Solutions to Accompany

Transcription

1 Solutions to Accompany Chapter 2 Engineering Economics Engineering economy problems require the calculation of various interest factors listed in Table 2.2. An alternative is to create an EES function library that can be easily accessed to calculate these factors. An example of such a user-written function library is shown on pages Subsequent problems in this chapter simply access the function library by calling the function name with the appropriate inputs. 13

2 INTERESTFACTORS Equations Single Payment Compound Amount Factor function F/P (i, n) F/P = (1 + i) n end Present Worth Factor function P/F (i, n) end P/F = 1 (1 + i) n Uniform Series Sinking Fund Factor function A/F (i, n) end A/F = i (1 + i) n 1 Compound Amount Factor function F/A (i, n) end F/A = (1 + i)n 1 i Capital Recovery Factor function A/P (i, n) end A/P = i (1 + i) n (1 + i) n 1 Uniform Series Present Worth Factor function P/A (i, n) 14

3 end P/A = (1 + i)n 1 i (1 + i) n Gradient Present Worth Factor function P/G (i, n) end P/G = (1 + i)n 1 (i 2 (1 + i) n ) n i (1 + i) n 15

4 PROBLEM 2.1 Interest Factor Calculator USER INPUTS 7.00% Nominal annual interest rate 4 Compounding periods per year 6 Number of years i = Calculated interest per period n = 24 Total number of periods Interest Factors P to F F/P F to P P/F F to A A/F A to F F/A P to A A/P A to P P/A G to P P/G

5 PROBLEM 2.02 Equations Problem 2.2 Interest Factor Calculator using the EES Diagram Window for I/O Input information Calculations These calculations access the user-written functions in InterestFactors.LIB F/P = F P (i/n yr, n n yr ) P/F = P F (i/n yr, n n yr ) A/F = AF (i/n yr, n n yr ) F/A = F A (i/n yr, n n yr ) A/P = AP (i/n yr, n n yr ) P/A = P A (i/n yr, n n yr ) P/G = P G (i/n yr, n n yr ) 17

6 PROBLEM 2.03 Equations PROBLEM 2.3 GIVEN: A credit card advertising a nominal interest rate i = FIND: The effective rate SOLUTION: i eff % = F/P 1 Interest Factor F/P = (1 + i/12) 12 Solution Variables in Main program F/P = i = i eff = [%] Key Variables i eff = [%] effective annual rate 18

7 PROBLEM 2.04 Equations PROBLEM 2.4 GIVEN: College fund establishment i nom = 0.04 F 1 = [] Withdrawl for first year of college (age 18) F 2 = [] Withdrawl for second year of college (age 19) F 3 = [] Withdrawl for third year of college (age 20) F 4 = [] Withdrawl for fourth year of college (age 21) FIND: Monthly deposit required to obtain this goal SOLUTION: There are several ways to do this. In this solution, all money will be moved to time zero. A P/A (i, n 21 ) = F 1 P F (i, n 18 ) + F 2 P F (i, n 19 ) + F 3 P F (i, n 20 ) + F 4 P F (i, n 21 ) The interest rate must be expressed on a monthly basis since interest is compounded monthly, but the analysis is considering years 18, 19, 20, and 21, i = i nom /12 The number of periods n n 21 corresponds to the number of months before a withdrawl is taken n 18 = n 19 = n 20 = n 21 = Solution Variables in Main program 19

8 A = [] F 1 = [] F 2 = [] F 3 = [] F 4 = [] i = i nom = 0.04 n 18 = 216 n 19 = 228 n 20 = 240 n 21 = 252 Key Variables A = [] monthly deposit required to meet goal 20

9 PROBLEM 2.05 Equations PROBLEM 2.5 GIVEN: A small scale silver mine A P = [] n = 15 i = 0.15 FIND: The amount of money that can be used to purchase the mine SOLUTION: Bringing all the mone back to the present, P = A P P/A (i, n) Solution Variables in Main program A P = [] i = 0.15 n = 15 P = [] Key Variables P = [] Maximum bid to purchase the mine 21

10 PROBLEM 2.06 Equations PROBLEM 2.6 GIVEN: Analysis of a short-term storage facility P = [] OM = 900 [] A = 3600 [] i = 0.10 FIND: How many years must the warehouse last? SOLUTION: There are several ways to solve this problem. Here, all costs will be converted to an annual series, P A/P (i, n) + OM = A Solution Variables in Main program A = 3600 [] i = 0.1 n = OM = 900 [] P = [] Key Variables n = Length of time the building must last (yrs) 22

