Example -9 (Using data of Example-4)

A material testing laboratory has two alternatives for purchasing a compression testing machine which will be used for determining the compressive strength of different construction materials. The alternatives are from two different manufacturing companies. The cash flow details of the alternatives are as follows;

Alternative-1: Initial purchase price = Rs.1000000, Annual operating cost = Rs.10000, Expected annual income to be generated from testing of different construction materials = Rs.175000, Expected salvage value = Rs.200000, Useful life = 10 years.

Alternative-2: Initial purchase price = Rs.700000, Annual operating cost = Rs.15000, Expected annual income to be generated from testing of different construction materials = Rs.165000, Expected salvage value = Rs.250000, Useful life = 5 years.

Find out the most economical alternative at interest rate of 10% per year by future worth method.

Solution:

As the alternatives have different life spans i.e. 10 years and 5 years, the comparison will be made over a time period equal to the least common multiple of the life spans of the alternatives i.e. 10 years. Thus the cash flow of Alternative-1 is analyzed for one cycle (duration of 10 years) whereas that of cash flow of Alternative-2 is analyzed for two cycles of duration 5 years each (already mentioned in Example-4).

The cash flow diagram of Alternative-1 is shown here again for ready reference.

Fig. 2.11 Cash flow diagram of Alternative-1 (shown for ready reference)

The equivalent future worth FW 1 (in Rs.) of Alternative-1 is determined as follows;

Putting the values of different compound interest factors in the above expression;

FW₁ = Rs.235971