 So, in the previous class I had actually drawn the transition matrix after step 1 for any switch in a state j and from there of course, then I also wrote the equation. So, probability that after step 1 you will be in state i in state j in time slot tau k in this time slot this will be given by if you know after 0th step or just before beginning of first step or the third step of previous time slot after that whatever is your state. So, you have to get that probability. So, if you are in state m then we need to actually multiply by all transition probability basically it is a row basically the whole column. So, this will be t now this is the same matrix which I had written earlier this is for state j for transitioning from m to i. So, for whichever things these are t j 1 m i is 0 those transitions are not permitted actually. So, only whatever are the valid those have to be used you will know this probability and you can always estimate this one and this summation has to be over m is equal to 1 to 14. This computation will be required sir if I need to converge to some value. So, what is the convergence condition and which value we had done this is from p 0 you will find out one I am going to come to convergence condition actually that is a steady state operation technically steady state is stable operation that is only we need to identify. So, now, let us come to the second matrix when we will go from state 1 to state 2 sorry whatever is there at the end of step 1 we have to go and find out what will happen after step 2. So, what is the transition probabilities in that step 2 actually. So, we will actually put that and that will be written as t 2 j. So, only thing the superscript will change from 1 to 2 because a transition which is going to happen in the second step. So, let me build up a matrix here. So, I am just going to put a 1 by 1. So, if you are in state 1 for example, remember in step 2 what happens only the packet from input to output actually that moment happens. Now, there is no packet at the input. So, no packet can go to the output. So, next state has to be only 1 and that will happen all the time. So, I have to put only 1 here rest everything will be 0. So, I will be I am just putting the entries which are existing all other entries will be 0 in the matrix. Now, look at the state 2 state 2 there is no packet which can go from input to output. So, after step 2 just before after step 1 if you are in state 2 step 2 will not make any difference you will remain only 2. So, there will be 1 here in this case look at 3 now. So, same is true for 3 if you will remain in state 3 only look at a state 4 now here. If you are in this state just after step 1 certainly this packet is going to move and once it moves you will go into state 2 and that will happen with the surety. So, you are going to put a 1 here all the entries will be only 1 and that is the important thing there is nothing like half or something because there are no exit and incoming probabilities p j bar and p a tilde are not participating here in a step 2 actually. So, the 5th one. So, 5th one will remain in 5th. So, 4 nothing will be coming 5th 1 there will be a 1 here 6th yeah there will be a movement out and you will end up in state number 3 state 3 I think you that is where you will come from 6 state 7 nothing will happen you will remain in 7 state 8 you will come to state 5 actually if you are in state 9 both of the packets will move out you will come to state 3 state 10 nothing will happen you will remain there in state 10 only state 11 you come to 7 actually right state 12 is also 7 state 13 you will remain in 13 state 14 will remain in 14. So, you have to just put somewhere a 14 here I do not have this space. So, I can put a 14 beneath 1 this will be a transition matrix for after step and I can again write down an equation. So, what will happen after step 2 you will be in a state i in a state j in time slot k will be given by whatever was the probability after step 1 transition matrix 2 in a state j from this has to be m. So, this will be a try see probability that you will be in state 1 is given by this and there is a transition probability you multiply. So, in one how you can come in what always for all other values it will be 0. So, probability of being in a state 2 even if it is there, but these transition probabilities are it does not matter. Similarly, look at what is the chance that you will be in a state 3 for example, state 3 is here i is 3. So, 1 2 3 transition is 0. So, only think 3 2 3 you will come. So, then there is a what was a after step 1 what is the probability of being in a step this is state 3 that has to come in because that comes that way actually. If you know the probability of being in a state this multiplied by transition probability you can be in this state. So, you can be in this state through this route through this route and through this route. So, sum of each probability into transition probability into transition probability sum of all these these are all mutually exclusive events will give you probability of being in a state 3 after step 2. Third one is similarly we will have third actually transition matrix this is this happens because of step 3 which is arrival into your incoming ports or incoming buffers. So, I will just draw it on this side. So, I call I actually have a superscript 3 in this transition matrix. So, if you have to go to 1 to 1 state. So, 1 to 1 remember it is because of the arrival process. If you are in state 1 there are possibility that 1 packet will arrive there is a possibility 2 packets will arrive at both the lines and there is a possibility