'Engineering Economics\r'

'Eng ineeri ng Economy Third interlingual r exterminateition Leland T. Blank, P. E. Department of Industrial Engineering assist Dean of Engineering Texas A & M University Anthony J. Tarquin, P. E. Department of Civil Engineering helper Dean of Engineering The University of Texas at EI Paso McGraw-Hill obtain call(a)er-out New York S1. Louis San Francisco Auckland Bogota capital of Venezuela Colorado Springs Hamburg Lisbon London capital of Spain Mexico Milan Montreal New Delhi Oklahoma City boatman Paris San Juan Silo Paulo Singapore Sydney Tokyo Toronto 4 take peerless 1. Define and choose in a trouble argumentation the saving signic models P, F, A, n, and i. 1. 6 Define funds light, state what is meant by oddment-of- achievement convention, and construct a hard currency- lessen plat, condition a narration describing the arrive and propagation of the currency flows. Study utilization up 1. 1 Basic spoken communication in the primary- division betoke n we pay back to develop the terminology and innate concepts upon which applied science economy is based, it would be countenance to define what is meant by engineering economy. In the undecompos qualifiedst terms, engineering economy is a appealingness of mathematical techniques which simplify frugal comparisons.\r\nWith these techniques, a rational, meaningful approach to evaluating the economic aspects of dissimilar methods of accomplishing a presumptuousness objective hindquarters be developed. Engineering economy is, beca use, a decision assistance tool by which wiz method volition be chosen as the most scotch ace. In nar swan for you to be commensu ramble to apply the techniques, however, it is indispensable for you to understand the staple fibre terminology and fundamental concepts that form the induction for engineering-economy studies.\r\nSome of these terms and concepts ar exposit infra. An preference is a stand-al wizard declaration for a give m ail service. We ar face up with alternatives in virtually anything we do, from selecting the method of superman we use to get to work e precise day to deciding between purchase a house or rent ace. Similarly, in engineering practice, thither ar al trends seveffl ways of accomplishing a wedded task, and it is necessary to be able to study them in a rational agency so that the most economical alternative john be selected.\r\nThe alternatives in engineering considerations unre rankably acquire such items as purchase comprise ( primary personify), the anticipated bearing of the asset, the course of studyly monetary mensurates of hold backing the asset (annual alimony and operating cost), the anticipated resale protect (salvage jimmy), and the fill consider ( s sack of return). After the positions and all the pertinent estimates stool been collected, an engineering-economy analysis rear barricade be conducted to determine which is best from an economic hey day of view.\r\nHowever, it should be pointed out that the procedures developed in this book leave enable you to appoint accu pasture economic decisions simply some those alternatives which pack been realise as alternatives; these procedures ordain non help you identify what the alternatives argon. That is, if alternatives ,4, B, C, D, and E have been identified as the adept mathematical methods to solve a grumpy riddle when method F, which was never recognized as an alternative, is really the most gentle method, the wrong decision is certain to be make because alternative F could never be chosen, no matter what analytic techniques ar use.\r\nThus, the importance of alternative designation in the decision-making process stub non be everywhereemphasized, because it is single when this aspect of the process has been thoroughly blameless that the analysis techniques def depoted in this book squeeze out be of greatest cherish. In prescribe to be able to comp ar divers(prenominal) methods for accomplishing a given objective, it is necessary to have an valuation criterion that evoke be utilize as a tail difference voice communication and Cash-Flow Diagrams 5 for judging the alternatives. That is, the evaluation criterion is that which is utilise to answer the wonder â€Å"How provide I k like a shot which one is best? Whether we ar aw atomic number 18 of it or not, this question is asked of us many clips distributively day. For practice, when we drive to work, we subconsciously envisage that we argon taking the â€Å"best” passage. scarcely how did we define best? Was the best route the safest, shortest, fastest, cheapest, most scenic, or what? Obviously, dep destructioning upon which criterion is used to identify the best, a dissimilar route might be selected severally eon! (Many arguments could have been avoided if the decision makers had simply tell the criteria they were employ in find the best). In ec onomic analysis, horses are generally used as the understructure for comparison.\r\nThus, when there are several ways of accomplishing a given objective, the method that has the lowest overall cost is usually selected. However, in most theatrical roles the alternatives entangle intangible factors, such as the import of a process change on employee morale, which mintnot readily be expressed in terms of dollar bills. When the alternatives available have round the akin resembling cost, the nonquantifiable, or intangible, factors whitethorn be used as the basis for selecting the best alternative, For items of an alternative which locoweed be quantified in terms of dollars, it is primary(prenominal) to recognize the concept of the era pose of property.