Question

You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for manufacturing high-pressure hydraulic hoses. The parameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an interest rate of 15% per year? Why is yours the correct choice?

Alternative X Y
First Cost $-45,000 $-58,000
Maintenance cost, per Year $-8000 $-4000
Salvage Value $2,000 $12,000
Life 5 years 5 years

The present worth of alternative X is( ______$ )and that of alternative Y is(_______ $ ).

Answer 1

Present value of Project X is $41990.6

Present value of Project Y is $34605.2

Step-by-step explanation:

PROJECT X

PARTICULARS YEAR Cost/Value Present Value factor 15% Present Value

Initial Cost 0 45000 1 45000

Maintenance 1- 5 8000 3.352 26816

cost

Annual Depreciation 1-5 (8600) 3.352 (28827.2)

Salvage Value 5 (2000) 0.4971 (994.2)

Present value of cash outflows 41,990.6

PROJECT Y

PARTICULARS YEAR COST/VALUE Present value factor 15% PRESENT VALUE

Initial Cost 0 58000 1 58000

Maintenance 1- 5 4000 3.352 13409

cost

Annual Depreciation 1-5 (9200) 3.352 (30838.4)

Salvage Value 5 12000 0.4971 (5965.2)

Present Value of cash outflow 34605.2

Note: Figures in parenthesis denote cash inflow

Working Notes

Depreciation for project X =
`(45000\ -\ 2000)/(5)` = $8600 p.a

Depreciation for project Y =
`(58000\ -\ 12000)/(5)` = $9200 p.a

Decision: Since present value of cash outflows is lesser for Project Y, it should be taken up.