Question

It is proposed to install thermal insulation in a residence to save on the summer-monsoon season air-conditioning costs. The estimated yearly saving is 20 thousand rupees. The cost of installation of the insulation is 150 thousand rupees. The life of the insulation is 12 years. For a compound interest rate of 9% per annum, the minimum salvage value of the insulation for which the proposal is competitive _________ thousand rupees (rounded off to nearest integer).

Hint:

To calculate the minimum salvage value, use the NPV of savings and add it to the initial cost of installation.

Answer 1

Step 1: Given Data

• Annual Savings, \( A = \text{₹}20{,}000 = 20 \, \text{thousand rupees} \)
• Installation Cost, \( C = \text{₹}150{,}000 = 150 \, \text{thousand rupees} \)
• Interest Rate, \( i = 9\% = 0.09 \)
• Life of Insulation, \( n = 12 \, \text{years} \)

Step 2: Present Worth of Savings (\( PW_{\text{savings}} \))
\[ PW_{\text{savings}} = A \times \left( \frac{(1 + i)^n - 1}{i(1 + i)^n} \right) \] \[ PW_{\text{savings}} = 20 \times \left( \frac{(1.09)^{12} - 1}{0.09 \times (1.09)^{12}} \right) \] Compute \( (1.09)^{12} \approx 2.8127 \): \[ PW_{\text{savings}} = 20 \times \left( \frac{2.8127 - 1}{0.09 \times 2.8127} \right) = 20 \times \left( \frac{1.8127}{0.2531} \right) \approx 20 \times 7.162 = 143.24 \, \text{thousand rupees} \] Step 3: Present Worth of Salvage Value (\( PW_{\text{salvage}} \))
\[ PW_{\text{salvage}} = S \times \left( \frac{1}{(1.09)^{12}} \right) = S \times 0.3555 \] Step 4: Net Present Worth (NPW)
The investment is justified if: \[ NPW = PW_{\text{savings}} + PW_{\text{salvage}} - C \geq 0 \] \[ 143.24 + 0.3555S - 150 \geq 0 \Rightarrow 0.3555S \geq 6.76 \Rightarrow S \geq \frac{6.76}{0.3555} \approx 19.01 \] Final Answer:
The minimum salvage value that makes the proposal competitive is \( \boxed{19} \) thousand rupees.