for questions (a) to (h),thanks! 4. (Go-back-N congestion control) Consider the
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for questions (a) to (h),thanks!
4. (Go-back-N congestion control) Consider the following graph, where a source is com- municating with a destination, using the go-back N along with the fast-retransmit and fast recovery scheme (i.e., TCP-Reno). Suppose each packet from the source is two bytes and ACK from the destination is one byte. Suppose sequence number of the first packet from the source is 100 and the sequence number of the first ACK fron the destination is 500. Set the initial value ssthresh 10 and cwnd 6. Suppose that only the pkt2 gets lost during the communication. pki2 pkus pkLe (a) What is the ACK number of the first ACK from the destination? (b) What is the sequence number and ACK umber of the pkt2? (c) What is the ACK number of the ACK for pkt3:7 (d) When will pkt2 be retransmitted (1). (2), (3), (1), or (5)? (e) Following the above question, what is the cwnd when it is retransmitted? (f) Following the above question, what is the ssthresh when it is retransmitted? (g) What is cund at time (1)? (h) What is the cwnd at time (5)?Explanation / Answer
In Go-Back-N congestion control method, for every packet sent by sender, there is a sequence number associated with it. This number is used at receiving end to place the packets in order and also to identify any loss of packets. Once it receives the packet, the receiver will send the acknowledgement to sender with an acknowledgement number. This number will be the next packet in sequence sent by sender.
Suppose if the packet sequence number sent by sender is 1, then upon receiving this packet, the receiver will send acknowledgement with acknowledgement(ACK) number of 2(next packet in sequence to be received by receiver in order).
a. The ACK number of first ACK from destination is: 102
This is because, the sequence number of first packet sent by sender is 100. The packet size is two bytes. Therefore the next packet to be received in order by receiver after receiving the first packet is 102(100+2). That number is used as the ACK number by the receiver to send it to sender, upon receiving the first packet with sequence number 100.
b. Packet 2 is the next packet to be sent after first packet(pkt1). As mentioned above, the sequence number of pkt1 is 100, then the next packet will have next sequence number in order. Since the packet size is 2 bytes it is added with sequence number.
Sequence number of pkt2: 102
ACK number of pkt2: 104
c. Pkt2 is lost. Whenever the receiver receives packets in out of order, then receiver will send an acknowledgement with ACK number which is missed out(which packet is expected by receiver for the packets to be in order). Thus when it receives first packet, it sends acknowledge(ACK 102) to receive the second packet. But the second packet is lost and 3rd packet is received. Hence packet 3 is not in order and receiver expected packet 2. Therefore it again sends acknowledgement with the number(ACK 102) for packet 2, so that sender will come to know packet 2 has not reached the destination.
Therefore ACK number is 102 for pkt3.
d. In fast retransmit and fast recovery, when a receiver receives out of order packets, for each packet it will send acknowledgements(duplicate ACKs) with an ACK number for missed out packet. When a sender receivers 3 duplicate ACKs, it retransmits the lost packet. Thus in this situation, upon receiving pkt1, receiver send ACK 102 so that sender knows pkt1 has reached. When pkt2 is lost, upon receiving pkt3, receiver sends the same ACK 102(1st duplicate ACK), so that sender knows the pkt2 has not reached. Again upon receiving pkt4, receiver sends the same ACK 102(2nd duplicate ACK), so that sender knows the pkt2 has not reached. Again upon receiving pkt5, receiver sends the same ACK 102(3rd duplicate ACK), so that sender knows the pkt2 has not reached.
After receiving 3 duplicate ACKs, the pkt2 is retransmitted by sender at (4)
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