### Comparison by Rate of Return Answers

1.

The information is not sufficient. You must do an incremental analysis.

1) Find cash flow of Project B—A over a common 10 year analysis period.

Initial Investment \$5,000

Annual Benefit 450

Salvage Value 500

2) Express incremental cash flow in the form of equations or using compound interest factors.

Set the NPV of the incremental cash flows equal to 0.

NPV = 0 = (-5,000) + 450(P/A, i%, 10) + 500(P/F, i%, 10)

Since the total cash out equals the total cash in, this ROR for the incremental investment is obviously 0%. Reject B-A and select A.

2.

Call the three sizes ranked in order of investment A, B, and C.

A alone:

5000(A/P, i, 5) = 25,000¥0.05 = 1250

(A/P, i, 5) = 0.25

7% < i < 8%. Therefore, reject A.

B alone:

10,000(A/P, i, 5) = 75,000¥0.05 = 3750

(A/P, i, 5) = 0.375

25% < i < 35%.. Therefore, accept B.

Look at the extra investment of C over B

(18,000 - 10,000)(A/P, i, 5) = 150,000¥0.04 - 75,000¥0.05 = 6000 - 3750 = 2250

(A/P, i, 5) = 0.28125

12% < i < 15%. Therefore, reject C => B is best choice.

3.

Rank the alternatives in order of investment: C, B, A

The incumbent is C, the challenger is B.

Extra investment B over C is 100 million, the savings per year is 8 million.

The ROR of the extra investment is

100(A/P, i, 20) = 8 or (A/P, i, 20) = 0.08

This is about 5%. Reject extra investment of B over C.

The incumbent is C, the challenger is A

Extra investment B over C is 150 million, the savings per year is 15 million.

The ROR of the extra investment is

150(A/P, i, 20) = 15 or (A/P, i, 20) = 0.10

This is about 8%. Reject extra investment of A over C.

The best selection is the smallest investment, that is the natural gas plant.

4a.

The investment yields exactly the MARR. The woman should accept the investment.

4b.

You must do an incremental analysis. A, with the smallest investment, is acceptable. Comparing B to A, we note that the ROR of the extra investment of B over A yields 10.5%. This is acceptable so A is rejected. Comparing C to B we note that the ROR of the extra investment of C over B is less than the MARR, so reject C. The remaining incumbent is B, so accept B.