no packet will arrive if 1 packet arrives. So, p j bar and q j bar you have to use there 2 possible ways. So, first thing p j bar let the packet be packet be there packet arrives actually p j bar square both packets will arrive. So, what will be the state both packets will arrive it is only arrival process no departure no transmission from input to output. So, you will end up in a state 9 and 8 both with equal probability remember 8 and 9 both will equal probability because they may be directed to any one of those situations. So, you have to write now. So, 8 and 9. So, this will be p j bar square p j bar square half and half when both packets are arriving when only one of them is arriving then what will happen you will always get in end up in 4. So, here you have to write 2 p j q j bar and if none of the packets arrive you remain in the same state. Similarly, state 2 you can have no packet arriving you will remain in a state 2 if 1 packet arriving 5 and 6 it will be either 5 or 6. So, 1 packet arriving probability is 2 p j q j both bars I am making half half of that and next one both packets will arrive both packet will arrive. So, there is one packet here. So, both packet will arrive one possibility they both are directed here they both are directed here one is directed here one is directed here one is directed here and one is directed here. So, this one is corresponds to which state 10 and next one corresponds to write this will correspond to 12 this one will correspond to 11 the next one also corresponds to 11 only. So, the probability for this is both are coming p j square is 1 by 4 p j bar square 1 by 4 p j bar square p j bar square. So, this what all actually has to be put here it will be 1 by 4 11 will be 1 by half. Similarly, for item 3 I think now should I leave or should I verify everything I can write down I think all the things and you please verify that will save time I think now you know the method how this has to be done that is more important. So, look at item 3 I think whatever is clear I will just keep on putting wherever there is some explanation required I will do that or actually it is not required it is possible 3 to 3 q j square bar 7 and then 13 and 14 this is from state 3 that is the way I think I can join them 8 and 9 now everything else will be diagonally unity actually. So, only initially you have to do some calculation rest everything is pretty simple after that. So, this will be the third transition matrix and from here you will actually get the third equation which will be what will happen after and this will be nothing but equal to p 0 i j k plus 1 sir how is going from 6 to 13 sir with half p j. This 6 to 13 6 is only one packet can arrive yes sir. If there is no packet arriving you will remain in state 6 13 sir 6 to 13 it is only arrival process no going from input to output port remember ok. So, you will remain in 6 if a packet arrives then this packet can be directed to any one of the ports outgoing port anyway arriving packet direction is also important. So, there will be two packets at the input and one at output. So, you might actually end up in choosing either this packet coming either this will be there or this will be there. See this is a situation 6 is this actually. So, if a new packet comes this packet can come here it has to come here only there is no other option and once this packet is here this can either be directed here or there can be an equal probability of this being directed here. There is no other possibility and this is nothing but if you twist these ports, but 11. There are 13 entries my answer for 6. Just hold on 12 and 13 yeah it is 11 and 12 you are right. I have made a wrong thing it is 11 and 12 rightly picked I think it is correct perfect you are right. So, this is clear now important thing is that when I am looking at what is going to be the state probability after step 3 that is nothing but after step 0 of k plus 1. So, step 3 in kth slot or tau k is nothing but step 0 in tau k plus 1. So, now, I have got the transient state probabilities for the next time slot through 3 iterations. Now, the method which actually is followed we also define something more here this is there in the paper actually you can even just make a copy from there. So, you cannot note it down you can just so we define another probability pj. So, we have defined pj bar pj tilde and now pj the last one this is a probability that packet exist an output link of a switch in stage j at a time tk I am not talking about interval this is a time tk because after time tk the first step will start then step 2 step 3 then tk plus 1 comes. So, that is why I was using all the time in earlier definitions interval tau k and I was defining a step because step can finish at any point of time. But, here it is this will be only happening at the slot interval k always because there can be situation your see analysis wise or what we will computational model can only be created for when t select and t pass either one of them is 0 that is why we have done the computational thing when you take for example, t pass is equal to half of t delay and t select also half of t delay is going to be slightly complicated in that case. So, paper actually only does a simulation thing for those cases and computational model is used only for the two extreme cases actually result will be somewhere midway in between the two extremes. So, this probability will be given by you can look into the states and find out step 1 sorry state 1 does not come into picture there is no packet if you are in state 2 with half probability you will have a packet at an output. Remember I am