11 PROBLEM 2.07 Equations PROBLEM 2.7 GIVEN: A cash flow diagram as shown P 0 = [] G = 2800 [] A = 3500 [] A 1 = [] A 10 = 4000 [] C 6 = [] Cost at year 6 I 8 = [] Income at year 8 i = n = 10 FIND: (a) the present value (b) the future value (c) the uniform annual value SOLUTION: Moving the cash flows to any point in the time line is relatively easy, except for the positive A values. They decrease linearly from 110,000 to 4,000. Calculating the gradient that causes this decrease, G A = A 1 A 10 n 1 23

12 Present value: DEBIT S 0 = P 0 + C 6 P/F (i, 6) + G P G (i, n) + A F A (i, 4) P F (i, 10) INCOMES 0 = A 1 P/A (i, 10) G A P G (i, 10) + I 8 P F (i, 8) P V = INCOMES 0 DEBIT S 0 Future value: F V = P V F/P (i, 10) Uniform annual value: AV = P V A/P (i, 10) Solution Variables in Main program A = 3500 [] AV = 8664 [] A 1 = [] A 10 = 4000 [] C 6 = [] DEBIT S 0 = [] F V = [] G = 2800 [] G A = [] i = INCOMES 0 = [] I 8 = [] n = 10 P V = [] P 0 = [] Key Variables P V = [] F V = [] AV = 8664 [] equivalent present value (yr 0) of the cash flow diagram equivalent future value (yr 10) of the cash flow diagram equivalent uniform annual series value of the cash flow diagram 24

13 PROBLEM 2.08 Equations PROBLEM 2.8 GIVEN: Investment in a new machine for a production facility P = [] A P = [] F S = [] i = 0.25 n = 20 Initial cost of the machine Annual increase in profits Salvage value of the machine Minimum required rate of return Life of the machine FIND: The annual acceptable maintenance expenditures SOLUTION: There are several ways to solve this problem. Converting all cash to an annual basis, P A/P (i, n) + A M,1 = F S AF (i, n) + A P Converting all the cash to a present value, P + A M,2 P/A (i, n) = A P P A (i, n) + F S P F (i, n) Converting all the cash to a future value, P F/P (i, n) + A M,3 F A (i, n) = A P F A (i, n) + F s Solution Variables in Main program 25

14 A M,1 = 3152 [] A M,2 = 3152 [] A M,3 = 3152 [] A P = [] F S = [] i = 0.25 n = 20 P = [] Key Variables A M,3 = 3152 [] A M,1 = 3152 [] A M,2 = 3152 [] Converting to a future value Converting to an annual series Converting to a present value 26

15 PROBLEM 2.09 Equations PROBLEM 2.9 GIVEN: A machine with the following economic data P = 9000 [] O = 750 [] M 1 = 800 [] M 2 = 1000 [] M 3 = 1400 [] M 4 = 1800 [] M 5 = 3000 [] i = 0.06 n = 5 FIND: The initial investment required to pay for the machine SOLUTION: Since the operating costs are given on an annual basis, and the compounding period is monthly, this problem can be solved on an annual basis using the effective interest rate i eff = F/P (i/12, 12) 1 Moving all the cash to time zero... P inv = P + O P/A (i eff, n) + M 1 P F (i eff, 1) + M 2 P F (i eff, 2) + M 3 P/F (i eff, 3) + M 4 P F (i eff, 4) + M 5 P F (i eff, 5) 27

16 Solution Variables in Main program i = 0.06 i eff = M 1 = 800 [] M 2 = 1000 [] M 3 = 1400 [] M 4 = 1800 [] M 5 = 3000 [] n = 5 O = 750 [] P = 9000 [] P inv = [] Key Variables P inv = [] Investment required to pay for the machine and its maintenance costs 28

17 PROBLEM 2.10 Equations PROBLEM 2.10 GIVEN: A chemical processing plant P = [] A P = [] A M = 0.03 A O Initial cost of the piping system Profits realized by the piping system Annual maintenance costs C e = 0.12 [/kwh] Cost of energy to run the system n = 15 Life of the system ρ = 48.5 [ lbm/ft 3] Density of the cyclohexane P = 6.4 [ψ] Pressure drop through the system V ol = 1200 [gpm] Volumetric flow rate of the cyclohexane FIND: The rate of return SOLUTION: The annual cost of energy used to move the fluid through the system is, ( Ẇ = ṁ P ) kw ρ ψ ft 3 /hr The mass flow rate of the cyclohexane through the system is, ṁ = ρ V ol ft3 /hr gpm 29

