If a stop-and-wait protocol is used, what is the fraction of time, in seconds, that the sender is busy sending bits into the channel (the utilization), given that the round-trip-time (RTT) for a router is 20ms when sending 2000 Byte packets ( a byte is 8 bits) at 500Mega bits per second (Mbps)

This question is incomplete, here is the complete question;

If a stop-and-wait protocol is used, what is the fraction of time, in seconds, that the sender is busy sending bits into the channel (the utilization), given that the round-trip-time (RTT) for a router is 20ms when sending 2000 Byte packets ( a byte is 8 bits) at 500Mega bits per second (Mbps)? Answer to the nearest tenth of a microsecond. Recall that the formula for utilization U is L/R / (L/R + RTT). Be careful of your units. Provide the answer in microseconds with one decimal place. Do not label your answer with the units, just provide the number .

Explanation:

The Answer to the question is given with proper step by step solution.

Given that : the round-trip-time (RTT in abbreviation) = 20ms = 0.02 micro seconds

L = 2000

Byte = 2000*8

bits = 16000 bits

R = 500Mbps = 500*106 bps

Now,calculate L/R = 16000 / 500*106

= 32 / 106

= 32 micro seconds

The Utilization (U) = fraction when the time sender is busy sending

= (L/R) / (L/R + RTT)

= (32) / (32 + 0.02)

= 32 / 32.02

= 0.99937539 seconds

= 999375.39 micro seconds

= 999375.4 (nearest tenth)

answerhard disks;