Answer to Question #122414 in Financial Math for Ishal

Question

Suppose that ABC Ltd is considering purchasing one of three new processing machines. Either machine would make it possible for the company to produce its products more efficiently.
Estimates regarding each machine are provided below:

Machine A Machine B Machine C
Original cost $79,000 $110,000 $244,000
Estimated life 7 years 8 years 10 years
Salvage value Nil Nil $30,000
Estimated annual cash inflows $30,000 $ 60,000 $58,500
Estimated annual cash outflows $ 7,000 $ 35,000 $18,500



A. If the projects cannot be repeated, which machine should ABC Ltd choose based on the NPV criteria at an 8% cost of capital?


B. If the projects can be repeated, which machine should ABC Ltd choose based on the NPV criteria at an 8% cost of capital?

Answer 1

a) $NPV_a=-79,000+(30,000-7,000)*\frac{(\frac{1}{1.08})^7-1}{\frac{1}{1.08}-1}=50,326.23$

$NPV_b=-110,000+(60,000-35,000)*\frac{(\frac{1}{1.08})^8-1}{\frac{1}{1.08}-1}=45,159.25$

$NPV_c=-244,000+(58,500-18,500)*\frac{(\frac{1}{1.08})^{10}-1}{\frac{1}{1.08}-1}-\frac{30,000}{1.08^{10}}=31,979.72$

 ABC Ltd should choose project A

b) For the comparable assessment, the same amount of time is needed. It is 14 years

A project will repeate 2 times; B project will repeate 1.75 times; C project will repeate 1.4 times

$NPV_a=-79,000-\frac{79,000}{1.08^7}+(30,000-7,000)*\frac{(\frac{1}{1.08})^{14}-1}{\frac{1}{1.08}-1}=79,691.11$

$NPV_b=-110,000-\frac{110,000*0.75}{1.08^8}+(60,000-35,000)*\frac{(\frac{1}{1.08})^{14}-1}{\frac{1}{1.08}-1}=68,022.22$

$NPV_c=-244,000-\frac{244,000*0.4}{1.08^{10}}(58,500-18,500)*\frac{(\frac{1}{1.08})^{14}-1}{\frac{1}{1.08}-1}-\frac{30,000}{1.08^{14}}=56,738.48$

The project A still the most effective