Network Engineering Asked on September 30, 2021
I’m new to networking sorry if my question sounds dumb, I was reading a textbook which says:
the outbound link for the packet is the one that leads to router B. A packet can be transmitted on a link only if there is no other packet currently being transmitted on the link and if there are no other packets preceding it in the queue; if the link is currently busy or if there are other packets already queued for the link, the newly arriving packet will then join the queue.
I’m confused here, let’s say the link between A and B is called L, so currently a packet called p1 is being transmitted in L and there is no queued packets in L, now a packet p2 arrives and at the same time p1 is still on the half way in L, so does it mean that p2 needs to wait until p1 complete its journey? if yes then isn’t it very unefficient?
The description in the book is a very general description, so specific media may vary in the details. When it says being transmitted, they mean router A is sending the electrical signals on L. Once A is finished, it can send p2. It doesn't have to wait for p1 to reach router B.
The time it takes for the electrical signals to reach B is the propagation delay.
Correct answer by Ron Trunk on September 30, 2021
One thing that is often unintuitive when starting networking is how significant the times to actually get the bits on the wire can be, and consequently how large packets are physically.
Say you have a 100Mbps transmission on fibre. The speed of light in fibre is roughly 2x10^8 m/s, so each bit is roughly 2m long. A thousand byte packet is roughly 2mx8x1000=16km long.
You might find a train to be a useful metaphor for a packet, to remember it's a thing that doesn't leave at a particular time, it's a thing that starts leaving at a particular time.
Answered by richardb on September 30, 2021
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