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# newnansm10 - Chapter 10 Uncertainty in Future Events 10-1(a...

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Chapter 10: Uncertainty in Future Events 10-1 (a) Some reasons why a pole might be removed from useful service: 1. The pole has deteriorated and can no longer perform its function of safely supporting the telephone lines 2. The telephone lines are removed from the pole and put underground. The poles, no longer being needed, are removed. 3. Poles are destroyed by damage from fire, automobiles, etc. 4. The street is widened and the pole no longer is in a suitable street location. 5. The pole is where someone wants to construct a driveway. (b) Telephone poles face varying weather and soil conditions; hence there may be large variations in their useful lives. Typical values for Pacific Telephone Co. in California are: Optimistic Life: 59 years Most Likely Life: 28 years Pessimistic Life: 2.5 years Recognizing there is a mortality dispersion it would be possible, but impractical, to define optimistic life as the point where the last one from a large group of telephone poles is removed (for Pacific Telephone this would be 83.5 years). This is not the accepted practice. Instead, the optimum life is where only a small percentage (often 5%) of the group remains in service. Similarly, pessimistic life is when, say, 5% of the original group of poles have been removed from the group. 10-2 If 16,000 km per year, then fuel cost = oil/tires/repair = \$990/year, and salvage value = 9,000 - 5x16,000x.05 = 9,000 – 4,000 = 5,000 EUAC 16,000 = 9,000( A / P ,8%,5) + 2x990 – 5,000( A / F ,8%,5) = 9,000x.2505 + 1,980 – 5,000x.1705 = 2,254.5 + 1,980 - 852.5 = \$3,382 Increasing annual mileage to 24,000 is a 50% increase so it increases operating costs by 50%. The salvage value drops by 5x8,000x.05 = 2,000 EUAC 24,000 = 9,000( A / P ,8%,5) + 2x1.5x990 – 3,000( A / F ,8%,5) = 9,000x.2505 + 1.5x1,980 - 3000x.1705 = 2,254.5 + 2,970 - 511.5 = \$4,713 Decreasing annual mileage to 8,000 is a 50% decrease so it decreases operating costs by 50%. The salvage value increases by 5x8,000x.05 = 2,000 EUAC 8,000 = 9,000( A / P ,8%,5) + 2x.5x990 – 7,000( A / F ,8%,5) = 9,000x.2505 + .5x1,980 – 7,000x.1705 = 2,254.5 + 990 - 1193.5 = \$2,051

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10-3 Mean Life = (12 + 4 x 5 + 4)/6 = 6 years PW of Cost = PW of Benefits \$80,000 = \$20,000 (P/A, i%, 6) Rate of Return is between 12% and 15% Rate of Return ≈ 13% 10-4 Since the pessimistic and optimistic answers are symmetric about the most likely value of 16,000, the weighted average is 16,000 km. If 16,000 km per year, then fuel cost = oil/tires/repair = \$990/year, and salvage value = 8,000 - 5x16,000x.05 = 9,000 – 4,000 = 5,000 EUAC 16,000 = 9,000( A / P ,8%,5) + 2x990 – 5,000( A / F ,8%,5) = 9,000x.2505 + 1980 – 5,000x.1705 = 2,254.5 + 1,980 - 852.5 = \$3,382 10-5 Probability of rolling a 7 or 11 = (6 + 2)/36 = 8/36 10-6 Since the P s must sum to 1: P (30K) = 1 - .2 - .3 = .5 E (savings) = .3(20K) + .5(30K) + .2(40K) = \$29K
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newnansm10 - Chapter 10 Uncertainty in Future Events 10-1(a...

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