Question

4. (15 points) Which option should be selected based on a present worth comparison at an interest rate of 10%? First cost, $ Annual operating & maintenance cost, $/year Salvage value, $ Life, in years Option X -230,000 -9,000 12,000 Option Y -380,000 -12,000 140,000 6 a. Draw the cash flow diagram for Option X. b. Calculate the present worth for Option X. c. Which option should be selected? Why?

Answer 1

	Option X	Option Y
First Cost	-230,000	-380,000
Annual O&M Cost	-9,000	-12,000
Salvage Value	12,000	140,000
Life	3 years	6 years

Interest Rate = 10%

The life of the alternatives is not equal. Hence, we need to apply the common multiple method and convert the unequal life into equal life. The LCM of 3 and 6 years is 6 years. Hence, the common life will be 6 years and the Option X is to be repeated two times.

a. Cash Flow Diagram of Option X

12000 12000 9000 9000 230000 230000

b. PW of Option X

PW = -230,000 – 230,000 (P/F, 10%, 3) – 9,000 (P/A, 10%, 6) + 12,000 (P/F, 10%, 3) + 12,000 (P/F, 10%, 6)

PW = -230,000 – 230,000 (0.75131) – 9,000 (4.35526) + 12,000 (0.75131) + 12,000 (0.56447)

PW = -426,209.28

c. Which option should be selected?

For this we need to calculate PW of Option Y

PW = -380,000 – 12,000 (P/A, 10%, 6) + 140,000 (P/F, 10%, 6)

PW = -380,000 – 12,000 (4.35526) + 140,000 (0.56447) = 353,237.32

On the basis of PW method, Option Y is to be selected because of less cost.

Select – Option Y