Droop for capacity sharing: 250 and 500 MW

To share a load change in proportion to capacity, the change on each unit must be the same fraction of its rating. Then from $R = (\Delta f/f_0)/(\Delta P/P_r)$, both droops come out equal.

Sharing by capacity: total capacity $= 750$ MW, $\Delta P_L = 100$ MW.

$$\begin{aligned} \Delta P_1 &= 100 \times \frac{250}{750} = 33.33\ \text{MW} \\ \Delta P_2 &= 100 \times \frac{500}{750} = 66.67\ \text{MW} \end{aligned}$$

Frequency change: $\Delta f = 59.5 - 60 = -0.5$ Hz, so $\Delta f/f_0 = 0.5/60 = 0.008333$ pu.

Droop of each unit (pu on own rating):

$$\begin{aligned} R_1 &= \frac{\Delta f/f_0}{\Delta P_1/P_{r1}} = \frac{0.008333}{33.33/250} = \frac{0.008333}{0.1333} = 0.0625 \\ R_2 &= \frac{\Delta f/f_0}{\Delta P_2/P_{r2}} = \frac{0.008333}{66.67/500} = \frac{0.008333}{0.1333} = 0.0625 \end{aligned}$$

Answer: Each generator needs a droop of 6.25 % on its own rating. G1 then goes from 125 to 158.33 MW and G2 from 250 to 316.67 MW, both at 63.3 % load, at 59.5 Hz.