year. The annual operating cost ($) for Alternative-1 = ($11/1m3 )(52 m3 /1hour)(y hours/year ) = $572y Now the equivalent uniform annual worth ($) of Alternative-1 is given by; AW1 = -4865000(A/P,i,n) -572y + 1250000(A/F,i,n) AW1 = -4865000(A/P,10.5%,12) -572y + 1250000(A/F,10.5%,12) AW1 = -4865000x0.1504 -572y + 1250000x0.0454 = - 674946 – 572y The annual operating cost ($) for Alternative-2 =($8/1m3 )(60 m3 /1hour)(y hours/year ) = $480y The equivalent uniform annual worth ($) of Alternative-2 is given by; AW2= -5350000(A/P,i,n) -480y + 1410000(A/F,i,n) AW2 = -5350000(A/P,10.5%,12) -480y + 1410000(A/F,10.5%,12) AW2 = -5350000x0.1504 -480y + 1410000x0.0454 = -740626 – 480y Now equating equivalent uniform annual worth Alternative-1 to that of Alternative-2; AW1 = AW2 , then - 674946 – 572y = -740626 – 480y gives y= 713.9 hours. Thus the breakeven value of number of operating hours per year is 713.9 hours, at which the equivalent uniform annual worth of Alternative-1 is equal to that of Alternative-2. The equivalent uniform annual worth of both the alternatives are determined at different values of annual operating hours and are shown below.

As shown the above Fig. , the line representing the equivalent uniform annual cost of Alternative-1 has greater slope than Alternative-2 as observed from this figure. In other words Alternative-1 has higher annual variable cost ($572y) as compared to Alternative-2 ($480y). Similarly Alternative-2 has higher constant equivalent annual cost ($740626) than Alternative-1 ($674946), as observed from expressions of equivalent uniform annual worth of both the alternatives and also from the above figure. If the expected annual operating hours are less than the breakeven value (i.e. 713.9 hours), then the construction company should select Alternatives-1 as its equivalent annual cost is less than that of Alternative-2 (as evident from above Fig.). Similarly if the expected annual operating hours are greater than breakeven value, then Alternative-2 should be selected as it shows lower equivalent annual cost as compared to Alternative-1.

8.2 Payback Analysis 8.2.1 Introduction Payback analysis is used to determine the amount of time, usually expressed in years, required to recover the first cost of an asset or project. The payback period, also called payback or payout period . There are two types of payback analysis as determined by the required return: • No return; i =0%: Also called simple payback, this is the recovery of only the initial investment. • Discounted payback; i >0%: The time value of money is considered in that some return , for example, 10% per year, must be realized in addition to recovering the initial investment. To calculate the payback period , np , for i = 0% or i> 0%, the following equations are used :

Where P is the initial investment and NCFt is the Net Cash Flow at year t = cash inflows - cash outflows Example 7 The board of directors of Halliburton International has just approved an $18 million worldwide engineering construction design contract. The services are expected to generate new annual net cash fl ows of $3 million. The contract has a potentially gained repayment clause to Halliburton of $3 million at any time that the contract is canceled by either party during the 10 years of the contract period. (a) If i = 15%, compute the payback period. (b) Determine the no-return payback period and compare it with the answer for i = 15%. This is an initial check to determine if the board made a good economic decision. Solution (a) The net cash flow each year is $3 million. For i>0 , the above equation is modified to include the single $3 million payment (call it CV for Cancellation Valuethat could be received at any time within the 10-year contract Period as follow : 0 = -P + NCF(P/A,i,nP) + CV(P/F,i,nP) In $1,000,000 units, 0 = -18 + 3(P/A,15%,nP) + 3(P/F,15%,nP) , by trail and error The 15% payback period is np = 15.3 years . This means that , during the period of 10 years, the contract will not deliver the required return. (b) If Halliburton requires absolutely no return on its $18 million investment,i.e i=0 , then 6 6 6 6 6 0= -P- 3 x nP -3 =-18x10 -3 x10 x nP -3 x10 =15 x10 - 3 x10 nP , therefore nP =5 There is a very significant difference in np for 15% and 0%. At 15% this contract would have to be in force for 15.3 years, while the no-return payback period requires only 5 years. A longer time is always required for i> 0% for the obvious reason that the time value of money is considered.

