Computer Networking: - TCP operation on a channel with loss Question 4- TCP Oper
ID: 3720940 • Letter: C
Question
Computer Networking:
- TCP operation on a channel with loss
Question 4- TCP Operation on a channel with Loss (20pt) send base nextseqnum already ack'ed usable, not yet sent 02002 sent, not yet ack'ed not usable 4 window size_ ?00 Let's consider a TCP sender at a Slow start with "send base" at 1000 with congestion window N-8. Assume MSS is 1000 Consider four TCP packets have been sent with sequence numbers 1000, 2000, 3000 and 4000 with 1000 bytes of data respectively. Then, a packet is received with TCP ACK number3000 with rwnd (receiver window) size of 5,000. Answer the following questions with an assumption that another ACK is not received yet a. (4pt) What is the new value of SendBase? b. (4pt) What is the last byte (number) that could be sent considering available rwnd size?b c. (4pt) What is the new congestion window value, N-? d. (4pt) What is the last byte (number) that could be sent if rwnd is infinite? C. d. Now, let's assume the sender sent five more TCP packet when it had received the TCP ACK number 3000 from the receiver before, and receives the sam e. (4pt) What is the next TCP packet's Sequence Number that this sender needs to transmit? e f. (4pt) What would have been the following TCP packet's Sequence Number of the sender, if all the previous packets had been received at the receiver successfully? CP packets with ACK number 3000 with the same rwnd 5000Explanation / Answer
Answer to a:
Sender received TCP ACK number 3000 with rwnd=5000.
This means receiver has successfully received the packets with sequence number 1000 and 2000 and is wainting for the packet with sequence number 3000. Also the reciver can receive 5000 bytes more. Therefore sender will move the send_base pointer to 3000.
Thus the new value of send_base = 3000
Answer to b:
The receiver is expecting the packet with sequence number 3000 and can receive 5000 bytes.
Thus the last byte that could be send considering this rwnd = 3000+4999 = 7999
Answer to c:
The congestion window before the ACK received was N=8
So the congestion window after the recive of the ACK packet is N=16 (Exponantial : 1, 2, 4, 8, 16, 32, ....)
Answer to d:
if rwnd is infinite, Window size = minimum(N, rwnd/MSS) = 16
Thus the last byte that could be send = 3000+ 15999 = 18999
Answer to e:
When sender recives 3 or more duplicate ACK, then sender send the Packet with sequence number in the ACK without waiting for its retransmission timeout.
Thus TCP packet's sequence number that the sender needs to transmit = 3000
Answer to f:
Receiver receives all the 5 packets successfully starting from sequence number 3000 with MSS=1000
Thus the TCP packet's sequence number = 3000+ 5x1000 = 8000.
However sender waits for the window to slide
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