 So, what is the Q discipline in output? Output is the Q discipline at output will be first and first output. Now, we are when any pattern is something WM at a delay after we are calculating W 2 W 2 and this W 2 is coming batch size I we are considering out of batch size we are. With a batch you can replace anywhere. Anyway, but we are selecting randomly. Yeah, it is a with equal probability. Your time packet with equal probability can be anywhere in the batch size of I, but anything with the earlier batches has to always go first. But, there is a first and first out among batches, within a batch it is a randomly random placement. And when we are doing the tagging. Tagging is just a concept. Tagging is a concept actually because actual tagging is not happening in the batch. That is true because we are calculating the probability tag packet coming batch size of I. So, we are assuming that the total size is I, but already we are selected. So, N minus 1 we are selecting C minus I, I minus 5. It is not clear. So, we are calculating the probability tag packet coming in batch size of I. And the output whether the batch size of I is coming or not what is the probability of that. Out of N we are saying we are taking the maximum batch size is N, minimum is 1. What is the chance that a given slot at an output both a batch of I will come. I will come, but one packet is already tagged with this particular packet. Out of this I, one is my packet. My packet. And it has the batch size has to be actually more greater than or equal to k plus 1 or a W2 is equal to k. That is true, but if I vary it between N minus 1 C I minus 1. One is my already tagged packet which is coming from one of the port. Now, from remaining N minus 1 I minus 1 should also come. Then only batch of I will be there. Tag packet anyway is part of it. That is the reason why it was done. So, ultimately yesterday what we got was probability that your tagged packet will be there and it will face suffer a delay of k. This is what we had estimated and this was nothing but summation of A I by T. I goes from k plus 1 to that is what we actually have got. And ultimately from here I can find out what is going to be the z transform of this which is a probability generating function E g s. So, I have to do summation for k going from 0 to N because there is that is the maximum which is possible the probability. Your batch the W2 delay cannot be more than N because your batch size cannot be more than N. The maximum is what we get is actually N minus 1 delay. It is not N but N minus 1. So, you have to do z raise to power k and then of course, whatever is this probability that W2 is equal to k has to be put here which in turn will give you. So, this actually can be solved to find out what is going to be your probability generating function W2 z actually. So, this is that is what we are interested in. So, you can actually now solve for it. So, this one is pretty straight forward ones I will give the technique. So, 1 over P I can always take out and then of course, A I we already know and I can solve. So, when k is equal to 0 for that there will be a series multiplied by z raise to power 0 for k is equal to 1 there will be a series number of terms will start reducing as my k increases. So, I have to actually basically write all these terms and I will just reorganize them, but there are many ways of doing it I will just do it this way. That is the way I have learnt it. So, for k is equal to 0 what is the series I am going to write in the first row. So, this will be nothing but n c 1 this is for the k i is equal to 1 and then of course, you will have a series I am not going to write all the terms, but that corresponds to when k is equal to 0. When k will be equal to 1 in that case all terms will be multiplied by z raise to power 1. I will have actually have because k is 1. So, I has to be 2 minimum and so on. So, I have z raise to power n minus 1 n c n I can write similarly for z square I think by this time you must be understanding it is you can write any term by just close observation actually. It has to be z square. It has to be z square because I am sorry for that and so on till the last term which is going to be there that is for n last row will always actually have only one term. So, you can now combine them you take only this term you take the term corresponding to n c 2 corresponding to n c n and you will end up in getting nothing, but the first one will have only one there is no term. So, there is only one term which is 1 summation of that because in every series first time series only with one term second time series is 1 n z third time series 1 z z square 1 z z square z 4 the last series will contain 1 z z square till z square n. This series you can always put g p and then put the things and this will give you the complete thing. So, I will just write the final thing. So, you will have 1 over p will be out z minus 1 by z minus 1. So, last term will be z to the power n minus 1 problem. z square n c n minus 1 or n plus 1 z to the power n minus 1 and so on and of course you can write on the last term. So, the last term will be now n c n c by n that is where the term actually goes. So, I can actually take 1 over p z minus 1 out this now can be written as summation of from i is equal to 1 to n z over i a i minus this minus 1 is there this very well can be now written as 1 