Engineering Economic Analysis
13th Edition
ISBN: 9780190296902
Author: Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle
Publisher: Oxford University Press
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Question
Chapter 8, Problem 11P
To determine
i.
Choice Table for interest Rates from 0% to 25%.
Expert Solution
Answer to Problem 11P
Choice table for both specification plan has been made.
Explanation of Solution
Given:
Cost of resurfacing a road = $1.5 M
Annual Maintenance = $120K
Time period = 10 years.
Calculation:
Current Specification:
| Current Plan | Maintenance Cost | Initial Cost Annuity = 1500000/13.1808 | FV = A×(1.06^10)/0.6 | Total Cost | PV of Perpetuity = A/i | |
| 13.181 | ||||||
| 0.01 | 120000 | 113802 | 233802 | 23380200 | ||
| 0.02 | 120000 | 113802 | 233802 | 11690100 | ||
| 0.03 | 120000 | 113802 | 233802 | 7793400 | ||
| 0.04 | 120000 | 113802 | 233802 | 5845050 | ||
| 0.05 | 120000 | 113802 | 233802 | 4676040 | ||
| 0.06 | 120000 | 113802 | 233802 | 3896700 | ||
| 0.07 | 120000 | 113802 | 233802 | 3340029 | ||
| 0.08 | 120000 | 113802 | 233802 | 2922525 | ||
| 0.09 | 120000 | 113802 | 233802 | 2597800 | ||
| 0.1 | 120000 | 113802 | 233802 | 2338020 | ||
| 0.11 | 120000 | 113802 | 233802 | 2125473 | ||
| 0.12 | 120000 | 113802 | 233802 | 1948350 | ||
| 0.13 | 120000 | 113802 | 233802 | 1798477 | ||
| 0.14 | 120000 | 113802 | 233802 | 1670014 | ||
| 0.15 | 120000 | 113802 | 233802 | 1558680 | ||
| 0.16 | 120000 | 113802 | 233802 | 1461263 | ||
| 0.17 | 120000 | 113802 | 233802 | 1375306 | ||
| 0.18 | 120000 | 113802 | 233802 | 1298900 | ||
| 0.19 | 120000 | 113802 | 233802 | 1230537 | ||
| 0.2 | 120000 | 113802 | 233802 | 1169010 | ||
| 0.21 | 120000 | 113802 | 233802 | 1113343 | ||
| 0.22 | 120000 | 113802 | 233802 | 1062736 | ||
| 0.23 | 120000 | 113802 | 233802 | 1016530 | ||
| 0.24 | 120000 | 113802 | 233802 | 974175 | ||
| 0.25 | 120000 | 113802 | 233802 | 935208 |
Proposed Plan
| Proposed Plan | Maintenance Cost | Initial Cost Annuity = 2100000/23.276 | FV = A×(1.06^15-1)/0.6 | Total Cost | PV of Perpetuity = A/i |
| 23.276 | |||||
| 0.01 | 90000 | 90221.68758 | 180221.7 | 18022168.76 | |
| 0.02 | 90000 | 90221.68758 | 180221.7 | 9011084.38 | |
| 0.03 | 90000 | 90221.68758 | 180221.7 | 6007389.587 | |
| 0.04 | 90000 | 90221.68758 | 180221.7 | 4505542.19 | |
| 0.05 | 90000 | 90221.68758 | 180221.7 | 3604433.752 | |
| 0.06 | 90000 | 90221.68758 | 180221.7 | 3003694.793 | |
| 0.07 | 90000 | 90221.68758 | 180221.7 | 2574595.537 | |
| 0.08 | 90000 | 90221.68758 | 180221.7 | 2252771.095 | |
| 0.09 | 90000 | 90221.68758 | 180221.7 | 2002463.196 | |
| 0.1 | 90000 | 90221.68758 | 180221.7 | 1802216.876 | |
| 0.11 | 90000 | 90221.68758 | 180221.7 | 1638378.978 | |
| 0.12 | 90000 | 90221.68758 | 180221.7 | 1501847.397 | |
| 0.13 | 90000 | 90221.68758 | 180221.7 | 1386320.674 | |
| 0.14 | 90000 | 90221.68758 | 180221.7 | 1287297.769 | |
