Ask Question

Suppose two hosts, A and B, are separated by 7,500 kilometers and are connected by a direct link of R = 10 Mbps. Suppose the propagation speed over the link is 2.5 x 10 8 meters/sec. Consider sending a large packet of 500,000 bits from Host A to Host B. How many milliseconds (ms) does it take before the receiver has received the entire 500,000-bit file?

+5
Answers (1)
  1. 13 October, 06:36
    0
    50 ms (milliseconds) will be taken in total to take the entire file by user.

    Explanation:

    In Computer networks, propagation delay is defined as the time in which a packet is sent from sender to receiver completely. It is computed by taking the ratio of link length and propagation speed.

    We know that:

    BDP (in bits) = total bandwidth (in bits/sec) * trip time (in sec)

    Now according to given condition we have:

    Bandwidth = R = 10 Mbps = 10,000,000 bps

    BDP = 500,000 bits

    For finding Propagation Delay:

    Propagation Delay = BDP / R

    Propagation Delay = 500,000/10,000,000 sec

    Propagation Delay = 0.05 sec

    Converting in milliseconds:

    Propagation Delay = 50 ms

    Hence. the delay would be 0.05 seconds and in milliseconds they will be equal to 50 ms
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Suppose two hosts, A and B, are separated by 7,500 kilometers and are connected by a direct link of R = 10 Mbps. Suppose the propagation ...” in 📗 Computers & Technology if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers