Question 17.14: Your company is looking at purchasing a new front-end loader...
Your company is looking at purchasing a new front-end loader and has narrowed the choice down to four loaders. The useful life of the loaders is five years. The
| Table 17-8 Cash Flow for Example 17-14 | ||||
| Cash Flow | Loader A ($) | Loader B ($) | Loader C ($) | Loader D ($) |
| Purchase Price | 110,000 | 127,000 | 120,000 | 130,000 |
| Annual Profit | 37,000 | 43,000 | 40,000 | 44,000 |
| Salvage Value | 10,000 | 13,000 | 12,000 | 13,000 |
purchase price, annual profit, and salvage value at the end of five years for each of the loaders is found in Table 17-8 .
Which front-end loader should your company purchase based on the incremental rate of return and a MARR of 20%?
The first step is to rank the alternatives in order of initial capital outlay. The loaders are compared in the following order: Loader A, Loader C, Loader B, and Loader D. Because Loader A has the lowest purchase price, it is designated the current best alternative.
Next, compare Loader A to Loader C. The difference in the purchase price is $10,000 ($120,000 – $110,000). The difference in annual profit is $3,000 ($40,000 – $37,000). The difference in salvage value is $2,000 ($12,000 – $10,000).
The present value of the difference in annual profits is determined by using Eq. (15-9) as follows:
P=A\left[\frac{(1+i)^{n}-1}{i(1+i)^{n}}\right] (15-9)
P_{ AP }=\$ 3,000\left[\frac{(1+i)^{5}-1}{i(1+i)^{5}}\right]
The present value of the difference in salvage values is determined by using Eq. (15-3) as follows:
P=\frac{F}{(1+i)^{n}} (15-3)
P_{ SV }=\frac{\$ 2,000}{(1+i)^{5}}
The incremental rate of return for purchasing Loader C in lieu of Loader A is calculated as follows:
NPV =\$ 3,000\left[\frac{(1+i)^{5}-1}{i(1+i)^{5}}\right]+\frac{\$ 2,000}{(1+i)^{5}}+(-\$ 10,000)
Setting the NPV to zero and solving by trial and error, we find the incremental rate of return equals 19.05%. Because the incremental rate of return is less than the MARR, Loader A continues to be the current best alternative.
Next, we compare Loader A to Loader B, the loader with the next lowest cost. The difference in the purchase price is $17,000 ($127,000 – $110,000). The difference in annual profit is $6,000 ($43,000 – $37,000). The difference in salvage value is $3,000 ($13,000 – $10,000).
The present value of the difference in annual profits is determined by using Eq. (15-9) as follows:
P_{ AP }=\$ 6,000\left[\frac{(1+i)^{5}-1}{i(1+i)^{5}}\right]
The present value of the difference in salvage values is determined by using Eq. (15-3) as follows:
P_{ SV }=\frac{\$ 3,000}{(1+i)^{5}}
The incremental rate of return for purchasing Loader B in lieu of Loader A is calculated as follows:
NPV =\$ 6,000\left[\frac{(1+i)^{5}-1}{i(1+i)^{5}}\right]+\frac{\$ 3,000}{(1+i)^{5}}+(-\$ 17,000)
Setting the NPV to zero and solving by trial and error, we find the incremental rate of return equals 25.32%. Because the incremental rate of return is greater than the MARR, Loader B is the new current best alternative and Loader A is eliminated from comparison.
Next, we compare Loader B to Loader D, the only remaining loader not compared. The difference in the purchase price is $3,000 ($130,000 – $127,000). The difference in annual profit is $1,000 ($44,000 – $43,000). The difference in salvage value is zero ($13,000 – $13,000).
The present value of the difference in annual profits is determined by using Eq. (15-9) as follows:
P_{ AP }=\$ 1,000\left[\frac{(1+i)^{5}-1}{i(1+i)^{5}}\right]
The incremental rate of return for purchasing Loader D in lieu of Loader B is calculated as follows:
NPV =\$ 1,000\left[\frac{(1+i)^{5}-1}{i(1+i)^{5}}\right]+\$ 0+(-\$ 3,000)
Setting the NPV to zero and solving by trial and error, we find the incremental rate of return equals 19.86%. Because the incremental rate of return is less than the MARR, Loader B continues to be the current best alternative. With no other alternative to compare, Loader B is the selected alternative; therefore, your company should purchase Loader B.