Question

In comparing alternatives, I and J by the present worth method, the equation that yields the present worth of alternative Jis Alternativel Alternative Initial cost, -150,000 - 250.000 Annual Income, $ per year 20,000 40,000 Annual expenses, $ per year --9,000 -14,000 Salvage value, $ 25,000 35,000 Life, years The interest rate is 15% per year. (PWJ--250,000 + 40,000(PIA ,15%.6) + 35,000(P/F.15%,6 (PWJ --250,000 + 26,000(PIA ,15%,6) +35,000(P/F.15%,6 (PWJ - -250,000 - 26,000(PLA 15%,6) +35,000(P/F ,15%,6 (PW J--250,000 - 26,000(PIA 15%,6) - 35,000(P/F ,15%,6

Answer 1

We must know the notations first :-

(P/A i,n) is Present Value factor of constant annual future payments.

(P/F i,n) is Present value factor of a single future payment.

Here, i = 15% and n = 6 for J

We can see from above table for Alternative J :-

Initial cost = -250000

• As it's paid in first year so the present value is same as the initial cost.

Annual Income = 40,000

• As it will be paid every year so the present value will be = 40000(P/A 15%,6)

​​​​​​​Annual Expenses = -14000

• As this cost will be incurred every year so the present value = -16000(P/A 15%,6)

​​​​​​​Salvage value = 35000

• As the Salvage value is the residual final value of good at the end of the term. So it's like a single resale value. Present value = 35000(P/F 15%,6)

​​​​​​​Sum of all present values will give Present Worth (PW) of J :-

PW​​​J = -250,000 + 40,000(P/A 15%,6) - 14,000(P/A 15%,6) + 35,000(P/F 15%,6)

PW​​​J = -250,000 + (40000-14000)(P/A 15%,6)+ 35,000(P/F 15%,6)

PW​​​J = -250,000 + 26,000(P/A 15%,6)+ 35,000(P/F 15%,6)

So, Option B ​​​​​​​is correct