18 The total energy used per year is, W = Ẇ (24 [hr/day]) (7 [day/week]) (52 [week]) The annual cost of the energy is, A O = C e W Assuming annual compounding, ( P A/P i ) %, n + A O + A M = A P Solution Variables in Main program A M = [] A O = 3502 [] A P = [] C e = 0.12 [/kwh] P = 6.4 [psi] i = [%] ṁ = [lbm/hr] n = 15 P = [] ρ = 48.5 [ lbm/ft 3] V ol = 1200 [gpm] W = [kw-hr] Ẇ = [kw] Key Variables i = [%] Rate of return 30

19 PROBLEM 2.11 Equations PROBLEM 2.11 GIVEN: A machine with the following economic data P inv = [] A = 1250 [] G = 200 [] S = 0.05 P i = 0.08 n = 7 FIND: The initial investment required to pay for the machine SOLUTION: The interest is compounded quarterly, but the gradient is on a yearly basis. Therefore, this problem will be solved by analyzing the cash flow diagram on a yearly basis but using the effective interest rate rather than the nominal interest rate. P inv + S P/F (i eff, n) = P + A P A (i eff, n) + G P G (i eff, n) i eff = F/P (i/4, 4) 1 Solution Variables in Main program 31

20 A = 1250 [] G = 200 [] i = 0.08 i eff = n = 7 P = [] P inv = [] S = 7247 [] Key Variables P = [] Maximum amount of money that can be spent on the machine 32

21 PROBLEM 2.12 Equations PROBLEM 2.12 GIVEN: A proposed heat recovery system with the following economic parameters, P = [] OM = [] Initial cost of the system Annual operating and maintenance costs n = 25 Economic life S = [] A = [] i mgmt = 0.12 Salvage value Annual savings in fuel minimum required rate of return FIND: (a) the rate of return for this scenario (b) if management requires a rate of return of 12%, how much can be spent on the heat recovery system? SOLUTION: (a) This problem can be solved in several ways. Converting all the funds to an annual series, ( P A/P i ) ( %, n + OM = A + S AF i ) %, n (b) The same equations from part (a) can be used... substitute i mgmt for i P b A/P (i mgmt, n) + OM = A + S AF (i mgmt, n) Solution Variables in Main program 33

22 A = [] i = [%] i mgmt = 0.12 n = 25 OM = [] P = [] P b = [] S = [] Key Variables i = [%] (a) rate of return for the heat recovery system P b = [] (b) Maximum that can be spent at time 0 for a rate of return of 12 34

23 PROBLEM 2.13 Equations PROBLEM 2.13 GIVEN: Three alternatives for a compressor drive Drive 1..3 = Gasoline, Diesel, Electric IC 1..3 = [, [], []] S 1..3 = 3200 [, 5800 [], 2400 []] F C 1..3 = 0.48 [lbm/hp hr, 0.36 [lbm/hp hr], 0] Cost fuel,1..3 = 3.28 [/gal, 3.89 [/gal], 0] Cost elec,1..3 = 0, 0, 0.11 [/kwh ] ρ fuel,1..3 = 43.7 [lbm/ft 3, 46.7 [ lbm/ft 3], 1 [ lbm/ft 3] ] elec 1..3 = 0, 0, 1 switch to specify non-electric (0) or electric (1) drive n = 15 i = 0.18 Ẇ = 400 [hp] time = 8 [hr/day] 270 [day] annual operation hours FIND: Which drive is the best economic alternative SOLUTION: 35