using 0 now here stage j time step k you can also use 3 k plus 1 that is all the same 3 k minus 1 3 k minus 1 I am assuming that it is for k p j will keep on changing remember and I have to keep on computing till p j stabilizes because in the beginning I will be starting with the extreme thing cases. Then you have only fifth state where there will be only one packet just you will list all the states where there is only one packet at the outgoing port and then plus without half all the states where two packets are there at the outgoing port. So, 2 5 6 10 11 and 12 and then both of them are there it is 3 7 13 and 14. So, now the procedure actually of this what is the difference of tau k and t k tau k is the whole interval from t k if this is t k this is t k plus 1 this interval is tau k t k is an instant is a time instant. So, what will happen at t k all switches will be in a step 0 or just before the step 1 they all will execute a step 1 then they will all execute a step 2 step 3 1 by 1. So, you will actually can if you can visualize its kind of remember when you are running one here 2 is running in the back next stage 3. So, from output port it will be flowing backward actually. So, this will go to 1 then next one will go to 1. So, this will move to 2 then this will move to 1 this will be 2 this will be 3 and so on. And once the 3 is done then it will be 0 for the next step. So, there is no execution after that. So, in your computation you are just running like a wave actually of 3 steps from forward to backward direction. So, all 3 will be added to the next time slot will come computationally it will be that way actually. This is actual probability because the throughput performance will be computed from this P j tilde will not give P j tilde is a conditional probability. If the packet exists then it will go out that is a P j tilde. If a vacancy exists at the input the packet will come in that is again a conditional probability this is absolute probability because all state probabilities are nothing but absolute probabilities which we are handling. That is what denominator numerators were actually computed because they were conditional probabilities. This will not give absolute throughput P j dot P j tilde. So, this is a rate at which the packets will be going out from a port and this will be same because no packets are being lost. So, technically incoming rate has to become equal to outgoing rate this will be used as a test actually during all your iterations when this will become same for all stages you have stabilized. Now, things cannot change anymore in your computation at that time you just measure what are the state probabilities. And once you know all that state probabilities you know this also because from those state probabilities you are anyway computing P j 0 P j and P j tilde. So, everything is stabilized and now you know state now you know this you know P j tilde you can find out the throughput performance from here take any stage does not matter this will be same for all stages. This will become independent of j in the beginning it is not because there is a transient in the system. So, that will be the check for essentially convergence sorry sorry sorry sorry sorry sorry it has to be P j it has to be P j. So, let us come to the procedure I have verbally told it, but let me just write it down a step by step how it happens. So, what we have as the boundary condition as 1 or 1.0 whatever way P 0 bar is going to be 1 these are two boundary conditions which will be used initially when there is no packet in the system at time t is equal to 0. I can get some kind of a initial values are required for the state probabilities to that I can iterate. So, all the state probabilities at t is equal to 0 all of them will be in state 1 will be 1 for all j's for all stages there is no packet in the system for all j and m is greater than equal to 1 greater than equal to 2 and less than equal to 14 that is initial condition. So, first you are going to find out and for all j this is the first step which we will be doing you will find out this thing for all stages which is viable and then you will compute for all stages. So, with these conditions you will find out then P 2 remember transition matrix every time you are computing transition matrix based on the probabilities you know the earlier things. So, you can always find out using those equations the next step from there you will get now again for this and this will be nothing but will become equal to P 0 1 m j go back again iterate keep on iterating this thing and you will keep on iterating till you will find that your P j dot P j tilde alternatively for all j's this should become same some constant k it will just give some constant k value this is a throughput part port actually is a fractional utilization of the port. Sir just repeat we have done this iteration sir we keep on doing it. Keep on doing it after some iteration you will you all the time observing this value also every time T k you will estimate what is the value of P j and P j tilde. But in this iteration you are not calculating P j tilde. P j tilde is dependent on the state probabilities remember. So, every step you will be computing you are going to compute the matrix you are going to compute the transition probability matrix and those P j tilde and P j bar for all stages this has to done iteratively every time and this you will keep on doing till you find this value is turns out to be constant for certain number of computations. So, paper actually does it for about 100 iterations it should see the same value actually it will be always converging converging