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1 tcp sequence numbers relevant lectures: lecture 14-16 suppose a and b…

Question

1 tcp sequence numbers

relevant lectures: lecture 14-16

suppose a and b create a tcp connection with initial sequence numbers 5000 and 80000, respectively, and an initial window of 4000 bytes. the table below depicts the flow of the connection, which has 3 main events:

  1. a sends three 200-byte segments, (which we will name dataa1, dataa2, and dataa3), and b sends acks for each.
  2. between segments dataa2 and dataa3, the application on b calls read() on the socket associated with this connection, which returns 400 bytes.
  3. b sends a 200-byte segment datab1 to a and begins the connection termination process with a fin.

in the table, fill in the seq, ack, and win fields for each packet shown, given the initial sequence numbers and window sizes.

hint: try to create a similar connection flow using the tcp reference, while looking at the packets sent in wireshark-this should allow you to view the changes in sequence numbers, and window sizes. another reference that may be useful is section 17.3 of the dordal textbook.

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$$\begin{tabular}{c|c|c} t & packets sent by a & packets sent by b \\\\ \\hline 0 & syn, seq=5000, win=4000 & \\\\ \\hline 1 & & syn,ack, seq=80000, ack=\\underline{\\quad\\quad}, win=4000 \\\\ \\hline 2 & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=4000 & \\\\ \\hline 3 & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad}, data=dataa1 & \\\\ \\hline 4 & & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad} \\\\ \\hline 5 & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad}, data=dataa2 & \\\\ \\hline 6 & & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad} \\\\ \\hline 7 & & b calls read(), which returns 400 bytes \\\\ \\hline 8 & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad}, data=dataa3 & \\\\ \\hline 9 & & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad} \\\\ \\hline 10 & & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad}, data=datab1 \\\\ \\hline 11 & ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad} & \\\\ \\hline 12 & & fin,ack, seq=\\underline{\\quad\\quad}, ack=\\underline{\\quad\\quad}, win=\\underline{\\quad\\quad} \\\\ \\hline 13 & & \\\\ \\end{tabular}$$

Explanation:

Analyze the initial handshake

We track the sequence numbers, acknowledgment numbers, and window sizes for both hosts, \(A\) and \(B\).

  • Host \(A\) initial sequence number (\(ISN_A\)): \(5000\).
  • Host \(B\) initial sequence number (\(ISN_B\)): \(80000\).
  • Both start with an initial window size of \(4000\) bytes.

At \(t=0\): \(A\) sends SYN, seq=5000, win=4000.

  • This consumes \(1\) sequence number.

At \(t=1\): \(B\) sends SYN, ACK.

  • Since \(A\)'s SYN had seq=5000, \(B\) acknowledges it with ack = 5001.
  • \(B\)'s sequence number is seq = 80000.
  • \(B\)'s window is win = 4000.
  • This SYN from \(B\) also consumes \(1\) sequence number.

At \(t=2\): \(A\) sends ACK to complete the handshake.

  • \(A\)'s sequence number is seq = 5001.
  • \(A\) acknowledges \(B\)'s SYN (seq=80000), so ack = 80001.
  • \(A\)'s window is win = 4000.

Track data transmission from A to B

At \(t=3\): \(A\) sends DataA1 (200 bytes).

  • seq = 5001.
  • ack = 80001 (no new data from \(B\) to acknowledge yet).
  • win = 4000.
  • This segment carries bytes \(5001\) to \(5200\).

At \(t=4\): \(B\) acknowledges DataA1.

  • \(B\)'s sequence number remains seq = 80001 (since \(B\) hasn't sent data yet).
  • \(B\) acknowledges up to byte \(5200\), so ack = 5201.
  • \(B\)'s buffer has received 200 bytes which are not yet read by the application. Thus, \(B\)'s advertised window decreases by 200: win = 4000 - 200 = 3800.

At \(t=5\): \(A\) sends DataA2 (200 bytes).

  • seq = 5201.
  • ack = 80001.
  • win = 4000.
  • This segment carries bytes \(5201\) to \(5400\).

At \(t=6\): \(B\) acknowledges DataA2.

  • seq = 80001.
  • \(B\) acknowledges up to byte \(5400\), so ack = 5401.
  • \(B\)'s buffer now holds 400 unread bytes (DataA1 and DataA2). Thus, win = 4000 - 400 = 3600.

Process the socket read event

At \(t=7\): The application on \(B\) calls read(), which returns 400 bytes.

  • This empties \(B\)'s receive buffer, restoring \(B\)'s available window size back to 4000.

At \(t=8\): \(A\) sends DataA3 (200 bytes).

  • seq = 5401.
  • ack = 80001.
  • win = 4000.
  • This segment carries bytes \(5401\) to \(5600\).

At \(t=9\): \(B\) acknowledges DataA3.

  • seq = 80001.
  • \(B\) acknowledges up to byte \(5600\), so ack = 5601.
  • \(B\)'s buffer now holds 200 unread bytes (DataA3). Thus, win = 4000 - 200 = 3800.

Track data transmission from B to A

At \(t=10\): \(B\) sends DataB1 (200 bytes).

  • seq = 80001.
  • ack = 5601.
  • win = 3800.
  • This segment carries bytes \(80001\) to \(80200\).

At \(t=11\): \(A\) acknowledges DataB1.

  • seq = 5601.
  • \(A\) acknowledges up to byte \(80200\), so ack = 80201.
  • \(A\)'s buffer has received 200 bytes. Since no read is mentioned on \(A\), \(A\)'s window decreases: win = 4000 - 200 = 3800.

Track connection termination

At \(t=12\): \(B\) sends FIN, ACK.

  • seq = 80201 (since 200 bytes of DataB1 were sent).
  • ack = 5601.
  • win = 3800.
  • The FIN flag consumes \(1\) sequence number.

At \(t=13\): \(A\) acknowledges \(B\)'s FIN.

  • seq = 5601.
  • \(A\) acknowledges \(B\)'s FIN (seq=80201),…

Answer:

  • t = 1: SYN,ACK, seq=80000, ack= <blank>8001</blank> , win=4000
  • t = 2: ACK, seq= <blank>5001</blank> , ack= <blank>80001</blank> , win=4000
  • t = 3: ACK, seq= <blank>5001</blank> , ack= <blank>80001</blank> , win= <blank>4000</blank> , data=DataA1
  • t = 4: ACK, seq= <blank>80001</blank> , ack= <blank>5201</blank> , win= <blank>3800</blank>
  • t = 5: ACK, seq= <blank>5201</blank> , ack= <blank>80001</blank> , win= <blank>4000</blank> , data=DataA2
  • t = 6: ACK, seq= <blank>80001</blank> , ack= <blank>5401</blank> , win= <blank>3600</blank>
  • t = 8: ACK, seq= <blank>5401</blank> , ack= <blank>80001</blank> , win= <blank>4000</blank> , data=DataA3
  • t = 9: ACK, seq= <blank>80001</blank> , ack= <blank>5601</blank> , win= <blank>3800</blank>
  • t = 10: ACK, seq= <blank>80001</blank> , ack= <blank>5601</blank> , win= <blank>3800</blank> , data=DataB1
  • t = 11: ACK, seq= <blank>5601</blank> , ack= <blank>80201</blank> , win= <blank>3800</blank>
  • t = 12: FIN,ACK, seq= <blank>80201</blank> , ack= <blank>5601</blank> , win= <blank>3800</blank>
  • t = 13: ACK, seq= <blank>5601</blank> , ack= <blank>80202</blank> , win= <blank>3800</blank>