Question

115 9 to A % he de en 2 M ts bu u ce fo th A 2 asl

Answer 1

A. Given Machine A initial cost = RM 51,000

Annual cash flow = RM 10,500

Salvage value, S = RM 3,000

Life, n = 12 years

The IRR can be calculated by trial and error method. Let us assume IRR = 15%

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

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

$\large P = \frac{10500}{0.15}[1-\frac{1}{(1+0.15)^{12}}]+\frac{3000}{(1+0.15)^{12}}$

P = 56,916.49 + 560.72

P = RM 57,477.21

The value is higher than RM 51,000 thus increase the IRR. Let us assume IRR = 20%

$\large P = \frac{10500}{0.2}[1-\frac{1}{(1+0.2)^{12}}]+\frac{3000}{(1+0.2)^{12}}$

P =46,611.77 + 336.47

P = RM 46,948.23

Now we can determine the IRR through linear interpolation as follows.

$\large IRR = 15 + (20-15)*\frac{57,477.21-51,000}{57,477.21-46,948.23}$

$\large IRR = 15 + 5*\frac{6,477.21}{10,528.98}$

IRR = 15 + 5*0.061517

IRR = 18.08%

Since, IRR is greater than MARR of 15%. Accept A.

Similarly for machine B we can calculate the IRR.

P = RM 27,000

A = RM 9,000

S = RM 900

N = 6 yrs

Let us assume IRR = 20%

$\large P = \frac{9000}{0.2}[1-\frac{1}{(1+0.2)^{6}}]+\frac{900}{(1+0.2)^{6}}$

P = 29,929.59 + 301.41

P = RM 30,231

Since it is greater than the value thus increase IRR. Let us assume IRR = 30%

$\large P = \frac{9000}{0.3}[1-\frac{1}{(1+0.3)^{6}}]+\frac{900}{(1+0.3)^{6}}$

P = 23,784.71 + 186.46

P = RM 23,971.16

Now using linear interpolation we can calculate IRR

$\large IRR = 20 + (30-20)*\frac{30,231-27,000}{30,231-23,971.16}$

IRR = 20 + 10*0.5161

IRR = 25.16% (Since, greater than MARR 15%) Accept.

Machine C.

When IRR = 20%

P = 9,000(PVIFA,_20%,₄) + 600(PVIF_20%,₄) = RM 23,987.96

Now assume IRR = 30%

P = 9000(PVIFA,30%,4)+600(PVIF,30%,4)

P = RM 19,706.24

IRR = 20 +(30-20)*(23,987.96-21,000)/(23,987.96-19,706.24)

= 20 + 10*2,987.96/4,281.72

= 20 + 10* 0.698

= 26.98% (Accept since greater than 15%)

Now, machine D.

The IRR = 0%

Reject machine D.

A. Machine C is most economical. Since its IRR is highest.

B. With a MARR of 20%

Machine A must be rejected. Since its IRR is less than MARR.

Machine B has an IRR of 25.07% Hence it can be selected.

Machine C has an IRR of 26.98% Hence it can be selected.

Machine D has IRR less than MARR so reject it.

Select machine C and B.

Please contact if having any query thank you.