Higher technological Institute Sheet 8 Engineering Economy MNG 102 Dr Hosny Abouzeid 1. A company can purchase a certain machine or rent one. If purchased, the machine will cost $15,000 and will have a 5-year life with a 10% salvage value. It’s operating cost will be $8000 per year. If the machine is rented, it will cost $400 per day. At an interest rate of 10% per year, calculate the minimum number of days the machine must be needed to justify its purchase . 2. A company for manufacturing plastic products has the following production data : Fixed Expenses = $ 90,000 Variable Costs per unit are : Direct Material = Rs. 5 , Direct Labor = $2 , Direct Overheads = 100% of Direct Labor ,and Selling Price per unit = $ 12. Calculate the break even quantity and the sales required to earn a profit of $450000. Calculate also the net profit for sales quantity 25% above the break even quantity . 3. Two methods of wild plant control in an irrigation canal are under consideration. Method A involves lining the canal at a cost of $30,000. The lining is expected to last 20 years. Maintenance with this method will cost $2 per mile per year. Method B involves spraying a chemical which costs $45 per gallon, with one gallon capable of treating 10 miles. Spraying equipment will cost $2,500 and will have a life of 3 years with no salvage value. At an interest rate of 10% per year, (a) how many miles of canal must require treatment in order for the two methods to breakeven, and (b) if 400 miles of canal must be treated each year, which method should be selected? 4. A small aerospace company is evaluating two alternatives: the purchase of an automatic feed machine and a manual feed machine for a finishing process. The auto feed machine has an initial cost of $23,000, an estimated salvage value of $4000, and a predicted life of 10 years. One person will operate the machine at a rate of $12 per hour. The expected output is 8 tons per hour. Annual maintenance and operating cost is expected to be $3500. The alternative manual feed machine has a first cost of $8000, no expected salvage value, a 5-year life, and an output of 6 tons per hour. However, three workers will be required at $8 per hour each. The machine will have an annual maintenance and operation cost of $1500 . All projects are expected to generate a return of 10% per year. How many tons per year must be finished to justify the higher purchase cost of the auto feed machine? 5. Two equivalent pieces of quality inspection equipment are being considered for purchase by Square D Electric. Machine 2 is expected to be versatile and technologically advanced enough to provide net income longer than machine 1.

Which of the two machines should be selected on the payback period method if the quality manager used a return of 15% per year .

ENGINEERING ECONOMY CONTENTS (MNG 102) Week Topic 1 Introduction to engineering economy(Objectives -terminology- Firm Objectives-Kinds of market structure – Types of economic systems) 2 Laws of supply and demands – Eng. Economy symbols - interest rate types, cash flow diagrams) – Balance sheet - Exercises 3 Present & future worth for single , uniform series factors and discrete payments calculations – Uses of interest tables - Exercises 4 Present & future worth and EUAS for uniform gradient payments - Calculation for unknown interest rate and number of years - Exercises 5 Geometric gradient payments calculation -Single project (or alternative) evaluation methods (Net present ,Internal rate of Return , ROR ) - Exercises 6 Selection methods between two or more alternatives of finite time periods - Exercises 7 Mid Term Exam. 8 Selection methods between two or more alternatives of infinite time periods (capitalized cost calculations) - Exercises 9 Benefit/cost ratio analysis & application for single and multiple alternatives- Exercises 10 Different Depreciation Methods - Exercises 11 Different Depreciation Methods - switching between models - Depletion calculations models for natural resources - Exercises 12 Breakeven Analysis & applications for single and two alternatives - Exercises 13 Payback Analysis models & applications - Exercises 14 Final Term Exam. References : 1. ENGINEERING ECONOMY , Leland Blank , P. E. and Anthony Tarquin , P. E. ,Seventh Edition , © 2012 by The McGraw-Hill Companies 2. ENGINEERING ECONOMIC ANALYSIS , NINTH EDITION , Donald G. Newnan and Ted G. , 2004 by Oxford University Press, Inc. 3. Fundamentals of engineering economics, Chans S. Park, 2004 by Pearson Education, Inc. , ISBN 0-13-030791-2 4. SCHAUM'S OUTLINE OF THEORY AND PROBLEMS ENGINEERING ECONOMICS , JOSE A. SEPULVEDA, Ph.D. , WILLIAM E. SOUDER, Ph.D. And BYRON S. GOTTFRIED, Ph.D. , SCHAUM'S OUTLINE SERIES McGRAW-HILL

The Higher Technological Institute (HTI) Department of Mechanical Engineering Subject: Engineering Economy Code: MNG10201

Dr.Hosny Abouzeid 90 min Examiner: Time: Mid Term Exam. May/Aug 2019 Exam Type: Term:

