Question

Question 4 Tiga jenis enjin hibrid sedang dipertimbangkan oleh syarikat yang berkesedaran tinggi terhadap alam sekitar untuk dilaburkan bagi digunakan oleh pengangkutan baru mereka. Anggaran ekonomi untuk setiap enjin ditunjukkan dalam Jadual S4 di Three types of hybrid engines are being considered by an environmentally conscious company to be invested for their new transporting equipment. The economic estimates for each engine are shown in Table Q4 below Jadual S4/Table Q4 Enjin Engine H-ll H-I Pelaburan permulaan Initial investment Penerimaan tahunan Annual receipts Perbelanjaan tahunan Annual expenses Jangka hayat (tahun) System life (years) RM 5,000 RM 20,000 RM 50,000 RM 2,000 RM 8,000 RM 12,000 RM 500 RM 2,500 RM 3,000 10 Berdasarkan Jadual S4 diatas, jawab soalan-soalan berikut: Based on Table Q4 above, answer the following questions a) Pilih enjin hibrid terbaik dengan menggunakan kaedah PW. MARR yang digunakan oleh syarikat tersebut adalah 15% setahun, dan enjin yang dipilih perlu berkhidmat selama 10 tahun. Anggapkan pengulangan. Jelaskan dengan nyata kenapa anda mencadangkan pilihan tersebut Select the best hybrid engine by using the PW method The MARR is 15% per year, and the engine chosen must provide service for 10 years. Assume repeatability. Clearly state the reason behind your recommendation (7 markah/marks) b) Sahkan cadangan yang anda perolehi dalam S4(a) dengan menggunakan kaedah IRR dalam menilai alternatif. Berikan alasan terhadap cadangan anda. Confirm your recommendation obtained in Q4(a) using the IRR method in evaluating alternatives. Justify your recommendation. (7 markah/marks) c) Apakah kelebihan utama menggunakan IRR dalam membuat keputusan bagi pemilihan projek? 4/6 KIX2002 What is the key advantage of using IRR in assessing project selections? (2 markah/marks)

Answer 1

The present worth can be determined using the following formula

$\large PW= -I -A(P/A,r\%,n)+S(P/F,r\%,n)$

Here the three machines have different useful life so in order to compare the PW take the LCM of their useful life and do the analysis for 10 years (LCM of 5,5 and 10).

Present worth of machine H1

I= -RM 5,000, Annual Reciepts = RM 2000,

Annual Expense = - RM 500

A = Reciepts + Expense = 2000 +(-500) = RM 1500

$\large PW= -2000 + 1500(P/A,15\%,10)-2000(P/F,15\%,5)$

$\large PW= -2000 + 1500*5.019-2000*0.2472$

$\large PW= -2000 + 7528.50- 494.40$

$\large PW= 5,034.10$

Machine H2

$\large PW= -20000 + 5500(P/A,15\%,10)-20000(P/F,15\%,5)$

$\large PW= -20000 + 5500*5.019-200000.2472$

$\large PW= -20000 + 27604.50 - 4944$

$\large PW= 2,660.50$

Machine H3

$\large PW= -50000 + 9000(P/A,15\%,10)$

$\large PW= -50000 + 9000*5.019$

$\large PW= -50000 + 45171$

$\large PW= -4,829$

Thus, select machine H 1. It has highest PW compared to other machines so H1 must be selected.

B. Determining IRR

Machine H1

$\large I = \frac{A}{1+r}+\frac{A}{(1+r)^2}+...+\frac{A}{(1+r)^5}$

$\large I = \frac{A}{r}[1-\frac{1}{(1+r)^5}]$

Let us assume r= 10%

$\large I = \frac{1500}{0.1}[1-\frac{1}{(1+0.1)^5}]$

$\large I = 5686.18$

Here, in case of H1 the initial investment, I = 5,000. Thus, r is greater than 10%. Let us assume r= 17%

$\large I = \frac{1500}{0.17}[1-\frac{1}{(1+0.17)^5}]$

$\large I = 4,799.019$

At a rate of 17% the initial investment is less than 5000. Thus r lies in between 10 to 17%. We can calculate IRR using linear interpolation technique as follows

$\large IRR =10\% +(17-10)\frac{5686.18-5000}{5686.18 -4,799.019}$

$\large IRR =10\% +7*\frac{686.18}{887.161}$

$\large IRR =10\% +7*0.773456$

$\large IRR =10\% +5.414192\%$

$\large IRR =15.42\%$

IRR of machine H1 = 15.42%

Similarly, determine IRR of machine H2

At a rate of 10%

$\large I = \frac{5500}{0.1}[1-\frac{1}{(1+0.1)^5}]$

$\large I = 20,849.32$

Here, the value is greater than 20,000 thus r must be greater than 10%. Let us assume r=15%

$\large I = \frac{5500}{0.15}[1-\frac{1}{(1+0.15)^5}]$

$\large I = 18,436.85$

Now using linear interpolation technique as follows

$\large IRR = 10 + (15-10)\frac{20,849.36-20,000}{20849.36-18,436.85}$

$\large IRR = 10 + 5*0.352065$

$\large IRR = 10 + 1.76$

$\large IRR = 11.76\%$

IRR of machine H2 = 11.76%

Similarly calculate for machine H3

When, r = 10%

$\large I = \frac{9000}{0.1}[1-\frac{1}{(1+0.1)^{10}}]$

I = RM 55,301.10

The value is greater than 50,000 thus r must be greater than 10% let us assume when r=15%

$\large I = \frac{9000}{0.15}[1-\frac{1}{(1+0.15)^{10}}]$

I = RM 45,168.91

Now it is less than 50,000 thus r lies in between 10 to 15%. Using linear interpolation technique we can calculate IRR as follows

$\large IRR = 10+(15-10)*\frac{55,301.10-50,000}{55,301.10-45,168.91}$

$\large IRR = 10+5*0.523193$

$\large IRR = 12.62\%$

Irr of machine H3 = 12.62%

Thus, IRR of machine H1 is highest hence it must me selected.

C. IRR is a better technique since it represents the outcome in percent thus it is considered to be better than PW method.

1. It indicates rate of return
2. It is not required to calculate cost of capital
3. It calculates break even.

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