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. Aug 14 '17 at 16:12