Question

A company is considering several alternatives that are shown below. If the firm uses a 10% interest rate and uses the rate of return approach to make the selection, which alternative should be chosen?

	P	Q	R	S
Initial Cost	19,000	16,000	11,000	8,500
Annual Ben.	3,500	3,100	2,200	1,350
Life	10 yrs	10 yrs	10 yrs	10 yrs

Answer 1

Arranging alternative in increasing order of initial cost

S < R < Q < P

Incremental analysis between S & R

Incremental initial cost (R-S) = 11000 - 8500 = 2500

Incremental annual benefits (R-S) = 2200 - 1350 = 850

Let incremental IRR be i%, then

850 * (P/A,i%,10) = 2500

(P/A,i%,10) = 2500 / 850 = 2.941176

using trail and error method

When i = 31%, value of (P/A,i%,10) = 3.009073

When i = 32%, value of (P/A,i%,10) = 2.930414

using interpolation

i = 31% + (3.009073-2.941176) /(3.009073-2.930414)*(32%-31%)

i = 31% + 0.86% ~ 31.9%

As Incremental IRR > MARR, alternative R must be selected

Incremental analysis between R & Q

Incremental initial cost (Q-R) = 16000 - 11000 = 5000

Incremental annual benefits (Q-R) = 3100 - 2200 = 900

Let incremental IRR be i%, then

900 * (P/A,i%,10) = 5000

(P/A,i%,10) = 5000 / 900 = 5.555556

using trail and error method

When i = 12%, value of (P/A,i%,10) = 5.650223

When i = 13%, value of (P/A,i%,10) = 5.426243

using interpolation

i = 12% + (5.650223-5.555556) /(5.650223-5.426243)*(13%-12%)

i = 12% + 0.42% ~ 12.4%

As Incremental IRR > MARR, alternative Q must be selected

Incremental analysis between P & Q

Incremental initial cost (P-Q) = 19000 - 16000 = 3000

Incremental annual benefits (P-Q) = 3500 - 3100 = 400

Let incremental IRR be i%, then

400 * (P/A,i%,10) = 3000

(P/A,i%,10) = 3000 / 400 = 7.5

using trail and error method

When i = 5%, value of (P/A,i%,10) = 7.721735

When i = 6%, value of (P/A,i%,10) = 7.360087

using interpolation

i = 5% + (7.721735-7.5) /(7.721735- 7.360087)*(6%-5%)

i = 5% + 0.61% ~ 5.6%

As Incremental IRR < MARR, alternative Q must be selected

Finally Q alternative must be selected