Chapter 3, Problem P48P

a.

Explanation of Solution

Given:

Link capacity= 10 Gbps

TCP segment size= 1500 bytes

Finding maximum window size:

Let the variable “X” represent the maximum of window’s size in segments.

Consider the expression, $\text{X × MSS/RTT = 10 Gbps}$ if the maximum sending rate goes beyond the link capacity, then the sending packets get dropped.

Substitute the given values as follows:

$\text{X × 1500 × 8/0}{}^{}$

b.

Explanation of Solution

Given:

Link capacity= 10 Gbps

TCP segment size= 1500 bytes

Let the variable “X” represent the maximum of window’s size in segments.

Finding average throughput:

For example the congestion window size “cwnd” varies from “X/2” to “X”, then the average window size is as follows:

$\begin{array}{l}\text{X = 125000}\\ \text{0}\end{array}$

c.

Explanation of Solution

Given:

Link capacity= 10 Gbps

TCP segment size= 1500 bytes

Let the variable “X” represent the maximum of window’s size in segments.

Time taken for recovering from packet loss:

When there is a loss while sending packets, the average window size becomes “93750/2”...