Question

. Consider a sender has to send data packets indexing from p1 to p5, it sends all the data packets in order (from p1 to p5), but the receiver has only received p1 and the data packet p2 is lost somewhere in the network, then the receiver declines all the data packets after p2 ( i.e. p3, p4, p5 ) because the receiver is waiting for packet p2 and will not accept any other data packet than that. So, now as the time out time index of p2 expires, the sender goes back 3 packets and starts sending all the data packets from p2 to p5 again. We can see the process in the illustration below,

In this 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 is the number of packets that A will transmit for sending the message to B? 1) 12 2) 14 3) 16 4) 18

Answer 1

Answer:

14 packets

Explanation:

First 4 pkts transmitted only one. Rest 5 pkts transmitted twice.