__Concept:__The transmission loss is given as

\({P_L} = \mathop \sum \limits_m \mathop \sum \limits_n \;{P_m}{B_{mn}}{P_n}\;\)

For two plant system, m = 1, 2 and n = 1, 2

Then,

P_{L} = P_{1} B_{11} P_{1} + P_{1} B_{12} P_{2} + P_{2} B_{21} P_{1} + P_{2} B_{22} P_{2}

B_{12} = B_{21}

\( \Rightarrow {P_L} = P_1^2{B_{11}} + 2{P_1}{P_2}{B_{12}} + P_2^2{B_{22}}\)

Penalty factor for plant 1 is,

\({P_{f1}} = \frac{1}{{1 - \frac{{\partial {P_L}}}{{\partial {P_1}}}}}\)

\(\frac{{\partial {P_L}}}{{\partial {P_1}}} = 2\;{P_1}\;{B_{11}} + 2\;{P_2}\;{B_{12}}\)

Penalty factor for plant 2 is

\({P_{f2}} = \frac{1}{{1 - \frac{{\partial {P_L}}}{{\partial {P_2}}}}}\;\)

\(\frac{{\partial {P_L}}}{{\partial {P_2}}} = 2{P_1}\;{B_{12}} + 2\;{P_2}\;{B_{22}}\)

__Calculation:__

Given that,

P_{1} = 150 MW, P_{2} = 200 MW

B_{11} = 0.003 MW^{-2}

B_{22} = 0.001 MW^{-2}

B_{12} = -0.0004 MW^{-2}

\(\frac{{\partial {P_L}}}{{\partial {P_1}}} = 2\;{P_1}\;{B_{11}} + 2\;{P_2}\;{B_{12}}\)

= 2(150) (0.003) + 2(200) (-0.0004)

= 0.74

\({P_{f1}} = \frac{1}{{1 - 0.74}} = 3.846\)