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 • Use BARTER( (مقايضة system in trade! No money! • 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)

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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 structure

Number of firms

Nature of products Selling price control

Condition of market entry

Perfect competition Large Homogeneous Non- existent Easy

Monopolistic competition

Large Are close substitutes Non- existent(low)

Easy

Oligopoly Few May be homogeneous or close substitutes

Medium or high Difficult

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.

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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 shareholders' equity )حقوق المساهمين). 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 .

• Marketable securities: )أوراق مالية قابلة للتداول(

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 debts, bills (فواتير) , loans, bonds(ودائع, سندات) ,etc.. . Liabilities have two main classes : 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 • Bank indebtedness( مديونيات البنك) • Interest payable • Rent, tax, utilities • Wages payable • Customer prepayments • Dividends payable )أرباح مستحقة الدفع(and others .

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.

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• Bonds Payable سندات مستحقة : This account includes the sinking fund amount of any bonds the company has issued.

• Pension fund liability: the money a company is required to pay into its employees' retirement accounts

• Deferred tax liability: إلتزامات ضريبيه مؤجلةtaxes that have been accrued but will not be paid for another year .

• Mortgages رهونات عقاريه

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 dividends أرباح.

C. Preferred and common stocks سهم المفضلة والشائعةاأل 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

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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 :

ieffa = (1 + r/m)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.

Solution:

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 .

P=?

A1=500 L.E A2=750 L.E

0 1 2 3 4 5

i = 10 %

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?

F = P(1+ n1 x il)(l+ i2)n2 = $1000[1+ (4)(0.12)](1+ 0.08)10

= $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

0 1 2 3 n-1 n

P? i= …. %

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.

2.2 Derivation of Single – Payment formulas Factors: Compound Amount (𝐹

𝑃)

&present worth ( 𝑃

𝐹 ):

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= …%

Then F2= Final sum at end of year 2 = P*(𝟏 + 𝒊)𝟐, 0 1 2 n=… F3=

Final sum at end of year 3 =F2+F2*i F?

= F2(1+i)= P*(𝟏 + 𝒊)𝟑 , So, Fn =Final sum at end of year n =P*(𝟏 + 𝒊)𝒏

𝑭?

𝑷 = (𝟏 + 𝒊)𝒏 called single payment compound amount factor

𝑷?

𝑭 =

𝟏

(𝟏+𝒊)𝒏 called single payment present worth factor

2.3 Derivation of Uniform – Series Present Worth Factor ( 𝑷

𝑨 ):

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=P1+P2+P3+…..= Pn = Sum of all the present worth values

= 𝑨

𝟏+𝒊+

𝑨

(𝟏+𝒊)𝟐 +………

𝑨

(𝟏+𝒊)𝒏 A = …… L. E

Multiply both sides by 𝟏

𝟏+𝒊 we get :

𝑷

𝟏+𝒊 = A[

𝟏

(𝟏+𝒊)𝟐+

𝟏

(𝟏+𝒊)𝟑+ ⋯ +

𝟏

(𝟏+𝒊)𝒏+

𝟏

(𝟏+𝒊)𝒏+𝟏]

Then,

𝑷

𝟏+𝒊 -P=A[

𝟏

(𝟏+𝒊)𝟐+

𝟏

(𝟏+𝒊)𝟑+ ⋯ +

𝟏

(𝟏+𝒊)𝒏+

𝟏

(𝟏+𝒊)𝒏+𝟏− {

𝟏

𝟏+𝒊+

𝟏

(𝟏+𝒊)𝟐+

𝟏

(𝟏+𝒊)𝟑+ ⋯ +

𝟏

(𝟏+𝒊)𝒏}]

P [ −𝒊

𝟏+𝒊 ] = A (

𝟏

𝟏+𝒊) [ 𝟏

(𝟏+𝒊)𝒏 - 1] =

𝑨

𝟏+𝒊[

𝟏−(𝟏+𝒊)𝒏

(𝟏+𝒊)𝒏]

𝑷

𝑨 =

(𝟏+𝒊)𝒏−𝟏

𝒊(𝟏+𝒊)𝒏 , Called Series Present Worth Factor

𝑨

𝑷 =

𝒊(𝟏+𝒊)𝒏

(𝟏+𝒊)𝒏−𝟏 , which is called the Capital Recovery Factor “CRF” which

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)

