Q.Assume a one-way delay of 10 ms and a link with an available bandwidth of 1Gbps. Further assume an MSS (Maximum Segment Size) of 1460 bytes.
a.Without using the window scale factor, what’s the maximum possible utilization (S)?
b.If we are using the window scale factor, determine the smallest window scale factor, that when combined with 65535 in the windows field in the TCP header, will give you a utilization of 1 or 100%.
c. In an environment where TCP achieves a ultilization of 10% with window size 65535, what window scale factor should be used to improve the utlization to 80%? Explain.
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(b)Optimal window size is when 4W/RD =1. Therefore W=RD/4 = (10^9* 10 * 10^-3)/4 = 1250000 bytes = 1250000/1460 sized segments = 856.1 segments. Use 858 segments and the window size = 1252680 bytes
(c)1252680 /65535 = 19.11. If window scale factor is 5. Window size =65535 * 2^5= 2097120 bytes.