Recall that in go-back-$N$ protocol, the transmitting window size is $N$ and the receiver window size is $1.$ Consider a pipelined, reliable transport protocol that uses go-back-$N$ with cumulative acknowledgment. Assume that the timeouts trigger retransmissions (but note that duplicate acknowledgments do not). Further assume that the receiver does not maintain any receive buffer, the one-way delay between the sender and receiver is $50 \mathrm{~ms}$, and every packet is $10,000$ bits long. Suppose that the sender must be able to send at a steady rate of $1 \mathrm{~Gb} / \mathrm{s}$ (gigabit per second) under ideal conditions.
- What should be the window size to allow the steady rate mentioned above?
- Suppose that the expected number of packets lost per $100,000$ packets is $1.$ If the sender uses a timeout of $500 \mathrm{~ms}$ and a window size of $20,000$ packets, calculate the expected gap between two timeouts, when
- The bottleneck link rate is $1 \mathrm{~Gb} / \mathrm{s}$.
- The bottleneck link rate is $2 \mathrm{~Gb} / \mathrm{s}$.