Question

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?

Answer 1

Answer:

a. Utilization =  0.00039

b. Throughput = 50Kbps

Explanation:

Given Data:

Packet Size = L = 1kb = 8000 bits

Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps

RTT = 20 msec

To Find

a. Sender Utilization = ?

b. Throughput = ?

Solution

a. Sender Utilization

As Given

Packet Size =  L = 8000 bits

Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps

Transmission Time = L/R = 8000 bits / 1 x 10⁹ bps = 8 micro-sec

Utilization =  Transmission Time / RTT + Transmission Time

                 =  8 micro-sec/ 20 msec + 8 micro-sec

                 =  0.008 sec/ 20.008 sec

Utilization =  0.00039

b. Throughput

As Given

Packet Size = 1kb

RTT = 20ms = 20/100 sec = 0.02 sec

So,

Throughput = Packet Size/RTT = 1kb /0.02 = 50 kbps

So, the system has 50 kbps throughput over  1 Gbps Link.