over p. Now, this term can be written as 1 minus a 0. So, I can write this I can combine this whole term a 0 can be taken inside and I can make this and this is nothing but a probability generating function of a i a z. So, expression will be a z minus 1 divided by p z minus 1 that is w 2 z. So, I can now combine these two things together. In fact, there would have been another simpler way of doing it you multiply by z and then subtract the two terms whenever you want to find out that kind of series that is also another simpler way that z minus 1 would have come out up around would have been simply a z minus 1. So, other way is actually fine. So, now w is equal to w 1 plus w 2 which actually implies w of z is w 1 of z this is a standard result again function actually for probability generating functions. So, whatever is the addition of the variables will become nothing but multiplication and p g f problem. So, this actually means you will have q of z 1 minus a of z minus I have actually taken 1 minus z proper one is also 1 minus z. So, that will be the final p g f you take the derivative of this and make z is equal to 1 that will give you the mean value of w w bar. So, you take that thing. So, this value which you will get is nothing but q bar plus second movement around 0 and first movement around 0 first movement around 0 is mean and a bar is p because n into p by n will be p a t. So, average value will be a bar and of course a square bar also you can estimate this should be p plus these are again all the standard stuff that we have and from here you will get w bar equal to whatever was w 1 w 1 was n minus 1 by n I think e square 2 p minus 1 1 minus p a t. So, that is for w this q bar plus the second term now you can put from here this will turn out to be p by 2 n minus 1 by you solve for this particular a square you take from here this a bar you take from here put it here solve you should get this and this in turn will give you which is nothing but this is average delay for m d 1 q only factor which is multiplying factor is n minus 1 by n. So, it technically all output q system will actually work like a m d 1 q's and the beauty of it is if you want to actually plot with p with the maximum throughput. So, when n goes to 1 this term goes away we have p by 2 1 minus p. So, if you solve for it this is nothing like a saturation the delay also goes to infinity. So, earlier case the q length the all will go asymptotically when p is equal to 1. So, same is with the delay also it is so that you can actually operate at any value of p does not matter only at p is equal to 1 you go to infinity which anyway is unestable for any kind of Markovian arrival process q p is equal to 1 you will always be unestable we have seen that for m m 1 infinity q at some point of time earlier. So, output q is actually the best option if you can achieve that, but you need not be able to implement that all the time. So, what we can do is now move over to the input queuing system for comparison for this input queuing system I will never be able to achieve very close throughput when p actually increases throughput is increasing it will try to reach to basically as far as possible only the q will build up, but my throughput performance will always become very close to 1. So, one packet per slot I will be able to push at output only my buffer length requirement will be high because of the randomness and because of the contingency and if you somehow can ensure that there is no contingency throughput will be 1, but input queuing because there is a ahead of the line blocking we can actually again do that modeling and find out what is the approximate value. So, it is about 53 something percent or 0.53 something that is a maximum throughput which can be achieved if I am going to have ahead of the line blocking in the input queuing. Sir, in the previous analysis we are able only computing the q will behave like m d 1 q with some multiplied factor. Yeah, it is n minus 1 by m. In. That is the only difference which is coming because of the batch arrivals. Batch arrivals. Batch arrivals technically act like a Markovian's processor. So, let us come to the input queue system. Now, this is going to have many outputs and there are n inputs. Again I am not worried about that n minus 1 only will be packaged to you. I am not bothered about that now. We have the buffers at the input. At any point of time I am not looking it for one particular output i. So, whatever is true for this i, output is also true for any other output. Say equivalent system. There is no bias. So, I can just take one output and analyze for that. Now, what will happen is the packets will keep on coming at the head. Once the packet will come. So, for this i thing there can be only look at the head how many packets are there. So, at the head for just for trying to the packets which are trying to reach to destination i, let those packets be numbered as BMI after nth time slot. One thing you should be sure that summation of BMI over all i's all possible output combinations. So, one to n they should be equal to how much? This only says that for i how many packets were queued up in the front of the queue? For i is equal to how many are queued up? Total number of queued up packet which is there in the head of the queues has to be equal to n only. This cannot be less or more. In worst case it is n. In