Full transcript of “Delays in Network and Throughput - Part 02 | Computer Networks | Prof. Neminath, IITI”. Click a timestamp to open the video at that moment. Need another video? Use the YouTube transcript generator.
0:03 [music] We understood different kinds of the delays experienced by packet or transmission. So mainly the transmission, processing, queuing and the propagation delay. So now let's define what is the overall delay. Overall delay at a particular node uh all these four kinds of the delays are incurred by a packet. So it's actually the sum of the all the four types of the delays what we call it as the nodal delay one node delay. So it's a nodal delay. D nodal is the
0:50 propagation delay the transmission delay queuing delay and the processing delay. all put together is the we call it as the well delay out of all the four. So we can see that some of them are actually kind of the pix quantities. So for example the propagation time is depending on the link characteristics per particular link. It's actually very unique. So it only varies based on the size of the packet. If the size is uniform then there is no so the propagation delay is going to be constant uh constant in the sense it is taking same moment of the time and even the uh transmission time is also kind of depending upon the the capacity of the
1:36 link that is also going to be constant. So if you assume >> [clears throat] >> uh the routing table has got a fixed number of the entries then in the worst case how much time it takes for the processing or doing the consultation with the routing table can also be uh you can derive a bound. So that is also going to be kind of uh a constant. So what is variable is uh the Q because uh depending upon because we are using the multiplex transmission. So depending upon how many number of the packets are already there inside the queue. So based on that the queueing time the time spent by a particular packet actually varies. So that is one variable kind of the delay. So we can roughly divide the
2:21 these three delays into fixed and variable. So the other than the queueing delay all the three are remaining three are fixed kind of the delays and the queuing delay is actually variable and uh if assuming if the uh packets are not experiencing any queuing delay in any of the routers all the cues from the source to destination in every half of all the routers are free. So immediately as soon as the packet comes to to a particular router it is actually processed and scheduled for transmission. Then the total end to end delay for the uh transmission for that particular packet is n times that is
3:08 number of the links that exist between the source and destination by times the sum of the processing transmission and the propagation delays. So if you assume all the links have got uniform propagation delay and uniform passing delay and transmission delay. So you can sum them up to per one node and then calculate the overall delay for the end to end transmission. So let's understand this with a particular uh example. So here is a case suppose he host A wants to send a two megabyte file. This is MB capital MB megabytes file to host B over a single link. There
3:54 is a link. The link has the following characteristics. The transmission rate is 4 mabits/s and the distance between A and B is 2,000 kilm and the propagation speed is 2 into two into 10 ^ of 8 m/s and the processing delay at each of the router is 2 milliseconds and there is one router sitting between A and B and the queueing delay experienced by that particular packet the average queuing Delay is is saying that five milliseconds. The file is actually divided into chunks of,000 bytes per packet. Now is asking to
4:39 calculate the following. Calculate the transmission delay per packet at a particular node. How much time it takes to do the transmission. Calculate the propagation delay. How much time it takes to send the data on the link. Once it's sent, how much time it takes to arrive. Calculate the total processing delay per particular packet and calculate the total queuing delay experienced by the packet and what is the end to end delay and approximately how much time it takes to do the uh transmission. So these are the questions he has asked. Let's try to uh build up. So what he's saying A is sending uh some data to B and this is actually separated
5:25 by a link and there is one router let's call that router as R that is what he has said this total distance between A and B is uh 2,000 kilometer and you can think of now 2,000 km link is because there is a router in between is divided into two parts So he has not given what is the ratio. So we can roughly assume 1,000 kilometers on the left hand side and 1,000 kilometer on the right hand side. And the total data that is supposed to be transferred is 4 mgabytes and the processing delay and the giving delay at the router is also given. So let's go one by one and try to answer
6:13 those questions. So first is is asking the transmission delay for one particular node per one packet. So that is the so L by R. So L is each packet is saying of 1,000 bytes. And now you require because the denominator is in the bits per second. So you require bits total that I multiplied 1,000 into 8 is 8,000 divided by the rate of transmission is 4 Mbps. So that is 4 into 10 ^ of 6. So that roughly turns out to be 2 milliseconds. 2 milliseconds time is spent per packet at every node. [snorts] Now the second question is
