Load reduction: 500 MW and 250 MW units

When load falls, both units move up their droop lines; the frequency rises and each unit sheds load in proportion to $P_r/R$.

Method: a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by

$$\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f$$

where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

Data: G1: 500 MW, $R_1 = 0.04$; G2: 250 MW, $R_2 = 0.02$; both fully loaded at 50 Hz (total 750 MW). Load reduced: $\Delta P_L = -100$ MW. Rated frequency 50 Hz.

Stiffness:

$$\begin{aligned} K_1 &= \frac{500}{0.04 \times 50} = 250\ \text{MW/Hz} \\ K_2 &= \frac{250}{0.02 \times 50} = 250\ \text{MW/Hz} \end{aligned}$$

Frequency deviation:

$$\Delta f = -\frac{\Delta P_L}{K_1 + K_2} = -\frac{-100}{500} = +0.2\ \text{Hz}$$

New frequency $= 50.2$ Hz.

Load reduction on each unit:

$$\Delta P_1 = -250 \times 0.2 = -50\ \text{MW}, \qquad \Delta P_2 = -250 \times 0.2 = -50\ \text{MW}$$

Answer: Frequency rises by 0.2 Hz (to 50.2 Hz). G1 now supplies 450 MW and G2 200 MW; both have equal stiffness, so they shed the load equally. The 11 kV rating does not affect the result.