Question

Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Investment Operating Costs $4,000 at end of year 1 and increasing by $400 per year thereafter $4,000 every 5 years 20 years $8,000 at end of year 1 and increasing by $800 per year thereafter None Overhaul Costs Life 10 years Salvage Value $8,000 if just overhauled negligible Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year The PW of Alternative A is $ 105606. (Round to the nearest dollar.) The PW of Alternative B is S. (Round to the nearest dollar.)

Answer 1

Initial capital investment is given and the annual cost is increasing every year.
Alternative A has overhaul cost every 5 years which must be added in the 5th, 10th, 15th and 20th year.
It also has a salvage value which needs to be subtracted.

Alternative B has a life of 10 years but it is assumed to be repeated so the initial capital investment is added in the 11th year.

	Alt-A		Alt-B
Year	Cash Flow	PW @ 10%	Cash Flow	PW @ 10%
0	45000	45000	15000	15000
1	4000	3636.36	8000	7272.73
2	4400	3636.36	8800	7272.73
3	4800	3606.31	9600	7212.62
4	5200	3551.67	10400	7103.34
5	9600	5960.84	11200	6954.32
6	6000	3386.84	12000	6773.69
7	6400	3284.21	12800	6568.42
8	6800	3172.25	13600	6344.50
9	7200	3053.50	14400	6107.01
10	11600	4472.30	15200	5860.26
11	8000	2803.95	31000	10865.31
12	8400	2676.50	16800	5353.00
13	8800	2549.05	17600	5098.09
14	9200	2422.65	18400	4845.30
15	13600	3255.73	19200	4596.33
16	10000	2176.29	20000	4352.58
17	10400	2057.58	20800	4115.17
18	10800	1942.47	21600	3884.95
19	11200	1831.29	22400	3662.58
20	7600	1129.69	23200	3448.53
		105605.87		132691.45

PW of A = -105606

PW of B = -132691