Question

You can use 2 methods to make a certain part. Method 1 has an initial cost of $ 20,000 and a residual value of $ 5,000 after 3; years, that is, its lifespan. Annual operating costs amount to $ 8,500 / year. Method 2 initially costs $ 15,000 and has a lifespan of 2 years. Its residual value is $ 3,000 and the annual operating costs are $ 7,000 / year. Using a 2-year planning horizon and the residual value method implied to find F2, choose the method there that is less expensive using the equivalent annual value (A) at a rate of 15%. ps: Show all your calculations to arrive at the results and make the cash flow diagrams

Answer 1

Method 1:

i =15%, n = 3 years, annual operating costs = 8500, first cost = 20,000, salvage value = 5000

So, EUAW (Equivalent Uniform Annual Worth) = First cost(A/P, i, n) + annual costs - salvage value(A/F, i,n)

The values of A/P and A/F are determined from compound interest tables for a given value of i and n. A/P is the capital recovery factor and A/F is the sinking fund factor.

Take a look at the cash flow diagram in fig 1: (downward arrowes imply cash outflow and upward arrows imply inflows)

So, EUAW = 20,000 * 0.4380 + 8500 - 5000 * 0.2880 = 8760 + 8500 - 1440 = $15,820

Now, for method 2:

i =15%, n = 2 years, annual operating costs = 7000, first cost = 15,000, salvage value = 3000

So, EUAW (Equivalent Uniform Annual Worth) = First cost(A/P, i, n) + annual costs - salvage value(A/F, i,n)

Take a look at the cash flow diagram in fig 2:

So, EUAW = 15,000 *0.6151  + 7000 - 3000 * 0.4651 = 9226.5 + 7000 - 1395.3 = $14,831.2

Now, as we have taken costs to be positive and income to be negative, explains why the first cost and annual operatinng costs are with positive sign and salvage value is negative, so the lesser EUAW is better.

In this case, $14,831.2<15820, and so method 2 is less expensive as cost incurred is lesser.