I am now working through book "Computer Networking - A Top Down Approach", and there is one question among others after the first chapter:

"Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B.
e)Suppose dprop is greater than dtrans . At time t=dtrans, where is the first bit of the packet?"

Here dprop is propagation delay (m/s), and dtrans is transmission delay (L/R).

I have looked through some pages on the Internet and all of them say that at time t=dtrans the first bit of the packet will be on the link already, but not in the Host B yet. My question is - why is it not possible for first bit to be in Host B in this scenario? Isn't this a question of ratio between dtrans and dprop and size of the packet (L) as well? For example, if there are 100 bits in packet and dprop equals 1.0001*dtrans, won't the first bit be in Host B already, while Host A will be trying to "push" the bits remained to the link?


1 Answer 1


"Transmission delay" is ambiguous and shouldn't be used when you need precision.

Each bit has a propagation delay of tp = distance / velocity.

Since the bits are serialized - one sent after the other -, there's a serialization delay for an entire frame or packet of ts = number of bits / bitrate. You only need to calculate ts once, since serialization, transmission and deserialization all overlap.

So, the first bit reaches Host B after tp, the last bit leaves Host A after ts and reaches Host B after ts + tp.

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