Eng ineeri ng Economy Third Edition Leland T. Blank, P. E. Department of Industrial Engineering Assistant Dean of Engineering Texas A & M University Anthony J. Tarquin, P. E. Department of Civil Engineering Assistant Dean of Engineering The University of Texas at EI Paso McGraw-Hill Book Company New York S1. Louis San Francisco Auckland Bogota Caracas Colorado Springs Hamburg Lisbon London Madrid Mexico Milan Montreal New Delhi Oklahoma City Panama Paris San Juan Silo Paulo Singapore Sydney Tokyo Toronto 4 Level One 1. Define and recognize in a problem statement the economy symbols P, F, A, n, and i. 1. 6 Define cash flow, state what is meant by end-of-period convention, and construct a cash-flow diagram, given a statement describing the amount and times of the cash flows. Study Guide 1. 1 Basic Terminology Before we begin to develop the terminology and fundamental concepts upon which engineering economy is based, it would be appropriate to define what is meant by engineering economy. In the simplest terms, engineering economy is a collection of mathematical techniques which simplify economic comparisons.

With these techniques, a rational, meaningful approach to evaluating the economic aspects of different methods of accomplishing a given objective can be developed. Engineering economy is, therefore, a decision assistance tool by which one method will be chosen as the most economical one. In order for you to be able to apply the techniques, however, it is necessary for you to understand the basic terminology and fundamental concepts that form the foundation for engineering-economy studies.

Some of these terms and concepts are described below. An alternative is a stand-alone solution for a give situation. We are faced with alternatives in virtually everything we do, from selecting the method of transportation we use to get to work every day to deciding between buying a house or renting one. Similarly, in engineering practice, there are always seveffl ways of accomplishing a given task, and it is necessary to be able to compare them in a rational manner so that the most economical alternative can be selected.

The alternatives in engineering considerations usually involve such items as purchase cost (first cost), the anticipated life of the asset, the yearly costs of maintaining the asset (annual maintenance and operating cost), the anticipated resale value (salvage value), and the interest rate (rate of return). After the facts and all the relevant estimates have been collected, an engineering-economy analysis can be conducted to determine which is best from an economic point of view.

However, it should be pointed out that the procedures developed in this book will enable you to make accurate economic decisions only about those alternatives which have been recognized as alternatives; these procedures will not help you identify what the alternatives are. That is, if alternatives ,4, B, C, D, and E have been identified as the only possible methods to solve a Particular problem when method F, which was never recognized as an alternative, is really the most attractive method, the wrong decision is certain to be made because alternative F could never be chosen, no matter what analytical techniques are used.

Thus, the importance of alternative identification in the decision-making process cannot be overemphasized, because it is only when this aspect of the process has been thoroughly completed that the analysis techniques presented in this book can be of greatest value. In order to be able to compare different methods for accomplishing a given objective, it is necessary to have an evaluation criterion that can be used as a basis Terminology and Cash-Flow Diagrams 5 for judging the alternatives. That is, the evaluation criterion is that which is used

Thus, when there are several ways of accomplishing a given objective, the method that has the lowest overall cost is usually selected. However, in most cases the alternatives involve intangible factors, such as the effect of a process change on employee morale, which cannot readily be expressed in terms of dollars. When the alternatives available have approximately the same equivalent cost, the nonquantifiable, or intangible, factors may be used as the basis for selecting the best alternative, For items of an alternative which can be quantified in terms of dollars, it is important to recognize the concept of the time value of money.

It is often said that money makes money. The statement is indeed true, for if we elect to invest money today (for example, in a bank or savings and loan association), by tomorrow we will have accumulated more money than we had originally invested. This change in the amount of money over a given time period is called the time value of money; it is the most important concept in engineering economy. You should also realize that if a person or company finds it necessary to borrow money today, by tomorrow more money than the original loan will be owed. This fact is also explained by the time value of money.