24 The lives are the same for all three alternatives. Therefore, the AC or PW method will work. Both soultions are shown here. duplicate k = 1, 3 AC k = IC k A/P (i, n) O fuel,k O elec,k + S k AF (i, n) P W k = IC k (O fuel,k + O elec,k ) P/A (i, n) + S k P F (i, n) Annual Cost Present Worth O fuel,k = gallons k Cost fuel,k Annual operating cost - fuel end O elec,k = kw h k Cost elec,k Annual operating cost - electricity ( ( )) F Ck gallons k = Ẇ time gal ρ fuel,k ft 3 gallons of fuel used per year ( ) kw h k = Ẇ kw hp time elec k kwh used per year Solution i = 0.18 n = 15 time = 2160 [hr] Ẇ = 400 [hp] Arrays Table: M ain Row Drive i elec i IC i S i F C i Cost fuel,i Cost elec,i [ ρ fuel,i gallons i [] [] [lbm/hp-hr] [/gal] [/kwh] lbm/ft 3 [gal] 1 Gasoline Diesel Electric Row kw h i O fuel,i O elec,i AC i P W i [kwh] [] [] [] [] E Report The electric motor is the best economic alternative. 36

25 PROBLEM 2.14 Equations Problem 2.14 GIVEN: Two possible heat treatment processes Process [1] P 1 = [] A o,1 = [] A p,1 = [] F 1 = [] Process [2] P 2 = [] A o,2 = [] A p,2 = [] F 2 = 7000 [] Tax information N = 8 t = 0.52 Depreciation is SYD FIND: Which process should be selected based on the rate of return analysis? 37

26 SOLUTION: A DUPLICATE loop will be used to calculate the interest rate for each alternative duplicate k = 1, 2 Calculation of the gradient D 1,k = 2 (P k F k ) D 2,k = 2 (P k F k ) A k = A p,k A o,k N N (N + 1) N N (N + 1) A AT,1,k = A k t (A k D 1,k ) A AT,2,k = A k t (A k D 2,k ) G k = A AT,1,k A AT,2,k Determination of the interest rate P k = A AT,1,k P/A (i k, N) G k P G (i k, N) + F k P F (i k, N) end Solution N = 8 t = 0.52 Arrays Table: M ain Row P i F i A o,i A p,i A i D 1,i D 2,i A AT,1,i A AT,2,i G i i i [] [] [] [] [] [] [] [] [] [] Report Based on this analysis, Process 1 has the higher rate of return. Therefore, select heat treatment process 1. 38

27 PROBLEM 2.15 Equations PROBLEM 2.15 GIVEN: Purchase of a computer system IC = [] S = 5000 [] O = [] A BT = [] n = 5 t = 0.50 FIND: The rate of return (a) before taxes (b) after taxes with no depreciation (c) after taxes using SLD (d) after taxes using SYD SOLUTION: Scenario Scenario 1 = no tax Scenario 2 = after tax no depreciation Scenario 3 = after tax with SLD Scenario 4 = after tax with SYD Scenarios [1], [2], and [3] duplicate k = 1, 3 IC = A AT,k P/A (i k, n) + S P F (i k, n) end 39

28 After tax cash flows A AT,1 = A BT O A AT,2 = (A BT O) (1 t) A AT,3 = (A BT O) t ((A BT O) D SLD ) Depreciation scheme - SLD D SLD = IC S n Scenario [4]: SYD IC + G P/G (i 4, n) = A AT,4 P A (i 4, n) + S P F (i 4, n) Convert interest rates to percent duplicate k = 1, 4 end i pct,k = i k 100 % Calculated the gradient and after tax cash flow at year 1 using the first two years of the scenario duplicate k = 1, 2 end D k = 2 (IC S) n k + 1 n (n + 1) A AT,SY D,k = (A BT O) t ((A BT O) D k ) G = A AT,SY D,1 A AT,SY D,2 A AT,4 = A AT,SY D,1 Solution A BT = [] D SLD = 2600 [] G = [] IC = [] n = 5 O = [] S = 5000 [] t = 0.5 Arrays Table: M ain Row Scenario i A AT,i i i i pct,i D i A AT,SY D,i [] [%] [] [] 1 no tax after tax no depreciation after tax with SLD after tax with SYD

29 Courtesy of CRC Press/Taylor & Francis Group 5,000 7, ,000 20,000 FIGURE 2.1 Economic cash flow diagram. 002x001.eps

30 Courtesy of CRC Press/Taylor & Francis Group 7,000 4, ,000 FIGURE 2.2 An equivalent version of the cash flow diagram shown in Figure x002.eps

31 Courtesy of CRC Press/Taylor & Francis Group F FIGURE 2.3 Cash flow diagram showing the equivalence between a uniform series, A, and a future sum, F. A 002x003.eps

32 Courtesy of CRC Press/Taylor & Francis Group P A FIGURE 2.4 Cash flow diagram showing the equivalence between a uniform series, A, and a present value, P. 002x004.eps

