Chapter 2

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MGMT74115

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Industrial Engineering

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Dec 6, 2023

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Chapter Factors: How Time and Interest Affect Money Royalty-Fraa/COREIS n Chapter | we leamed the basic concepts of engineering economy and their role in decision making. The cash flow is fundamental to every economic study. Cash flows occur in many configurations and amounts—isolated single values, series that are uniform, and series that increase or decrease by constant amounts or constant percentages. This chapter develops the commonly used engineering economy factors that consider the time value of money for cash flows. The application of factors is illustrated using their mathematical forms and a standard notation format. Spreadsheet and calculator functions are illustrated.
Purpose: Use tabulated factors or spreadsheet/calculator functions to account for the time value of money. LEARNING QUTCOMES . Use the compound amount factor and present worth factor for F/P and P/F factors single payments. ' . Use the uniform series factors. | P/A, A/P, F/A and A/F factors 3. Use the arithmetic gradient factors and the geometric gradient Gradients formula. - . Use uniform series and gradient factors when cash flows are Shifted cash flows shifted. . Use a spreadsheet or calculator to make equivalency Spreadsheets/Calculators calculations.
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.
P = given Chapter 2 Factors: How Time and Interest Affect Money Fod == F - i = given I i=given } I ! 1 l i : ] l | 1 n-2 n-1 ‘u F=17 F = given {a) (5) FIGURE 2.1 Cash flow diagrams for single-payment factors: (a) find F and (b) find P. A standard notation has been adopted for all factors. Tt is always in the general form (X/Y¥.in). The letter X represents what is sought, while the letter ¥ represents what is given. For example, F/P means find F when given P The i is the interest rate in percent, and n represents the number of periods involved. Thus, (F/P.6%.20) represents the factor that is used to calculate the future amount F accumulated in 20 penods if the interest rate is 6% per period. The P is given. The standard nota tion, simpler to use than formulas and factor names, will be used hereafter. Table 2.1 summarizes the standard notation and equations for the F/P and P/F factors. To simplify routine engineering economy calculations, tables of factor values have been prepared for a wide range of interest rates and time periods from 1 to large n values, depending on the i value. These tables are found at the end of this book. For a given factor, interest rate, and time, the correct factor value 1s found at the intersection of the factor name and n. For example, the value of the factor (P/F.5%.10) is found in the P/F column of Table 10 at period 10 as 0.6139. When it is necessary to locate a factor value for an § or n that is not in the interest tables, the desired value can be obtained in one of several ways: (1) by using the formulas derived in Sections 2.1 to 2.3, (2) by linearly interpolating between the tabulated values. or (3) by using a spreadsheet or calculator function as discussed in Section 2.5. For many cash flow series or for the sake of speed, a spreadsheet function may be used in lieu of the tabulated factors or their equations. For single-payment series, no annual series A is present and three of the four values of F, P, i, and n are known. When solving for a future worth by spreadsheet, the F value is calculated TABLE 2.1 F/P and P/F Factors: Notation, Equation and Function Factor Standard Notation Equation with Spreadsheet Calculator Notation Name Equation Factor Formula Function Function (F/P.in) Single-payment F = P(F{P.in) F=mMl+ F¥V(i%.n, Py FV{inAP) compound amount (P/F.i.n) Single-payment P = Fi(P{F.in) P = F1/1 + 07 PVii%.n . F) PV(inAF) |JIL.'-_‘iL.',Il1 worth
2.1 Single-Payment Formiulas (F/P and P/F) by the FV function, and the present worth P is determined using the PV function. The formats are included in Table 2.1. (Refer to Section 2.5 and Appendix A for more information on the FV and PV spreadsheet functions.) The calculator formats for FV and PV functions detailed in Table 2.1 include the annual series A, which 15 entered as () when only single-amounts are involved. In standard notation form, the relation used by calculators to solve for any one of the parameters P, F, A, i, or n is A(P/Ai%.n) + F(P/Fi%n) + P =10 Using the factor eguations, this relation expresses the present worth of uniform series, future values and present worth values as: (1= (1 +i/100)" _ o -’t( e )+ F1+ (/100" +P=10 [2.3] i/ 100 F ' When only single payments are present, the first term is (0. 37 An engineer received a bonus of 512,000 that he will invest now. He wants to calculate the equivalent value after 24 years, when he plans to use all the result ing money as the down payment on an island vacation home. Assume a rate of retum of 8% per year for each of the 24 years. Find the amount he can pay down, using the tabulated factor, the factor formula, a spreadsheet function, and a calculator function. Solution The symbols and their values are P = 512,000 F=1 i = 8% per year n = 24 years The cash flow diagram is the same as that in Figure 2.1a. Tabulated: Determine F, using the F/P factor for 8% and 24 years. Table 13 provides the factor value. F = P(F/P.in) = 12,000(F /P 8% 24) = 12,00006.3412) = $76,004.40 Formula: Apply Equation [2.1] to calculate the future worth F. F = P(l + 0" = 12,000(1 + 0.08)* = 120000634 1181) = $76.094.17 Spreadsheet: Use the function = FV(i%,n,A,P). The cell entry is = FV(8%, 24..12000). The F wvalue displayed is ($76.094.17) in red or —376,094.17 in black to indicate a cash outflow. EXAMPLE 2.1
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.
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