Networks and services

To finish with this set of examples, the last thing we will see is, as a continuation of the last example, the transmission of a file. In fact, on the attached slide, there is as well the information of this file and what they are asking in this case is: How long does it take to transfer this file from A to C? We are going to draw the network again, thus we will be able to contextualize everything. In this case, we are sending packets from A to C, and it responds with an ACK, with a packet that recognizes that it has received last packet, they are data packets. We have a file here, which has a FL = 1*10^9 Bytes size. Remember that one Byte is nothing but 8 bits, therefore this file’s size in bits will be 8Gbits, and they are asking: How long will it take to transfer all this file from A to C?, we have to take into account that for each packet C receives, it sends an ACK, and when A receives this ACK, it sends this packet.

To solve this problem, for instance, we will make the drawing again, just like we did before, and we will try to see how the system works. As before, we have node A, we have the Router, and we have node C. We draw the timing in a vertical shaft and we take a reference value where we send the first packet. In this case, this packet will be sent to link 1, this would be link 1, and this, link 2. It is processed on the Router and it is sent to C. What does C do once the packet has been received? We have said that it responds with an ACK, therefore on the opposite direction it would send this recognition packet, as you can see on the slides, if this packet was 12000 bits, this packet here will be only 120 bits, to simply recognize that this packet has been received correctly. This ACK is sent on the opposite direction: it first passes through the Router, where it is also processed, and then it is sent again to A, where it is processed as well. Therefore, on this point right here, we have that A is aware that the packet has been sent been sent before, it is correctly received, and it can start transferring a second packet. If this was packet 1, now it would start transferring packet 2, and so on.

Therefore, to answer how long it would take to transfer this whole file, we need two things. First of all, we need to know how long we need for each packet, and in this case, the time is from this point here until A knows that the packet has been sent and correctly received. What would we have to do?: add this term right here and this one here, this is the propagation time on link 1, the transmission time on link 1, processing time on the Router, transmission time on link 2 (the data), propagation time on link 2, processing time on link 2 and repeat the same on ACK. ACK transmission time, ACK propagation time, ACK processing time on the Router, ACK transmission time on link 1, ACK propagation time on the link and node A processing time.

Notice that the propagation times are independent of the packet size, therefore, in this case, this value right here will be exactly equal to the value here, in other words, this range right here would be the same as the one here. And in the same way, this range right here will be the same as the one here. The different terms are written but basically you have to repeat the same we have seen before, but for the packet as well as for ACK. In any case, if we have the timing of a packet and they are asking for the time needed to transfer the whole file, another thing we need to know is how many packets have to be sent. This would be the timing to send a packet. And we can refer as N to the number of packets that are yet to be sent. If the image has 1Gbit, and each packet has a length of 12000 bits, from which 200 are headers, this means that only 10000 are data, the packet number is calculated like the file size divided by the data size of each packet. Here, a typical mistake is to divide by the total size, just be careful and be aware that it has to be divided only by the data size. If we do this division, simple in this case, they are 10000 bits by the data size. 8 Gbit would be 8 thousand million bits of the file size. We can simplify this and we would have 800 thousand packets. Notice that 8 Gbits can be the size of a movie or any other file of this kind, and this implies transferring 8000-bits data packets that we would have to transfer in 800000. Therefore, how will we calculate the total timing of the transmission?: well, just multiplying the delay of a packet by the number of packets we have.

Imagine this delay here is around half a second, this means that multiplied by this value right here, might produce several thousands of seconds and, in practice, this means several hours. Why can it be such a high value when we calculate this? Basically, notice that this link has a really low capacity, only 250 Kbits, really low for the current values, but maybe 10 years ago it was a normal value, and if, today, downloading one of these movies can take about one or two hours, maybe 10 years ago it took 12 hours or more, and the result of this calculation is around those lines.