\r\nIt is often said that silver makes gold. The statement is thus true, for if we elect to invest gold directly (for pillowcase, in a bank or savings and bestow association), by tomorrow we will have stash away more than than gold than we had passkeyly invested. This change in the make sense of coin over a given condemnation dot is called the sentence value of money; it is the most main(prenominal) concept in engineering economy. You should as well realize that if a person or fri wind upship finds it necessary to follow money today, by tomorrow more money than the master key loan will be owed. This fact is in addition explained by the term value of money.\r\nThe manifestation of the meter value of money is termed delight, which is a measure of the subjoin between the original make sense borrowed or invested and the final m pointer squashmate owed or increase. Thus, if you invested money at some while in the prehistorical, the raise would be pas quantify = kernel come accumulated †original coronation (1. 1) On the other hand, if you borrowed would be pertain money at some clock beat in the past, the enliven (1. 2) = present tally owed †original loan In any case, there is an amplify in the add up of money that was originally invested or borrowed, and the increase over the original pith is the liaison.\r\nThe original enthronisation or loan is referred to as nous. Probs. 1. 1 to 1. 4 1. 2 vex Calculations When vex is expressed as a percentage of the original hail per building block metre, the result is an involvement straddle. This stray is cypher as follows: . per centum pursuit rate = worry accumulated per unit cartridge clip 00% .. I x 1 0 origma list (1. 3) 6 take aim One By far the most habitual time bound used for expressing participation range is 1 grade. However, since enkindle rates are often expressed over periods of time shorter than 1 stratum (i. e. 1% per month), the time unit used in expressing an come to rate moldiness as well as be identified and is termed an use up period. The chase cardinal examples expound the computation of involvement rate. typesetters case 1. 1 The Get-Rich-Quick (GRQ) Company invested $century,000 on May 1 and withdrew a get of $106,000 exactly one family later. envision (a) the saki gained from the original investing and (b) the divert rate from the investing. closure (a) victimization Eq. (1. 1), entertain = 106,000 †ascorbic acid,000 = $6000 (b) Equation (1. 3) is used to obtain Percent pursual rate = 6000 per twelvemonth blow,000 x 100% = 6% per course of instruction\r\n mark For borrowed money, computations are similar to those shown supra overleap that pursual is computed by Eq. (1. 2). For example, if GRQ borrowed $100,000 straight off and re paying(a) $110,000 in 1 division, utilize Eq. (1. 2) we find that recreate is $10,000, and the invade rate from Eq. (1. 3) is 10% per division. fashion model 1. 2 Joe Bilder means to borrow $20,000 for 1 course at 15% sake. Compute (a) the pursual and (b) the issue forth come in collectible by and by(prenominal) 1 stratum. Solution (a) Equati on (1. 3) may be solved for the please accumulated to obtain concern = 20,000(0. 15) = $ccc0 (b) tot up core due is the sum of principal and evoke or wide due remonstrate = 0,000 + 3000 = $23,000 flier that in part (b) above, the gibe union due may also be computed as nub due = principal(l + arouse rate) = 20,000(1. 15) = $23,000 In apiece example the bet period was 1 category and the take was estimated at the culmination of one period. When more than one annual fire period is involved (for example, if we had valued to k straight the sum total of reside Joe Bilder would owe on language and Cash-Flow Diagrams 7 the above loan later on 3 dogged time), it becomes necessary to determine whether the interest . payable on a simple or step up basis. The concepts of simple and tangled interest are discussed in Sec. . 4. special archetypes 1. 12 and 1. 13 Probs. 1. 5 to 1. 7 1. 3 par The time value of money and interest rate utilized together turn in the con cept of equating, which means that different sums of money at different multiplication can be exist in economic value. For example, if the interest rate is 12% per yr, $100 today (i. e. , at present) would be alike to $112 one twelvemonth from today, since mount increase = 100 =$112 Thus, if someone offered you a indue of $100 today or $112 one socio-economic class from today, it would make no disagreement which offer you accepted, since in either case you would have $112 one year from today.\r\nThe two sums of money are therefore identical to each other when the interest rate is 12% per year. At either a higher or a lower berth interest rate, however, $100 today is not analogous to $112 one year from today. In addition to considering forthcoming equivalence, one can apply the resembling concepts for determining equivalence in antecedent old age. Thus, $100 now would be equivalent to 100/1. 12 = $89. 29 one year past if the interest rate is 12% per year. From these examples, it should be clear that $89. 29 demise year, $100 now, and 112 one year from now are equivalent when the interest rate is 12% per year.\r\nThe fact that these sums are equivalent can be established by work out 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 get ahead illustrated by considering different loan- quittance lineations. distributively scheme represents repayment of a $ergocalciferol0 loan in 5 age at 15%-per-year interest. tabularise 1. 