The manifestation of the time value of money is termed interest, which is a measure of the increase between the original sum borrowed or invested and the final amount owed or accrued. Thus, if you invested money at some time in the past, the interest would be Interest = total amount accumulated – original investment (1. 1) On the other hand, if you borrowed would be Interest money at some time in the past, the interest (1. 2) = present amount owed – original loan In either case, there is an increase in the amount of money that was originally invested or borrowed, and the increase over the original amount is the interest.

The original investment or loan is referred to as principal. Probs. 1. 1 to 1. 4 1. 2 Interest Calculations When interest is expressed as a percentage of the original amount per unit time, the result is an interest rate. This rate is calculated as follows: . Percent interest rate = interest accrued per unit time 00% .. I x 1 0 origma amount (1. 3) 6 Level One By far the most common time period used for expressing interest rates is 1 year. However, since interest rates are often expressed over periods of time shorter than 1 year (i. e. 1% per month), the time unit used in expressing an interest rate must also be identified and is termed an interest period. The following two examples illustrate the computation of interest rate. Example 1. 1 The Get-Rich-Quick (GRQ) Company invested $100,000 on May 1 and withdrew a total of $106,000 exactly one year later. Compute (a) the interest gained from the original investment and (b) the interest rate from the investment. Solution (a) Using Eq. (1. 1), Interest = 106,000 – 100,000 = $6000 (b) Equation (1. 3) is used to obtain Percent interest rate = 6000 per year 100,000 x 100% = 6% per year

Comment For borrowed money, computations are similar to those shown above except that interest is computed by Eq. (1. 2). For example, if GRQ borrowed $100,000 now and repaid $110,000 in 1 year, using Eq. (1. 2) we find that interest is $10,000, and the interest rate from Eq. (1. 3) is 10% per year. Example 1. 2 Joe Bilder plans to borrow $20,000 for 1 year at 15% interest. Compute (a) the interest and (b) the total amount due after 1 year. Solution (a) Equation (1. 3) may be solved for the interest accrued to obtain Interest = 20,000(0. 15) = $3000 (b) Total amount due is the sum of principal and interest or Total due Comment = 0,000 + 3000 = $23,000 Note that in part (b) above, the total amount due may also be computed as Total due = principal(l + interest rate) = 20,000(1. 15) = $23,000 In each example the interest period was 1 year and the interest was calculated at the end of one period. When more than one yearly interest period is involved (for example, if we had wanted to know the amount of interest Joe Bilder would owe on Terminology and Cash-Flow Diagrams 7 the above loan after 3 years), it becomes necessary to determine whether the interest . payable on a simple or compound basis. The concepts of simple and compound interest are discussed in Sec. . 4. Additional Examples 1. 12 and 1. 13 Probs. 1. 5 to 1. 7 1. 3 Equivalence The time value of money and interest rate utilized together generate the concept of equivalence, which means that different sums of money at different times can be equal in economic value. For example, if the interest rate is 12% per year, $100 today (i. e. , at present) would be equivalent to $112 one year from today, since mount accrued = 100 =$112 Thus, if someone offered you a gift of $100 today or $112 one year from today, it would make no difference which offer you accepted, since in either case you would have $112 one year from today.

The two sums of money are therefore equivalent to each other when the interest rate is 12% per year. At either a higher or a lower interest rate, however, $100 today is not equivalent to $112 one year from today. In addition to considering future equivalence, one can apply the same concepts for determining equivalence in previous years. Thus, $100 now would be equivalent to 100/1. 12 = $89. 29 one year ago if the interest rate is 12% per year. From these examples, it should be clear that $89. 29 last year, $100 now, and 112 one year from now are equivalent when the interest rate is 12% per year.

