Question

The fuel cost functions in rupees/hour for two 600 MW thermal power plants are given by \[ \text{Plant 1: } C_1 = 350 + 6P_1 + 0.004P_1^2 \] \[ \text{Plant 2: } C_2 = 450 + aP_2 + 0.003P_2^2 \] where \( P_1 \) and \( P_2 \) are the power generated by plant 1 and plant 2, respectively, in MW and \( a \) is constant. The incremental cost of power (\( \lambda \)) is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are

• A. 200, 350
• B. 250, 300
• C. 325, 225
• D. 350, 200

Hint:

To determine the optimal power generation for each plant, set the incremental cost equal for both plants and solve the system of equations.

Answer 1

The Correct Option is B

Step 1: Incremental cost condition.
The incremental cost (\( \lambda \)) is the rate of change of the cost function with respect to power generated: \[ \frac{dC_1}{dP_1} = \lambda, \quad \frac{dC_2}{dP_2} = \lambda. \] For Plant 1: \[ \frac{dC_1}{dP_1} = 6 + 0.008P_1 \] For Plant 2: \[ \frac{dC_2}{dP_2} = a + 0.006P_2 \] Since \( \lambda = 8 \) rupees per MWh: \[ 6 + 0.008P_1 = 8 \quad \text{and} \quad a + 0.006P_2 = 8. \] Step 2: Solving for \( P_1 \) and \( P_2 \).
From the equation for Plant 1: \[ 0.008P_1 = 2 \quad \Rightarrow \quad P_1 = 250 \, \text{MW}. \] From the equation for Plant 2: \[ a + 0.006P_2 = 8 \quad \Rightarrow \quad P_2 = 300 \, \text{MW}. \] Step 3: Conclusion.
The optimal generation for Plant 1 and Plant 2 are 250 MW and 300 MW, respectively.