0
$\begingroup$

This is a problem from a MOOC on computer networking:

We wish to send a message of size $150,000$ bytes over the network. There are four hops, each of length $20$ km and running at $100$ Mb/s. However, before sending we split the message into $1500$ byte packets. What is the end-to-end delay of the message? Use speed of light in copper $c = 2 * 10^8$ m/s, and round your answer to the nearest integer millisecond.

HINT: Break the problem into two parts: the end-to-end delay of one packet and the delay of the rest of the message across the slowest link.

After struggling for a bit, I obtained the answer by following the hint as follows:

$$ 4\left(\frac{1500 \ B * 8 \ (b/B)}{100 \ Mb/s} + \frac{20 \ km}{2 * 10^8 \ m/s}\right) + \left(\frac{150000}{1500} - 1\right)\left(\frac{1500 \ B * 8 \ (b/B)}{100 \ Mb/s}\right) = 12.76 \ ms $$ The answer is correct but I don't understand why the packetization delay only was considered for the rest of the message. Why wasn't the propagation delay also considered for the remaining packets?

$\endgroup$
  • 1
    $\begingroup$ The reason is due to the pipeline nature of the solution: the second packet doesn't need to wait until the first one reaches its destination, it can be transmitted just one hop behind. Google latency vs throughput to get more intuition. $\endgroup$ – Ran G. Aug 14 '17 at 16:12
0
$\begingroup$

In this case the propagation delay is more than the transmission delay, the propagation delay acts as a bottleneck for the whole network therefore the total time depend on it.

Consider a pizza shop, it takes 10 min to prepare pizza but 5 min to transport it, having infinite labour to transport, calculate how much time will it take to finsh job for 100 pizza.

Your ans might be correct but equation is wrong, you should draw the transmission diagram for better understanding.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.