Question

Two planes are being evaluated, with the projected life of each being 10 years. The cash flow for each plane are below. Use MARR of 15%. Based on the rate of return which is the most better.

	A	B
First Cost	$350 000	$415 000
M & O Costs	15 000	10 000
Annual Benefit	90 000	125 000
Salvage Value	35 000	45 000

Answer 1

Plane A

The present value of cash flow can be written as follows

$\large P = \frac{A}{1+IRR}+...\frac{A}{(1+IRR)^n}+\frac{S}{(1+IRR)^n}$

For plan A

A = 90000 - 15000 = $ 75000 per year

Salvage cost, S = $ 35000

We can calculate IRR using Trial and error method

Let us assume IRR = 20%

P_A  = 75,000(PVIFA,20%,10) + 35,000(PVIF,20%,10)

= 75,000*4.192 +35,000*0.1615

= 314,400 + 5652.70

= $ 320,052.70

Now assume IRR = 10%

P_A = 75,000*6.1445 + 35,000*0.3855

= 460,842.5 + 13,494.01

= $ 474,336.51

Thus the IRR of plan A lies in between 10 to 20%. We can determine it using linear interpolation technique as follows

$\large IRR_A = 10 + (20-10) *\frac{474,336.51-350,000}{474,336.51 - 320,052.7}$

$\large IRR_A = 10 + (20-10) *\frac{124,336.51}{154,283.8}$

$\large IRR_A = 10 + 10 *0.8059$

$\large IRR_A = 10 + 8.059$

$\large IRR_A = 18.059\%$

Similarly calculate for plan B

Let us assume when IRR is 20%

P_B = 115,000(PVIFA,20%,10) +45,000(PVIF,20%,10)

= 115,000*4.192 + 45,000*0.1615

= $ 489,347.50

Since the initial investment is less than the calculated value thus increase the rate

Let us assume IRR = 30%

P_B = 115,000*3.10 + 45,000*0.072538

= 355,527 + 3264.22

= $ 358,791.20

Linear interpolation technique as follows

$\large IRR_B= 20 + (30-20) *\frac{489,347.5 -415,000}{489,347.5 - 358,791.2}$

$\large IRR_B= 20 + 10*\frac{74,347.5}{130,556.2}$

$\large IRR_B= 20 + 5.69$

IRR_B = 25.69%

Both the plans can be selected since marr is less PV.

Select plan B

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