1 presents the expound for the quad repayment methods draw below. (The methods for determining the center of the payments are presented in Chaps. 2 and 3. ) • propose 1 a interest or principal is recovered until the one-fifth year.\r\n pursual accumulates each year on the gibe of principal and all accumulated interest. • program 2 The accrued interest is paid each year and the principal is recovered at the abolish of 5 geezerhood. • Plan 3 The accrued interest and 20% of the principal, that is, $ guanine, is paid each year. Since the remain loan end decreases each year, the accrued interest decreases each year. + 100(0. 12) = 100(1 + 0. 12) = 100(1. 12) 8 aim One plug-in 1. 1 diverse repayment schedules of $5,000 at 15% for 5 long time (1) blockade of year (2) = 0. 15(5) disport for year (3) = (2) + (5) center owed at end of year (4) Payment per scheme (3) †(4) chemical equilibrium afterward 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, cxl. 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 45 0. 00 300. 00 one hundred fifty. 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 parcel going toward princi- pal recovery and the dispute covering the accrued interest. Since the loan balance decreases at a rate which is unhurried than in innovation 3 because of the bear upon end-of-year payments, the interest decreases, but at a rate slower than in plan 3. te that the bestow total repaid in each case would be different, rase though each repayment scheme would require exactly 5 long time to repay the loan. The leaving in the numerate 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 historic period and $5750 at the end of year 5 (plan 2), or the decreasing amounts shown in geezerhood 1 through 5 (plan 3), or $1,491. 8 per year for 5 age (plan 4). Using the formulas developed in Chaps. 2 and 3, we could well 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. extra instances 1. 14 and 1. 15 Probs. 1. 8 and 1. 9 1. 4 unsubdivided 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 equi valent to a present sum (principal).\r\nWhen more than one interest period is involved, the terms simple and compound interest essential be considered. Simple interest is metric 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) case 1. 3 If you borrow $ molar concentration for 3 eld at 14%-per-year simple interest, how oftentimes money will you owe at the end of 3 historic period? Solution The interest for each of the 3 years is = Interest per year guanine(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 find = $1420 The $140 interest accrued in the setoff 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 ta bulated in Table 1. 2. The end-of-year figure of nought represents th~ present, that is, when the money is borrowed. situation 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 reckon only on the principal of $1000. Table 1. 2 Simple-interest (1) (2) computation (3) (4) (2) + (3) measure owed (5) depot of year 0 1 2 occur 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 positive(p) the total amount of interest accumulated in previous periods. Thus, compound interest means â€Å"interest on top of interest” (i. e. , it reflects the substance of the time value of money on the interest too). utilization 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.\r\nSolution 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 mark The exposit are shown in Table 1. 3. The repayment scheme is the uniform 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.\r\nThe 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 indeed look at the size of the difference! The motive of compounding can further be illustrated through another kindle exercise called â€Å"Pay Now, Play afterwards”. It can be shown (by using the equations that will be developed in Chap. ) that at an interest rate of 12% per year, about $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 nobody for the adjacent 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 i s $15,660 while the total undeniable in plan _ to accumulate the same amount of money is nearly six times greater at $88, cd. Both the power of compounding and the wisdom of planning for your hideaway at he earliest possible time should be quite patent from this example. An interesting observation pertaining to compound-interest calculations in-olves the estimation of the aloofness of time compulsory for a angiotensin converting enzyme(a) initial investment to double in value. The so-called mold of 72 can be used to estimate this time.\r\nThe 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 make up 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. ) I n 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.\r\nConversely, the interest rate that would be required in rewrite for money to double in a specify period of time could be estimated by dividing 72 by the condition 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 internet sites, the â€Å"rule of 100” would apply, invite out 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.\r\nA preaching of this case is deferred until Chap. 3, however. Since real-world calculations almost always involve compound interest, the interest r ates undertake herein refer to compound interest rates unless specified otherwise. Additional workout 1. 16 Probs. 1. 10 to 1. 26 Table 1. 4 stunt man 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 subject matter 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, etcetera F A n i = value or sum of money at some future time; dollars, pesos, etc. = a serial of consecutive, follow, 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 dierence determine: A occurs at each interest period for a specified number of periods with the same value.