Question
In: Economics
A large standby electricity generator in a hospital operating room has a first cost of $69,250...
A large standby electricity generator in a hospital operating room has a first cost of $69,250 and may be used for a maximum of 6 years. Its salvage value, which decreases by 15% per year, is described by the equation S = 69,250(1 − 0.15)n, where n is the number of years after purchase. The operating cost of the generator will be constant at $6,750 per year, and the interest rate is 12% per year. Determine the economic service life and associated AW values.
The economic service life is 6 years and the AW value is $−______ .
Solutions
Expert Solution
Using Excel for ESL analysis
| Year | Discount factor | O&M cost | PV (O&M) | Cumulative (O&M) | Cumulative (O&M) + Initial Cost | Salvage value | PV (Salvage value) | NPV | (A/P,15%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+69250 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | 0.86957 | 6750 | 5870 | 5870 | 75120 | 58863 | 51185 | 23935 | 1.150000 | 27525.00 |
| 2 | 0.75614 | 6750 | 5104 | 10974 | 80224 | 50033 | 37832 | 42391 | 0.615116 | 26075.58 |
| 3 | 0.65752 | 6750 | 4438 | 15412 | 84662 | 42528 | 27963 | 56699 | 0.437977 | 24832.78 |
| 4 | 0.57175 | 6750 | 3859 | 19271 | 88521 | 36149 | 20668 | 67853 | 0.350265 | 23766.50 |
| 5 | 0.49718 | 6750 | 3356 | 22627 | 91877 | 30727 | 15277 | 76600 | 0.298316 | 22851.12 |
| 6 | 0.43233 | 6750 | 2918 | 25545 | 94795 | 26118 | 11291 | 83504 | 0.264237 | 22064.81 |
| Discount factor | 1/(1+0.15)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
As minimum EUAC is in 6th yr, therefore
ESL = 6 yrs
AW = -22064.81 ~ -22065
Showing formula in excel
| Year | Discount factor | O&M cost | PV (O&M) | Cumulative (O&M) | Cumulative (O&M) + Initial Cost | Salvage value | PV (Salvage value) | NPV | (A/P,15%,n) | EUAC |
| A | B | C | D=C*B | E | F=E+69250 | G | H=G*B | I=F-H | J | K = I*J |
| 1 | =1/(1.15)^A3 | 6750 | =C3*B3 | =D3 | =69250+E3 | =69250*(1-0.15)^A3 | =G3*B3 | =F3-H3 | =0.15*((1 + 0.15)^A3)/((1 + 0.15)^A3-1) | =I3*J3 |
| 2 | =1/(1.15)^A4 | 6750 | =C4*B4 | =E3+D4 | =69250+E4 | =69250*(1-0.15)^A4 | =G4*B4 | =F4-H4 | =0.15*((1 + 0.15)^A4)/((1 + 0.15)^A4-1) | =I4*J4 |
| 3 | =1/(1.15)^A5 | 6750 | =C5*B5 | =E4+D5 | =69250+E5 | =69250*(1-0.15)^A5 | =G5*B5 | =F5-H5 | =0.15*((1 + 0.15)^A5)/((1 + 0.15)^A5-1) | =I5*J5 |
| 4 | =1/(1.15)^A6 | 6750 | =C6*B6 | =E5+D6 | =69250+E6 | =69250*(1-0.15)^A6 | =G6*B6 | =F6-H6 | =0.15*((1 + 0.15)^A6)/((1 + 0.15)^A6-1) | =I6*J6 |
| 5 | =1/(1.15)^A7 | 6750 | =C7*B7 | =E6+D7 | =69250+E7 | =69250*(1-0.15)^A7 | =G7*B7 | =F7-H7 | =0.15*((1 + 0.15)^A7)/((1 + 0.15)^A7-1) | =I7*J7 |
| 6 | =1/(1.15)^A8 | 6750 | =C8*B8 | =E7+D8 | =69250+E8 | =69250*(1-0.15)^A8 | =G8*B8 | =F8-H8 | =0.15*((1 + 0.15)^A8)/((1 + 0.15)^A8-1) | =I8*J8 |
| Discount factor | 1/(1+0.15)^n | |||||||||
| (A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) | |||||||||
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