Question

For purchasing a batch reactor, three alternatives \( P \), \( Q \), and \( R \) have emerged, as summarized in the table below. For a compound interest rate of \( 10\% \) per annum, choose the correct option that arranges the alternatives, in order, from the least expensive to the most expensive. \begin{table}[h!] \centering \begin{tabular}{|c|c|c|c|} \hline & P & Q & R
\hline Installed Cost (lakh rupees) & 15 & 25 & 35
\hline Equipment Life (years) & 3 & 5 & 7
\hline Maintenance Cost (lakh rupees per year) & 4 & 3 & 2
\hline \end{tabular} \end{table}

• A. P, Q, R
• B. R, P, Q
• C. R, Q, P
• D. Q, R, P

Hint:

To compare alternatives with different lifetimes, use the Equivalent Annual Cost (EAC) method, which accounts for both initial cost and recurring maintenance costs.

Answer 1

The Correct Option is C

Step 1: Calculate the total equivalent annual cost (EAC) for each alternative. The EAC is calculated as: \[ \text{EAC} = \frac{C \cdot i}{1 - (1 + i)^{-n}} + \text{Annual Maintenance Cost}, \] where: - \( C \) = Installed cost, - \( i = 0.10 \) (interest rate), - \( n \) = Equipment life (years). Step 2: Calculate EAC for each alternative. Alternative P: The total capitalized cost (\(TCC\)) is calculated using the formula: \[ TCC = c_0 + \frac{c_0 - c_s}{(1 + i)^n - 1} + \frac{c_m}{i}, \] where: \[ c_0 = \text{Installed cost}, \quad i = \text{Interest per year}, \quad n = \text{Equipment life}, \quad c_m = \text{Maintenance cost (annual expense)}, \quad c_s = \text{Salvage value}. \] Substituting the values for Alternative P: \[ c_0 = 15 \, \text{lakh}, \quad c_s = 0, \quad n = 3, \quad i = 0.1, \quad c_m = 4 \, \text{lakh}. \] \[ (TCC)_P = 15 + \frac{15 - 0}{(1.1)^3 - 1} + \frac{4}{0.1}. \] Simplify: \[ (TCC)_P = 15 + \frac{15}{1.1^3 - 1} + 40 = 100.317 \, \text{lakh}. \] Alternative Q: Substituting the values for Alternative Q: \[ c_0 = 25 \, \text{lakh}, \quad c_s = 0, \quad n = 5, \quad i = 0.1, \quad c_m = 3 \, \text{lakh}. \] \[ (TCC)_Q = 25 + \frac{25 - 0}{(1.1)^5 - 1} + \frac{3}{0.1}. \] Simplify: \[ (TCC)_Q = 25 + \frac{25}{1.1^5 - 1} + 30 = 95.944 \, \text{lakh}. \] Alternative R: Substituting the values for Alternative R: \[ c_0 = 35 \, \text{lakh}, \quad c_s = 0, \quad n = 7, \quad i = 0.1, \quad c_m = 2 \, \text{lakh}. \] \[ (TCC)_R = 35 + \frac{35 - 0}{(1.1)^7 - 1} + \frac{2}{0.1}. \] Simplify: \[ (TCC)_R = 35 + \frac{35}{1.1^7 - 1} + 20 = 91.89 \, \text{lakh}. \] Concept: The method with the minimum total capitalized cost (\(TCC\)) is the least expensive. \[ (TCC)_R<(TCC)_Q<(TCC)_P. \] Conclusion: The least to most expensive methods are: \[ \text{R, Q, P}. \]