| 0.15 | 90000 | 90221.68758 | 180221.7 | 1201477.917 | |
| 0.16 | 90000 | 90221.68758 | 180221.7 | 1126385.548 | |
| 0.17 | 90000 | 90221.68758 | 180221.7 | 1060127.574 | |
| 0.18 | 90000 | 90221.68758 | 180221.7 | 1001231.598 | |
| 0.19 | 90000 | 90221.68758 | 180221.7 | 948535.1979 | |
| 0.2 | 90000 | 90221.68758 | 180221.7 | 901108.438 | |
| 0.21 | 90000 | 90221.68758 | 180221.7 | 858198.5124 | |
| 0.22 | 90000 | 90221.68758 | 180221.7 | 819189.4891 | |
| 0.23 | 90000 | 90221.68758 | 180221.7 | 783572.5548 | |
| 0.24 | 90000 | 90221.68758 | 180221.7 | 750923.6983 | |
| 0.25 | 90000 | 90221.68758 | 180221.7 | 720886.7504 |
Conclusion:
Choice table for both specification plan has been made.
To determine
ii.
Specification to be preferred when interest rate is 6%.
Expert Solution
Answer to Problem 11P
As the Present value of
Explanation of Solution
Given:
Cost of resurfacing a road = $1.5 M
Annual Maintenance = $120K
Time period = 10 years.
Calculation:
| Current Specification | ||||
| Figures in Million $ | ||||
| Year | Particulars | Amount | P.V Factor @ 6% | Total |
| 0 | Resurface cost | 1.5 | 1 | 1.50 |
| 1 | Annual maintenance cost | 0.12 | 0.904159 | 0.11 |
| 2 | Annual maintenance cost | 0.12 | 0.85298 | 0.10 |
| 3 | Annual maintenance cost | 0.12 | 0.804698 | 0.10 |
| 4 | Annual maintenance cost | 0.12 | 0.759149 | 0.09 |
| 5 | Annual maintenance cost | 0.12 | 0.716179 | 0.09 |
| 6 | Annual maintenance cost | 0.12 | 0.67564 | 0.08 |
| 7 | Annual maintenance cost | 0.12 | 0.637397 | 0.08 |
| 8 | Annual maintenance cost | 0.12 | 0.601317 | 0.07 |
| 9 | Annual maintenance cost | 0.12 | 0.567281 | 0.07 |
| 10 | Annual maintenance cost | 0.12 | 0.53517 | 0.06 |
| Total | 2.35 | |||
| Proposed Specification | ||||
| Figures in Million $ | ||||
| Year | Particulars | Amount | P.V Factor @ 6% | Total |
| 0 | Resurface cost | 2.1 | 1 | 2.10 |
| 1 | Annual maintenance cost | 0.09 | 0.904159 | 0.08 |
| 2 | Annual maintenance cost | 0.09 | 0.85298 | 0.08 |
| 3 | Annual maintenance cost | 0.09 | 0.804698 | 0.07 |
| 4 | Annual maintenance cost | 0.09 | 0.759149 | 0.07 |
| 5 | Annual maintenance cost | 0.09 | 0.716179 | 0.06 |
| 6 | Annual maintenance cost | 0.09 | 0.67564 | 0.06 |
| 7 | Annual maintenance cost | 0.09 | 0.637397 | 0.06 |
| 8 | Annual maintenance cost | 0.09 | 0.601317 | 0.05 |
| 9 | Annual maintenance cost | 0.09 | 0.567281 | 0.05 |
| 10 | Annual maintenance cost | 0.09 | 0.53517 | 0.05 |
| 11 | Annual maintenance cost | 0.09 | 0.504878 | 0.05 |
| 12 | Annual maintenance cost | 0.09 | 0.4763 | 0.04 |
| 13 | Annual maintenance cost | 0.09 | 0.449339 | 0.04 |
| 14 | Annual maintenance cost | 0.09 | 0.423905 | 0.04 |
| 15 | Annual maintenance cost | 0.09 | 0.39991 | 0.04 |
| Total | 2.94 | |||
Conclusion:
As the Present value of cash outflow is less in current specification, the current specification will be selected.
