Question 10.P.9: It is necessary to pump twice as much water as can be handle......
It is necessary to pump twice as much water as can be handled by the existing small pump, which is now 5 years old. This pump can be sold now for $1200 or kept for 5 years, after which it will have zero salvage value. Operating expenses are $3000 per year. If the pump is kept, a similar one must be purchased for $3500, with operating costs of $2500 per year; its salvage values after 5 and 10 years are the same as for the original pump. A large pump, equal in capacity to the two small pumps, costs $6000, with operating expenses of $4500 per year. New machines have economic lives of 10 years, and zero salvage values at that date. Analyze the situation, if the MARR is 10%.
There are 3 possible alternatives: (1) to replace the present pump by a new, large pump; (2) to buy a new small pump now and every fifth year hereafter; (3) to buy a new small pump now, and after 5 years to sell it and install a single large pump. We calculate the present-worth costs for a 10-year study period.
plan 1 PW_1 = $6000 + $4500(P/A, 10%, 10) = $33 651
plan 2 PW_2 = $1200 + $3000(P/A, 10%, 5) + $3500 + $2500(P/A, 10%, 10)
+ [$3500(A/P, 10%, 10) + $25001 (P/A, 10%, 5) (P/F, 10%, 5)
= $38 658
plan 3 PW_3 = $1200 + $3000(P/A, 10%, 5)+ $3500 + $2500(P/A, 10%, 5)- $1200(P/F, 10%, 5)
+ [$6000(A/P, 10%, 10) + $45001 (P/A, 10%, 5) (P/F, 10%, 5)
= $37 694
Note, in the calculations for plans 2 and 3, how certain costs are spread out over ten years and then are partially brought back into the study period.
The conclusion is that plan 1 is best, with plan 3 being slightly superior to plan 2.