Station A needs to send a message consisting of 9 packets to Station B using a sliding window (window size 3) and go-back-n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that A transmits gets lost (but no acks from B ever get lost), then what isthe number of packets that A will transmit for sending the message to B?

To determine the number of packets Station A will transmit to Station B, given the sliding window protocol with a window size of 3, and a go-back-n error control strategy where every 5th packet that A transmits gets lost, let’s walk through the process.

The sliding window protocol allows Station A to send up to 3 packets before needing an acknowledgment (ACK) for the first packet in the window. With go-back-n, if a packet is lost, all packets sent after the lost packet must be retransmitted, even if they were received successfully.

Since every 5th packet gets lost:

1. The first four packets (1, 2, 3, 4) are sent successfully. Packet 5 is lost.
2. Station A realizes packet 5 is lost and hence retransmits packets 5, 6, and 7.
3. Packet 5 (on the second try, transmission number 6 overall) is successfully transmitted, but now the 10th packet from the start of the process, which is packet 8 this time (since we are counting retransmissions in our total), is lost.
4. Station A retransmits packets 8, 9, and receives ACKs without loss.

Let’s count the transmissions including retransmissions:

– Original sequence of transmissions: 1, 2, 3, 4, (5 lost), 6, 7, 8

– First re