Answer:
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)
