in a replacement analysis for a vaccum seal on a spacecraft the following data are known about the challenger, the initial investment is 7,000. there is no annual maintenance cost for the first three years, however, it will be 3,800 in each of years four and five, and then 4,900 in the sixth year and increasing by 3,300 each year thereafter. The salvage value is 0 at all times and MARR is 8% per year. What is the economic life of this challenger?
Please show every step.
Homework Answers
MARR = 8%
Using Excel
| Year | Discount factor | O&M cost | PV (O&M) | Cumulative (O&M) | Cumulative (O&M) + Initial Cost | (A/P,10%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+7000 | J | K = F*J |
| 1 | 0.92593 | 0.00 | 0.00 | 0.00 | 7000.00 | 1.08000 | 7560.00 |
| 2 | 0.85734 | 0.00 | 0.00 | 0.00 | 7000.00 | 0.56077 | 3925.38 |
| 3 | 0.79383 | 0.00 | 0.00 | 0.00 | 7000.00 | 0.38803 | 2716.23 |
| 4 | 0.73503 | 3800.00 | 2793.11 | 2793.11 | 9793.11 | 0.30192 | 2956.74 |
| 5 | 0.68058 | 3800.00 | 2586.22 | 5379.33 | 12379.33 | 0.25046 | 3100.48 |
| 6 | 0.63017 | 4900.00 | 3087.83 | 8467.16 | 15467.16 | 0.21632 | 3345.78 |
| 7 | 0.58349 | 8200.00 | 4784.62 | 13251.78 | 20251.78 | 0.19207 | 3889.81 |
| 8 | 0.54027 | 11500.00 | 6213.09 | 19464.87 | 26464.87 | 0.17401 | 4605.28 |
| 9 | 0.50025 | 14800.00 | 7403.68 | 26868.56 | 33868.56 | 0.16008 | 5421.67 |
| Discount factor | 1/(1+0.08)^n | ||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | ||||||
From Table, we see EUAC is lowest at 2716.23 at year 3
Therefore Economic life = 3 yrs
showing formula in excel
| Year | Discount factor | O&M cost | PV (O&M) | Cumulative (O&M) | Cumulative (O&M) + Initial Cost | (A/P,10%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+7000 | J | K = F*J |
| 1 | =1/(1.08)^N4 | 0 | =P4*O4 | =Q4 | =7000+R4 | =0.08*((1 + 0.08)^N4)/((1 + 0.08)^N4-1) | =S4*T4 |
| 2 | =1/(1.08)^N5 | 0 | =P5*O5 | =R4+Q5 | =7000+R5 | =0.08*((1 + 0.08)^N5)/((1 + 0.08)^N5-1) | =S5*T5 |
| 3 | =1/(1.08)^N6 | 0 | =P6*O6 | =R5+Q6 | =7000+R6 | =0.08*((1 + 0.08)^N6)/((1 + 0.08)^N6-1) | =S6*T6 |
| 4 | =1/(1.08)^N7 | 3800 | =P7*O7 | =R6+Q7 | =7000+R7 | =0.08*((1 + 0.08)^N7)/((1 + 0.08)^N7-1) | =S7*T7 |
| 5 | =1/(1.08)^N8 | 3800 | =P8*O8 | =R7+Q8 | =7000+R8 | =0.08*((1 + 0.08)^N8)/((1 + 0.08)^N8-1) | =S8*T8 |
| 6 | =1/(1.08)^N9 | 4900 | =P9*O9 | =R8+Q9 | =7000+R9 | =0.08*((1 + 0.08)^N9)/((1 + 0.08)^N9-1) | =S9*T9 |
| 7 | =1/(1.08)^N10 | =P9+3300 | =P10*O10 | =R9+Q10 | =7000+R10 | =0.08*((1 + 0.08)^N10)/((1 + 0.08)^N10-1) | =S10*T10 |
| 8 | =1/(1.08)^N11 | =P10+3300 | =P11*O11 | =R10+Q11 | =7000+R11 | =0.08*((1 + 0.08)^N11)/((1 + 0.08)^N11-1) | =S11*T11 |
| 9 | =1/(1.08)^N12 | =P11+3300 | =P12*O12 | =R11+Q12 | =7000+R12 | =0.08*((1 + 0.08)^N12)/((1 + 0.08)^N12-1) | =S12*T12 |
| Discount factor | 1/(1+0.08)^n | ||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | ||||||
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