In the Case of Free Space Propagation

Suppose the path loss is depicted by free space propagation model, then

$$32.4 + 20 \log \left ( d \right ) + 20 \log \left ( f_c \right )$$ L₀ =

$$32.4 + 20 \log \left ( 10^4 \right ) + 20 \log \left ( 0.9 \right )$$ =

= 111.5   dB (13)

W = P_t + G_t + G_r - L₀ - S / N -F +204

= 0 + 0 + 0 - 111.5 - 10 - 2 + 204

= 80.5   dB (14)

As we assume that the maximum data rate equals to the channel bandwidth, the maximum data rate can be acheived is R = W = 10^8.05   bps.