Question

Suppose a packet is 10K bits long, the channel transmission rate connecting a sender and receiver is 10 Mbps, and the round-trip propagation delay is 10 ms. What is the maximum channel utilization of a stop-and-wait protocol for this channel

Answer 1

Answer:

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)

Explanation:

um I dont know what to put for the explanation so yea