every slot if the packets are coming fully loaded condition. So, this will be equal to n for saturated condition. When there is always a packet available at the input in every time slot. In fact, a similar kind of equation which we had earlier is also whole truth. Whatever was the after the earlier time slot whatever was the number which was queued up one of them will be picked up and will be sent to the outgoing port. Remaining will remain there at the head only thing now the balancing conditions will be different. There will be new packets which can arrive in the nth time slot which I will again write as a m i for that output port. Now, remember these will be not out of n earlier case it was out of n actually out of n i packets are coming in the batches. Now it is whatever are the ports for which the packet has been pushed out to the outgoing port in the m minus first slot only on those lines new packets will be coming in and from those only I will get some packets which will be destined for i. So, I will not use n here I will use something else when I will make an estimation and one packet which will be going out. So, this relation is very similar. So, in fact a very similar result can be used here also except now that I will not be using n I will be using something else here. So, probability factor which was there will actually get changed that is only thing. So, the probability that your a m i will be k I will not use n earlier I was using n heads these are free ports on which the packets will arrive in the free head of the line cubes front of the cube which actually has become free in the next slot packet only will come on those this f m minus 1 is that number c i with equal probability I am looking at a saturated condition. So, p is equal to 1. So, I am trying to find out what is the upper bound on the throughput performance under saturated condition whatever you achieve you cannot achieve better than that unless we do some manipulation we will do that actually. So, it is 1 by this man has to be there 1 minus 1 over n what is this f m minus 1. So, this is by definition whatever is the blocked q's I sum up all of them for each outgoing port this value I have taken in worst case will be equal to n, but ideally speaking this should be always less than or equal to n the difference between them is what is f m minus 1. So, n minus the difference is this that should be f m minus 1 and of course f m minus 1 should be equal to that is where the new packets will come should also be equal to this arrivals these two conditions would always be satisfied and I can actually make an estimate of I can call something on f bar which is the many mean number of free inputs used after every time slot this is statistical average of this under saturated condition they let that be at bar. So, you can actually observe that f bar by n that which fraction of packets will actually will be going as a throughput that is the maximum which you can achieve what is happening is in the first switch in this case some packets remain some move some packet remains some moves in every time slot on an average the fraction of packets which are moving out divided by total lines that should be the fractional utilization of each outgoing link that is a load that is the number of packets going out, but no packet is being lost here remember it is a infinite buffer. So, this has to be equal to rho naught I am interested in finding out what is rho naught actually. Now, there is an approximation here this is a very simple judgment when n goes to infinity this A i under steady state condition E M i will tend to become Poisson's statistics. So, new arrivals which are coming in will be nothing they will be Poisson with parameter rho naught because that is a utilization every utilization coming. So, it should be Poisson's statistics with rho naught actually I should call it a steady number of packets for output i A i I am not writing M because this is a steady state number this is Poisson with parameter. Output utilization yes yes yes you are putting packet all the time, but you have to maintain a steady state. So, output utilization has to be equal to input utilization if input utilization is because this is a maximum which you can achieve under saturated condition you cannot have input utilization higher than this if you are going to have it your q will blow up it will not be stable q. So, stability condition does not come at rho naught is equal to 1 which was happening in the earlier case in output queuing system when p was utilization of every link when it is going to 1 then only you are going to instability here you will actually achieve your stability even before that. So, that is a bad thing about input queues. So, you will blow up to infinity only even before p actually goes equal to 1. So, p has to be maintained at lesser than 1 for input queues which to operate. So, there is a chances you may actually leave to instability because p is equal to 1 most likely you usually never operate with that p is equal to 1 fully saturated condition. So, what I am saying is under fully saturated condition your q is going to blow up, but there is a maximum utilization. So, if you keep input utilization equal to output utilization your q will be almost kind of stable just it will actually blow up at that point, but anything which is blow rho naught it is going to remain