Question

A tractor has a first cost of $40,000, a monthly op- erating cost of $1500, and a salvage value of $12,000 in 10 years. The MARR is 12% per year. An identical tractor can be rented for $3200 per month (operating cost not included). If n is the minimum number of months per year the tractor must be used in order to justify its purchase, the relation to find n is represented by:

Determine the number of month for breakeven

(a) -?40,000(A?P,1%,10) ?- 1500n ?+ 12,000(A?F,1%,10) =? 3200n
(b) -?40,000(A?P,12%,10) -? 1500n ?+ 12,000(A?F,12%,10) ?= 3200n
(c) ?-40,000(A?P,1%,120) ?- 1500n +? 12,000(A?F,1%,120) =? 3200n
(d) ?-40,000(A?P,11.4%,10) ?- 1500n ?+ 12,000(A?F,11.4%,10) ?= 3200n

Answer 1

C) -40,000(A/P, 1%,120) - 1500n + 12,000(A/F, 1%, 120) = 3200n

Since its monthly operating cost is 1500, we need the MARR in monthly format. Thus MARR = 12/12 = 1%

first cost = -40,000(A/P, 1%, 120)

Operating cost = Operating Cost*Number of months to use = -1500n

salvage value = 12,000(A/F, 1%,120)

Total annual worth = -40,000(A/P, 1%, 120) -1500n +12,000(A/F, 1%, 120)

Annual worth of rented cost = -3200 * number of months = -3200n

Total annual worth must be equal to annual worth of rented cost =

-40,000(A/P, 1%, 120) -1500n +12,000(A/F, 1%, 120) = 3200n