The fact that these sums are equivalent can be established by computing the interest rate as follows: 112 100 = 1. 12, or 12% per year and 8~~~9 = 1. 12, or 12% per year The concept of equivalence can be further illustrated by considering different loan-repayment schemes. Each scheme represents repayment of a $5000 loan in 5 years at 15%-per-year interest. Table 1. 1 presents the details for the four repayment methods described below. (The methods for determining the amount of the payments are presented in Chaps. 2 and 3. ) • Plan 1 a interest or principal is recovered until the fifth year.

Interest accumulates each year on the total of principal and all accumulated interest. • Plan 2 The accrued interest is paid each year and the principal is recovered at the end of 5 years. • Plan 3 The accrued interest and 20% of the principal, that is, $1000, is paid each year. Since the remaining loan balance decreases each year, the accrued interest decreases each year. + 100(0. 12) = 100(1 + 0. 12) = 100(1. 12) 8 Level One Table 1. 1 Different repayment schedules of $5,000 at 15% for 5 years (1) End of year (2) = 0. 15(5) Interest for year (3) = (2) + (5) Total owed at end of year (4) Payment per plan (3) – (4) Balance after payment (5) Plan 1 0 1 2 3 4 5 Plan 2 0 1 2 3 4 5 Plan 3 0 1 2 3 4 5 Plan 4 0 1 2 3 4 5 $ 750. 00 862. 50 991. 88 1,140. 66 1,311. 76 5,750. 00 6,612. 50 7,604. 38 8,745. 04 10,056. 80 0 0 0 0 10,056. 80 $10,056. 80 $ $5,000. 00 5,750. 00 6,612. 50 7,604. 38 8,745. 04 0 $750. 00 750. 00 750. 00 750. 00 750. 00 $5,750. 00 5,750. 00 5,750. 00 5,750. 00 5,750. 00 $ 750. 00 750. 00 750. 00 750. 00 5,750. 00 $8,750. 00 $5,000. 00 5,000. 00 5,000. 00 5,000. 00 5,000. 00 0 $750. 00 600. 00 450. 00 300. 00 150. 00 $5,750. 00 4,600. 00 3,450. 00 2,300. 00 1,150. 00 $1,750. 00 1,600. 00 1,450. 0 1,300. 00 1,150. 00 $7,250. 00 5,000. 00 4,000. 00 3,000. 00 2,000. 00 1,000. 00 0 $750. 00 638. 76 510. 84 363. 73 194. 57 $5,750. 00 4,897. 18 3,916. 44 2,788. 59 1,491. 58 $1,491. 58 1,491. 58 1,491. 58 1,491. 58 1,491. 58 $7,457. 90 $5,000. 00 4,258. 42 3,405. 60 2,424. 86 1,297. 01 0 • Plan 4 Equal payments are made each year with a portion going toward princi- pal recovery and the remainder covering the accrued interest. Since the loan balance decreases at a rate which is slower than in plan 3 because of the equal end-of-year payments, the interest decreases, but at a rate slower than in plan 3. te that the total amount repaid in each case would be different, even though each repayment scheme would require exactly 5 years to repay the loan. The difference in the total amounts repaid can of course be explained by the time value of money, since the amount of the payments is different for each plan. With respect to equivalence, the table shows that when the interest rate is 15% per year, $5000 at time 0 is equivalent to $10,056. 80 at the end of year 5 (plan 1), or $750 per year for 4 years and $5750 at the end of year 5 (plan 2), or the decreasing amounts shown in years 1 through 5 (plan 3), or $1,491. 8 per year for 5 years (plan 4). Using the formulas developed in Chaps. 2 and 3, we could easily show that if the payments in Terminology and Cash-Flow Diagrams 9 each plan (column 4) were reinvested at 15% per year when received, the total amount of money available at the end of year 5 would be $10,056. 80 from each repayment plan. Additional Examples 1. 14 and 1. 15 Probs. 1. 8 and 1. 9 1. 4 Simple and Compound Interest The concepts of interest and interest rate were introduced in Sees. 1. 1 and 1. 2 and ed in Sec. 1. 3 to calculate for one interest period past and future sums of money equivalent to a present sum (principal).

