Authors: |
James F. Kurose, Keith W. Ross |

ISBN: |
9780132856201 |

Edition: |
6 |

Chapter: |
Transport Layer |

Exercise: |
Problems |

Question: |
15 |

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 = 10^{9}bps.

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 *D*_{trans } 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.**