33 Courtesy of CRC Press/Taylor & Francis Group G 2G 3G 4G 5G 6G 7G 8G 9G P FIGURE 2.5 Cash flow diagram showing the equivalence of a uniform linearly increasing series and a present value. 002x005.eps

34 Courtesy of CRC Press/Taylor & Francis Group Example Using Microsoft Excel GIVEN: A_p = A_o = S = G = n = 50, Annual profit increase due to equipment 1, Annual operation cost 5, Salvage value of equipment at year n 1, Maintenance gradient 10 Number of periods in the analysis (years) FIND: COSTS: P + A o P,i,n A i = + G P = A_o*(P/A,i,n) = G*(P/F,i,1) = G*(P/F,i,2) = G*(P/F,i,3) = G*(P/F,i,4) = G*(P/F,i,5) = G*(P/F,i,6) = G*(P/F,i,7) = G*(P/F,i,8) = G*(P/F,i,9) = G*(P/F,i,10) = 15.62% P,i,n G 225, , , , , , , , , , PV COSTS = 246, Rate of return (interest rate) Calculated with Goal Seek P = A p,i,n A P + S,i,n F Initial cost of equipment Calculated with PV function Calculated with PV function Calculated with PV function Calculated with PV function Calculated with PV function Calculated with PV function Calculated with PV function Calculated with PV function Calculated with PV function Calculated with PV function Calculated with PV function INCOMES: A_p*(P/A,i,n) = 245, S*(P/F,i,n) = 1, PV INCOMES = 246, Calculated with PV function Calculated with PV function PV(COST - INCOME) = (0.00) FIGURE 2.6 Microsoft Excel solution of Example 2.5. Set to zero in Goal Seek 002x006.eps

35 Courtesy of CRC Press/Taylor & Francis Group P = 20, A = (6)(12) FIGURE E2.2

36 Courtesy of CRC Press/Taylor & Francis Group m Payment = Interest Principal Balance 20, , , , , , , FIGURE E , , (0.00)

37 Courtesy of CRC Press/Taylor & Francis Group A p = 50,000 S = 5, A o = 1,000 G = 1,000 P =? FIGURE E2.4

38 Courtesy of CRC Press/Taylor & Francis Group Machine A Machine B A A = 600 A B = 500 P A = 800 P B = 1,000 FIGURE E2.6

39 Courtesy of CRC Press/Taylor & Francis Group Machine A Machine B A A = 600 A B = 500 P A = 800 P B = 1,500 FIGURE E2.7A

40 Courtesy of CRC Press/Taylor & Francis Group A A = 600 A A = 600 P A0 = 800 P A5 = 800 FIGURE E2.7B

41 Courtesy of CRC Press/Taylor & Francis Group A A1 = 600 A A2 = 350 P A0 = 800 P A5 = 800 FIGURE E2.7C

42 Courtesy of CRC Press/Taylor & Francis Group A = 85, IC = 300,000 FIGURE E2.8A

43 Courtesy of CRC Press/Taylor & Francis Group A AT,k = A BT,k t(a BT,k D k ) A AT,k = 85,000 (0.51)(85,000 0) = 41, IC = 300,000 FIGURE E2.8B

44 Courtesy of CRC Press/Taylor & Francis Group A AT,k = A BT,k t(a BT,k D k ) A AT,k = 85,000 (0.51)(85,000 30,000) = 56, IC = 300,000 FIGURE E2.8C

45 Courtesy of CRC Press/Taylor & Francis Group G = 2,782 A AT,1 = 69, IC = 300,000 FIGURE E2.8D

46 Courtesy of CRC Press/Taylor & Francis Group A AT,1 = 69, G = 2,782 IC = 300,000 FIGURE E2.8E

47 Courtesy of CRC Press/Taylor & Francis Group A 1 = 110,000 i = 12.5% 12,000 A 10 = 4, G = 2,800 A = 3,500 P 0 = 200, ,000 FIGURE P2.7

48 Courtesy of CRC Press/Taylor & Francis Group 5,000 cfm 10 F Supply fan 5,000 cfm Heat exchanger 1 Flow control valve 5,000 cfm Pump Flow control damper Heat exchanger 2 Furnace Exhaust fan 10,000 cfm 85 F 10,000 cfm 75 F FIGURE P2.12