Question

In Go-Back-4 ARQ protocol, the sequence number is set [0−7], that is, [0,1,2,3,4,5,6,7] to avoid potential dilemma as in Selective-Repeat. Given following assumptions, answer questions. the transmission. - The packet transmission time is τ, and the propagation time Tₚ=τ. The transmission time of ACK is negligible (note that the propagation time is still applicable).

- Pkt-2 and ACK-5 are lost for the first 0-7 packets. - Assume that all packets and ACKs arrive at t ⁻.
- The window positions then can be determined by assuming that packets/ACKs are received at t⁺.
- Time out Tₒᵤₜ=RTT=W⋅τ=4τ (from the beginning of transmission).

Answer following questions.
Assume that Go-Back-N protocol is used.

Answer 1

In the Go-Back-N ARQ protocol, the sender's window size is limited by the sequence number range. If packets or ACKs are lost, the sender will retransmit them. The timeout for the Go-Back-N protocol is set to the RTT multiplied by the window size.

Step-by-step explanation:

In Go-Back-N ARQ protocol, the sender's window size is limited by the sequence number range. In this case, the sequence number is set from 0 to 7, which allows for a window size of 8. This means that the sender can have up to 8 unacknowledged packets in flight at any given time.

Assuming that packets 2 and ACK 5 are lost, the sender will continue sending packets 2 to 7. However, when the sender reaches packet 6 and doesn't receive an ACK for packet 2, it will assume that all packets from 2 to 5 are lost and will retransmit them. This is because the sender's window moves forward only when it receives an ACK for the oldest unacknowledged packet.

The timeout for the Go-Back-N protocol is set to the round-trip time (RTT) multiplied by the window size. In this case, the RTT is 4τ and the window size is 8, so the timeout is 32τ. If a packet or ACK is not received within the timeout period, the sender will retransmit the corresponding packet(s).