When comparing two or more alternatives by the rate of return method, the first step is to classify them as either mutually exclusive or independent, and second as either revenue or service-only alternatives, because their classification determines what they are to be compared against. There are only three types of alternatives based on these classifications as shown in Table 8.1. The way they are to be evaluated is also shown in the Table.
All alternatives will be considered mutually exclusive unless indicated otherwise. Independent alternatives are compared only against the do-nothing alternative wherein all alternatives which have a rate of return that exceeds the MARR are selected. For mutually exclusive alternatives, the do-nothing is a viable option when revenue alternatives are involved. This is essentially what was done in the previous chapter when there was only one alternative involved and its rate of return was determined. When there is more than one alternative involved, the alternatives must be compared against each other on an incremental basis as described below. An incremental analysis refers to an economic analysis of the difference in cash flow between two alternatives. Such an analysis is based on the fact that if the extra investment required in the alternative which has the higher initial investment does not earn at least the minimum attractive rate of return, then that increment of investment should not be made because that increment of money could be better invested elsewhere (where it would earn at least the MARR). The procedure for comparing mutually exclusive alternatives can be summarized as follows:
(1 (1) Rank the alternatives in terms of increasing initial investment cost. If the alternatives are revenue alternatives, then do-nothing is added and is the first alternative.
(2) Identify the first two alternatives as A and B
(3) Tabulate the difference in cash flow between the first two alternatives (i.e. the incremental cash flow) by subtracting the cash flow for alternative A from the cash flow for alternative B (i.e. B-A) over their least common multiple of lives.
(4) Find the rate of return on the incremental cash flow. If i* is ³ MARR, eliminate A, or vice versa.
(5) Compare the survivor with the next-in-line alternative per steps (3) and (4) above.
Continue steps (3) thru (5) until only one alternative remains
Select the best alternative from the four mutually exclusive alternatives shown below, using an MARR of 12% per year.
First rank the alternatives in terms of increasing initial investment, with do-nothing (DN) as the first one (because the alternatives are revenue alternatives): DN, 4, 2, 1, 3
Next, tabulate the difference in cash flow between the first two alternatives, DN and 4
Now, find the rate of return on the difference column:
0 = -20,000 + 8,000 (P/A, i, 4)
(P/A, i, 4) = 2.5000
i is between 20% and 22%
Since i > 12%, eliminate DN and tabulate the incremental cash flow between the survivor (i.e. alt 4) and the next alternative (i.e. 2)
Find the rate of return on the difference column:
0 = -30,000 + 9000 (P/A, i 4)
(P/A, i, 4) = 3.3333
i is between 7% and 8%
Since i < 12%, eliminate 2
Compare survivor (i.e. 4) against next alternative (i.e. 1): (The tabulation of cash flow is left to the student) The rate of return equation is:
0 = -60,000 + 25,000 (P/A, i, 4)
(P/A, i, 4) = 2.4000
i is between 24% and 25%
Since i > 12%, eliminate 4
Finally, tabulate the cash flow and compare the survivor, 1, with 3 over their least common multiple of years (i.e. 8)
Set-up and solve the rate of return equation on the difference column:
0 = -65,000 + 2000 (P/A, i, 8) + 80,000 (P/F, i, 4)
Solve for i by trial and error:
i = 10.1%, eliminate 3
Alternative 1 is the only one that remains. Therefore, it is the best one.
The rate of return equations that were set up in the previous example were set up on the basis of present worth (i.e. all cash flows were converted into present values). However, the cash flows could also have been set up in terms of their annual worth or future worth and the same i* values would have been obtained.