In: Economics
Compare the 2 sytems at MARR = 10% per year by using tabulated factors. What system costs less over the span of 8 years.
| Pull System | Push System | |
| Robot and Support equipement first cost, ($) | -1,500,000 | -2,250,000 | 
| Annual M&O cost, ($ per year) | -700,000 | -600,000 | 
| Rebuild cost in year 3, ($) | 0 | -500,000 | 
| Salvage Value, ($) | 100,000 | 50,000 | 
| Estimated life, (years) | 8 | 8 | 
| Alternative Pull System will cost less: | ||||
| Explanation as follows: | ||||
| Alternative-Pull | ||||
| Year | Cash flows | PVF @ 10% | Present value | |
| 0 | 1500000 | 1 | 1500000 | |
| 1 | 700000 | 0.909091 | 636363.6 | |
| 2 | 700000 | 0.826446 | 578512.4 | |
| 3 | 700000 | 0.751315 | 525920.4 | |
| 4 | 700000 | 0.683013 | 478109.4 | |
| 5 | 700000 | 0.620921 | 434644.9 | |
| 6 | 700000 | 0.564474 | 395131.8 | |
| 7 | 700000 | 0.513158 | 359210.7 | |
| 8 | 600000 | 0.466507 | 279904.4 | |
| Present value of cash outflows | 5187798 | |||
| Note: salvage value is deducted from Year-8 | ||||
| Alternative-Push | ||||
| Year | Cash flows | PVF @ 10% | Present value | |
| 0 | 2250000 | 1 | 2250000 | |
| 1 | 600000 | 0.909091 | 545454.5 | |
| 2 | 600000 | 0.826446 | 495867.8 | |
| 3 | 1100000 | 0.751315 | 826446.3 | |
| 4 | 600000 | 0.683013 | 409808.1 | |
| 5 | 600000 | 0.620921 | 372552.8 | |
| 6 | 600000 | 0.564474 | 338684.4 | |
| 7 | 600000 | 0.513158 | 307894.9 | |
| 8 | 550000 | 0.466507 | 256579.1 | |
| Present value of cash outflows | 5803288 | |||
| Note: Salvage value reduced from Year-8 | ||||
| Note: Rebuild cost is added in Year-3 | ||||