Question

Either of the cost alternatives shown below can be used in a chemical refining process. If the company’s MARR is 15% per year, determine which should be selected on the basis of an incremental ROR analysis.

	A	B
First cost ,$	− 40,000	− 61,000
Annual cost, $/year	− 25,000	− 19,000
Salvage value, $	8,000	11,000
Life, years 5	5	5

-       A.       B.       C.       D.       E.       F.	The incremental rate of return computed using a present worth analysis is nearest to is the incremental rate of return to be found
-       A.       B.       C.       D.       E.       F.	Which alternative is to be chosen?

A.	15%
B.	17%
C.	Alternative A
D.	16%
E.	14%
F.	Alternative B

Answer 1

MARR = 15%

Incremental cost (B-A) = 61000 - 40000 = 21000

Annual saving = -19000 + 25000 = 6000

Incremental salvage value = 11000 - 8000 = 3000

t = 5 yrs

Let incremental rate of return be i%, then PW at i% =0

PW = -21000 + 6000*(P/A,i%,5) + 3000*(P/F,i%,5) = 0

6000*(P/A,i%,5) + 3000*(P/F,i%,5) = 21000

dividing by 3000

2*(P/A,i%,5) + (P/F,i%,5) = 7

Using trail and error method

When i = 15%, 2*(P/A,i%,5) + (P/F,i%,5) = 7.20149

When i = 16%, 2*(P/A,i%,5) + (P/F,i%,5) = 7.02470

When i = 17%, 2*(P/A,i%,5) + (P/F,i%,5) = 6.85480

using interpolation

i = 16% + (7.02470-7) / (7.02470-6.85480) *(17-16)

i = 16% + 0.14537

i = 16.15% (Approx) (option D)

As Incremental IRR is greater than MARR, Project B should be selected (option F)