Chapter 2

2.1 Single-Payment Formulas (F/P and P/F) 2.1 SINGLE-PAYMENT FORMULAS (F/P AND P/F) The most fundamental equation in engineering economy is the one that determines the amount of money F accumulated after n years {or periods) from a single pres ent worth P, with interest compounded one time per year (or period). Recall that compound interest refers to interest paid on top of interest. Therefore, if an amount P is invested at time ¢ = (0, the amount F; accumulated | year hence at an interest rate of i percent per year will be Fi =P+ Pi = P(1 + i) where the interest rate is expressed in decimal form. At the end of the second year, the amount accumulated F, is the amount after year | plus the interest from the end of year | to the end of year 2 on the entire F,. F,=F, + Fji =P+ + P+ The amount £, can be expressed as Fo=P(l +i+i+i) =P(1 +2i+ %) =Pl +i)* Similarly, the amount of money accumulated at the end of year 3 will be Fy=P(l +i) By mathematical induction, the future worth F can be calculated for n years using F=PFP1+i" [2.1] The term (1 + i) is called a factor and is known as the single-payment compound amount factor (SPCAF), but it is wsually referred to as the F/P factor This is the conversion factor that yields the future amount F of an initial amount P after n years at interest rate i. The cash flow diagram is seen in Figure 2.la. Reverse the situation to determine the P value for a stated amount £ Simply solve Equation [2.1] for P. . 1 " 'LL',I + :"]"‘ The expression in brackets is known as the single-payment present worth factor (SPPWF), or the P/F factor. This expression determines the present worth P of a given future amount F after n years at interest rate i. The cash flow diagram is shown in Figure 2.15. Note that the two factors derived here are for single payments; that is, they are used to find the present or future amount when only one payment or receipt is involved.

Chapter 2 Factors: How Time and Interest Affect Money Caleulator: Use the TVM function FV{in.A.F). The numerical values are FVIR.24,0,120000), which displays the future worth value $—76,094.17, which indicates it is a cash outflow. The slight difference in answers between tabulated, formula, and calculator solutions 15 due to round-off error and how different methods perform equiva lence calculations. An equivalence imterpretation of this result is that $12,000 today 15 worth 376,094 after 24 years of growth at 8% per year compounded annually. - EXAMPLE 2.2 Hewlett-Packard has completed a study indicating that $50,000 in reduced main tenance this year (i.e., year zero) on one of its processing lines resulted from improved wireless monitoring technology. a. If Hewlett-Packard considers these types of savings worth 209 per year, find the equivalent value of this result after 5 years. b. If the $50,000 maintenance savings occurs now, find its eguivalent value 3 years earlier with interest at 20% per year. Solution a. The cash flow diagram appears as in Figure 2.1a. The symbols and their vilues are P = 350,000 F=1 i = 20% per year n = 5 years Use the F/P factor to determine F after 5 years. F = P(F/Pin) = $50,000(F/P.20%.5) = 50,000(2.4883) — §124,415.00 The function = FV(209%.5,,50000) also provides the answer. See Figure 2.2. . VN Y "N NN - SN NN -3 - S Y N N N _— 1 | I I —— I 3 Example 2.2a Example 2.2b 4 5 F= -$124,416 P = -$28,935 6 [ ) i3] . [ | | | | ] =5 Spreadsheet function with } | Spreadsheet function with | | A omitted: | A omitted: 1':' = FV(20%5,.50000) = PV(20% 3, 50000) 12 13 FIGURE 2.2 Use of single-cell spreadsheet functions to find F and P values, Example 2.2.

