Ask Question
3 October, 08:49

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)

+1
Answers (1)
  1. 3 October, 09:07
    0
    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)
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Computers & Technology if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers