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I was reading "Computer Networking; A Top-Down Approach" by "Kurose Ross" that I bumped in a question. Consider a 2 Link Network with a router as the following picture: enter image description here

by definition:

The instantaneous throughput at any instant of time is the rate (in bits/sec) at which Host B is receiving the file

and

the average throughput of the file transfer is F/T bits/sec.

Consider we have file with size XL in which X is the number of packets and L is the size of packet. So as we see in the picture, there is Rs bits/sec Link bottleneck, so (X-1) packets will be transferred to client after the first packet gets to the router. the first packet needs L/Rs seconds to get to the router. It means transferring the whole file needed XL/Rc + L/Rs seconds, Which by definition gives the following throughput:

enter image description here

but in book, the answer of the same question is:

Having determined the throughput, we can now approximate the time it takes to transfer a large file of F bits from server to client as F/min{Rs, Rc}. For a specific example, suppose you are downloading an MP3 file of F = 32 million bits, the server has a transmission rate of Rs = 2 Mbps, and you have an access link of Rc = 1 Mbps. The time needed to transfer the file is then 32 seconds.

I can not understand why the book doesn't care about the last packet. What is the reason?

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I think the reason is :

Having determined the throughput, we can now approximate the time it takes...

The last packet is received in about a ms, and this doesn't change much the calculated value.

In other word, it is negligible.

Note that, anyway, actual communication in a network use various protocols, each of which has some kind of overhead that will impact the actual throughput.

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I can not understand why the book doesn't care about the last packet. What is the reason?

Because you are referencing two different examples that are slightly different. The first example which is above uses Rs as the bottleneck (link with lowest available bandwidth).

In the second example, Rc is the bottleneck. This results in a slightly different way to calculate the time needed to transfer the data.

Edit: Also re-reading the posted question with more care, the first example appears to be talking about calculating the average throughput where the second example appears to be calculating the average time it takes to transfer a file. Not having the full context of the book in front of me, it is likely that wording on the two questions is slightly different.

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  • My mistake, but it's not the case. It won't change the conclusion. I changed the question to the correct form. – Aidin.T May 4 '17 at 23:28
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    @Aidin.T, the edits you made to the example/description of the formula cause it to no longer make sense. You also twice flipped Rs and Rc in the last example you were asking about, so we are still in a state where one example has Rs as the bottleneck and the other has Rc as the bottleneck. Looking at the original post, you mixed up Rs/Rc in the formula description, but everything else there should have remained untouched. For example, it should have read "L/Rs" and "XL/Rc + L/Rs" to match your formula graphic. – YLearn May 4 '17 at 23:50
  • I think now everything's fine, and about your edit in answer: the time that is calculated as the "time needed to transfer the file" used in the aforementioned "throughput". as in the text we have "Having determined the throughput, we can now approximate the time it takes to transfer a large file of F bits from server to client as F/min{Rs, Rc}" which by the definition means that "min{Rs, Rc}" is the throughput. – Aidin.T May 5 '17 at 0:16
  • @Aidin.T, when you changed from to client before the last packet gets to the router to read to client after the first packet gets to the router AND from When the last packet gets to the router, all other packets has been transferred, so the last packet needs L/Rc seconds to read the first packet needs L/Rs seconds you changed the entire meaning of that section and makes it inaccurate to the provided formula. Thinking that is fine indicates you may not understand things correctly. Also finding "throughput" and finding "time to transfer" are two ENTIRELY different questions/exercises. – YLearn May 5 '17 at 3:38

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