# Calculating the end-to-end delay of a message sent over a network

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?

• 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. – Ran G. Aug 14 '17 at 16:12