2.2 Uniform Seres Formulas (PAA, AP, AIF, FIA) 41 TABLE 2.2 P/A and A/P Factors: Notation, Equation and Function Factor Facior Standard Notation Spreadsheet Calculator Notation MName Formula Equation Function Function (1+ =1 (PIAL L) Uniform-series ? P = A(P/A,in) PV(i%.n AF) PVI(inAF) present worth A d ] ) il + )" . . ) , (AP i 1) Capital recovery “*'—]I“l A = P{A/P,in) PMT( % nP.F) PMTiinP.F) [ i In standard factor notation, the equations are P = A(P/A, i, n) and A = P(A/P, i, n), respectively. It is important to remember that in these equations, the P and the first A value are separated by one interest period. That is, the present worth P is always located one interest period prior to the first A value. It is also important to remember that the n is always egual o the number of A values. The factors and their use to find P and A are summarized in Table 2.2. The spreadsheet and calculator functions shown in Table 2.2 are capable of deter mining both P and A values in lieu of applying the P/A and A/P factors. The PV function calculates the P value for a given A over n years, and a separate F value in year n, if present. The format is = P¥V(i% n,A,F) Similarly, the A value is determined using the PMT function for a given P value in year () and a separate F, if present. The format is = PMT(i%.n,P,F) In addition to { and n, the calculator functions shown in Table 2.2 include all three cash flow parameters—F, F, and A. The function uses Equation [2.3] to solve for one of the five parameters, given values for the remaining four. How much money should you be willing to pay now for a guaranteed $600 per EXAMPLE 2.4 year for 9 years starting next year, at a rate of return of 16% per year? Solution The cash flow diagram (Figure 2.6) fits the P/A factor. The present worth is: P = 6(P/A16%.9) = 60IN4.6065) = 32763.90 The spreadsheet PV function = —PV(16%,9,600) entered into a single spreadsheet cell will display the answer P = 32763.93. Similarly. the calculator function PV(16.9.600,0) results in P = $—2763.93.

Chapter 2 Factors: How Time and Interest Affect Money When a problem involves finding § or » (instead of P, F, or A), the solution may require trial and error. Spreadsheet or calculator functions can be used to find [ OT n in MOsL cases. 2.3 GRADIENT FORMULAS FIGURE 2.10 Conventional arithmetic gradient series without the base amount. The previous four equations involved cash flows of the same magnitude A in each interest period. Sometimes the cash flows that occur in consecutive interest periods are not the same amount (not an A value), but they do change in a predictable way. These cash flows are known as gradienrs, and there are two general types: anthmetic and geometric. An arithmetic gradient is one wherein the cash flow changes (increases or decreases) by the same amount in each period. For example, if the cash flow in penod I ix 3800 and in perind 2 it 1s 3900, with amounts increasing by 3100 in each subse guent interest period, this is an arithmetic gradient &, with a value of $100. The eguation that represents the present worth of an arithmetic gradient series is: _E[“ +i)" =1 n il +0" (1+ " . [2.4] i Equation [2.4] is denved from the cash flow diagram in Figure 2.10 by using the PF factor to find the equivalent P in year (0 of each cash fow. Standard factor notation for the present worth of an arithmetic gradient is P = G{P/G,i%.n). This equation finds the present worth in year () of the gradient only (the $100 increases mentioned earlier starting in year 2). It does nor include the base amount of money that the gra dient was built upon (3800 in the example). The base amount in time period 1 must be accounted for separately as a uniform cash flow senies. Thus, the general equation to find the present worth of an arithmetic gradient cash flow series is P = Present worth of base amount + present worth of gradient amount . = A(PIAi%.n) + G(PIG,i%.n) [2.5] where A = amount in period | (; = amount of change in cash flow between periods 1 and 2 n = number of periods from | through a of gradient cash flow i = interest rate per period 0 1 2 3 4 5 m—1 n ] 1 ] [L I T l 1] [ 27 1N A3 r in-2)G Y (n— 1)

2.4 Calculations for Cash Flows That Are Shifted 47 Solution ¢ The cash Aow at the end of year | 15 $250,000, increasing by g = 5% per year (Figure 2.11). The present worth is found using Equation [2.8]. . (l_nfi)‘5 P = 250,000 = o 0,12 — 0.05 = 250,000(3.94005) = $985.013 250,000 {1.05) = 53035, 876 250000 - o NeENY L] I I L] ] 1 2 3 4 5 i=12% = 5% F=1 FIGURE 2.11 Cash flow with g = 5%, Example 2.9 In summary, some basics for gradients are: s Arnthmetic gradients consist of two parts: a uniform series that has an A value equal to the amount of money in period |, and a gradient that has a value equal to the change in cash low between periods 1 and 2. # For arithmetic gradients, the gradient factor is preceded by a plus sign for in creasing gradients and a minus sign for decreasing gradients. & Conventional arithmetic and geometric cash flows start between periods | and 2, with the A value in each equation equal to the magnitude of the cash flow in period 1 and the P value in year 0. s Geometric gradients are handled with Equation [2.8] or [2.9], which yield the present worth of all the cash flows. 2.4 CALCULATIONS FOR CASH FLOWS THAT ARE SHIFTED When a uniform series begins at a time other than at the end of period 1, it is called a shifted series. In this case several methods based on factor equations or tabulated values can be used to find the equivalent present worth P. For example, P of the

