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How long does it take a packet of length L to propagate over a link of distance D, propagation speed S, and transmission rate R bps. Does this delay depend on transmission rate?

Answer given is No.

But I don't understand how it is possible. Until the packet is transmitted , it can't be propagate over the link.
asked in Computer Networks by Active (1k points) | 136 views
propagation delay = D / S

here you can see that it is depending only on distance and speed. This is not depending on any other factor. So answer is NO

1 Answer

+1 vote
Best answer
It won't. Propagation delay will depend only on distance and speed.

Understand it with this analogy. Imagine a conveyor belt. It is of some length (Distance) and is rolling with some speed (propagation speed). Now, at one end of the belt, you are putting packets (You are performing transmission). As soon as you put the packet, it is left for delivery. You put another packet it is left for delivery. The time required would depend on length and speed of the belt. Hence, Propagation time depends on distance and speed. Note that at a time there can be many packets on the belt towards the same destination.

Now the time taken to put n packets will depend on two factors. How big n is and at what rate you can put packets on the belt. Say you have very good machinery to put packets. Now, it burns down to how many packets can fit on the belt at a moment. You have very good machinery doesn't mean you will put all the packets at the same time. The belt will crash. Say the belt is wide enough to accommodate 5 packets at a time (say second). This is the bandwidth of that medium. Hence, transmission rate will depend on length and bandwidth.

With this analogy, you can even know how the capacity of a medium is defined. Capacity depends on bandwidth and propagation delay. The capacity of the belt would be at that moment, at max, how many packets can be present on that entire belt. If the bandwidth is 5 packets per second and delay is 10s, at a time it can have 50 packets on it.

Hope you get it :)
answered by Loyal (7.2k points)
selected by
Thanks for explaining it with relevant example.

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