stable. I mean output bokeh is better. This obvious obviously commonsense tells it has to be actually better. So, once this is rho naught you look at this expression and this expression this is very similar to what we did in output queuing I had a very similar expression this was a Poisson statistics for n is equal to infinity and for binomial for finite number of n and we could directly estimate what was the statistics of output size it was m d 1 q thing. So, I can use the same result here, but n is going to infinity. So, n minus 1 by n is 1 actually. So, what ultimately you will get is the q length I do not have to put b i bar this value will be now it is rho naught I am not going to put p it is rho naught that is actually equal to p p is also equalization, but I am going to rho naught here that is what it will be and of course, now I can use this expression under steady state condition m actually can be removed this value will be nothing, but n by n becomes 1. So, b i I know already from here. So, this 1 by n can be taken out. So, now solve for this upper one. So, this is nothing, but rho naught understood in steady state conditions 1 minus and you will have 1 over n. So, for each of these components summation I am taking bar actually on top of it both I am taking expectation on both sides. So, this will be nothing, but n number of times you have to add b i bar which is nothing, but rho naught something. So, this should be because I am actually having n terms all same values is multiplied by n this can be cancelled with this solve for this rho naught. So, you should get minus rho naught is equal to rho naught square 2 plus rho naught square minus there also has to be a 2. So, this rho naught square I can make it equal to 0 this is the equation that we will get. So, for it 2 minus root 2 yeah. So, you will get actually these two solutions for rho naught rho naught can never be greater than 1 obviously. So, 2 plus root 2 cannot be the solution. So, only possible solution is 2 minus root 2. So, that is the maximum throughput performance which you can have out of a input output switch it cannot be better than this and this number I think turns out to be 0.586 58.6 percent. So, this actually means at the input side number of packets which are coming on every incoming port on an average per unit slot has to be less than 0.586 for stability. So, because what is happening is as p changes as p changes your throughput performance rho naught actually will change with this and it will saturate roughly about 0.58. So, throughput is saturating, but my input is high somewhere. So, if the input is higher and output is low what will happen something has to be stored flow is not balanced and where it will be stored in that input queues. So, queues will blow up ultimately which is not good actually. So, you have to always operate below 0.586. Now, question is can I do better than this is it possible I think it is possible to do better than this. So, what we do is this actually means we are going to operate if there is a input queues switch in 2 possible ways. So, that I can improve my throughput to about 80 percent actually instead of limiting myself to 0.586, 0.88 something I can really should be able to achieve I cannot just go to 1. So, what you have to do is very simple rule. So far my p is less than 0.586 I will keep on operating here. Suppose what I do is instead of this whatever packets I can transmit I transmit if I cannot transmit I do not keep them I do not keep head of the line blocking I simply drop those packets let the new packets coming. So, now and that will improve the throughput performance remember there are two phenomena which are happening here one is the packets which are arriving at the input when you are actually working with lower p lower arrival rate then less number of packets are coming. So, what happens is even if you keep head of the line blocking operational does not matter because packets most likely will not come and you should be able to transmit. So, you achieve a higher throughput performance after certain threshold if you do not drop the head of the line packets that blocking itself now will cause you the loss because sufficient packets are anyway arriving so that you can achieve higher throughput performance. So, from there this actually idea came. So, if you start dropping the packets how you will estimate a throughput every time there is a fresh lot coming in the front of the queue. So, what is the going to be performance what is going to be the throughput performance of the I output. So, far even if I get what is the probability that I will get one or more than one packet destined for I out of this end that will be the throughput performance that is the probability that you will have a packet in the outgoing slot. If you have one or more than one being direct destined for this if it is more than one then only one will be going if it is less than one or it is 0 then nothing will be going. So, 1 and 2, 3, 4, 5 for all of them only one packet will be going. So, you just estimate the probability that I will be greater than or equal to 1 that is nothing but your throughput performance no and this probability is 1 minus n c 0 p by m this is nothing but 1 minus. So, that will be your rho naught or the throughput performance and when n goes to infinity then what will happen this rho naught will become 