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a. In case of rdt 3.0 Stop and Wait, suppose we send a packet of 1KB...

a. In case of rdt 3.0 Stop and Wait, suppose we send a packet of 1KB through 1 Gbps link and RTT=20 msec. Find the sender utilization? Find the sender throughput?

b. Now if the sender is allowed to send five packets without waiting for acknowledgement of first packet (pipelined operation), Find utilization and throughput of sender.
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Answer #1

a)

Given length of a packet(L) : 1KB

Link bandwidth(B) : 1Gbps.

RTT : 20msec

Transmission Time: (L/ B) = (8 Kilobits)/(109) = 8 nanoseconds = 0.008 micro sections

So, utilization of sender = (Transmission time)/(Transmission Time + RTT) = 0.008/(0.008 + 20) = 3.99 * 10-4 = 0.0399 % utilization:

b)

Now the sender utilization with 5 packets pipelining: ( 5 *Transmission time)/(Transmission Time + RTT) = (5 * 0.008)/(0.008 + 20) = 0.199 % utilization.

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