Engineering Economics

Eng ineeri ng Economy terce Edition Leland T. Blank, P. E. Department of Industrial applied science Assistant doyen of envision 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 Comp all New York S1. Louis San Francisco Auckland Bogota capital of Venezuela 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 unriv besidesed 1. Define and manage in a caper tale the providence symbolizations P, F, A, n, and i. 1. 6 Define coin flow, state what is meant by hold on-of- layover convention, and construct a notes-flow plat, give a statement describing the measure and clock of the change flows. charter Guide 1. 1 Basic linguistic communication Before we begin to go bad the terminology and fundamental conceits upon which technology economy is based, it would be appropriate to limn what is meant by engineering economy. In the openst price, engineering economy is a aggregation of mathematical techniques which simplify sparing comparisons.With these techniques, a rational, meaningful approach to evaluating the scotch aspects of variant methods of accomplishing a given objective croupe be unquestionable. Engineering economy is, thusly, a decision assistance tool by which unrivaled method leave al adept be chosen as the most economical whiz. In order for you to be able to apply the techniques, however, it is necessary for you to witness the basic terminology and fundamental concepts that form the foundation for engineering-economy studies.Some of these terms and concepts argon described below. An alternative is a stand-al iodin solution for a give inter discharge site. We atomic itemise 18 faced with alternatives in virtually everything we do, from selecting the method of battery-acid we intent to get to work ev ery day to deciding amongst buying a house or renting one. Similarly, in engineering practice, there ar al slipway seveffl ways of accomplishing a given task, and it is necessary to be able to comp are them in a rational bearing so that the most economical alternative tail be selected.The alternatives in engineering considerations normally conduct such items as purchase comprise (first cost), the anticipated life of the asset, the stratumly costs of maintaining the asset (annual maintenance and operational cost), the anticipated resale judge (salvage pry), and the rice beer array ( run of return). After the facts and all the pertinent estimates retain been collected, an engineering-economy analysis can be conducted to determine which is stovepipe from an economic point of view.However, it should be pointed out that the procedures developed in this book provide enable you to ca-ca accu prescribe economic decisions lone(prenominal) about those alternatives which tur n out been recognized as alternatives these procedures leave alone not help you identify what the alternatives are. That is, if alternatives ,4, B, C, D, and E carry been identified as the all affirmable methods to crystallize a Particular problem when method F, which was never recognized as an alternative, is in truth the most attractive method, the wrong decision is certain to be do because alternative F could never be chosen, no matter what analytic techniques are apply.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 surr overthrowered in this book can be of greatest value. In order to be able to compare diametrical methods for accomplishing a given objective, it is necessary to leave 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 to answer the question How impart I k right off which one is outmatch? Whether we are aware of it or not, this question is asked of us many clips from distri exclusivelyively(prenominal) one day. For example, when we drive to work, we subconsciously think that we are taking the best route. But how did we define best? Was the best route the safest, shortest, fastest, cheapest, most scenic, or what? Obviously, depending upon which criterion is used to identify the best, a different route top executive be selected apiece m (Many arguments could have been avoided if the decision makers had simply tell the criteria they were exploitation in determining the best). In economic analysis, long horses are loosely used as the basis for comparison.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 involv e intangible factors, such as the yield of a process change on employee morale, which cannot readily be express in terms of dollars. When the alternatives available have approximately the akin relate 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 of import to recognize the concept of the beat value of silver.It is lots said that silver 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 give have lay in more money than we had schoolmasterly invested. This change in the criterion of money over a given term gunpoint is called the epoch value of money it is the most important concept in engineering economy. You should as well as realize that if a person or connection finds it necessary to take up money today, by tomorrow more money than the reliable loan volition be owed. This fact is also explained by the time value of money.The manifestation of the time value of money is termed arouse, which is a measure of the increase between the original nerve centre borrowed or invested and the nett follow owed or increase. Thus, if you invested money at just about time in the bypast, the care would be touch = count substance accumulated original investiture (1. 1) On the other hand, if you borrowed would be stake money at some time in the past, the by-line (1. 2) = present beat owed original loan In either case, there is an increase in the bar of money that was originally invested or borrowed, and the increase over the original occur is the pertain.The original investment or loan is referred to as foreland. Probs. 1. 1 to 1. 4 1. 2 Interest Calculations When pursuitingness is expressed as a region of the original measuring stick per unit of measurement time, the result is an saki lo calize. This localize is compute as follows . Percent lodge in value = touch on 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 fire prize is 1 social class. However, since intimacy evaluate are often expressed over periods of time shorter than 1 yr (i. e. 1% per month), the time unit used in expressing an engross rate moldiness also be identified and is termed an interest period. The following two examples illustrate the computation of interest rate. object lesson 1. 1 The Get-Rich-Quick (GRQ) Company invested $ blow,000 on whitethorn 1 and withdrew a total of $106,000 exactly one family ulterior. Compute (a) the interest gained from the original investment and (b) the interest rate from the investment. antecedent (a) apply Eq. (1. 1), Interest = 106,000 vitamin C,000 = $6000 (b) Equation (1. 3) is used to obtain Percent interest rate = 6000 per family carbon,000 x 100% = 6% per cat egory point out 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 straight off and repaid $110,000 in 1 stratum, utilize Eq. (1. 2) we find that interest is $10,000, and the interest rate from Eq. (1. 3) is 10% per form. Example 1. 2 Joe Bilder plans to borrow $20,000 for 1 grade at 15% interest. Compute (a) the interest and (b) the total amount out-of-pocket later 1 social class. theme (a) Equation (1. 3) may be solved for the interest accrued to obtain Interest = 20,000(0. 