Question

For a 1Mbits/sec channel with 10ms propagation delay, what is the channel utilization when sending 2KByte segments if Go-Back-N (Window size 5) protocol is used? What is the throughput? Show your work.

Answer #1

First lets calculat transmission delay.

Transmission delay is given by Packet size/Bandwidth of channel.

Transmission delay = 2KB/1Mb

= 2K * 8 bits /1M Bits

= 16K * 10-6

=16 * 10-3

= 16 ms

Here transmission delay=16 ms and propagation delay = 10 ms.

Channel utilization = N/(1+2a)

Here a = Tp/Tt

= 10/16 = 0.625

So utilization = 5/1+0.625

= 5/1.625

=3.07

So here channel utilization = 100%

Throughput = Channel Utilization * Bandwidth = 1*1Mbps

= 1Mbps

Here since utilization is greater than 1 , utilization is 100%

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