Economic Minimum Life Service
| For 12 years | ||||||||||||||
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | ||
| splitter | -5800 | 600 | ||||||||||||
| operating cost | -1000 | -1100 | -1210 | -1331 | -1464.1 | -1610.51 | -1771.56 | -1948.72 | -2143.59 | -2357.95 | -2593.74 | -2853.12 | ||
| loan | 5400 | -270 | -270 | -5670 | ||||||||||
| total | -400 | -1270 | -1370 | -6880 | -1331 | -1464.1 | -1610.51 | -1771.56 | -1948.72 | -2143.59 | -2357.95 | -2593.74 | -2253.12 | |
| pv @ 7% | 1 | 0.934579 | 0.873439 | 0.816298 | 0.762895 | 0.712986 | 0.666342 | 0.62275 | 0.582009 | 0.543934 | 0.508349 | 0.475093 | 0.444012 | 8.942686 |
| pv | -400 | -1186.92 | -1196.61 | -5616.13 | -1015.41 | -1043.88 | -1073.15 | -1103.24 | -1134.17 | -1165.97 | -1198.66 | -1232.27 | -1000.41 | -18366.8 |
| equivalent annual cost | 2053.854584 | 2053.855 | ||||||||||||
| For 6 years | ||||||||||||||
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | ||||||||
| splitter | -5800 | 2200 | ||||||||||||
| operating cost | -1000 | -1100 | -1210 | -1331 | -1464.1 | -1610.51 | ||||||||
| loan | 5400 | -270 | -270 | -5670 | ||||||||||
| total | -400 | -1270 | -1370 | -6880 | -1331 | -1464.1 | 589.49 | |||||||
| pv @ 7% | 1 | 0.934579 | 0.873439 | 0.816298 | 0.762895 | 0.712986 | 0.666342 | 5.76654 | ||||||
| PV | -400 | -1186.92 | -1196.61 | -5616.13 | -1015.41 | -1043.88 | 392.8021 | -10066.2 | ||||||
| equivalent annual cost | 1745.622158 | 1745.622 | ||||||||||||
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Question 
Individual Computer Spreadsheet Problem with Executive Summary Details
The following are suggestions and useful information to assist you with completing the Individual Computer Spreadsheet Problem with Executive Summary. As always, I welcome your questions.
Where:
- DP is the down payment
- SV is the salvage value
- PW is the Present Worth of…
- AOC is the annual operating cost
- LA is the loan amount
Suggestions:
Economic Minimum Life Service
- The Economic Service Life (ESL) that this problem is based upon can be found discussed in the Text in Chapter 11 Section 2, page 308.
- The equation above is adapted especially for this problem and is based upon the Equation 11.3 in the Text.
- The ESL topic is also specifically discussed in Video Lecture 12 Continued, “More on Replacement Analysis” at the 11 minute and 25 second time stop. And you can find it in the Viewing Guide starting on page 205.
- Please be aware that “Economic Service Life (ESL)” is synonymous with “Minimum Cost Life”.
- You can find the exact CFD for this problem at the very end of the Video Lecture #15 Continued, “Taxation” at the 30 minute and 52 second time stop.
- You will need to apply the above equation for each year of the 12 year study. I suggest working it by hand then doing the spreadsheet. If you have difficulty with the built in functions of excel, then just write in the interest factor formulas and let the spreadsheet compute the answers for each year. Or you can use a combination of excel functions and interest formulas. Your choice.
- This individual computer spreadsheet problem has some similarity to homework Practice Problem 14 and Group Homework Problem 11.15.
Individual Computer Spreadsheet Problem: A self-employed worker operates a firewood-splitting service. He purchased a commercial-grade wood splitter for $5800. He used $400 of business capital and financed the balance at 5% per year for 3 years. The estimated values of the splitter for the next 6 years are $2200 after the first year of ownership, decreasing by $400 per year to year 5, after which the resale value remains at $600. Annual operating costs are expected to be $1000 the first year, increasing by 10% each year thereafter. He considers keeping the splitter at least 6 years. If money is worth 7% per year, for how many years should the splitter be retained? (Perform the Economic Minimum Life analysis for at least 12 years.)