𝑭

𝑨 =

𝑭

𝑷∗

𝑷

𝑨= (𝟏 + 𝒊)𝒏 ∗

(𝟏+𝒊)𝒏−𝟏

𝒊(𝟏+𝒊)𝒏 =

(𝟏+𝒊)𝒏−𝟏

𝒊= 𝑼𝑺𝑪𝑨𝑭

∴𝑨

𝑭=

𝒊

(𝟏+𝒊)𝒏−𝟏 = SFF which gives the value of A in terms of F, i, n.

In these cases A starts at year 1 and ends at year n where F is existed as shown.

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 𝑭

𝑷 at i= 6% ; n=20. To get value of

F; multiply this factor by P i.e F = P*𝑭

𝑷 or F = P (F/P, 6%, 20)

The following table summarize the standard factors notation used in the interest tables .

F 0 1 2 3 …………. n-1 n

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

N

Single Payment Uniform Series Payments

Compound

amount F/P

Present worth

P/F

Sinking Fund A/F

Compound amount

F/A

Capital Recovery

A/P

Present Worth

P/A

1 2 3 4 5 . . . . .

100

1.005

1.0151

0.995

0.9851

1.0

0.33167

1.0

3.015

1.005

0.33667

0.995

2.9702

1 2 3 . . . . . . .

100

Examples: • (F/A, 0.5%, 3) = 3.015 from table A-1 at i=0.5%; n=3

Or (F/A, 0.5%, 3) = (𝟏+𝒊)𝒏−𝟏

𝒊 =

(𝟏+𝟎.𝟎𝟎𝟓)𝟑−𝟏

𝟎.𝟎𝟎𝟓= 𝟑. 𝟎𝟏𝟓𝟎𝟐𝟓

• (P/A, 5%, 10) = 7.7217 from table A-7 at i=5%; n=10

Or (P/A, 5%, 10) = (𝟏+𝒊)𝒏−𝟏

𝒊(𝟏+𝒊)𝒏 =

(𝟏.𝟎𝟓)𝟏𝟎−𝟏

𝟎.𝟎𝟓(𝟏.𝟎𝟓)𝟏𝟎 = 7.7217

• (P/F, 25%, 35) = 0.0004 from table A-25

Or (P/F, 25%, 35) = 𝒊

(𝟏+𝒊)𝒏=

𝒊

(𝟏.𝟐𝟓)𝟑𝟓 = 0.0004

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:

𝑭𝒊 = 𝑭𝒊𝟏 + 𝒊−𝒊𝟏

𝒊𝟐−𝒊𝟏(𝑭𝒊𝟐 − 𝑭𝒊𝟏), where 𝑭𝒊 corresponds to F/P or P/F or P/A, A/p, ….. etc

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

𝒊𝟏 = 𝟕%

i=?

𝒊𝟐=8%

10

10

10

0.14238 = 𝑭𝒊𝟏

F=?

0.14903=𝑭𝒊𝟐

∴𝑨

𝑷|

𝟕.𝟑= 𝟎. 𝟏𝟒𝟐𝟑𝟖 +

𝟕. 𝟑 − 𝟕

𝟖 − 𝟕(𝟎. 𝟏𝟒𝟗𝟎𝟑 − 𝟎. 𝟏𝟒𝟐𝟑𝟖) = 𝟎. 𝟏𝟒𝟒𝟑𝟕𝟓

∴ 𝑨 = 𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝒖𝒏𝒊𝒇𝒐𝒓𝒎 𝒂𝒏𝒏𝒖𝒂𝒍 𝒔𝒆𝒓𝒊𝒆𝒔

=𝑷 ∗𝑨

𝑷|

𝟕.𝟑= 𝟏𝟎𝟎, 𝟎𝟎𝟎 ∗ 𝟎. 𝟏𝟒𝟒𝟑𝟕𝟓 = 𝟏𝟒𝟒𝟑𝟕𝟓 𝑳. 𝑬

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:

With a gradient G G =25 in this case the value of G is obtained as follow :