When more than one interest period is involved, the terms simple and compound interest must be considered. Simple interest is calculated using the principal only, ignoring any interest that was accrued in preceding interest periods. The total interest can be computed using the relation Interest = (principal)(number of periods)(interest rate) = Pni (1. 4) Example 1. 3 If you borrow $1000 for 3 years at 14%-per-year simple interest, how much money will you owe at the end of 3 years? Solution The interest for each of the 3 years is = Interest per year 1000(0. 14) = $140 Total interest for 3 years from Eq. (1. 4) is Total interest = 1000(3)(0. 4)= $420 Finally, the amount due after 3 years is 1000 + 420 Comment = $1420 The $140 interest accrued in the first year and the $140 accrued in the second year did not earn interest. The interest due was calculated on the principal only. The results of this loan are tabulated in Table 1. 2. The end-of-year figure of zero represents th~ present, that is, when the money is borrowed. Note that no payment is made by the borrower until the end of year 3. Thus, the amount owed each year increases uniformly by $140, since interest is figured only on the principal of $1000. Table 1. 2 Simple-interest (1) (2) computation (3) (4) (2) + (3) Amount owed (5) End of year 0 1 2 Amount borrowed $1,000 Interest Amount paid 3 $140 140 140 $1,140 1,280 1,420 $ 0 0 1,420 10 Level One In calculations of compound interest, the interest for an interest period is calculated on the principal plus the total amount of interest accumulated in previous periods. Thus, compound interest means “interest on top of interest” (i. e. , it reflects the effect of the time value of money on the interest too). Example 1. 4 If you borrow $1000 at 14%-per-year compound interest, instead of simple interest as in the preceding example, compute the total amount due after a 3-year period.

Solution The interest and total amount due for each year is computed as follows: Interest, year 1 = 1000(0. 14) = $140 Total amount due after year 1 = 1000 + 140 = $1140 Interest, year 2 = 1140(0. 14) = $159. 60 Total amount due after year 2 = 1140 + 159. 60 = $1299. 60 Interest, year 3 = 1299. 60(0. 14)= $181. 94 Total amount due after year 3 = 1299. 60 + 181. 94 = $1481. 54 Comment The details are shown in Table 1. 3. The repayment scheme is the same as that for the simple-interest example; that is, no amount is repaid until the principal plus all interest is due at the end of year 3.

The time value of money is especially recognized in compound interest. Thus, with compound interest, the original $1000 would accumulate an extra $1481. 54 – $1420 = $61. 54 compared with simple interest in the 3-year period. If $61. 54 does not seem like a significant difference, remember that the beginning amount here was only $1000. Make these same calculations for an initial amount of $10 million, and then look at the size of the difference! The power of compounding can further be illustrated through another interesting exercise called “Pay Now, Play Later”. It can be shown (by using the equations that will be developed in Chap. ) that at an interest rate of 12% per year, approximately $1,000,000 will be accumulated at the end of a 40-year time period by either of the Table 1. 3 Compound-interest (1) (2) computation (3) (4) = (2) + (3) (5) End of year 0 1 2 3 Amount borrowed $1,000 Interest Amount owed $1,140. 00 1,299. 60 1,481. 54 Amount paid $140. 00 159. 60 181. 94 $ 0 0 1,481. 54 Terminology and Cash-Flow Diagrams 11 – llowing investment schemes: • Plan 1 Invest $2610 each year for the first 6 years and then nothing for the next 34 years, or • Plan 2 Invest nothing for the first 6 years, and then $2600 each year for the next 34 years!! ‘ote that the total investment in plan 1 is $15,660 while the total required in plan _ to accumulate the same amount of money is nearly six times greater at $88,400. Both the power of compounding and the wisdom of planning for your retirement at he earliest possible time should be quite evident from this example. An interesting observation pertaining to compound-interest calculations in-olves the estimation of the length of time required for a single initial investment to double in value. The so-called rule of 72 can be used to estimate this time.

