1
$\begingroup$

I have recently started reading the book Computer Networking: A Top-down Approach in hopes to get introduced to computer networks. When attempting one of the questions from the book for practice, I got stuck and spent the whole day today trying to make some progress.

The questions is:

Consider sending a large file of F bits from Host A to Host B. There are three links (and two switches) between A and B, and the links are uncongested (that is, no queuing delays). Host A segments the file into segments of S bits each and adds 80 bits of header to each segment, forming packets of $L = 80 + S$ bits. Each link has a transmission rate of $R$ bps. Find the value of $S$ that minimizes the delay of moving the file from Host A to Host B. Disregard propagation delay.

I was able to identify that the number of packets we have is $\frac{F}{S}$, and the delay for the first packet will be $N \cdot \frac{L}{R} = 3 \cdot \frac{80+S}{R} seconds$ where $N$ is the number of links. This is all I could come up with and I don't know what to do from here. A detailed and newbie-friendly explanation to this question would be greatly appreciated!

$\endgroup$

1 Answer 1

0
$\begingroup$

You are on the right track. The value of $S$ that minimizes the delay of moving the file from host A to host B is $S = \sqrt{40 F}$, which can be shown as follows.

The first packet reaches the destination after a delay of $3 \times d_{trans}$, where $d_{trans} = \frac{80+S}{R}$ is the transmission delay, as you’ve already shown. Each of the subsequent $\frac{F}{S}-1$ packets that follow this first packet will take another $d_{trans}$ time to reach the destination. Hence, the total delay for all the packets is $d_{trans} \times \left(3 + \frac{F}{S} - 1\right) = d_{trans} \left( \frac{F}{S} +2\right) = \left(\frac{80+S}{R}\right)\left(\frac{F}{S}+2\right) =: f(S)$.

We have that $$f(S) = \frac{80F}{RS} + \frac{160}{R} + \frac{F}{R} + \frac{2S}{R}.$$

To find the minimum value of $f(S)$, we set the derivative $f’(S)$ to equal $0$, and we obtain

$$\frac{80F}{R} \left(\frac{-1}{S^2}\right) + \frac{2}{R} = 0,$$ i.e. $S = \sqrt{40F}.$ At this value of $S$, $f’’ > 0$, and so this is a local minimum.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.