1. Answer ONE of The Following Questions : [2 Marks] 1.1 Mention the Objectives and functions of a firm . 1.2 Mention the Kinds of Market structure indicating two main characteristics of each 2. Select the correct Answer from the following Questions : :[2 Marks] 2.1 An economic system in which private property is almost totally restricted is called ... A. A mixed economy B. Competition C. Free enterprise D. A centrally planned economy 2.2 ABC Co. has current assets of $50,000 and total assets of $150,000. ABC has current liabilities of $30,000 and total liabilities of $80,000. What is the amount of ABC's owner's equity? A. $20,000 B. $30,000 C. $70,000 D. $120,000 3. Answer The Following Questions Using Either (√ ) OR ( χ ) Only: :[2 Marks] 3.1 In supply curve , the supplied quantity increases with increasing the price . 3.2 If current assets > current liabilities ,the then company does not has enough cash to run the business 4. Answer The Following Problems: [24 Marks] 4.1 Assume that you have been offered an investment opportunity in which you may invest $5000 at 6%per year simple interest for 3 years or you may invest the same $5000 at 5% per year compound interest for 3 years. Which investment offer you accept? Why ?

4.2 Calculate the present worth and the future worth of an expenditure of 51000 L.E per year for 6 years starting 4 years from now if the interest rate if 15% per year . Draw the cash flow diagram and list the values of the engineering economy symbols used in this problem. Calculate the equivalent annual expenditure if it starts from first year up to the end of the interest period.

4.3 Determine the present worth of a loan which will be repaid in annual payments starting one year from now and increase gradually by 1000 L.E per year starting from year two up to year ten where the final payments will be 10000 L.E with interest rate of 15 % per year compounded quarterly .

4.4 The fixed cost of a machine maintenance will be 7000 L.E per year ; while the remaining cost will be 400 L.E starting two year from now and increase gradually and uniformly by 400 L.E per year . Calculate the expected present worth ; future worth ; and the equivalent uniform annual amounts for the total cost during the next ten years if the interest rate will be 14.4 % yearly .

Good Luck إسم الطالب : رقم الطالب : HIGHER TECHANOLOGICAL INSTITUTE 10th of Ramadan City Subject: Engineering Economy,G1 Quiz1: May/Aug. 2019 Examiner: Dr. Hosny Abouzeid Time: 25 min. ANSWER TWO ONLY OF THE FOLOWING QUESTIONS : 1. A new plant to produce steel tubing requires an initial investment of $10 million. It is expected that after four years of operation an additional investment of $5 million will be required; and after seven years of operation, another investment of $3 million. Annual operating costs will be $2 million and annual revenues will be $7 million. The life of the plant is 10 years. If the interest rate is 14% per year, Determine the economic acceptability of this plant . 2.Select one of the two machines below on the basis of their present worth , using an interest rate Of 15% per year . Machine P Machine Q First cost $290000 $370000 Salvage Value $40000 $50000 Annual maintenance cost $30000 $35000 Overhaul every 3 years $37000 $20000 Life , years 4 6 3. Compare between the following two machines on the bases of their capitalized cost with interest rate of 14 % per year . Machine First Cost AOC Salvage Value Over haul Lift ,years every 6 years X 75000 L.E 124000 L.E 15000 L.E ------8 Y 300000 L.E 24000 L.E 0 7500 L.E ∞

The Higher Technological Institute (HTI) Department of Mechanical Engineering Subject: Engineering Economy Code: MNG10201

Examiner: Dr.Hosny Abouzeid Time: 90 min Exam Final Exam. May/Aug 2019 Term:

Type:

