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( $ ).

Present value of Project X is $41990.6

Present value of Project Y is $34605.2

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  =  = $8600 p.a

Depreciation for project Y =  = $9200 p.a

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

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