An electric switch manufacturing company is try­ing to decide between three different assembly methods. Method A has an estimated first cost of $40,000, an annual operating cost (AOC) of $9000, and a service life of 2 years. Method B will cost $80,000 to buy and will have an AOC of $6000 over its 4-year service life. Method C costs $130,000 initially with an AOC of $4000 over its 8-year life. Methods A and B will have no salvage value, but Method C will have equipment worth 10% of its first cost. Perform both (a) future worth, and (b) present worth analyses to select the method at i = 10% per year.

Question

What will be an ideal response?

Answer 1

(a) FWA = -40,000[(F/P,10%,8) + (F/P,10%,6) + (F/P,10%,4) + (F/P,10%,2)]
- 9000(F/A,10%,8)
= -40,000[(2.1436) + (1.7716) + (1.4641) + (1.2100)] – 9000(11.4359)
= $-366,495

FWB = -80,000[(F/P,10%,8) + (F/P,10%,4)] - 6000(F/A,10%,8)
= -80,000[(2.1436) + (1.4641)] - 6000(11.4359)
= $-357,231

FWC = -130,000(F/P,10%,8) - 4000(F/A,10%,8) + 13,000
= -130,000(2.1436) - 4000(11.4359) + 13,000
= $-311,412

Select method C
(b) Find the PW values from the FW values with n = 8 years. Select method C

PWA = FWA(P/F,10%,8) = -366,495(0.4665) = $-170,970

PWB = FWB(P/F,10%,8) = $-166,648

PWC = FWC(P/F,10%,8) = $-145,274