The rule i based on the fact that the time required for an initial lump-sum investment to double in value when interest is compounded is approximately equal to 72 divided by the interest rate that applies. For example, at an interest rate of 5% per year, it would take approximately 14. 4 years (i. e. , 72/5 = 14. 4) for an initial sum of money to double in value. (The actual time required is 14. 3 years, as will be shown in Chap. 2. ) In Table 1. 4, the times estimated from the rule of 72 are compared to the actual times required for doubling at various interest rates and, as you can see, very good estimates are obtained.

Conversely, the interest rate that would be required in order for money to double in a specified period of time could be estimated by dividing 72 by the specified time period. Thus, in order for money to double in a time period of 12 years, an interest rate of approximately 6% per year would be required (i. e. , 72/12 = 6). It should be obvious that for simple-interest situations, the “rule of 100” would apply, except that the answers obtained will always be exact. In Chap. 2, formulas are developed which simplify compound-interest calculations. The same concepts are involved when the interest period is less than a year.

A discussion of this case is deferred until Chap. 3, however. Since real-world calculations almost always involve compound interest, the interest rates specified herein refer to compound interest rates unless specified otherwise. Additional Example 1. 16 Probs. 1. 10 to 1. 26 Table 1. 4 Doubling time estimated actual time from rule of 72 versus Doubling lime, no. of periods Interest rate, % per period 1 Estimated from rule 72 Actual 70 35. 3 14. 3 7. 5 2 5 10 20 40 36 14. 4 7. 2 3. 6 1. 8 3. 9 2. 0 12 Level One 1. 5 Symbols and Their Meaning The mathematical symbols: relations sed in engmeenng economy employ the following P = value or sum of money at a time denoted as the present; dollars, pesos, etc. F A n i = value or sum of money at some future time; dollars, pesos, etc. = a series of consecutive, equal, end-of-period month, dollars per year, etc. amounts of money; dollars per = number of interest periods; months, years, etc. = interest rate per interest period; percent per month, percent per year, etc. The symbols P and F represent single-time occurrence values: A occurs at each interest period for a specified number of periods with the same value.

It should be understood that a present sum P represents a single sum of money at some time prior to a future sum or uniform series amount and therefore does not necessarily have to be located at time t = O. Example 1. 11 shows a P value at a time other than t = O. The units of the symbols aid in clarifying their meaning. The present sum P and future sum F are expressed in dollars; A is referred to in dollars per interest period. It is important to note here that in order for a series to be represented by the symbol A, it must be uniform (i. e. the dollar value must be the same for each period) and the uniform dollar amounts must extend through consecutive interest periods. Both conditions must exist before the dollar value can be represented by A. Since n is commonly expressed in years or months, A is usually expressed in units of dollars per year or dollars per month, respectively. The compound-interest rate i is expressed in percent per interest period, for example, 5% per year. Except where noted otherwise, this rate applies throughout the entire n years or n interest periods. The i value is often the minimum attractive rate of return (MARR).

All engineering-economy problems must involve at least four of the symbols listed above, with at least three of the values known. The following four examples illustrate the use of the symbols. Example 1. 5. If you borrow $2000 now and must repay the loan plus interest at a rate of 12% per year in 5 years, what is the total amount you must pay? List the values of P, F, n, and i. Solution In this situation P and F, but not A, are involved, since all transactions are single payments. The values are as follows: P = $2000 Example 1. 6 i = 12% per year n = 5 years

If you borrow $2000 now at 17% per year for 5 years and must repay the loan in equal yearly payments, what will you be required to pay? Determine the value of the symbols involved. Terminology and Cash-Flow Diagrams 13 ~- ution = S2000 = ? per year for 5 years = 17% per year = 5 years – ere is no F value involved. – 1 In both examples, the P value of $2000 is a receipt and F or A is a disbursement. equally correct to use these symbols in reverse roles, as in the examples below. Example 1. 7 T you deposit $500 into an account on May 1, 1988, which pays interest at 17% per year, hat annual amount can you withdraw for the following 10 years?

