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I am stuck with this homework question. I honestly have no clue what it is talking about.

The question asked: You are transmitting 100 bit packets over a long distance link of 40 km. I assume that the data rate is 1 Mbps and the propagation delay on the link is 2 * 10^8 m/sec.
How long will it take the packet to reach the destination from the source?

  1. .1 ms
  2. .3 ms
  3. .4 ms
  4. .2 ms

Will it be .4 ms?

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Thanks for asking your homework question appropriately and letting us know that it's for school! Just a word of advice from a fellow student, you might want to go in and talk to your professor if you've having trouble. While you can get help from the internet, believe me from experience that if you just go in and have a chat with your prof, it'll make like much easier in the long run :) – nhinkle May 7 '11 at 19:10

Given a link speed of c and a bandwidth of b, the one-way time T to transfer a packet of s length is:

size of packet / bandwidth + link distance / link speed

or to be short:

s / b + d / c

The math, you can do yourself!

Why does this formula give you the time needed to transmit? The whole thing is explainable as:

  • The first part is the time needed to push all Bits onto the link. The link has a certain bandwidth (bits per second that can be pushed on it). You're dividing Bits through Bits/sec, which ultimately gives you seconds.
  • The second part is the time needed to transfer the whole thing. The link distance is meters, divided by the speed in meters/sec, which again gives us seconds.

We can then just add both components and get the whole time.

Note: This formula is assuming that there is no constant routing delay, including the IP stack, and no variable queueing delay in the links, which you would have to simply add in more realistic calculations.

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