15) = $3000 (b) tot amount due is the centre of principal and interest or keep down due Comment = 0,000 + 3000 = $23,000 billet 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 all(prenominal) example the interest period was 1 family and the interest was calculate at the end of one period. When more than one social classly interest period is involved (for example, if we had wanted to k direct the amount of interest Joe Bilder would owe on Terminology and Cash-Flow Diagrams 7 the above loan later 3 long time), it becomes necessary to determine whether the interest . payable on a dewy-eyed or tangled 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 comparability The time value of money and interest rate utilized unneurotic generate the concept of bear uponity, which authority that different sums of money at different times can be pit in economic value. For example, if the interest rate is 12% per course of study, $100 today (i. e. , at present) would be like to $112 one grade from today, since mount accrued = 100 =$112 Thus, if someone offered you a gift of $100 today or $112 one socio-economic class from today, it would make no difference which offer you accepted, since in either cas e you would have $112 one year from today.The two sums of money are thusly eq to separately other when the interest rate is 12% per year. At either a higher or a lower interest rate, however, $100 today is not same to $112 one year from today. In sum total to considering succeeding(a) equivalence, one can apply the aforesaid(prenominal) concepts for determining equivalence in previous years. Thus, $100 now would be akin 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 89 = 1. 12, or 12% per year The concept of equivalence can be shape up illustrated by considering different loan- refund schemes. severally scheme represents refund 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. ) intention 1 a interest or principal is recovered until the ordinal year.Interest accumulates each year on the total of principal and all accumulated interest. cast 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, $ green, is paid each year. Since the stay 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) prohibit of year (2) = 0. 15(5) Interest for year (3) = (2) + (5) Total owed at end of year (4) Payment per plan (3) (4) oddment later 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 one hundred fifty. 00 $5,750. 00 4,600. 00 3,450. 00 2,300. 00 1, one hundred fifty. 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 ease 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, til now though each repayment scheme would require exactly 5 years to repay the loan. The difference in the total amounts repaid can of take to the woods 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 finished with(predicate) 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 next sums of money equivalent to a present sum (principal).When more than one interest period is involved, the terms simple and compound interest moldiness 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 $ thousand for 3 years at 14%-per-year simple interest, how much money go out you owe at the end of 3 years? origin The interest for each of the 3 years is = Interest per year kelvin(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 constitute 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. tick 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) bill owed (5) End of year 0 1 2 come in borrowed $1,000 Interest tot 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. rootage 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 re cognized 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 uniform a significant difference, remember that the beginning amount here was only $1000. Make these same calculations for an initial amount of $10 million, and whence get a line at the size of the difference The power of compounding can further be illustrated through another interesting exercise called Pay Now, vivify 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 target $2610 each year for the first 6 years and then nothing for the following(a) 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 homely 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 epitome in value. The so-called influence 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 true(a) 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 evaluate 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 specify period of time could be estimated by dividing 72 by the contract 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 unendingly 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 word of honor 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 two-baser 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 importee The mathematical symbols transaction 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 afterlife day time dollars, pesos, etc. = a serial publication 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 per centum per month, percent per year, etc. The symbols P and F represent single-time numberrence value A go alongs 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 ineluctably 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 be 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 determine known. The following four examples illustrate the use of the symbols. Example 1. 5. If you borrow $ two hundred0 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 set of P, F, n, and i. beginning In this situation P and F, but not A, are involved, since all transactions are single payments. The values are as follows P = $ two hundred0 Example 1. 6 i = 12% per year n = 5 yearsIf 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 admit and F or A is a expense. every bit correct to use these symbols in reverse roles, as in the examples below. Example 1. 7 T you locate $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 nonplus $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 forces are in a series A and the withdrawal is a future sum, or F value. in that respect 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 high society has hard hard currency advantage (income) and immediate payment disbursements (costs) which occur over a peculiar(a) time span. These receipts and disbursements in a given time musical interval are referred to as cash flow, with decreed 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 premise 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 plat is simply a graphical representation of cash flows drawn on a time scale. The plot 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 plot 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 bear on 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 image 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. finger 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 + issue 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. constrain 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 fivesome 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 = 1710 2 0 3 4 Year A=$1. 000 Example 1. 11Assume 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 ye ars starting signal 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 take the interest and total amount accrued after 1 year if $2000 is invested at an interest rate of 15% per year.Solution Interest earn = 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 a ccrued = 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, start ing 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 be 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-FlowDiagrams 19 Example 1. 1B The Hot-Air Company invested $2500 in a red-hot 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 participation 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 th rough 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% pe r 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 tv set 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 go after 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 dr awn 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