Total packets sent till now from sender side = 6Īfter receiving the acknowledgement for packet-4, sender slides its window and sends packet-7. Total packets sent till now from sender side = 5Īfter receiving the acknowledgement for packet-3, sender slides its window and sends packet-6. Total packets sent till now from sender side = 4Īfter receiving the acknowledgement for packet-2, sender slides its window and sends packet-5. Total packets sent till now from sender side = 3Īfter receiving the acknowledgement for packet-1, sender slides its window and sends packet-4. Since sender window size is 3, so sender sends 3 packets (1, 2, 3). 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? All packets are ready and immediately available for transmission. 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. Here, we have assumed that the acknowledgement for all the packets are sent independently.Therefore, the possible values of acknowledgement number ranges from (total n values).This is because sender’s window size is ‘n’.At any time, maximum number of outstanding packets can be ‘n’.For the (K-2) th packet, acknowledgement number would be ‘K-1’ and so on.For the (K-1) th packet, acknowledgement number would be ‘K’.It means it has received the packets ranging from 0 to K-1 whose acknowledgements are are on the way.Receiver expects the packet having sequence number ‘K’ at time ‘t’.Acknowledgement number is the next expected sequence number by the receiver.Therefore, last packet in sender’s window will have sequence number K+n-1.So, outstanding packets in sender’s window waiting for the acknowledgement starts from K.It is given that sender has received the acknowledgement for all these packets.It means it has processed all the packets ranging from 0 to K-1.
Assume the sender has already received the ACKs. What are the possible sets of sequence numbers inside the sender’s window at time ‘t’. Assume that the medium does not reorder messages. Suppose that at time ‘t’, the next inorder packet the receiver is expecting has a sequence number of ‘K’. = (100 x 8 bits) / (20 x 10 3 bits per sec)Ĭonsider the Go back N protocol with a sender’s window size of ‘n’.