F
,IN3701 Assignment 1 (COMPLETE ANSWERS) Semester 2 2024
(232195) - DUE 20 August 2024 ; 100%
QUESTION 1 [20 marks] Batlokwa Industries wishes to select
one of three possible machines, each of which is expected to
satisfy the firm’s ongoing need for additional aluminium
extrusion capacity. The three machines, A, B and C, are equally
risky. The firm plans to use a 12% cost of capital to evaluate
each of them. The initial investment and annual cash inflows
over the life of each machine are shown in the following table:
Year Machine A Machine B Machine C 0 (R92 000) (R65 000)
(R100 500) 1 R12 000 R10 000 R30 000 2 R12 000 R20 000 R30
000 3 R12 000 R30 000 R30 000 4 R12 000 R40 000 R13 000 5
R12 000 - R30 000 6 R12 000 - REQUIRED: 1.1 Calculate the NPV
for each of the three projects. (9 marks) 1.2 Calculate the
annualised net present value (ANPV) of each machine. (9
marks) 1.3 Based on the NPV and IRR calculated above, would
you advise Batlokwa (Pty) Ltd to invest their funds in the
replacement? Give a reason for your answer. (2 marks)
1.1 Calculate the NPV for each of the three projects
The NPV (Net Present Value) formula is:
NPV=∑t=0nCt(1+r)t\text{NPV} = \sum_{t=0}^{n} \frac{C_t}
{(1+r)^t}NPV=∑t=0n(1+r)tCt
where:
CtC_tCt = Cash flow at time t
, rrr = Discount rate (cost of capital), which is 12% in this
case
nnn = Number of periods
For Machine A, B, and C, we need to discount each cash flow
back to present value and then sum them up.
Machine A
NPVA=∑t=06Ct(1+0.12)t\text{NPV}_A = \sum_{t=0}^{6} \
frac{C_t}{(1+0.12)^t}NPVA=t=0∑6(1+0.12)tCt
Cash flows:
C0=−92,000C1=12,000C2=12,000C3=12,000C4=12,000C5=12,00
0C6=12,000\begin{align*} C_0 &= -92,000 \\ C_1 &= 12,000 \\
C_2 &= 12,000 \\ C_3 &= 12,000 \\ C_4 &= 12,000 \\ C_5 &=
12,000 \\ C_6 &= 12,000 \\ \end{align*}C0C1C2C3C4C5C6
=−92,000=12,000=12,000=12,000=12,000=12,000=12,000
Machine B
NPVB=∑t=04Ct(1+0.12)t\text{NPV}_B = \sum_{t=0}^{4} \
frac{C_t}{(1+0.12)^t}NPVB=t=0∑4(1+0.12)tCt
Cash flows:
C0=−65,000C1=10,000C2=20,000C3=30,000C4=40,000\
begin{align*} C_0 &= -65,000 \\ C_1 &= 10,000 \\ C_2 &=
20,000 \\ C_3 &= 30,000 \\ C_4 &= 40,000 \\ \end{align*}C0C1
C2C3C4=−65,000=10,000=20,000=30,000=40,000
Machine C