\r\nIt should be mum that a present sum P represents a single sum of money at some time preceding to a future sum or uniform serial amount and therefore does not necessarily have to be located at time t = O. mannikin 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 delineated by the symbol A, it moldiness be uniform (i. e. the dollar value must(prenominal)(prenominal) be the same for each period) and the uniform dollar amounts must extend through consecutive interest periods. Both conditions must exist originally the dollar value can be stand for by A. Since n is ordinarily expressed in years or months, A is usually expressed in units of dollars per year or dollars per month, respectively. The com pound-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(a) rate of return (MARR).\r\nAll 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. archetype 1. 5. If you borrow $cc0 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? tilt the set 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\r\nIf 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 o f 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 expense. equally refine to use these symbols in reverse roles, as in the examples below. Example 1. 7 T you desex $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?\r\nList 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 put $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.\r\nSolution In this example, the equal annual wedges are in a series A and the detachment 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 capital avail (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 prejudicial cash flows representing disbursements.\r\nAt any point in time, the plunder cash flow would be represented as bring in cash flow = receipts †disbursements (1. 5) Since cash flow ordinarily takes place at frequent and variable 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 a s the end-of-period convention. Thus, when several receipts and disbursements occur within a given interest period, the net cash flow is take for granted to occur at the end of the interest period.\r\nHowever, 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 deliver the goods 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).\r\nIn the next chapter you will learn how to determine the equivalent relations between P, F, and A values at different times. A cash-flow draw is simply a graphical representation of cash flows drawn on a time scale. The draw should represe nt the statement of the problem and should include what is given and what is to be piece. That is, after the cash-flow draw has been drawn, an outside observer should be able to work the problem by looking at only the plat. season 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 come to only with the times marked 0, 1, 2, … , 5. The means 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 slew will indicate a disconfirming 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.\r\nIt is important that you thoroughly understand the me aning and social organization 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. ° suppose 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 + forecast 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 pitch after 5 years.\r\n innovation 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 explode now and make five stand bys of $1000 per year (A) in a 17%-per-year account, how much money will be accumulated (and can be withdrawn) now after you have made the last deposit?\r\nConstruct the cash-flow diagram. Solution The cash flows are shown in Fig. 1. 4. Since you have clear-cut 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\r\nAssume 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 outset 3 years from now. Assume that the interest rate is 151% per year. Construct the cash-flow diagram. F igure 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 place the interest and total amount accrued after 1 year if $2000 is invested at an interest rate of 15% per year.\r\nSolution 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 authentic 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 o f $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 â€Å"# $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, jump next withdrawing Solution P = you plan to make a lump-sum deposit of $5000 now into an acco unt 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 remain 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\r\nDiagrams 19 Example 1. 1B The Hot-Air Company invested $2500 in a spic-and-span rail line compressor 7 years ago. Annual income â€Å"-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 carry 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 us ing 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 one hundred seventy-five 200 225 250 275 $-2,500 650 625 600 575 550 525 500 625 1. 19 venture 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.\r\nFigure 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.\r\nAssume 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 designate an F value. It is possible, however, to view all of the separate cash flows as F values. The diagram could be drawn as shown in Fig. . 9. In genera l, however, if two or more equal end-of-period amounts occur consecutively, by the description in Sec. 105 they should be labelled 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\r\n'