1 minus 0 power minus p. Now, how many packets are being lost per line your incoming rate is p, p number of packets are coming per line per unit slot while your outgoing is rho naught. So, how many packets are being lost p minus rho naught packets are lost per line per packet. So, this is a per line whatever is being lost per line per slot per line per slot this many packets are lost. So, p are arriving rho naught is going out p minus rho naught is going to be lost and if you know how many packets are coming you divide by that. So, this should be your probability that a packet a tagged packet is going to be lost. So, this will become that probability. So, this is the probability that a tagged packet will be lost and then of course, now your rho naught is 1 minus e raise power minus p for a given p this is what is the value in this case when this is going to be greater than or equal to whatever I have solved for fully saturated conditions 2 minus root 2 find out the values of p for which this is going to happen. So, till 0.586 if my p is till that point it is fine after that I want to actually have higher throughput. So, what is the value of p? So, from here you should be able to get a value of p which is equal to. So, this is the crossover point after which you should start dropping the packets. Which one? And you solve for it actually it will let me solve for that this is very simple. So, if you write these two equations this will turn out to be root 2 minus 1 root 2 you on that put on that side. This 2 minus 1 is 1 1 minus root 2 minus e raise power minus p. So, root 2 minus 1 I will change the sign that is the way to come and then of course, this is nothing but you multiply by root 2 plus 1 divide by root 2 plus 1. This will turn out to be 1 actually 2 minus 1 is 1 and then you take the log. So, p has to be greater than equal to natural log of root 2 plus 1. So, the moment you crossover this particular value. So, you have to estimate what is this value you will find that your power this is remember is this throughput is dependent on p at certain probability this will be better than the situation throughput at that particular point of p you should switch over and then when you put p is equal to 1 here that is the maximum which you can achieve you cannot achieve anything better than that you put p is equal to 1. So, 1 minus 1 over e which is roughly I think 0.88 0.88 2. Is it we are applying the hybrid approach? No what is this value which will come I think this is 0.88 2 when you start dropping the packets, but you cannot actually operate you have to start dropping the packets when it crosses 0.886 and that is the maximum which you can achieve and anyway when you are dropping your packets your buffer will never be filled up everything which is coming afresh all the time. So, you have to start dropping the packet on the tail side of it also. Now, when you are dropping there is no there is only one buffer required for input so, this actually means you can actually operate with input switch, but you have to have operating regime. So, whenever your input arrival rate is less than 0.586 it is fine the movement create 0.586 is going to explode is better to switch over to dropping the packets and it will keep on operating and in this zone you will find out you are dropping this thing actually will always give a better performance for dropping the packet situation. So, this is actually the another threshold because of this. So, one is because of the stability region I have to switch over so 0.586 whenever it is there. So, till this point I will be operating with ordinary strategy ahead of the line blocking will be happening at this point actually stability happens. So, I have to switch over to dropping the situation, but I may not be still getting only getting 0.586 performance it will not be better it will only become better when I will cross over nature and log of 1 plus 2. Somebody has a calculator to find out the value I think it is 0.88 it should not be 0.88 should be the same value 0.88 till this after this point you will start getting the benefit. So, below this you are operating below without saturation region you are operating in saturation region here where you will only operate the throughput of 0.586 and after that you will get you start dropping. So, even if you had drop or does not drop does not matter because even if you are dropping your performance will be poorer than 0.586. So, q will remain unstable that is only problem in this zone. So, what you do is at the input buffer you limit your amount length of the q. So, the q get fully filled up you start dropping the packet that is only rate because you do not have infinite buffer you require infinite buffer for this operating region. So, usually buffers will always be finite once they fill up you start dropping you do not actually store any incoming new packet unless head of the line gets clear. At this point you start dropping from the head of the line because then throughput performance will be better than 0.586 because if you drop from head of the line here you will not get a better performance your row not will be lower than 0.586 at least you are operating at 0.586 which is maximum possible. So, at this point you will start becoming better. So, I think it should be something like this that of the line and then here it will become better and to reach