GATE 2026 EE Q.58: solution and answer

GATE 2026 · EEQ.58 · 2 marks · NAT

A system with two generators G1 and G2 (without generator limits) is shown. The total load on the system is 1184 MW. The expressions for the cost of generation (C_1 and C_2) and real power loss (P_Loss) are as follows: C_1(P_G1) = 1000 + 50 P_G1 + 0.01 (P_G1)^2 Rs/MWh; C_2(P_G2) = 2000 + 50 P_G2 + 0.001 (P_G2)^2 Rs/MWh; P_Loss = 0.001 (P_G2 - 50)^2 MW. When the generators are operating at their optimal generation, meeting the total load requirement, the real power loss in the system is ________ MW (Round off to one decimal place). Consider the Lagrange multiplier lambda = 70.25 for optimal generation.

GATE 2026 EE question 58: Single-line diagram: generator G1 on the left feeds a bus with power P_G1 flowing in (arrow to the right) and a load of… — Figure from the GATE 2026 EE paper. It is part of the question.

Answer and solution

ANSWER

The real power loss at optimal generation is about 20.0 MW.

Official answer: marks to all candidatesSource: Official answer key, GATE 2026 EE (IIT Guwahati)

Set up the Lagrangian with losses: $L = C_{1} + C_{2} + λ (P_{D} + P_{L oss} - P_{G 1} - P_{G 2})$ with $P_{D} = 1184 MW$ . The loss depends only on $P_{G 2}$ , so $\partial P_{L oss} / \partial P_{G 1} = 0$ and the condition for G1 is simply $\frac{d C _{1}}{d P _{G 1}} = λ$ .
$50 + 0.02 P_{G 1} = 70.25 \Rightarrow P_{G 1} = \frac{20.25}{0.02} = 1012.5 MW$ .
Power balance (generation = load + loss): $P_{G 1} + P_{G 2} = 1184 + 0.001 (P_{G 2} - 50)^{2}$ , so $P_{G 2} = 171.5 + 0.001 (P_{G 2} - 50)^{2}$ .
Expand: $0.001 (P_{G 2}^{2} - 100 P_{G 2} + 2500) = 0.001 P_{G 2}^{2} - 0.1 P_{G 2} + 2.5$ , giving $0.001 P_{G 2}^{2} - 1.1 P_{G 2} + 174 = 0 \Rightarrow P_{G 2}^{2} - 1100 P_{G 2} + 174000 = 0$ .
$P_{G 2} = \frac{1100 - 110 0 ^{2} - 4 ( 174000 )}{2} = \frac{1100 - 514000}{2} = \frac{1100 - 716.94}{2} \approx 191.53 MW$ . The other root, about $908.5 MW$ , would mean a loss of about $737 MW$ and makes $1 - \partial P_{L oss} / \partial P_{G 2}$ negative, so it is not a valid operating point.
Loss: $P_{L oss} = 0.001 (191.53 - 50)^{2} = 0.001 (141.53)^{2} \approx 20.03 MW$ .
Check with G2's coordination equation: $\frac{d C _{2} / d P _{G 2}}{1 - \partial P _{L oss} / \partial P _{G 2}} = \frac{50 + 0.002 ( 191.53 )}{1 - 0.002 ( 141.53 )} = \frac{50.38}{0.717} \approx 70.3 \approx λ$ , consistent. (Solving G2's equation on its own, $50 + 0.002 P_{G 2} = 70.25 [1 - 0.002 (P_{G 2} - 50)]$ gives $P_{G 2} \approx 191.4 MW$ and $P_{L oss} \approx 19.99 MW$ , so either route rounds to $20.0 MW$ .)

KEY IDEA

With losses, each generator's incremental cost equals $λ (1 - \partial P_{L oss} / \partial P_{G i})$ . G1 does not appear in the loss formula, so $d C_{1} / d P_{G 1} = λ$ gives $P_{G 1}$ directly. The power balance including the loss then fixes $P_{G 2}$ , and with it the loss.

WATCH OUT FOR

Leaving the loss out of the power balance (taking $P_{G 2} = 1184 - 1012.5 = 171.5$ MW gives a loss of 14.8 MW), or applying $d C_{2} / d P_{G 2} = λ$ to G2 without the penalty factor. Another mistake is keeping the larger root of the quadratic (about 908 MW), which is not physical.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 9 Dispatch & control questions, with notes

GATE 2026 EE Q.58: solution and answer

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