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!


1 Answer 1


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.


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