# Given two alternatives: Data A B First Cost $4,000$6,000 Annual Cost $1,000$500 Annual...

## Question:

Given two alternatives:

Data A B
First Cost $4,000$6,000
Annual Cost $1,000$500
Annual Benefit $2,000$2,200
Life, Years 4 10
Salvage Value $3,000$1,000

Assuming that alternatives are replaced at the end of their useful life, determine the better alternative using annual cash flow analysis at an interest rate of 10%

## Annual cash flow analysis

Based on the infinite analysis period assumption, annual cash flow analysis is one of the major methods of resolving alternatives into comparable values. When initial cost P and salvage value S are given, the equivalent uniform annual cost (EUAC) can be calculated.

Answer:B is the better alternative based on annual cash flow analysis.

Solution:

There are three ways to compute equivalent uniform annual cost:

{eq}(1): EUAC=P(A / P, i, n)-S(A / F, i, n) {/eq}

{eq}(2): E U A C=(P-S)(A / F, i, n)+P i {/eq}

{eq}(3): E U A C=(P-S)(A / P, i, n)+Fi {/eq}

where:

(A / P, i, n) is to find A given P.

(A / F, i, n) is to find A given F.

In this case, let's apply way (2) to calculate EUAC for both alternatives.

The formula of (A / P, i, n) is as follow:

{eq}(A / P, i, n)=\left[\frac{i(1+i)^{n}}{(1+i)^{n}-1}\right] {/eq}

Plug in numbers:

For alternative A:

{eq}E U A C_A=(P-S)(A / P, i, n)+Fi=(P-S)\left[\frac{i(1+i)^{n}}{(1+i)^{n}-1}\right]+Fi=(4000-3000)\left[\frac{10\%(1+10\%)^{4}}{(1+10\%)^{4}-1}\right]+3000*10\%=$615.47 {/eq} {eq}E U A C_B=(P-S)(A / P, i, n)+Fi=(P-S)\left[\frac{i(1+i)^{n}}{(1+i)^{n}-1}\right]+Fi=(6000-1000)\left[\frac{10\%(1+10\%)^{10}}{(1+10\%)^{10}-1}\right]+1000*10\%=$913.73 {/eq}

Thus equivalent uniform annual worth (EUAW) for both alternatives are as follow:

{eq}E U A W_A=2000-1000-615.47=$384.53 {/eq} {eq}E U A W_B=2200-500-913.73=$786.27 {/eq}

{eq}E U A W_B>E U A W_A {/eq}

Thus, B is the better alternative based on annual cash flow analysis.