Julia must choose between two different designs for a safety closure, which will be in use indefinitely. Model A has a life of three years, a first cost of $8000, and maintenance of $1000 per year. Model B will last four years, has a first cost of $10,000, and has maintenance of $800 per year. A salvage value can be estimated for Model A using a depreciation rate of 40% and declining-balance depreciation, while a salvage value for Model B can be estimated using straight-line depreciation and the knowledge that after one year its salvage value will be $7500. Interest is at 14%. Using a present worth analysis, which design is better?
Depreciation rate= 40%
Year Depreciation WDV
1 3,200 =40.%*8000 4,800 =8000-3200
2 1,920 =40.%*4800 2,880 =4800-1920
3 1,152 =40.%*2880 1,728 =2880-1152
Salvage value at the end of 3 years= 1,728
Cash flow for Model A
Year Cost Maintenance Salvage Total
0 -8,000 -8,000
1 -1,000 -1,000
2 -1,000 -1,000
3 -1,000 1,728 728
Cash inflow is positive, cash outflow is negative
Let us find the Present value (PV) of cash flows
Discount rate used is interest rate = 14%
Year Cash flow Discount factor @ Discounted cash ...
Using the present worth analysis of two different designs for a safety closure, which will be in use indefinitely, the better design is identified.