𝑮 = 𝒇𝒊𝒏𝒂𝒍 𝒑𝒂𝒚𝒎𝒆𝒏𝒕𝒔 𝒐𝒇 𝒖𝒏𝒊𝒇𝒐𝒓𝒎 𝒈𝒓𝒂𝒅𝒊𝒆𝒏𝒕 − 𝑩𝒂𝒔𝒆 𝒑𝒂𝒚𝒎𝒆𝒏𝒕

𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒚𝒆𝒂𝒓 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒖𝒏𝒊𝒇𝒐𝒓𝒎 𝒈𝒓𝒂𝒅𝒊𝒆𝒏𝒕 𝒑𝒆𝒓𝒊𝒐𝒅 − 𝟏

∴ 𝑮 = 𝑨𝒇 − 𝑨𝒃

𝑵 − 𝟏

400+ (n-1)25

Ab+ (n-1) G

400+ (n-2)25

Ab+ (n-2) G

450

Ab

+2G

425

Ab

+G

400

Ab

0 1 2 3 …………. n-1 n

Cash Flow Diagram For A Uniform -

Gradient

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 𝑷

𝑮 (UGPWF)

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)

P = G[ 𝟏

(𝟏+𝒊)𝟐+

𝟐

(𝟏+𝒊)𝟑+ ⋯ +

𝒏−𝟐

(𝟏+𝒊)𝒏−𝟏+

𝒏−𝟏

(𝟏+𝒊)𝒏 ] ………….. (A)

Multiply both sides by (1+i) we get:

P(1+i) = G[ 𝟏

𝟏+𝐢+

𝟐

(𝟏+𝐢)𝟐+

𝟑

(𝟏+𝐢)𝟑+ ⋯ +

𝐧−𝟐

(𝟏+𝐢)𝐧−𝟐+

𝒏−𝟏

(𝟏+𝒊)𝒏−𝟏 ]…. (B)

(B) – (A) will give:

P[ (1+i) -1] = G [𝟏

𝟏+𝒊+

𝟐−𝟏

(𝟏+𝒊)𝟐+

𝟑−𝟐

(𝟏+𝒊)𝟑+ ⋯ +

(𝒏−𝟏)−(𝒏−𝟐)

(𝟏+𝒊)𝒏−𝟏−

𝒏−𝟏

(𝟏+𝒊)𝒏]

Pi = G {[𝟏

𝟏+𝒊+

𝟏

(𝟏+𝒊)𝟐+

𝟏

(𝟏+𝒊)𝟑+ ⋯ +

𝟏

(𝟏+𝒊)𝒏−𝟏] −

𝒏−𝟏

(𝟏+𝒊)𝒏 }

P= 𝑮

𝒊 [ (

𝑷

𝑨 -

𝟏

(𝟏+𝒊)𝒏 ) -

𝒏−𝟏

(𝟏+𝒊)𝒏 ]

= 𝑮

𝒊 [

(𝟏+𝒊)𝒏−𝟏

𝒊(𝟏+𝒊)𝒏 -

𝒏

(𝟏+𝒊)𝒏 ]

∴ 𝑷

𝑮 =

𝟏

𝒊 [

(𝟏+𝒊)𝒏−𝟏

𝒊(𝟏+𝒊)𝒏 -

𝒏

(𝟏+𝒊)𝒏 ]

2.8.2 Uniform gradient annual series factor 𝑨

𝑮 (UGASF):

This factor is used to convert from uniform gradient to an equivalent annual series as shown below:

i=….%

0 1 2 3 …………. n-1 n

P? G=……

Ab G

i=….%

0 1 2 …………. n-1 n

A

i=….%

0 1 2 3 …………. n-1 n

It’s value is obtained as follow: 𝑨

𝑮=

𝑨

𝑷∗

𝑷

𝑮 =

𝒊(𝟏+𝒊)𝒏

(𝟏+𝒊)𝒏−𝟏∗

𝟏

𝒊 [

(𝟏+𝒊)𝒏−𝟏

𝒊(𝟏+𝒊)𝒏 -

𝒏

(𝟏+𝒊)𝒏 ] =

𝟏

𝒊 -

𝒏

(𝟏+𝒊)𝒏−𝟏

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:

(P/G, 4%, 6) = 𝟏

𝟎.𝟎𝟒 [

(𝟏+𝟎.𝟎𝟒)𝟔−𝟏

𝟎.𝟎𝟒(𝟏+𝟎.𝟎𝟒)𝟔 -

𝟔

(𝟏+𝟎.𝟎𝟒)𝟔 ] = 12.50624

Or from the table:

N 1% 2% 3% 4% 5% 6% N

2 3 . . . . 6

14.321

13.68

13.076

12.506

11.968

11.459

2 . . . . . 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)

Present Worth Gradient Factors

(P/G)

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

= 60000(F/P, 5%, 10) + 30000(F/P, 5%, 8) + 40000(F/P, 5%, 5)

= 60000(1.6289) + 30000(1.4775) + 40000(1.2763)

= 193111 L.E

Or to avoid mistakes in compute n:

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)

= 702605 L.E.

EUAS = Ab + G(A/G, 5%, 10) = 50000

+ 10000 (4.099) = 90990

Or

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

(P/F, i%, 5) = 𝟔𝟎𝟎𝟎𝟎

𝟏𝟎𝟎𝟎𝟎𝟎 = 0.6

From tables

F?

i= 5% n=10

0 1 2 3 4 5

60000

0

30000 40000

0 1 2 3 …………. n = 10

50000

G=10000 L.E

i=5%

i?

0 1 2 3 4 n=5

P=600000 L.E

F =

100000

i N P/F

𝟏𝟎

?

𝟏𝟏

5

5

0.6209

0.6

0.5935

Using the equation: (P/F, i%, 5) = 0.6 = 𝟏

(𝟏+𝒊)𝟓

∴ 𝒊 = 𝟏

𝟎.𝟔𝟎.𝟐− 𝟏 = 𝟏. 𝟏𝟎𝟕𝟔 − 𝟏 = 𝟎. 𝟏𝟎𝟕𝟔 = 𝟏𝟎. 𝟕𝟔%

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.

100000 = 10000(P/A, 15%, n) + 1000(P/G, 15%, n)

= PT

n P/A P/G PT Notes 5 15 20 22 24

3.3533 5.8474 6.2543 6.3587 6.4338

5.775 26.693 33.582 35.615 37.302

39297 85167 96175 99202

101638

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 :

(P/G, 5.5%, 6) = 𝟏

𝟎.𝟎𝟓𝟓 [

(𝟏+𝟎.𝟎𝟓𝟓)𝟔−𝟏

𝟎.𝟎𝟓𝟓(𝟏+𝟎.𝟎𝟓𝟓)𝟔 -

𝟔

(𝟏+𝟎.𝟎𝟓𝟓)𝟔 ] = 11.71

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

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

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 .

0 1 2 3 4 5 6 7 8

EUAS

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 .

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% .

G1 = (8000 – 4000)/(5-1) = 1000 L.E ; G2 = (2000 -10000)/(5-1) = - 2000 L.E

PG1 =present worth of increasing gradient or deposits

0 1 2 3 4 5 6 7 8 i=6%

0 1 2 3 4 5 6 7 8

PT

7500

PI 0 1 2 3 4 5 6

EUAS

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

4000

8000

10000 G2

G1

2000

= 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 :

An = A1 / (1+g) n-1 , 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

= An / (1+i) n = A1 ((1+g) n-1 / (1+i) n

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

P = A1 [ 𝟏−(𝟏+𝒈)𝒏(𝟏+𝒊)−𝒏

𝒊−𝒈 ] , where i not equal g

= 100.0 [ 𝟏−(𝟏.𝟏)𝟓(𝟏.𝟎𝟖)−𝟓

−𝟎.𝟎𝟐 ] = $480.42

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

The present worth value is comprised of three components The present modification cost = $8,000 The present value of the future salvage value The 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:

F=P(1+i)n = A1 [ 𝟏−(𝟏+𝒈)𝒏(𝟏+𝒊)−𝒏

𝒊−𝒈 ](𝟏 + 𝐢) n

F = A1[((1 + i)n - (1 + g)n)/(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.

F = nA1(1 + i)n-1 , 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 . Solution: 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:

14000 L.E

0 1 2 3 4 5

28000 L.E

140000

8000 L.E

0 1 2 3 4 5

36000 L.E

100000

0

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