To determine
iii.
Economic Difference between both specifications.
Expert Solution
Answer to Problem 11P
As Proposed Specification has High present Value compared to Current specification and there is relatively high difference between the Cost involvements in Current plan compared to proposed plan, so it leads to economic difference.
Explanation of Solution
Given:
Cost of resurfacing a road = $1.5 M
Annual Maintenance = $120K
Time period = 10 years.
Concept used:
As Proposed Specification has High present Value compared to Current specification and there is relatively high difference between the Cost involvements in Current plan compared to proposed plan, so it leads to economic difference.
Conclusion:
As Proposed Specification has High present Value compared to Current specification and there is relatively high difference between the Cost involvements in Current plan compared to proposed plan, so it leads to economic difference.
To determine
iv.
Impact on environment due to road surfacing.
Expert Solution
Answer to Problem 11P
- Alteration of landscape which leads to negative impact on wildlife in a number of deleterious ways.
- Mortality of wildlife due to destruction of their habitats.
- The debris from the tires of vehicles results to land pollution.
- Road noise affects birds negatively. Bird population declines due to proximity of road areas.
Explanation of Solution
Given:
Cost of resurfacing a road = $1.5 M
Annual Maintenance = $120K
Time period = 10 years.
Concept used:
- Alteration of landscape which leads to negative impact on wildlife in a number of deleterious ways.
- Mortality of wildlife due to destruction of their habitats.
- The debris from the tires of vehicles results to land pollution.
- Road noise affects birds negatively. Bird population declines due to proximity of road areas.
Conclusion:
- Alteration of landscape which leads to negative impact on wildlife in a number of deleterious ways.
- Mortality of wildlife due to destruction of their habitats.
- The debris from the tires of vehicles results to land pollution.
- Road noise affects birds negatively. Bird population declines due to proximity of road areas.
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Chapter 8 Solutions
Engineering Economic Analysis
Ch. 8 - Prob. 1QTCCh. 8 - Prob. 2QTCCh. 8 - Prob. 3QTCCh. 8 - Prob. 4QTCCh. 8 - Prob. 5QTCCh. 8 - Prob. 1PCh. 8 - Prob. 2PCh. 8 - Prob. 3PCh. 8 - Prob. 4PCh. 8 - Prob. 5P
Ch. 8 - Prob. 6PCh. 8 - Prob. 7PCh. 8 - Prob. 8PCh. 8 - Prob. 9PCh. 8 - Prob. 10PCh. 8 - Prob. 11PCh. 8 - Prob. 12PCh. 8 - Prob. 13PCh. 8 - Prob. 14PCh. 8 - Prob. 15PCh. 8 - Prob. 16PCh. 8 - Prob. 17PCh. 8 - Prob. 18PCh. 8 - Prob. 19PCh. 8 - Prob. 20PCh. 8 - Prob. 21PCh. 8 - Prob. 22PCh. 8 - Prob. 23PCh. 8 - Prob. 24PCh. 8 - Prob. 25PCh. 8 - Prob. 26PCh. 8 - Prob. 27PCh. 8 - Prob. 28PCh. 8 - Prob. 29PCh. 8 - Prob. 30PCh. 8 - Prob. 31PCh. 8 - Prob. 32PCh. 8 - Prob. 33PCh. 8 - Prob. 34PCh. 8 - Prob. 35PCh. 8 - Prob. 36PCh. 8 - Prob. 37PCh. 8 - Prob. 38PCh. 8 - Prob. 39PCh. 8 - Prob. 40PCh. 8 - Prob. 41PCh. 8 - Prob. 42PCh. 8 - Prob. 43PCh. 8 - Prob. 44PCh. 8 - Prob. 45P
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