converging all the time minor correction. But if it is going to happening at the tenth decimal place you do not bother further actually. With time actually if you plot you will stabilize that is a convergence of the algorithm and once you converge you have achieved the values now you know all the state probabilities you can do whatever you wish now with it. An important thing these initial values are fine this P n minus 2 tilde is 1 if you change P j bar is equal to 1 from there to some other value you can change the load factor incoming load factor. So, this for maximum loading condition when it is 1 you can make it half also and then start doing the computation then also you will get a throughput. So, throughput versus load also can be done by getting the various points that is not maximum loading is when P 0 bar is 1. So, why you change only 100. That is a paper this guy must have observed after doing computation for 100 it remains stable it does not matter then mostly it is it is basically by observation. 100 for a 1 way to walk n. n there is no n here. Then and the state is why it is changing. See size of switches does not matter now size of the switches now being reflected in terms of number of stages. The number of ports you do not worry is the number of stages what matters and since it is a delta is a buffered delta buffered banyan actually whatever it is because does not matter whether it is delta or not a delta here. I have not used anywhere anything related to routing what I am stating is there are two inputs both inputs should be independent of each other that is the assumption which we have used this input arrival at this input and arrival at this input are independent departure at this and departure at this are independent and this will happen if they are always connecting to always disjoint sets which is true in I think banyan network by design actually that is true if there are no loops there is exactly one path between any input and output port this will be satisfied. So, this result is true irrespective of the topology. So, this is what has to be done for all j's this has to become constant and after this you can actually find out the throughput. So, throughput will be what? So, throughput will be now remember is p j dot p j delta that is what is going to be constant, but p j might be different across the switches p j will be different across the switch across the stages because if the p j is higher you will find p j delta will be lower as you go across p j will be going down and p j delta will be increasing as your j increases actually, but this relatively will become product will become constant that is a flow rate. So, whatever is incoming has to go out that is what the condition basically. So, there is no storage in the switch which is stabilized now in a stabilized condition because you are starting from your initial state is all 0's there is no packet in the system. So, it will start storing packets actually because they are buffers inside. So, once the storage is stabilized this condition will be satisfied in that case. So, it is like if you have for example, a water is flowing you have a tank and then tank is going out. So, this buffer if it tank is empty the rate here will be higher rate here will be lower you can build up a cascade of this technically is nothing, but something similar another tank another thing. So, find out flow rate here find out flow rate here find out flow rate here you will be in steady state when all these three flow rates are same and this is nothing, but flow rate. Flow rate at the output port of every stage has to be same and flow rate is same at all ports does not mean that your P j will be same at all ports P j's probability that packet exist. So, this will always be higher for the initial stage or the input side and will reduce as you move to the output side and P j tilde will be reverse. So, they both will stabilize. So, one will be actually decreasing other one will be increasing this product will become constant. There is no relation P j bar in relation P j bar P j bar also if you want you can push what will be the flow rate for P j bar it will be I will say Q j Q j bar. So, the P j is the packet Q j packet is not there and packet arrives that is incoming port side flow rate. This has to be constant across all j's that is also fine this also is the flow rate for all j's if this is going to also become constant then also you have converged that is a equivalent condition symmetrically and you can similarly look at what happens Q j will be smaller on the input side stage. Stage is a closer to input as you move towards the output side this Q j will start increasing while P j bar will be smaller in the input side and it increases it decreases actually as you move it will be higher it will be decreasing as you move towards the output side. So, I can write there it is P j bar or P j in this side this is nothing, but P j tilde or is a pairing basically. Sir, what is the guaranteed. So, if you want to take you either take this as a pair or you take this as a pair under stable conditions their products will become constant you are asking some question. Sir, what is the guarantee of convergence although we say that one is increasing. Mostly it is a first order system. But take any consideration the rate at which one increases or decreases. This is only a intuition actually honestly speaking I do not have the proof whether it will converge not, but since it seems to be like a first order system there is no second order. So, if you make a difference equation you do not get a second order difference equation you get