6:59 asking is to calculate the propagation delay per link. So per link is remember we have divided that into two places A and uh uh between A and this router and between router and the B and the distance is 2,000. So we assumed it to be half of that. So now per link we can think of this as exactly there are two links and per link distance is 1,000 km that is in terms of the meters it is 10 ^ 6 m. So d by s. So d is 10 ^ of 6 and the s is is said that it is 2 into 10 ^ of 8 m/s. So that turns out to be 5
7:46 millisecond is the propagation delay per link. So from at 5 milliseconds once the data transmission is done from the A side after the 5 millconds the data arrives at the router and similarly another 5 millconds time will go for the from the R to B as well and the processing delay because there is only one router in between and only one time the packet is inspected the header is inspected and only one time the errors are checked in the middle. So that is 2 milliseconds you're saying. So the processing delay is exactly 2 milliseconds. So if there are multiple routers then you need to account for that. So the total queuing delay is asking to
8:33 calculate again there is only one router only one time the packet is queued up and the average queuing delay is saying 5 milliseconds. So the total queuing delay is also going to be 5 milliseconds. Again if there are multiple routers so you need to account for multiple routers as well. In this case the end to end delay is the time taken. It is a sum of again all the four kinds of the delays. So transmission per one link it is 2 milliseconds. So there are two links it is two times the 2 milliseconds and then the propagation time is saying 5 milliseconds and then then the there are two uh links and the two times the data is transmitted. So it is two times 5
9:19 milliseconds and now the queueing happens one time the average Q and 5 milliseconds and the processing time is 2 milliseconds only one time the packet is getting processed. So exactly after 21 milliseconds of the time. So the first packet is actually arrived at the destination B. So the last thing that he is asking is to calculate the total uh time for the transmission of the two megabytes to the data. So the we need to calculate the total number of the packets that are generated each of the size 1,000 bytes. So 2 into 10 ^ 6 megabytes is the uh number of the bytes of the total file and then the each file each packet is of
10:06 the,000 bytes. So exactly 2,000 packets are actually generated. Now the time taken for the entire transmission need to be understood in the context of the pipeline transmission. So we calculated that the first packet arrives at the exactly after 21 milliseconds and then the once the first packet arrives for every 2 milliseconds subsequently there is a new packet arriving at the destination. How many of them? Exactly uh 1999 number of the packets 2,000 minus one the first packet and each one at the interval of the 2 millisecond arriving.
10:51 So that turns out to be the initial this is the first packet transmission time but for the remaining 1999 packets it's exactly 2 milliseconds. Why this is happening like this? So this is how the transmission is actually happening in the pipeline transmission. So when the P1 arrives now you can see this P2 arrives and immediately next P3 arrives. So like this if you have multiple 2,000 packets for every 2 milliseconds that is the exactly the time taken for this links how much time it takes to reach the other end. So everything is rest of the things whatever to the left of this everything is happening in the p parallelly in the pipeline fashion. So something is
11:38 getting processed here parallelly something is getting processed here something is getting transmitted from here to here and something is parallely getting transmitted from this router to this router like that and even the source is also sending the data from the uh uh source to the first top router and something like that. So everything is actually getting pipelined. So in that context, so in that context, so for 1999 packets, it is exactly every 2 milliseconds the package is arriving. So sum of these two quantities is exactly approximately 4.02 seconds is the time taken for the entire transmission. So that's how you work out the uh reasons. So this is uh
12:24 assuming the pipeline operations. So if you do non-pipeline operations again the things would uh change that brings us to the uh last portion of the today's discussion which is something called as the throughput. So throughput is basically the rate or the uh time uh time taken for the transmission. So at a particular interval how much of the data can I send from the source to destination or from one hop to another hop is the uh throughput. So normally we measure the throughput from the end to end perspective. There are two kinds of the throughputs that I can measure. One is called the instantaneous that is the at this point of time what is the transmission rate uh and the second one
13:11 is the average transmission rate. So why this do continuous varying because some of the delays experienced by the packets particularly the queueing delay is actually varying and at a if you have multiple number of routers in between. So again multiple queuing delays it is experiencing. So I cannot really determine the exact uh the que delay for the entire transmission. The first packet may encounter problems. second packet may not necessarily encounter that much of giving delay and so forth. That's why the we usually calculate the average uh uh throughput that you get over a particular link. Let's say I have
