Compute the sender utilization for this protocol in the case the roundtrip time between sender and receiver is 7 ms, the network throughput is 1Mbps and each packet has a size of 1500 bytes

Given Information:

Round-trip time = RTT = 7 ms

Network throughput = 1 Mbps

Packet size = 1500 bytes

Required Information:

Sender utilization = U = ?

Answer:

Sender utilization = 63.16%

Explanation:

The transmission delay is given by

T = Packet size/Network throughput

T = 1500*8 bits/1x10⁶ bits/s

T = 12x10⁻³ sec

The Propagation delay is given by

Pd = RTT/2

Where RTT is the round trip time

Pd = 7x10⁻³/2

Pd = 3.5x10⁻³ sec

The sender utilization is

U = 1 / (1 + 2 * (Pd/T))

U = 1 / (1 + 2 * (3.5x10⁻³ / 12x10⁻³))

U = 0.6316

U = 63.16%

Therefore, the sender utilization or efficiency for this protocol is 63.16% which is basically the average traffic over this link or in other words the link has been used for 63.16% of the time.