Answer :

Solution :

Total delivering power is

P₁ + P₂ = 260 MW .....(1)

The incremental cost of Generator-1 is

$I{C}_1=\frac{d{F}_1}{d{P}_1}=40+0.2P_1$

The incremental cost of Generator-2 is

$I{C}_2=\frac{d{F}_2}{d{P}_2}=32+0.4P_2$

For the optimum cost, the incremental cost of each generator must be equal.

IC₁ = IC₂

⇒ 40 + 0.2 P₁ = 32 + 0.4 P2

⇒ 0.2 P1 - 0.4 P₂ = - 8 .....(2)

By solving equations (1) and (2), we get

P₁ = 160 MW

P₂ = 100 MW