TCP

Question 18

While opening a TCP connection, the initial sequence number is to be derived using a time-of-day (ToD) clock that keeps running even when the host is down. The low order 32 bits of the counter of the ToD clock is to be used for the initial sequence numbers. The clock counter increments once per millisecond. The maximum packet lifetime is given to be

Which one of the choices given below is closest to the minimum permissible rate at which sequence numbers used for packets of a connection can increase?

A	0.015/s
B	0.064/s
C	0.135/s
D	0.327/s

Computer-NetworksTCPGATE 2009

Question 18 Explanation:

32 bits are used to represent a sequence number. So we have 2³² different sequence number.
The maximum packet lifetime is given is given 64s.
Maximum data rate possible(bandwidth) to avoid the wraparound = 2³²/64 = 2²⁶ Byte/sec.
The clock counter increments once per milliseconds = That means when then counter increments next possible sequence number is generated. The packet lifetime is 64 seconds and after this 64 seconds next sequence number is come. So that means in this 64 seconds only 1 sequence number is generated.
Hence the minimum rate is = 1/64 = 0.015/sec.

Correct Answer: A

Question 18 Explanation:

32 bits are used to represent a sequence number. So we have 2³² different sequence number.
The maximum packet lifetime is given is given 64s.
Maximum data rate possible(bandwidth) to avoid the wraparound = 2³²/64 = 2²⁶ Byte/sec.
The clock counter increments once per milliseconds = That means when then counter increments next possible sequence number is generated. The packet lifetime is 64 seconds and after this 64 seconds next sequence number is come. So that means in this 64 seconds only 1 sequence number is generated.
Hence the minimum rate is = 1/64 = 0.015/sec.

Computer-Networks

TCP

Computer-Networks

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