ISYE 3025 - Week 4
ISYE 3025 - Week 4 ISYE 3025
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This 7 page Class Notes was uploaded by Moriah Mattson on Saturday September 24, 2016. The Class Notes belongs to ISYE 3025 at Georgia Institute of Technology taught by Kelly Bartlett in Fall 2016. Since its upload, it has received 8 views. For similar materials see Engineering Economy in Industrial Engineering at Georgia Institute of Technology.
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Date Created: 09/24/16
ISYE 3025 Engineering Economics Week 4 Video Fundamentals of an Economics Decision Goals: To illustrate the fundamental choice between Benefits and Costs Develop a formal approach and notation to determining the most economical decision Introduce the basic decision criterions of Prospective Benefit and Prospective Cost Introducing Economics Decisions Assume you can either get a 1year magazine subscription for $15 or a 2year magazine subscription for $25. And, as a special offer these prices are guaranteed for the next four years. Which option would you prefer? In order to determine this, we should base this decision on the differences! Then, we must break down the differences into two components. (Benefit and cost series) Difference = Benefit Cost The benefit and cost series represent two vectors. These funds (F anB ) caCbe invested. And, we can determine which series would accumulate more money by an equivalence calculation. Formal Representation And economic decision criterion is hot we make economics decisions Measure of Worth: expression of monetary value associated with an investment alternative Decision Rule: judges if the monetary value is an economical option or not Minimum Attractive Rate of Return (MARR) We can look at the special interest rate (i) known as the MARR to determine if the opportunity is economical or not The MARR looks at those opportunities readily available to us to invest our funds This is the minimum standard to compare alternatives to Also known as, marginal growth rate, discount rate, cutoff rate, hurdle rate, and yield Example Suppose we invest $100 at an MARR, i = 5%. Is this an economical decision? We cannot say if it is an economical decision without a standard to compare it to. The MARR sets this standard... If MARR < 5%, it would be a good investment If MARR > 5%, it would not be a good investment Values of the MARR 1) For individuals, the MARR typically represents the minimum attractive opportunities to invest in money markets 2) For corporations, the MARR typically represents the minimum attractive opportunities to invest in the company Notation Ajt net cash flow j = index on opportunities A > 0 is a net receipt jt Ajt 0 is a net expense Ajt 0 for t<0 and t>N A = B C jt jt jt Bjt A jt A jt, else B = 0 jt Cjt A jtA jt, else C = 0 jt Measure of Worth N N B (i) = ∑ B (1+i) T−t and C (i) = ∑ C (1+i) T−t , where i = MARR j t = 0 j j t = 0 j Typically, T = 0 or T = N Decision Rule Accept j if B (i) jt (i), ojtrwise reject j Example A manufacturer is considering buying 10 new machines, which cost $100,000 a piece and have a life expectancy of 5 years. The cost to install all machines is $25,000. Each machine will reduce labor costs by $30,000, but will increase energy costs by $8000. If the MARR = 8%/year, are these machines economical? Break down into A = B C Rt Rt Rt Now, we determine which series would accumulate more money. By accepting the machines, you would be poorer by relative comparison. Every decision has benefits and costs, and the question is “Do the Benefits exceed the Cost?” Video Future Worth, Present Worth, Annual Worth There are three classic economic criteria based on worth: Future Worth, FW(i) j Present Worth, PW(i) j Future Worth, FW(i) j Future Worth Measure of Worth: FW(ij= B(i)j C(i)jwhere i = MARR N N N = ∑ B (1+i) T−t − ∑ C (1+i) T−t = ∑ [B −C ](1+i) T−t t = 0 jt t = 0 jt t = 0 jt jt N = ∑ A (1+i) T−t , typically T = N t = 0 jt Decision Rule: Accept j if FW(i) >j, otherwise reject Future Worth Example Referring back to the machine example from the previous videos. We can use these equations for Future Worth to determine whether it is a good investment. FW(ij= B(i) jC(i) j$1,290,652 $1,506,061 = $215,409 < 0, OR FW(0.08) = $1,025,000(F/P, 0.08, 5) + $220,000(F/A,0.08,5) = $215,409 < 0 j Present Worth Measure of Worth: N PW(ij= FW(i)(1+j , where i = MARR N N−t −N = [ ∑ A (1+jt ](1+i) t = 0 N −t = ∑ A (1jt) t = 0 Decision Rule: Accept j if PW(i) > j otherwise reject Present Worth Example Consider an engineer that is evaluating 3 alternative routes for a power line. Use an i = 0.05/year and N = 15 years. Below is the relevant data: Then, identify the cash flows. We will compare PW (i) to PWR1i). R2 PW (0.05) = 72K + 24.9K(P/A, 0.05, 14) + 54.9K(P/F, 0.05, 15) = +200.8K R1 PW (R25) = 110K + 28.6K(P/A, 0.05, 14) + 48.6K(P/F, 0.05, 15) = +196.5K > Can we say Route 1 is the best choice because it is the highest number? No, we cannot right now. We do not know if we can claim this yet, we must evaluate the differences Since PW (i) R2W (i) = R1K < 0, Route 1 is the more economical option. Now, we will compare PW (i) to PW (i). R1 R3 PW (R15) = 72K + 24.9K(P/A, 0.05, 14) + 54.9K(P/F, 0.05, 15) = +200.8K PW (R35) = 64K + 27.3K(P/A, 0.05, 14) + 51.3K(P/F, 0.05, 15) = +230.9K Since PW (i) R3W (i) = +R11K > 0, Route 3 is the most economical option. An Equivalent Decision Rule For any two alternatives j = A1, A2: PW (i) = PW (i) PW (i) A1A2 A1 A2 Thus, equivalent decision rule is: prefer A1 over A2 when PW (i) > PW (i) A1en there is A2capital rationing (which will be talked about later) Annual Worth Measure of Worth: AW(ij= PW(i)(Aj, i, N), where i = MARR = FW(i)j/F, i, N) Decision Rule: Accept j if AW(i) j0, otherwise reject AW (R15) = +19.3K > 0 AW (R25) = +18.9K > 0 AW (0.05) = +22.2K > 0 R3 AW R2R1(0.05) = 0.4K < 0 AW (0.05) = +2.9K > 0 R3R1
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