Question

problem: make a present - worth comparison of the equal - service machines for which the costs are shown below, if i = 10% per year.

	first cost, p	annual operating cost, aoc	salvage value, sv	life, years
type b	$3,500	700	350	5

Explanation:

Step1: Define present worth formula

The present worth (PW) formula for a machine is:
$$PW = -P - AOC \times \frac{(1+i)^n - 1}{i(1+i)^n} + SV \times \frac{1}{(1+i)^n}$$
where $P$ = first cost, $AOC$ = annual operating cost, $SV$ = salvage value, $i$ = interest rate, $n$ = life in years.

Step2: Calculate PW for Type A

Substitute $P=\$2500$, $AOC=\$900$, $SV=\$200$, $i=0.10$, $n=5$:
First, compute the present worth factor for AOC: $\frac{(1+0.10)^5 - 1}{0.10(1+0.10)^5} = \frac{1.61051 - 1}{0.10 \times 1.61051} = 3.79079$
Salvage value factor: $\frac{1}{(1+0.10)^5} = 0.62092$
$$PW_A = -2500 - 900 \times 3.79079 + 200 \times 0.62092$$
$$PW_A = -2500 - 3411.71 + 124.18$$

Step3: Calculate PW for Type B

Substitute $P=\$3500$, $AOC=\$700$, $SV=\$350$, $i=0.10$, $n=5$:
$$PW_B = -3500 - 700 \times 3.79079 + 350 \times 0.62092$$
$$PW_B = -3500 - 2653.55 + 217.32$$

Answer:

Present Worth of Type A: $\$-5787.53$
Present Worth of Type B: $\$-5936.23$
Since $PW_A > PW_B$, Type A is the more economical machine.