|Authors:||James F. Kurose, Keith W. Ross|
Consider the cross-country example shown in Figure 3.17. How big would the window size have to be for the channel utilization to be greater than 98 percent? Suppose that the size of a packet is 1,500 bytes, including both header fields and data.
Consider two systems A and B.
The round-trip propagation delay between A and B (RTT) = 30 ms.
Transmission rate of the link between A and B (R) = 1Gbps = 109bps.
The size of the data packet (L)= 1500 bytes (1500x8 bits)
The following is the formula to calculate the time required to transmit the packet over the 1 Gbps link:
where Dtrans is the time required.
Calculate the window size such that the channel utilization is greater than 98%.
Therefore, to utilize 98% of the channel, the window size should be 2451 packets approximately.