# Chapter 4-5

ENGINEERING ECONOMY Chapter 4-5 Comparison of Alternatives Annual Worth Analysis DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses

7. Select the preferred investment DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Annual Worth Analysis Single Alternative DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Annual Worth Method converts all cash flows to a uniform annual series over the planning horizon using i=MARR

a popular DCF method n i (1 i ) t AW (i %) At (1 i ) n t 0 (1 i ) 1 AW (i %) PW (i %) ( A | P i %, n) n DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY

SMP Investment A \$500,000 investment in a surface mount placement machine is being considered. Over a 10-year planning horizon, it is estimated the SMP machine will produce net annual savings of \$92,500. At the end of 10 years, it is estimated the SMP machine will have a \$50,000 salvage value. Based on a 10% MARR and annual worth analysis, should the investment be made? AW(10%) = -\$500K(A|P 10%,10) + \$92.5K + \$50K(A|F 10%,10) = \$14,262.50 =PMT(10%,10,500000,-50000)+92500 = \$14,264.57 DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY

SMP Investment How does annual worth change over the life of the investment? How does annual worth change when the salvage value decreases geometrically and as a gradient series? DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM use SOLVER to determine the DPBP using AW analysis. ENGINEERINGLets ECONOMY

DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM

ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Annual Worth Analysis Multiple Alternatives

DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Example 7.4 Recall the example involving two alternative designs for a new ride at a theme park: Alt. A costs \$300,000, has net annual after-tax revenue of \$55,000, and has a negligible salvage value at the end of the 10-year planning horizon; Alt. B costs \$450,000, has revenue of \$80,000/yr., and has a negligible salvage value. Based on an AW analysis and a 10% MARR, which is preferred? AWA(10%) = -\$300,000(A|P 10%,10) + \$55,000 = -\$300,000(0.16275) + \$55,000 = \$6175.00 =PMT(10%,10,300000)+55000 = \$6176.38 AWB(10%) = -\$450,000(A|P 10%,10) + \$80,000 = -\$450,000(0.16275) + \$80,000 = \$6762.50

=PMT(10%,10,450000)+80000 = \$6764.57 Analyze the impact on AW based on salvage values decreasing geometrically to 1 after 10 years. DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM

ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY AW(A) AW(B) \$40,000 \$30,000

AW(A) = AW(B) when MARR = 10.56% \$20,000 \$10,000 -\$20,000 -\$30,000 MARR DR. MAISARA MOHYELDIN GASIM 20% 18% 16%

14% 12% 10% 8% 6% 4% 2% -\$10,000

0% \$0 ENGINEERING ECONOMY Example 7.5 For The Scream Machine alternatives (A costing \$300,000, saving \$55,000, and having a negligible salvage value at the end of the 10year planning horizon; B costing \$450,000, saving \$80,000, and having a negligible salvage value), using an incremental AW analysis and a 10% MARR, which is preferred? AWA(10%) = -\$300,000(A|P 10%,10) + \$55,000 = -\$300,000(0.16275) + \$55,000 = \$6175.00 =PMT(10%,10,300000)+55000 = \$6176.38 > \$0 (A is better than do nothing) AWB-A(10%) = -\$150,000(A|P 10%,10) + \$25,000 = -\$150,000(0.16275) + \$25,000 = \$587.50

=PMT(10%,10,150000)+25000 = \$588.19 > \$0 (B is better than A) Prefer B DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Example 7.6 If an investors MARR is 12%, which mutually exclusive investment alternative maximizes the investors future worth, given the parameters shown below? EOY 0 1 2 3 4

5 6 CF(1) -\$10,000 \$5,000 \$5,000 \$10,000 CF(2) -\$15,000 \$5,000 \$5,000 \$5,000 \$5,000 \$5,000 \$7,500

CF(3) -\$20,000 \$0 \$3,000 \$6,000 \$9,000 \$12,000 \$15,000 Consider 3 scenarios: individual life cycles; least common multiple of lives; and one-shot investments DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Example 7.6 (Continued)