actually first order difference equation that is why I think it should converge. But that is my intuition I have not verified that anything which is first order usually will converge only second order will actually show the oscillatory behavior or resonance conditions and under which the convergence cannot happen only important yes. If it is a first order your decaying coefficient has to be negative if it is positive then it will explode then also it is unstable. But instability cannot happen because the probabilities are bounded from 0 to 1 because what happens in this case this is what we call total things are conserved. There is a principle of conservation actually being used here honestly speaking total sum of all possible events which can happen mutually exclusive sum of probabilities will always be 1. So, once probability increase others one will go down. So, you cannot get that case of exploding to infinity in first order system. So, this first order system also cannot oscillate because since it is a first order. So, only possibility existed this first order will always converge. So, intuitively I know this is the way it is going to happen. But I think exact way is you build up actual differential equation as time evolution happens and based on that you make a judgment that will be the more sophisticated and better approach to prove it paper does not talk about it. I think again the paper the author has done it through this intuition itself and since the intuition they are satisfied and they have seen it happening they assume their intuition is correct they never tried verifying it formula. So, once you have this you can now build up what we call the throughput performance. Let me write down the throughput is what it will be. So, what are the chances that packet will be there at the n minus second output port of the n minus 2 stage and if this packet exist this packet will go out immediately this we know because p j tilde of n minus 2 is 1. So, that is why we have taken actually I have to always find out this flow rate take this flow rate. This is what is going to be transmitted in 1 t delay and that is what gives you the throughput and total number of ports will be 2 raised power n divide by t delay. Alternatively if you want to put in terms of j it will be under steady state condition this will become independent for all j's this value will turn out to be same because this is constant all across and p j p j tilde is nothing for j is equal to n minus 2 this will be nothing but p of n minus 2. So, this is what I am using and this what gives you the throughput performance and what is a maximum throughput which we can get 2 raised power n packets per t delay which can be moved out. So, that is the maximum which you can get when p n minus 2 is 1 actually. So, you divide by the maximum possible value that gives a normalized throughput. So, normalized throughput is that is basically fractional utilization of the link this is normalized throughput and turn around time is turn around time is the time required for the packet to move from input to outgoing port delay part. So, that also similarly can be estimated here. So, what you can do is t delay multiplied by summation of in each stage probability of it is not going out is this happens for k minus 1 times then it goes out with this probability. So, it means a packet is going to in a stage suffer k delay this is the average value I am looking at the average estimate and k can go from 1 to infinity. So, this average once you find out this I am leaving it to you I think this string theory theme principle you can always use to find out this average value. This will turn out to be p j minus 1 sorry this has to be j minus 1 j minus 1 because I am looking at in the j th stage at the input how much delay is suffered. So, that will be in terms of the previous stage what is the probability from it is output buffer the packet goes out that is why it has to be j minus 1. So, this is what will be your average value and of course, this is the one stage. So, total amount of delay will be nothing, but t delay summation over 1 over p j minus 1 tilde go from j 1 to 0 to n minus 1 I think total delay in that transmission from 0 th stage to n minus first stage all stages I have taken I have done summed up actually and what is the maximum delay possible is total number of stages into t delay. So, you can divide by that. So, your normalized delay will be 1 over p j minus 1 1 over n that is a normalized delay that is what will be suffered. So, I think with that I close here the buffer delta analysis only thing is the results which are need to be seen, but my idea was not to actually discuss the result result will always be better if you use buffer delta. An important outcome of this paper was that with one buffer between two stages or two buffers between two stages after that does not matter 3 or 4 or 5 does not have any improvement in the performance. So, 2 is what supposed to be optimum value which gives the best performance. The performance is better than unbuffered delta surely we slight increase in delay which happens and this almost become as good as cross bar. See cross bar is better performing thing than delta because of the blocking cross bar does not have any blocking. So, you still use a delta, but by using buffering you are able to reach to become as good as a cross bar. So, that was essentially the outcome of the paper, but I was more interested in teaching the methodology which was involved because this is a very generic thing can be used at lot of other places also. So, I think with that we close on the buffer delta system.