Chapter 2 Factors: How Time and Interest Affect Money actual year 2, not year 3. Also, n = 6, not &, for the P/A factor. First find the value of F; of the shifted series. B = $5000 P/A 8% 6) Since Py 15 located in year 2, now find P, i year (). P, = F{(P/F.8%.2) The total present worth is determined by adding P, and the initial payment P in year (). Pr="FP,+ P, = 5000 + 500{P/A 8% 6)(P/F8%.2) = 5000 + 50004.6229)(0.8573) = $6981.60 To determine the present worth for a cash flow that includes both uniform series and single amounts at specific times, use the P/F factor for the single amounts and the PfA factor for the series. To caleulate A for the cash flows, first convert every thing to a P value in year (), or an F value in the last year Then obtain the A value using the A/P or A/F factor, where n is the total number of years over which the A is desired. Many of the considerations that apply to shifted uniform series apply to anth metic gradient series as well. Recall that a conventional gradient series starts between periods | and 2 of the cash flow sequence. A gradient starting at any other time is called a shifted gradient. The n value in the P/ and A/G factors for the shifted gradient is determined by renumbering the time scale. The period in which the gradient first appears is labeled period 2. The n value for the factor is deter mined by the renumbered period where the last gradient increase occurs. The P/G factor values and placement of the gradient series present worth P for the shifted arithmetic gradients in Figure 2.16 are indicated. It is important to note that the A/ factor cannot be used to find an equiva lent A wvalue in periods | through n for cash flows involving a shifted gradient Consider the cash flow diagram of Figure 2166 To find the equivalent annual series in yvears | through 10 for the arithmetic gradient series only, first find the present worth of the gradient in year 5, take this present worth back to year (), and then annualize the present worth for 10 years with the A/P factor. If you apply the annual series gradient factor (A/G.i.5) directly, the gradient is converted into an equivalent annual series over years 6 through 10 only. Remember, Lo find the equivalent A series of a shifted gradient through all of the periods, first find the present worth of the gradient at actual time 0, then apply the (A/Pd.n) factor. If the cash flow series involves a geometric gradient and the gradient starts at a time other than between periods 1 and 2, it is a shified gradient. The P, is located in 2 manner similar to that for P above, and Equation [2.8] is the factor formula.

2.5 Using Spreadsheets and Calculators P = 33000 deposit now and F = 35000 returmed n = 10 years hence, the function = —PMT(5%,10,—3000,5000) will display A = %9. This is the same as using the A/P and A/F factors to find the equivalent net A = 39 per year between the deposit now and retum 100 years later. A = —3000(A/P5%, 107} + 5000 (A/F,5%,10) = —389 + 398 = 59 Number of periods n: Use the NPER function = NPER{i% AP F) if A 1s exactly the same for each of n years; either P or F can be omitted, but not both. For example, for P = 525,000 deposit now and A = 33000 per year return, the function = NPER(5%, 3000, —25000) will display n = 11.05 years to recover P at 5% per year. This is the same as using trial and error to find »# in the relation (0 = —25,000 + 3,000(P/A.5% n). When cash flows vary im amount or tming, it is usually necessary to enter them on a spreadsheet, including all zero amounts, and utilize other functions for P, F, or A values. All spreadsheet functions allow another function to be embed ded in them, thus reducing the time necessary to get final answers. Example 2.12 illustrates these functions and the embedding capability. Example 2.13 demon strates how easily spreadsheets handle arithmetic and percentage gradients and how the IRR (rate of retum) function works. Carol just entered college and her grandparents have offered her one of two gifts. They promised to give her 325,000 toward a new car if she graduates in 4 years. Altermatively, if she takex 5 years to graduate, they offered her $5000 each year starting after her second year is complete and an extra $5000 when she graduates. Draw the cash flow diagrams first. Then, use { = 8% per year o show Carol how o use spreadsheet functions and her financial caleulator TVM functions to determine the following for each gift offered by her grandparents. a. Present worth P now b. Future worth F five years from now c. Eguivalent annual amount A over a total of 5 years d. Number of years it would take Carol to have $25,000 in hand for the new car if she were able to save $5(00 each year starting next year. Solution Spreadsheet: The two cash How series, labeled Gift A (lump sum) and Gift B (spread out), are in Figure 2.18. The spreadsheet in Figure 2.19a lists the cash flows (don’t forget to enter the $0 cash flows so the NPV function can be used), and answers to each part using the PV, NPV, FV, or PMT functions as explained below. In some cases, there are alternative ways Lo obtain the answer. Figure 2.196 shows the function formulas with some comments. Refer to Appendix A for a complete description of how each function operates. Remem ber that the PV, FV, and PMT functions will return an answer with the opposite EXAMPLE 2.12