Scenario 1: individual life cycles AW1(12%) = -\$10,000(A|P 12%,3) + \$5000 + \$5000(A|F 12%,3) = \$2318.25 =PMT(12%,3,10000,-5000)+5000 = \$2318.26 AW2(12%) = -\$15,000(A|P 12%,6) + \$5000 +\$2500(A|F 5%,6) = \$1659.63 =PMT(12%,6,15000,-2500)+5000 = \$1659.68 AW3(12%) = -\$20,000(A|P 12%,6) + \$3000(A|G 12%,6) = \$1651.55 =PMT(12%,6,-1000*NPV(12%,0,3,6,9,12,15)+20000) = \$1651.63 DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Example 7.6 (Continued) Scenario 2: least common multiple of lives

EOY 0 1 2 3 4 5 6 CF(1') -\$10,000 \$5,000 \$5,000 \$0 \$5,000 \$5,000 \$10,000

CF(2) -\$15,000 \$5,000 \$5,000 \$5,000 \$5,000 \$5,000 \$7,500 CF(3) -\$20,000 \$0 \$3,000 \$6,000 \$9,000 \$12,000

\$15,000 AW1(12%) = -\$10,000(A|P 12%,6) + \$5000 + \$5000(A|F 12%,6) - \$5000(A|P 12%,3)(A|P 12%,6) = \$2318.22 =PMT(12%,6,10000,-5000)+5000 +PMT(12%,6,PV(12%,3,,-5000))= \$2318.26 AW2(12%) = \$1473.17 = \$1473.23 Identical results! AW3(12%) = \$1651.55 = \$1651.63 DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Example 7.6 (Continued) Scenario 3: one-shot investments EOY

0 1 2 3 4 5 6 CF(1) -\$10,000 \$5,000 \$5,000 \$10,000 \$0 \$0 \$0

CF(2) -\$14,500 \$5,000 \$5,000 \$5,000 \$5,000 \$5,000 \$5,000 CF(3) -\$20,000 \$0 \$3,000 \$6,000 \$9,000 \$12,000 \$15,000

AW1(12%) = {-\$10,000 + [\$5000(P|A 12%,3) + \$5000(P|F 12%,3)]}(A|P 12%,6) = \$1354.32 =PMT(12%,6,10000-PV(12%,3,-5000,-5000)) = \$1354.29 AW2(12%) = \$1473.17 = \$1473.23 AW3(12%) = \$1651.55 = \$1651.63 DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Considering scenarios 1 and 2, is it reasonable to assume an investment alternative equivalent to Alt. 1 will be available in 3 years? If so, why was the MARR set equal to 12%?

DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Example 7.7 Three industrial mowers (Small, Medium, and Large) are being evaluated by a company that provides lawn care service. Determine the economic choice, based on the following cost and performance parameters: First Cost: Operating Cost/Hr Revenue/Hr Hrs/Yr Useful Life (Yrs) Small \$1,500

\$35 \$55 1,000 2 Medium \$2,000 \$50 \$75 1,100 3 Large \$5,000 \$76 \$100 1,200

5 Use AW analysis to determine the preferred mower, based on a MARR of 12%. DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Example 7.7 (Continued) AWsmall = -\$1500(A|P 12%,2) + \$20(1000) = \$19,112.45 =PMT(12%,2,1500)+20*1000 = \$19,112.45 AWmed = -\$2000(A|P 12%,3) + \$25(1100) = \$26,667.30 =PMT(12%,3,2000)+25*1100 = \$26,667.30 AWlarge = -\$5000(A|P 12%,5) + \$24(1200) = \$27,412.95 =PMT(12%,5,5000)+24*1200 = \$27,412.95 What did we assume when solving the example?

DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Example 7.8 If a 5-year planning horizon were used, what salvage values are required to have the same AW as before? The small mower will be replaced at the end of year 4; the medium mower will be replaced at the end of year 3. One year of service of the small mower will have the following cash flows: SVsmall = \$19,112.45(F|A 12%,1) - \$20,000(F|A 12%,1) + \$1500(F|P 12%,1) = \$792.45 =FV(12%,1,-19112.45)-FV(12%,1,-20000,1500) = \$792.45 SVmed = \$26,667.30(F|A 12%,2) - \$27,500(F|A 12%,2) + \$2000(F|P 12%,2) = \$743.48

