Question

A small construction project having a useful life of 5 years has five mutually exclusive alternatives. With an MARR of 6% and using incremental IRR which alternative should be selected? Note: First solve for the IRR for each alternative. You may use a spreadsheet to iterate to solve, but solve for the IRR for Alternative II 'by hand' by iterating between the interest rate tables in the back of the book. Alternatives III $400 $90 Initial Cost Uniform Annual Benefit $800 $210 $200 $60 IV $1000 $250 V $300 $60

Answer 1

In the question it is being asked to solve alternative II maunally thus i am soling it first manually.

The net present worth of the alternative can be mathematically written as

NPW11 = -200+ 60(P/A, 1%,5)

We can determine the IRR using trial and error method. Assume rate = 15%

N PWH = -200 + 60(PA, 15%.5)

NPW11 = -200 + 60 x 1-1.15-5 0.15

NPW111 = –200+60 x 3.352155

NPW111 = –200+ 201.129305

NPW111 = $ 1.129305

At IRR the NPW =0, at 15% the NPW is 1.129305 therefore, we have to increase the rate. Assume i = 16%

N PW111 = –200+ 60(P/A, 16%,5)

NPW116 = -200 + 60 x 1-1.16-5 0.16

On solving above equation we get

NPW1116% = -$ 3.54238

Now we can determine the IRR using linear interpolation.

NPW115% IRRJ = 15+ (16 – 15) * NPW - N PW1116%

1.129305 IRRI = 15 + 1129305 - (-3.54238)

S IRR = 15+ 0.24173

IRR (alternative II)= 15.24%

Refer the attached picture of excel screen shot for the IRR

Year V -800 -200 -400 -300 IV -1000 250 250 60 210 210 60 60 60 90 90 2 | 60 210 90 250 60 210 60 90 250 60 210 60 15.24% 90 4.06% 250 | 7.93% 60 0.00% IRR 9.81% |

As we can see some of the alternatives have IRR greater than MARR. Therefore, reject the do nothing option.

First select the alternative with the lowest initial investment as the base alternative then determine the incremental cash flow by subtracting the cash flow of base alternative from the next costly alternative.

Here alternative II is cheapest and Alternative V is second cheapest. But as we can see the IRR of alternative V is 0.00%. Thus, reject alternative V.

Now compare the next costlier alternative. That is alternative III. Subrat alternative II from III, we get

​​​​​​

|- || -200 30 90 30 -400 -200 90 60 60 90 60 60 90 60 4.06% | 15.24% 30 30 90 30 -8.88%

From above picture we can say if the IRR of any alternative is less than MARR, then reject the alternative.

Now accept alternative II and reject IIi.

Again the same process refer the attached picture

| 60 -800 -200 210 210 60 210 60 210 60 | 210 | 60 9.81% | 15.24% | |- || -600 150 | 150 150 150 150 7.93% | |

The incremental IRR greater than marr therefore, accept alternative I and reject II.

Now compare alternative I with alternative IV.

IV. -1000 250 250 250 250 250 7.93% -800 210 210 210 |- || -200 40 40 40 210 40 210 9.81% 40 | 0.00%

The incremental IRR between alternative IV and I. The incremental rate of return is 0%. Thus, reject alternative IV and select Alternative I.

Accept alternative I.

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