The following information refers to three mutually exclusive alternatives. The decision maker wishes to choose the right machine but is uncertain what MARR to use. Create a choice table that will help the decision maker to make the correct economic decision.

Question

Machine Z	Machine Y	Machine X
$4250	$7000	$2000	Initial cost
1000	1100	650	Uniform annual benefit
8	12	6	Useful life, in years

Accepted Answer

In this example, the lives of the three alternatives are different. As we saw in Chapter 6 and Examples 8–3 and 8–4, when the service period is expected to be continuous and the assumption of identical replacement reasonable, we can assume a series of replacements and compare annual worth values just as we did with present worth values.

We will make these assumptions in this instance and plot EUAW. This was done in a spreadsheet using the Excel function = benefit + PMT (interest rate, life, initial cost), and the result is Figure 8–5.

The EUAW graph shows that for low values of MARR, Y is the correct choice. Then, as MARR increases, Z and X become the preferred machines. If the “do nothing” alternative is available, then it becomes the best choice for higher MARR values.

In this case, it is first necessary to solve for the EUAC of the initial cost. This is subtracted from the annual benefit of each machine to calculate the EUAWs. In the top block of rows, each interest rate is set equal to the highlighted interest rate in cell B5. Copying the formula =$B$5 is easiest. In the lower block of rows, each interest rate is set equal to the highlighted interest rate in cell B13. Then two GOAL SEEKS are used to find the highlighted interest rates that make their respective differences in EUAWs equal to zero. Notice that Figure 8–5 has been used to determine that finding the incremental or breakeven rates between Y & Z, Z & X, and X & doing nothing is necessary. The breakeven rate between X and doing nothing equals [latex]IRR_{X}[/latex], and it is found using the RATE function.

	A	B	C	D	E	F	G	H	I	J
1	Machine	i	n	PMT	PV	FV	Solve for	EUAC	Ann. Benefit	Benefit - cost
2	X	3.1%	6		-2000	0	PMT	371	650	279
3	Y	3.1%	12		-7000	0	PMT	709	1100	391
4	Z	3.1%	8		-4250	0	PMT	609	1000	391
5		3.1%					Difference in EUAW for Y & Z			0
6	3.1%						Difference in EUAW for Z & X			112
7
8	Machine	i	n	PMT	PV	FV	Solve for	EUAC	Ann. Benefit	Benefit - cost
9	X	10.8%	6		-2000	0	PMT	470	650	180
10	Y	10.8%	12		-7000	0	PMT	1068	1100	32
11	Z	10.8%	8		-4250	0	PMT	820	1000	180
12		10.8%					Difference in EUAW for Y & Z			148
13	10.8%						Difference in EUAW for Z & X			0
14
15	Machine	i	n	PMT	PV	FV	Solve for	IRR
16	X		6	650	-2000	0	i	23.20%

If MARR ≥23.2% do nothing

If 23.2% ≥ MARR ≥ 10.8% choose X

If 10.8% ≥ MARR ≥ 3.1% choose Z

If 3.1% ≥ MARR ≥ 0% choose Y

If the “do nothing” alternative is not available, then the table reads

If MARR ≥ 10.8% choose X

If 10.8% ≥ MARR ≥ 3.1% choose Z

If 3.1% ≥ MARR ≥ 0% choose Y

The choice now is back in the decision maker’s hands. There is still the need to determine MARR, but if the uncertainty was, for example, that MARR was some value in the range 12 to 18%, it can be seen that for this problem it doesn’t matter. The answer is Machine X in any event. If, however, the uncertainty of MARR were in the 7 to 13% range, two things are clear. First, the decision maker would have to determine MARR with greater accuracy to maximize the EUAW. Second, there is little practical difference between the EUAWS for machines X and Y.

Question 8.5: The following information refers to three mutually exclusive...

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