Find the link utilization in stop and wait protocol if the bandwidth of the line $2 kbps$, round trip time is $20$ second and the packet size is $2000 bytes$.
Answer given is $20\%$. Because they are considering only Round Trip time in calculation and not taking TT.
According to me it should be
$\frac{1}{1+2*\frac{10}{8}}= \frac{1}{3.5} = 0.2857 = 28.57%$
