TCP

Question

You are hired to design a reliable byte stream protocol that uses a sliding window protocol like TCP .The protocol will run over 1Gbps n/w. the RTT is 140 ns and maximum segment lifetime (MSL)is 60sec.

 How many bits would you require in the Advertise window field of TCP header to keep pipe full??

Answer 1

 

q1) Max. amount of data that can be send by the sender = 140 ns * 1 Gbps = 140 b = 140/8 B

No. of bits needed to represent 140/8 B of data < = No. of bits needed to represent 2^8/2^3 B of data

No. of bits needed to represent 140/8 B of data is at most 5 bits.

Advertised window field needs 5 bit so that it can send 140/8 B of data to keep the pipe full.

 

q2) Data send in MSL = 1Gbps * 60 sec = 60 Gb = 60/8 GB

   Max. no. of bits in sequence no. field of TCP header to prevent wrap around of data = log(base 2) (60 * 10^9/8) = 33 bits