List the symbol values. Solution p = $500 A =? per year i = 17% per year n= 10 years Comment The value for the $500 disbursement P and receipt A are given the same symbol names as before, but they are considered in a different context. Thus, a P value may be a receipt (Examples 1. 5 and 1. 6) or a disbursement (this example). Example 1. 8 If you deposit $100 into an account each year for 7 years at an interest rate of 16% per year, what single amount will you be able to withdraw after 7 years? Define the symbols and their roles.

Solution In this example, the equal annual deposits are in a series A and the withdrawal is a future sum, or F value. There is no P value here. A = $100 per year for 7 years F =? i = 16% per year n = 7 years Additional Example 1. 17 Probs. 1. 27 to 1. 29 14 Level One 1. 6 Cash-Flow Diagrams Every person or company has cash receipts (income) and cash disbursements (costs) which occur over a particular time span. These receipts and disbursements in a given time interval are referred to as cash flow, with positive cash flows usually representing receipts and negative cash flows representing disbursements.

At any point in time, the net cash flow would be represented as Net cash flow = receipts – disbursements (1. 5) Since cash flow normally takes place at frequent and varying time intervals within an interest period, a simplifying assumption is made that all cash flow occurs at the end of the interest period. This is known as the end-of-period convention. Thus, when several receipts and disbursements occur within a given interest period, the net cash flow is assumed to occur at the end of the interest period.

However, it should be understood that although the dollar amounts of F or A are always considered to occur at the end of the interest period, this does not mean that the end of the period is December 31. In the situation of Example 1. 7, since investment took place on May 1, 1988, the withdrawals will take place on May 1, 1989 and each succeeding May 1 for 10 years (the last withdrawal will be on May 1, 1998, not 1999). Thus, end of the period means one time period from the date of the transaction (whether it be receipt or disbursement).

In the next chapter you will learn how to determine the equivalent relations between P, F, and A values at different times. A cash-flow diagram is simply a graphical representation of cash flows drawn on a time scale. The diagram should represent the statement of the problem and should include what is given and what is to be found. That is, after the cash-flow diagram has been drawn, an outside observer should be able to work the problem by looking at only the diagram. Time is considered to be the present and time 1 is the end of time period 1. (We will assume that the periods are in years until Chap. . ) The time scale of Fig. 1. 1 is set up for 5 years. Since it is assumed that cash flows occur only at the end of the year, we will be concerned only with the times marked 0, 1, 2, … , 5. The direction of the arrows on the cash-flow diagram is important to problem solution. Therefore, in this text, a vertical arrow pointing up will indicate a positive cash flow. Conversely, an arrow pointing down will indicate a negative cash flow. The cash-flow diagram in Fig. 1. 2 illustrates a receipt (income) at the end of year 1 and a disbursement at the end of year 2.

It is important that you thoroughly understand the meaning and construction of the cash-flow diagram, since it is a valuable tool in problem solution. The three examples below illustrate the construction of cash-flow diagrams. ° Figure 1. 1 A typical cash-flow time scale. Year 1 Year 5 r=;:;; r+;:;. I 1 2 Time o I I 3 4 I 5 Terminology and Cash-Flow Diagrams 15 + Figure 1. 2 Example of positive and negative cash flows. 2 3 Time Example 1. 9 Consider the situation presented in Example 1. 5, where P = $2000 is borrowed and F is to be found after 5 years.

