10. The table below shows some of the contents of packet exchanged for a TCP con
ID: 3729501 • Letter: 1
Question
10. The table below shows some of the contents of packet exchanged for a TCP connection. All numbers are in decimal, not hex. (3 pts) Type of packet Packet Number 192.167.1.247 192.167.1.253 Seq.no. 17768656 Ack.no. 0 LEN = 0 bytes SYN Seq.no. 82980009 Ack.no. 17768657 LEN = 0 bytes Seq.no. 17768657 Ack.no. 82980010 LEN = 0 bytes Seq.no. 17768657 Ack.no. 82980010 LEN = 72 bytes of data ACK 4 Seq.no. 82980010 Ack No: LEN = 60 bytes of data Seq. no: Ack.No: LEN = m. The IP address of the client is: n. The type of packet for Packet Number 2 is: o. The value of the ack field in Packet Number 5 is p. q. (don't make a simple error here) Complete Packet 6 assuming it's just an ACK and has no Payload Why does Packet Number 1 have a Sequence Number of 17768656Explanation / Answer
m . IP address of the client is : 192.167.1.253
n. Packet number 2 is Non-SYN RST type
o. The value of the ack field in Packet number 5 : 177668658
p. Seq no. : 17768658, Ack no. : 82980011, LEN : 60 bytes of data
q. Because the ACK no. before the arrival of Packet no. 1 was same i.e. 17768656
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