=FV(12%,2,-26667.3)-FV(12%,2,-27500,2000) = \$743.48 SVlarge = \$0 DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Principle #8 Compare investment alternatives over a common period of time DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Fundamentals of Engineering Examination Even though you might not encounter a situation in your

professional practice that requires the least common multiple of lives assumption to be used, it is very likely you will have problems of this type on the FE Exam. Therefore, you need to be familiar with how to solve such problems. Specifically, on the FE Exam, unless instructions are given to do otherwise, calculate the annual worth for a life cycle of each alternative and recommend the one that has the greatest annual worth. DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Capital Recovery Cost DR. MAISARA MOHYELDIN GASIM

ENGINEERING ECONOMY \$F 0 1 2 End of Year n -1 n

0 1 2 \$CR \$CR n -1 \$CR \$P CFD for Capital Recovery Cost (CR).

DR. MAISARA MOHYELDIN GASIM n \$CR ENGINEERING ECONOMY Capital Recovery Cost Formulas CR = P(A|P i,n) F(A|F i, n) CR = (P-F)(A|F i, n) + Pi CR = (P-F)(A|P i,n) + Fi CR =PMT(i,n,-P,F) Example P = \$500,000 F = \$50,000 i = 10% n = 10 yrs CR = \$500,000(0.16275) - \$50,000(0.06275) = \$78,237.50 CR = \$450,000(0.06275) + \$500,000(0.10) = \$78,237.50

CR = \$450,000(0.16275) + \$50,000(0.10) = \$78,237.50 CR =PMT(10%,10,-500000,50000) = \$78,235.43 DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY Geometrically Decreasing Salvage Value Decreasing Gradient Salvage Value \$180,000 \$600,000 \$160,000 \$500,000 \$120,000

\$400,000 \$100,000 \$300,000 \$80,000 \$60,000 \$200,000 \$40,000 \$100,000 \$20,000 \$0 \$0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Age (Yrs) DR. MAISARA MOHYELDIN GASIM CR (decreasing gradient SV) CR(geometrically decreasing SV) \$140,000 ENGINEERING ECONOMY 1. 2. 3. 4. 5.

6. 7. 8. 9. 10. Pit Stop #7No Time to Coast True or False: Annual worth analysis is the most popular DCF measure of economic worth. True or False: Unless non-monetary considerations dictate otherwise, choose the mutually exclusive investment alternative that has the greatest annual worth over the planning horizon. True or False: The capital recovery cost is the uniform annual cost of the investment less the uniform annual worth of the salvage value. True or False: If AW > 0, then PW>0, and FW>0. True or False: If AW(A) > AW(B), then PW(A) > PW(B).

True or False: If AW (A) < AW(B), then AW(B-A) > 0. True or False: If AW(A) > AW(B), then CW(A) > CW(B) and DPBP(A) < DPBP(B). True or False: AW can be applied as either a ranking method or as an incremental method. True or False: To compute capital recovery cost using Excel, enter =PMT(i%,n,P,F) in any cell in a spreadsheet. True or False: When using annual worth analysis with mutually exclusive alternatives having unequal lives, always use a planning horizon equal to the least common multiple of lives. DR. MAISARA MOHYELDIN GASIM ENGINEERING ECONOMY 1. 2. 3. 4.

5. 6. 7. 8. 9. 10. Pit Stop #7No Time to Coast True or False: Annual worth analysis is the most popular DCF measure of economic worth. FALSE True or False: Unless non-monetary considerations dictate otherwise, choose the mutually exclusive investment alternative that has the greatest annual worth over the planning horizon. TRUE True or False: The capital recovery cost is the uniform annual cost of the investment less the uniform annual worth of the salvage value. TRUE True or False: If AW > 0, then PW > 0, and FW > 0. TRUE

True or False: If AW(A) > AW(B), then PW(A) > PW(B). TRUE True or False: If AW (A) < AW(B), then AW(B-A) > 0. TRUE True or False: If AW(A) > AW(B), then CW(A) > CW(B) and DPBP(A) < DPBP(B). FALSE True or False: AW can be applied as either a ranking method or as an incremental method. TRUE True or False: To compute capital recovery cost using Excel, enter =PMT(i%,n,P,F) in any cell in a spreadsheet. TRUE True or False: When using annual worth analysis with mutually exclusive alternatives having unequal lives, always use a planning horizon equal to the least common multiple of lives. FALSE DR. MAISARA MOHYELDIN GASIM

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