Aconsulting engineering firm is considering two models of suvs for the company principals. a gm model will have a first cost of $36,000, an operating cost of $4000, and a salvage value of $15,000 after 3 years. a ford model will have a first cost of $32,000, an operating cost of $4100, and also have a $15,000 resale value, but after 4 years. (a) at an interest rate of 12% per year, which model should the consulting firm buy? conduct an annual worth analysis. (b) what are the pw values for each vehicle?

Part (a): Annual worthy analysis
Annual worthy = -Annuities  of PV (cost) - Annual expenses + Annuities of FV (salvage cost)

Then,
For GM model:
Annual worthy = -36,000(A|P,12%,3) -4000 + 15,000(A|F,12%,3)
Where,
(A|P,12%,3) = {[1-(1+0.12)^-3]/0.12}^-1 = 0.41635;
(A|F,12%,3) = {[(1+0.12)^3-1]/0.12}^-1 = 0.29635
Therefore,
Annual worthy for GM model = -36,000(0.41635) - 4,000 + 15,000(0.29635) = -$50,543.35

For Ford model:
Annual worthy = -32,000(A|P,12%,4) - 4,100 + 15,000(A|F,12%,4)
Where,
(A|P,12%,4) = {[1-(1+0.12)^-4]/0.12}^-1 = 0.32923
(A|F,12%,2) = {[(1+0.12)^4-1]/0.12}^-1 = 0.20923
Therefore,
Annual worthy of Ford model = -32,000(0.32923) - 4,100 + 15,000 (0.20923) = -$11,496.91

It can be noted that Annual worth of Ford > Annual worthy of GM model and thus the best option is the Ford model

Part (b): Present Worth (PW)
PW of GM model = -36,000 - 4,000(P/A,12%,3) + 15,000(P/F,12%,3)
Where
(P/A,12%,3) = [1-(1+0.12)^-3]/0.12 = 2.40183
(P/F,12%,3) = 1/(1+0.12)^3 = 0.71178
Therefore,
PW of GM model = -36,000 - 4,000(2.40183) + 15,000 (0.71178) = -$34,930.62

PW of Ford model = -32,000 - 4,100(P/A,12%,4) + 15,000(P/F,12%,4)
Where,
(P/A,12%,4) = [1-(1+0.12)^-4]/0.12 = 3.03735
(P/F,12%,4) = 1/(1+0.12)^4 = 0.63552
Therefore,
PW of Ford model = -32,000 - 4,100(3.03735) + 15,000 (0.63552) = -$10,014.065

Based on PW, PW of Ford model > PW for GM model and thus Ford model is the better option.

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