Question

Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits
per second and 20 milliseconds propagation delay. Assume that the transmission time
for the acknowledgment and the processing time at nodes are negligible. Find the
minimum frame size in bytes to achieve a link utilization of at least 50%.​

Answer 1

Explanation:

Transmission or Link speed = 64 kb per sec

Propagation Delay = 20 milisec

Since stop and wait is used, a packet is sent only

when previous one is acknowledged.

Let x be size of packet, transmission time = x / 64 milisec

Since utilization is at least 50%, minimum possible total time

for one packet is twice of transmission delay, which means

x/64 * 2 = x/32

x/32 > x/64 + 2*20

x/64 > 40

x > 2560 bits = 320 bytes