to 1 over because that is what you get the maximum you cannot get anything better than this what is this value 1 over 1 minus e 0.38 0.38 no it should be high 0.63 0.63. So, how come this one values oh this is a row not that is p is equal to 1. So, it is 0.586 0.63 whatever you will be reaching here and you will saturate with that actually. So, 0.586 0. what ever is 16 which is this value that is where you will saturate with that you cannot get anything better than this is not possible when p is equal to 1. So, that is the maximum. Sir utilization is arrival rate by departure rate or departure rate. Utilization no it is not the ratio it is you look at the output line for how much fraction of the time the packet is there. So, you put a slotted system in every slot what is the probability of packet is there. So, if you observe for 100 hours. So, whatever is your utilization into 100 hours for those many times you will see the packet for remaining time you will not see the packet. Utilization how much you are utilizing the link. Generally in QC systems we will be taking lambda by nu lambda is arrival rate and nu is departure rate. No that is a load that is not utilization that is a load lambda by nu is a load factor that is nothing to be utilization you can compute utilization in those terms that is fine. Usually that will be that is known as load load is also technically equal to utilization under steady state conditions that also gives you the fraction of the time for which a link will remain utilization. Link will be having some packet. Here we are dropping the packet structure in Q and H 1. So, incoming utilization is higher than the outgoing utilization. Is there any relation between the next packet which is coming in the same. All are independent there is no correlation. Markovian arrival process actually means every packet is coming independently. I am only assuming a parameter called probability p. In a slot a packet is there or not there. Next slot if a packet is there or not there is independent. And how these lost packets are because how these lost packets. We do not actually what they have gone. That is a responsibility of higher layer. This is what is technically known as congestion. See a switch has the capacity but there is a congestion building up because too many people are trying to go through the same line. And you have to be dropped because it cannot go through. Congestion cannot be cured in the network in any switching node or any link. Only thing is that you can start reducing your traffic which is being pumped into. You go on to the road for example go to the city try to drive. There is a congestion what you have to do. Do you get a solution? Best is you can start improving utilization or some discipline. You can still improve the performance but if it is still there is a congestion you have to live with that. Best is do not go. So have a radio broadcast telling everybody there is a congestion on this road please do not drive through this. So incoming rate itself you are trying to reduce and that will then keep the congestion within limit. And that also what we do in networks. In TCP algorithm for example there is a TCP Reno. I remember it technically does this. It guesses when the congestion builds up it will start reducing its transmit window. So it will pump less number of packets per unit. But everybody should do it. It should not that there are 4 honest guys who are doing it. This guy says okay now whichever let me push through. So this kind of free riding can actually happen. This happens between UDP and TCP for example. UDP does not have congestion control. Last semester I have taught this. So if you are transmitting say audio video transmissions typically on UDP. When you are doing a file transfer network congestion happens your file transfer will slow down actually because of congestion control algorithm. UDP there is no congestion control. It will see a very good bandwidth it will actually even pump more. You can actually conduct this experiment try downloading a heavy file UGP and also try to play a video. Make sure it is a UDP transport this one is TCP. And let the link be congested. And you will find your video actually will still perform better. Not an issue. What will suffer is always the TCP traffic. But this is an observation we are still looking for better algorithms for doing UDP based congestion control. So in that field what is the axis corresponding to 5.586? This one this one is TCP. No sir on the y axis. Y axis is the utilization. No sir. The value corresponding to 0.586 of the y axis. This one. And roughly you will be increasing and there will be some formulation based on that. Sir what is that rho naught corresponding to? Rho naught is output port utilization corresponding to 0.586 the value. Maximum you can get is only 0.586 is this thing. Actually this will be lower this is not 0.586. But you will saturate that 0.586 this much I know. This 0.586 might be happening somewhere here. So your saturation will actually start happening before it. This is not the knee point. It is somewhere happening the 0.586. I have not done the exact calculation. But see by intuition important thing is by intuition and by understanding I should be able to predict. I should be able to generate hypothesis. So science is all about that. So gaining insight based on that predicting something.