Tcp

Question

Consider  a reliable byte stream protocol that use SWP running over 100 Mbps network .Propagation delay is 80 ms  and maximum segment  lifetime is 80 s

a) How many bits are needed for receiver window size in TCP?

A) 19  B)20 C)21 D)none

B) how many bits we need in sequence field in Tcp header?

A) 21 B)28 C)30 D)15

Comments

In 1rtt we can send 160ms*100Mbps/8 byes=2*10^6=2MB

This requires 21 bit

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Ans of A na?

Answer 1

B)

 

WAT = MAX SEQ NO. POSSIBLE / BANDWIDTH

HERE WAT =80 SEC(MSL)

and B.W. = 100 Mbps= 100MBps/8 (bit to byte)

 

now max seq no.possible= log (wat *bw) (base 2)

=log (80 * 100*10^6/8) (base 2)

=30 bits

=

Comments

can you please specify what is WAT?

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Wrap Around- The Process of using up all the seq no. and repeating a previously used seq no. is called wrap around and the time taken to wrap around is called WRAP AROUND TIME(WAT)