13:57 a pipe and uh this is the sender and this is the receiver. Oh, this is sender and there is a router in between. There is a pipe between the sender and the router and the second pipe between the router and the receiver. Now if these two have the different capacities now the end to end throughput that I get are going to be constrained by the capacity is of these two links. So there are two cases that I might have. The first link let's say there is the first link has a capacity of RS bits per second and the second link has got the capacity of RC bits per second and the first scenario is if the RS rate
14:44 is smaller than the RC which is the case in this one what is the rate at which effective rate at which I can actually send the data? it turns out to be the smaller of these two cases. So in this case even though I have a bigger pipe in the next uh line but the I cannot send more than what this the initial pipe actually can carry. So the overall rate at which you can send is the smaller of these two which turns out to be the RS. The second case when you have a bigger pipe in the first stage and a smaller pipe in the second stage. So pipe is in link. If you have higher capacity link in the first phase and a smaller capacity link in the second stage. Now
15:30 again the smaller of these two which is turns out to be RC bits per second that is the throughput that you get. So in a way the end to end throughput that you get is constrained by the bottlenecket link link capacity. So for example, if you have let's say S is the sender and R is the recipient and you have R1 R2 R3 and uh if let's say this is 1 kbps 2 kbps and this is 3 kbps 4 kbps. So smaller up all these four links which is the 1 kbps that is the effective rate of the transmission from 8 to SS to R. So that is how you calculate the uh
16:16 throughput. So here is a slightly different scenario and what is shown here is there are multiple senders. So this is the multiplexity transmission from a n number of the senders. So let's say there are uh like 10 senders and then there are 10 recipients and this there is a single link from this point to this point over which all these transmissions data is actually flowing and if you give let's say this capacity is R and if you divide the there are 10 parallel transmissions each capacity uh is an equal capacity is given to an individual transmission from uh one
17:03 sender to one recipient. So exactly R10th of the portion is given to the individual transmissions. Then what is the effective throughput of the individual the case how much of the uh this I get. So it turns out to be the minimum of these RS is the this capacity and the RC is the this rate and R by 10 is this rate. So the minimum of whatever is RC RS and R / 10 is the effective throughput of one transmission from one sender to one receiver. So that's how you calculate the uh end to end
17:48 throughput of a particular transmission depending upon the scenario whether the senders this link is having the higher capacity or R10 is having the lower capacity and RC is having the lower capacity. So whatever is the lowest capacity so that dictates the effective throughput that you get over the uh end to end transmission. So uh now let's take a simple example and try to find out what is the throughput that you get uh per this particular transmission. So what is asking is uh there is a host A which wants to send a file to host B and the path from the host A to B has three links uh of the rate 500 kbps, 2 Mbps
18:33 and 1 Mbps. Assuming there are no other topic in the network, what is the throughput for this file transfer is asking. Second one, suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how long will it take for a transfer of the file is asking. So let's uh try to understand. So let's say the sender is here and the recipient is here. uh let me call it as RC and uh you have the first link with the 500 kbps capacity and the second link with the 1 Mbps capacity and the third
19:20 one is is saying uh 2 Mbps capacity and as we said throughut is actually the minimum of all these quantities it is 500 300 kbps, one uh Mbps and two Mbps. So that turns out to be 500 kbps. So that the bottleneck link dictates the distance. And [snorts] the second thing that is asking is part B if the file is 4 million bytes. So 4 million bytes is the total size. The file size is 4 into 10 ^ 6 and 8
20:10 bits per second. This is the size and the total time taken for the transmission and is the effective throughput that you get. So it is just size divide by the throughput that you have. So this is 4 into 10 ^ 6 * 8 / the throughput is 500 kbps. So this is the bits now. So this is kilobits per second. So that turns out to be 4 into 10 ^ of 6 into 8 divided by 5 into 10 ^ of 5. So 10 ^ 5 and this was this is
20:55 actually uh 40 into 8 / 5. So this is 5 1 5 inside a this is 64 seconds is the total time taken by the uh per this particular transmission. So that's how you calculate the throughput of the uh sorry total time per transmission taking the throughput and the size of the file into account. So first you calculate the throughput and then you estimate the total time taken for the transmission. Thank you.
