Question

Consider a long distance TCP connection, where packets are pipelined to maximally use the channel, which has a bandwidth of 1Gbps. How large would the window size have to be, in terms of number of packets, for the channel utilization to be greater than 98 percent? Suppose that the RTT is 30 milliseconds and the size of a packet is 1,500 bytes, including both header fields and data.

Answer 1

To achieve channel utilization greater than 98% on a long distance TCP connection, the window size should be 2500 packets.

Step-by-step explanation:

To calculate the window size required for channel utilization of greater than 98%, we need to consider the bandwidth-delay product. The bandwidth-delay product represents the maximum amount of data that can be in transit on the network at any given time. It is calculated by multiplying the bandwidth by the round-trip time (RTT). In this case, the bandwidth is 1Gbps, which is equivalent to 1,000,000,000 bits per second. The RTT is given as 30 milliseconds, which is equivalent to 0.03 seconds. Therefore, the bandwidth-delay product is 1,000,000,000 * 0.03 = 30,000,000 bits.

Since the size of a packet is 1,500 bytes, we need to convert the bandwidth-delay product to packets. One packet is 1,500 bytes, which is equivalent to 12,000 bits (1 byte = 8 bits). Therefore, the window size required would be 30,000,000 / 12,000 = 2500 packets.