This is a question from text bookmy problem from part A is w

This is a question from text book,my problem from part A is what formula should we use and how to calculate the RTT and MSS,

Consider that only a single TCP (Reno) connection uses one 10Mbps link which does not buffer any data. Suppose that this link is the only congested link between the sending and receiving hosts. Assume that the TCP sender has a huge file to send to the receiver, and the receiver\'s receive buffer is much larger than the congestion window. We also make the following assumptions: each TCP segment size is 1,500 bytes; the two-way propagation delay of this connection is 150 msec; and this TCP connection is always in congestion avoidance phase, that is, ignore slow start. a. What is the maximum window size (in segments) that this TCP connection can achieve? b. What is the average window size (in segments) and average throughput (in bps) of this TCP connection? c. How long would it take for this TCP connection to reach its maximum window again after recovering from a packet loss?

Solution

Formula to calculate the window Size is:

You can calculate the window size as:

window Size = (RTT * Bandwidth) / 8

To calculate RTT

RTT = bits in flight / bandwidth

To calculate MSS

MSS = MTU - 40

MTU is always 576 so

MSS = 536