Construct the cash-flow diagram for this case, assuming an interest rate of 12% per year. Solution Figure 1. 3 presents the cash-flow diagram. Comment While it is not necessary to use an exact scale on the cash-flow axes, you will probably avoid errors later on if you make a neat diagram. Note also that the present sum P is a receipt at year 0 and the future sum F is a disbursement at the end of year 5. Example 1. 10 If you start now and make five deposits of $1000 per year (A) in a 17%-per-year account, how much money will be accumulated (and can be withdrawn) immediately after you have made the last deposit?

Construct the cash-flow diagram. Solution The cash flows are shown in Fig. 1. 4. Since you have decided to start now, the first deposit is at year 0 and the [lith Comment deposit and withdrawal occur at the end of year 4. Note that in this example, the amount accumulated after the fifth deposit is to be computed; thus, the future amount is represented by a question mark (i. e. , F = ? ) Figure 1. 3. Cash-flow diagram for Example 1. 9. + P = $2. 000 i = 12% o 2 3 4 5 Year F= ? 16 Figure 1. 4 Cashflow diagram for Example 1. 10. Level One F= ? i = 17″10 2 0 3 4 Year A=$1. 000 Example 1. 11

Assume that you want to deposit an amount P into an account 2 years from now in order to be able to withdraw $400 per year for 5 years starting 3 years from now. Assume that the interest rate is 151% per year. Construct the cash-flow diagram. Figure 1. 5 presents the cash flows, where P is to be found. Note that the diagram shows what was given and what is to be found and that a P value is not necessarily located at time t = O. Solution Additional Examples 1. 18 to 1. 20 Probs. 1. 30 to 1. 46 Additional Examples Example 1. 12 Calculate the interest and total amount accrued after 1 year if $2000 is invested at an interest rate of 15% per year.

Solution Interest earned = 2000(0. 15) = $300 Total amount accrued = 2000 + 2000(0. 15) = 2000(1 + 0. 15) = $2300 Figure 1. 5 Cashflow diagram for Example 1. 11. A = $400 o 2 3 4 5 6 7 Year p=? Terminology and Cash-Flow Diagrams 17 Example 1. 13 a) Calculate the amount of money that must have been deposited 1 year ago for you to have $lOQO now at an interest rate of 5% per year. b) Calculate the interest that was earned in the same time period. Solution a) Total amount accrued = original deposit + (original deposit)(interest rate). If X = original deposit, then 1000 = X + X(0. 5) = X(l + 0. 05) 1000 = 1. 05X 1000 X=-=952. 38 1. 05 Original deposit = $952. 38 (b) By using Eq. (1. 1), we have Interest = 1000 – 952. 38 = $47. 62 Example 1. 14 Calculate the amount of money that must have been deposited 1 year ago for the investment to earn $100 in interest in 1 year, if the interest rate is 6% Per year. Solution Let a = a = = total amount accrued and b = original deposit. Interest Since a Interest Interest b b + b (interest rate), interest can be expressed as + b (interest rate) b =b = b (interest rate) $100 = b(0. 06) b = 100 = $1666. 67 0. 06 Example 1. 5 Make the calculations necessary to show which of the statements below are true and which are false, if the interest rate is 5% per year: (a) $98 now is equivalent to $105. 60 one year from now. (b) $200 one year past is equivalent to $205 now. (c) $3000 now is equivalent to $3150 one year from now. (d) $3000 now is equivalent to $2887. 14 one year ago. (e) Interest accumulated in 1 year on an investment of $2000 is $100. Solution (a) Total amount accrued = 98(1. 05) = $102. 90 =P $105. 60; therefore false. Another way to solve this is as follows: Required investment = 105. 60/1. 05 = $100. 57 =P $9? Therefore false. b) Required investment = 205. 00/1. 05 = $195. 24 =p $200; therefore false. 18 Level One (e) Total amount accrued = 3000(1. 05) = $3150; therefore true. (d) Total amount accrued = 2887. 14(1. 05) = $3031. 50 “# $3000; therefore false. (e) Interest = 2000(0. 05) = $100; therefore true. Example 1. 16 Calculate the total amount due after 2 years if $2500 is borrowed now and the compoundinterest rate is 8% per year. Solution The results are presented in the table to obtain a total amount due of $2916. (1) (2) (3) (4) = (2) + (3) (5) End of year Amount borrowed $2,500 Interest Amount owed Amount paid o 1 2 Example 1. 17 $200 216 2,700 2,916 $0 2,916 Assume that 6% per year, starting next withdrawing Solution P = you plan to make a lump-sum deposit of $5000 now into an account that pays and you plan to withdraw an equal end-of-year amount of $1000 for 5 years year. At the end of the sixth year, you plan to close your account by the remaining money. Define the engineering-economy symbols involved. $5000 A = $1000 per year for 5 years F = ? at end of year 6 i = 6% per year n = 5 years for A Figure 1. 6 Cashflow diagram for Example 1. 18. $650 $625 $600 $575 $ 550 $525 $500 $625 t -7 -6 -5 -4 -3 -2 -1 t o Year P = $2,500 Terminology and Cash-Flow

