## Sunday, November 10, 2019

### Engineering Management

EEMT 5510 Homework #5 12-24. An improved design of a computerized piece of continuous quality measuring equipment used to control the thickness of rolled sheet products is being developed. It is estimated to sell for \$125,000 more than the current design.Based on present test data, however, the typical user has the following probabilities of achieving different performance results and cost savings (relative to the current unit) in the first year of operation (assume these annual cost savings would escalate 5% per year thereafter; a five-year analysis period is used; the MARR=18%, and the net market value after five years is 0): |Performance Results |Probability |Cost Savings in Year One | |Optimistic |0. 0 |\$60,000 | |Most likely |0. 55 |40,000 | |Pessimistic |0. 15 |18,000 | Based on the E(PW), is the new design preferable to the current unit? Based on a decision tree analysis, what is the EVPI? What does the EVPI tell you?Without information, the optimal decision is to take the new design, shown by the decision tree below |scenarios |Year 0 cost |Year 1 Saving |Year2 Saving | | |Results (j) |p(j) |Decision |Outcome | | |Optimistic |0. 30 |New |\$79,063 | | |Most Likely |0. 55 |New |11,042 | | |Pessimistic |0. 5 |Current | 0 | | | |Expected Value: |\$29,792 | EVPI = \$29,792 ? \$20,225 = \$9,567 Note:The EVPI is the maximum amount that ought to be spent to obtain additional information prior to making a decision. Suggested Exercises 1. A bridge is to be constructed now as part of a new road. An analysis has shown that traffic density on the new road will justify a two-lane bridge at the present time.Because of uncertainty regarding future use of the road, the time at which an extra two lanes will be required is currently being studied. The estimated probabilities of having to widen the bridge to four lanes at various times in the future are as follows: |Widen Bridge In |Probability | |3 years |0. 1 | |4 years |0. 2 | |5 years |0. | |6 years |0. 4 | The present esti mated cost of the two-lane bridge is \$2,100,000. If constructed now, the four-lane bridge will cost \$4,000,000. The future cost of widening a two-lane bridge will be an extra \$2,100,000 plus \$350,000 for every year that widening is delayed. If money can earn 12% per year, what would you recommend? Option 1: construct 4 lanes now: PW( cost ) = \$4,000,000 Option 2: widen to 4 lanes later with 0. 1 probability,PW( cost ) = 2100000 + (2100000+350000*3)(P/F, 12%, 3) = 4,342,108 with 0. 2 probability, PW( cost ) = 2100000 + (2100000+350000*4)(P/F, 12%, 4) = 4324313 with 0. 3 probability, PW( cost ) = 2100000 + (2100000+350000*5)(P/F, 12%, 5) = 4284593 with 0. 4 probability, PW( cost ) = 2100000 + (2100000+350000*6)(P/F, 12%, 6) = 4227851 Hence the expected PW(cost) = 4,275,592 > 4,000,000 So we recommend constructing 4 lanes now. 2. Suppose that a random variable (e. g. , market value for a piece of equipment) is normally distributed, with mean = \$180 and variance = 36\$2.What is the proba bility that the actual market value is at least \$176? Normally distributed random variable: E(X) = \$180, V(X) = 36 (\$)2 Pr{X ( 176} = ? Z = [pic] = ? 0. 67 Pr{X ( 176} = Pr{Z ( ? 0. 67} = 1 ? Pr{Z ( ? 0. 67} = 1 ? 0. 2514 = 0. 7486 3. A potential project has an initial capital investment of \$100,000. Net annual revenues minus expenses are estimated to be \$40,000 (A\$) in the first year and to increase at the rate of 6. 48% per year. The useful life of the primary equipment, however, is uncertain, as shown in the following table: Useful Life, Years(N) |p(N) | |1 |0. 03 | |2 |0. 10 | |3 |0. 30 | |4 |0. 30 | |5 |0. 17 | |6 |0. 10 |Assume that im = MARR = 15% per year and f = 4% per year. Based on this information, a. What are the E(PW) and SD(PW) for this project? b. What is the Pr{PW>0}? c. What is the E(AW) in R\$? a. |Year N |Profit in Year |PW (Profit in |Total PW (Profit |PW (with investment until |Probability | | |N |Year N) |until Year N) |Year N) | | |1 |40000 |\$34,782. 61 |\$34,7 82. 61 |(\$65,217. 39) |0. 03 | |2 |42592 |\$32,205. 7 |\$66,988. 28 |(\$33,011. 72) |0. 1 | |3 |45351. 96 |\$29,819. 65 |\$96,807. 93 |(\$3,192. 07) |0. 3 | |4 |48290. 77 |\$27,610. 40 |\$124,418. 33 |\$24,418. 33 |0. 3 | |5 |51420. 01 |\$25,564. 83 |\$149,983. 17 |\$49,983. 17 |0. 17 | |6 |54752. 03 |\$23,670. 81 |\$173,653. 98 |\$73,653. 98 |0. 1 |E[ PW ] = 16972,Var[PW] = 1097641387 SD[PW] = 33130 (Standard deviation is the square root of Var) b. Prob{ PW>0 } = 0. 3+0. 17+0. 1 = 0. 57 Note: This is a discrete random variable with known distribution. So we do not use the normal distribution table. c. Since f=4%, we have real MARR, ir = (15%-4%)/(1+4%) = 10. 577% use the equation:AWr (until Year N) = PW (until Year N) (A/P, 10. 577%, N) AWr(1) = (72115), AWr(2) = (19168), AWr(3) = (1296), AWr(4) = 7799, AWr (5)=13380, AWr (6) = 17198, so E[AWr] = 1865 The project is questionable though E[PW]>0. The probability of PW