ANSWER FIVE ONLY OF THE FOLOWING QUESTIONS :[5X8Marks] 1 . Facility has aging cooling system which currently runs 70% of the time the plant is open – Pump will only last 5 more years. As it deteriorates, the pump run time is expected to increase 6% per year . New cooling system of price $80000 would only run 50% of the time . Which of these two systems should be accepted , if : either pump uses 300 kWh, Electricity cost $0.2/KWh , Plant runs 270 days per year, 24 hours per day and interest rate is 15 % . 2. There are three sites for flood – control dams ( designated as sites A , B ,and C) . The construction costs are $15 millions , $ 18 millions , and $30 millions , and maintenance costs are expected to be $22500 , $ 30000 , and $34500 respectively , for sites A , B , and C . In addition , a $112500 expenditure will be required every 10 years at each site . The present cost of flood damage is $3 million per year . If only the dam at site A is constructed , the flood damage will be reduced to $2.4 million per year . If only the dam at site B is constructed , the flood damage will be $1.8 million per year . If only the dam at site c is built , the flood damage will be reduced to $1.16 million per year . If the interest rate is 5% per year , determine which ones , if any , should be built on the bases of their B /C ratios . Assume that the dams will be permanent . 3. A new machine costs $320 000, has a useful life of 12 years, and can be sold for $30000 at the end of its useful life. It is expected that $10000 will be spent to dismantle and remove the machine at the end of its useful life. Compute the depreciation rate , the book values for years 1&12 for this machine using: A. straight-line method . B. the sum-of-years'-digits method . 4. A coal mining company has owned a mine for the past 5 years . During this time the following tonnage of raw material has been removed each year : 80000 , 104000 , 116000,120000, and 112000 tons . The mine is estimated to contain a total of 2million tons of coal and the mine had an initial cost of $7 million . If the company had a gross income for this coal of $30 per ton for the first two years and $36 per ton for the last three years . Comput the depletion charges each year. Compute the percent of the initial cost that has been written off in these 5 years . 5. Two methods of wild plant control in an irrigation canal are under consideration. Method A involves lining the canal at a cost of $60,000. The lining is expected to last 20 years. Maintenance with this method will cost $4 per mile per year. Method B involves spraying a chemical which costs $90 per gallon, with one gallon capable of treating 10 miles. Spraying equipment will cost $5000 and will have a life of 3 years with no salvage value. At an interest rate of 10% per year, (a) how many miles of canal must require treatment in order for the two methods to breakeven, and (b) if 400 miles of canal must be treated each year, which method should be selected? 6. Two quality inspection equipment are being considered for purchase by Square D Electric. Machine 2 is expected to be versatile and technologically advanced enough to provide net income longer than machine 1. First cost,$ Annual NCF,$ Maximum life,years Machine A 24000 6000 8 Machine B 16000 2000(years 1-5) 15 6000(years 6-15) Which of the two machines should be selected on the payback period method if the quality manager used a return of 14% per year . Good Luck The Higher Technological Institute (HTI) Department of Mechanical Engineering

Subject: Engineering Economy Code: MNG102 Examiner: Dr.Hosny Abouzeid Time: 90 min

Final Exam. Exam Type: Term:

ANSWER FIVE ONLY OF THE FOLOWING QUESTIONS : 1. Facility has aging cooling system which currently runs 70% of the time the plant is open – Pump will only last 5 more years. As it deteriorates, the pump run time is expected to increase 6% per year . New cooling system of price $80000 would only run 50% of the time . Which of these two systems should be accepted , if : either pump uses 300 kWh, Electricity cost $0.2/KWh , Plant runs 270 days per year, 24 hours per day and interest rate is 15 % .

2. Compare between the following two machines on the bases of their capitalized cost with interest rate of 14 % per year . mmmMachinegSaff First CostStsts AOC Salvage ValuemmmOver haul every 6 ll6years ye life , years X X 150000 L.E 125000 L.E 30000 L.E ------2000 7 Y 2000Y 300000 L.E2000 50000 L.E2000 0 2000 20000 L.E 2000 ∞

3. Freeport-McMoRan Copper and Gold has purchased a new ore grading unit for $88,000. The unit has an anticipated life of 10 years and a salvage value of $11,000. Use the DB and DDB methods to compare the values of depreciation charge and book values at year 1 and 10 only.

4. A gold mine was purchased for $15 million. It has an anticipated gross income of $7.5 million per year for years 1 to 5 and $4.5 million per year after year 5. Compute annual depletion amounts for the mine. How long will it take to recover the initial investment at i = 0%? . Assume percentage gross income,PD , for gold equal 0.15 .

5. A company for manufacturing plastic products has the following production data : Fixed Expenses = $ 135,000 . Variable Costs per unit are : Direct Material = $7.5 , Direct Labour = $3 , Direct Overheads = 100% of Direct Labour . Selling Price per unit = $ 18. Calculate the breaheven quantity and the sales required to earn a profit of $675000. Calculate also the net profit for sales quantity 25% above the break even quantity .

6. Two equivalent pieces of quality inspection equipment are being considered for purchase by Square D Electric. Machine 2 is expected to be versatile and technologically advanced enough to provide net income longer than machine 1. First cost ,$ Annual NCF , $ Maximum life,years Machine 1 24000 3000 7 Machine 2 16000 2000(years 1-5) , 6000(years 6-14) 14 Which of the two machines should be selected on basis of the payback period method if the quality manager used a return of 15% per year . Good Luck