Diagrams 19 Example 1. 1B The Hot-Air Company invested $2500 in a new air compressor 7 years ago. Annual income “-om the compressor was $750. During the first year, $100 was spent on maintenance, _ cost that increased each year by $25. The company plans to sell the compressor for salvage at the end of next year for $150. Construct the cash-flow diagram for the piece f equipment. The income and cost for years – 7 through 1 (next year) are tabulated low with net cash flow computed using Eq. (1. 5). The cash flows are diagrammed . Fig. 1. 6. Solution End of year Net cash flow Income Cost -7 -6 -5 -4 -3 -2 -1 0 1 Example 0 750 750 750 750 750 750 750 750 + 150 $2,500 100 125 150 175 200 225 250 275 $-2,500 650 625 600 575 550 525 500 625 1. 19 Suppose that you want to make a deposit into your account now such that you can withdraw an equal annual amount of Ai = $200 per year for the first 5 years starting 1 year after your deposit and a different annual amount of A2 = $300 per year for the following 3 years. How would the cash-flow diagram appear if i is 14! % per year? Solution The cash flows would appear as shown in Fig. 1. 7. Comment The first withdrawal (positive cash flow) occurs at the end of year 1, exactly one year after P is deposited.

Figure 1. 7 Cash-flow diagram for two different A values, Example 1. 19. A2 = $300 A, = $200 0 1 2 3 4 i = 14+% 5 6 7 8 Year p=? 20 Level One p=? j = 12% per year Figure 1. 8 Cash-flow diagram for Example 1. 20. F2 1996 1995 A = $50 A = $150 = $50 F, = $900 Example 1. 20 If you buy a new television set in 1996 for $900,. maintain it for 3 years at a cost of $50 per year, and then sell it for $200, diagram your cash flows and label each arrow as P, F, or A with its respective dollar value so that you can find the single amount in 1995 that would be equivalent to all of the cash flows shown.

Assume an interest rate of 12% per year. Solution Comment Figure 1. 8 presents the cash-flow diagram. The two $50 negative cash flows form a series of two equal end-of-year values. As long as the dollar values are equal and in two or more consecutive periods, they can be represented by A, regardless of where they begin or end. However, the $150 positive cash flow in 1999 is a single-occurrence value in the future and is therefore labeled an F value. It is possible, however, to view all of the individual cash flows as F values. The diagram could be drawn as shown in Fig. . 9. In general, however, if two or more equal end-of-period amounts occur consecutively, by the definition in Sec. 105 they should be labeled A values because, as is described in Chap. 2, the use of A values when possible simplifies calculations considerably. Thus, the interpretation pictured by the diagram of Fig. 1. 9 is discouraged and will not generally be used further in this text. p=? j = 12% per year F. = $150 1. 9 A cash flow for Example 1. 20 considering all values as future sums. Figure 1996 1995 1997 1998 1999 F2 = $50 F3 = $50 F, = $900