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 is __________ 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

Correct Answer: 18

To determine the minimum salvage value (\(S\)) that makes the insulation installation competitive, we analyze the proposal using the Present Worth (PW) method with a compound interest rate of 9% per annum. Given: Yearly Savings (\(A\)) = \rupee20,000 Installation Cost (\(C\)) = \rupee150,000 Life of Insulation (\(n\)) = 12 years Interest Rate (\(i\)) = 9% = 0.09 Step 1: Present Worth of Savings (\(PW_{{savings}}\))
The yearly savings form a uniform annual series. The present worth is calculated using the Uniform Series Present Worth Factor (USPWF): \[ PW_{{savings}} = A \times \left( \frac{(1 + i)^n - 1}{i(1 + i)^n} \right) \] Substituting the values: \[ PW_{{savings}} = 20 \times \left( \frac{(1.09)^{12} - 1}{0.09 \times (1.09)^{12}} \right) \] Calculating the terms: \[ (1.09)^{12} = 2.8127 \] \[ PW_{{savings}} = 20 \times \left( \frac{2.8127 - 1}{0.09 \times 2.8127} \right) = 20 \times \left( \frac{1.8127}{0.2531} \right) = 20 \times 7.1604 = 143.208 { thousand rupees} \] Step 2: Present Worth of Salvage Value (\(PW_{{salvage}}\))
The salvage value is a single future amount. Its present worth is calculated using the Single Payment Present Worth Factor (SPPWF): \[ PW_{{salvage}} = S \times \left( \frac{1}{(1 + i)^n} \right) = S \times \left( \frac{1}{2.8127} \right) = S \times 0.3555 \] Step 3: Net Present Worth (NPW)
For the proposal to be competitive, the Net Present Worth (NPW) must be at least zero: \[ NPW = PW_{{savings}} + PW_{{salvage}} - C \geq 0 \] Substituting the known values: \[ 143.208 + 0.3555S - 150 \geq 0 \] Solving for \(S\): \[ 0.3555S \geq 6.792 \] \[ S \geq \frac{6.792}{0.3555} = 19.1 { thousand rupees} \] Rounding to the nearest integer: \[ S \geq 19 { thousand rupees} \] Final Answer: The minimum salvage value for which the proposal is competitive is \(\boxed{19}\) thousand rupees.