Question

A firm is considering three mutually exclusive alternatives as part of a production improvement program. The alternatives are:

	A	B	C
installed cost	$10,000	$15,000	$20,000
annual benefit	1,625	1,530	1,890
useful life (yrs)	10	20	20

The salvage value of each alternative is zero. At the end of 10 years, Alternative A could be replaced with another A with identical cost and benefits.

(a) which alternative should be selected if interest is 6%?

(b) 3%

(c) if there is a difference between parts (a) and (b) can you explain it?

Answer 1

As per the question Alternatives A,B and C are mutually exclusive

	Alternative A	Alternative B	Alternative C
Installed Cost (I)	$10000	$15000	$20000
Annual Benefit (A)	$1625	$1530	$1890
Useful life (N)	10	20	20

Salvage value (S) = 0

(A) As the question the rate of interest (i)= 6%

Though the lifecycle of A alternative is not equal with alternative B and C, so to convert the unequal lifecycle to equal lifecycle. The effective Life cycle of Alternative A has to repeat 1 times for estimating the net present worth (NPW)

Alternative A

NPW of Alternative A = 10000 + 10000(P/F,6%,10) + 1625(P/A,6%,20)

NPW of Alternative A = 10000 + 10000(0.5584) + 1625(11.4699)

NPW of Alternative A = 10000 + 5584 + 18638.59 = $34222.59

Alternative B

NPW of Alternative B = 15000 + 1530(P/A,6%,20) = 15000 + 1530(11.4699)

NPW of Alternative B = 15000 + 17548.95 = $32548.95

Alternative C

NPW of Alternative C = 20000 + 1890(P/A,6%,20) = 20000 + 1890(11.4699)

NPW of Alternative C = 20000 + 21678.11 = $41678.11 (Highest NPW)

Among three different alternatives, Alternative C should be selected for highest net present worth (NPW)

(B) As the question the rate of interest (i)= 3%

Though the lifecycle of A alternative is not equal with alternative B and C, so to convert the unequal lifecycle to equal lifecycle. The effective Life cycle of Alternative A has to repeat 1 times for estimating the net present worth (NPW)

Alternative A

NPW of Alternative A = 10000 + 10000(P/F,3%,10) + 1625(P/A,3%,20)

NPW of Alternative A = 10000 + 10000(0.7441) + 1625(14.8775)

NPW of Alternative A = 10000 + 7441 + 24175.94 = $41616.94

Alternative B

NPW of Alternative B = 15000 + 1530(P/A,3%,20) = 15000 + 1530(14.8775)

NPW of Alternative B = 15000 + 22762.57 = $37762.57

Alternative C

NPW of Alternative C = 20000 + 1890(P/A,3%,20) = 20000 + 1890(14.8775)

NPW of Alternative C = 20000 + 28118.47 = $48,118.47 (Highest NPW)

Among three different alternatives, Alternative C should be selected for highest net present worth (NPW)

(C) there is no difference between part (A) and part (B) for selecting the best alternative, because in both the part the net present worth of alternative C is highest, so alternative C has to be selected in the both part (A) and part (B).

	NPW of Alternative A	NPW of Alternative B	NPW of Alternative C
Rate of Interest (i) = 6%	$34222.59	$32548.95	$41678.11 (Highest)
Rate of Interest (i) = 3%	$41616.94	$37762.57	$48,118.47 (Highest)

Due to the change in the rate of interest the value of net present worth of alternatives changes which is given in the above table.