Question

A design engineer needs to purchase a membrane module (M) for a plant. Two options are available: M1 and M2. The overall plant life is 7 years. Interest rate is \(8%\) per annum, compounded annually. The table gives purchase cost (lakhs of rupees) and expected life (years): \[ \begin{array}{c|cc} & \text{M1} & \text{M2}\\ \hline \text{Purchase cost (lakhs)} & 10 & 5\\ \text{Expected life (years)} & 5 & 3 \end{array} \] Find the difference in the net present value (NPV) of these two options over the 7-year horizon (lakhs of rupees).

Hint:

For a finite project horizon, count each replacement at its purchase time and discount to present using \((1+i)^{-t}\).
If salvage is not specified, ignore it; otherwise include discounted salvage at the horizon.

Answer 1

Correct Answer: -4.8

Step 1: Over 7 years, replacements are needed to keep the module available. For M1 (life 5 y): buy at \(t=0\) and again at \(t=5\) y. For M2 (life 3 y): buy at \(t=0,3,6\) y. Salvage is ignored. Step 2: Present values with \(i=0.08\). \[ \text{NPV}_{\text{M1}}=10+\frac{10}{(1.08)^5}=10+\frac{10}{1.4693}=16.81. \] \[ \text{NPV}_{\text{M2}}=5+\frac{5}{(1.08)^3}+\frac{5}{(1.08)^6} =5+\frac{5}{1.2597}+\frac{5}{1.5869}=12.12. \] Step 3: Difference: \[ \Delta \text{NPV}=\text{NPV}_{\text{M1}}-\text{NPV}_{\text{M2}} =16.81-12.12=4.69\approx \boxed{4.7}. \]