Chapter 2 Factors: How Time and Interest Affect Money Lump-sum gift Spread-out gift F =525 (00 A F = 55000 A = £5000) T ] 1 2 3 4 5 0 1 ZT _"sl -1I 5T | | ] 1 ] l | I I I I 1 I T I I 1 1 Years Years FIGURE 2.18 Cash flows for Carol’s gift from her grandparents, Example 2.12. sign from that of the cash flow entries. The same sign 15 maintained by enter ing a minus before the function name. a. Rows 12 and 13: There are two ways to find P; either the PV or NPV func tion. NPV requires that the zeros be entered. (For Gift A, omitting zeros in years 1, 2, and 3 will give the incorrect answer of P = $23 148, because NPV assumes the 525,000 occurs in year | and discounts it only one year at 8%.) The single-cell PV is hard to use for Gift B since cash fAows do not start until year 2; using NPV is easier. bh. Rows 16 and 17: There are two ways to use the FV function to find F at the end of year 5. To develop FV comectly for Gift B in a single cell with out listing cash flows, add the extra 35000 in year 5 separate from the FV for the four A = 35000 values. Altematively, cell D17 embeds the NPV function for the P value into the FV function. This is a very convenient way to combine functions. . Rows 20 and 21: There are two ways o use the PMT function to find A for 5 years; find P separately and use a cell reference, or embed the NPV function into the PMT to find A in one operation. d. Row 24: Finding the years to accumulate $25,000 by depositing 35000 each year using the NPER function is independent of either plan. The entry = NPER(B%,—5000,,.25000) results in 4.3719 years. This can be con firmed by caleulating S000(F/A.8% 43719 = 5000(5.0000) = $25.000 (The 4.37 years is about the time it will take Carol to finish college. Of course, this assumes she can actually save 35000 a vear while working on the degree.) Calculator: Table 2.4 shows the format and completed calculator function for each gift, followed by the numerncal answer below it. Minus signs on final answers have been changed to plus as needed to reflect the same sense as that in the spreadsheet solution. When calculating the values for Gift B, the func tions can be performed separately, as shown, or embedded in the same way as the spreadsheet functions are embedded in Figure 2.19. In all cases, the answers are identical for the spreadsheet and calculator solutions.

2.5 Using Spreadsheets and Calculators e () || extra 35000} in year 5 as Function recognizes F value; no cash flow listing needed Embedded NPV tinds P in year 0 |l A 'y ———, LEy Cash flow, § 2 Year Gift A GitB 3] 0 3 . 4 1 of Ly x| 2 of 5,000 L5 — — a o 5,000 3| Be sure to 4 25,000 5,000 i cnicr cach zero 5 + o 10,000/ i cash flow i | Function applied 2| [ PV {single call § $18,378 | 13 | - | NP §iarar | . 15 | 18 = FW [single call) 527,000 527,531 | b. Fuiun rorthz | ar | sk sl | FV with embedded NPV s27,531 | 18 il - 20 . = PMT (reference P) §4 502 4,683 == c. Annmual worth; years 1-5 1 1 | L FMT with embedded NEV w602 84,683 22 i d. Years to 525,000 MNPER Tor bath gins 437 437 28 o (a) S I T T— ELN | Cash flaw. § 2| Year Gif & [ GiftB A a LS n 1] o 5| 2 o 5000 E | |13 o 5000 el |4 25000 5000 [ |8 1 10000 5] [ ] | Furetion applied a2 Present | | P wilhin = A = P, A 28000) 13 | * ) s —— [ | HEY = MFV{E%,C4CH " i) i | h. Future worth; year 5 l ! L) = FVE, T5008) = FUINtE LS00 ¢ BN0D______+4 7] | | P wits sssident My = W% B NPV DL-DE) | ] ] | I .' t ] | . o] | PT (e P s PTG ECEE) = PRI 5.0 21 |E' A 5 years 1-51 | T it vt MY = TS, B NFVIRS, CACCRJ| = FMTES S NPV DA OEy | = = | | [ _g_ | d. Years to $25.000 [ | NPER {sawe for botly) = MPER{I 5500, 35800 = NPERE% 5000, 25300) Functions reference P determined using PV or NPV function FIGURE 2.19 (a) Use of several spreadsheet functions to find P, F, A, and n values, and (b) format of functions to obtain values, Example 2.12.