 So, we will start from where we left in the previous lecture. So, I will now give how you actually iterate. This is a particular specific scenario of how you solve Markov chain. Technically, it is actually solution of a Markov chain by doing iterative computation actually, because sometimes when this number of states are extremely large, it is very difficult to get a solution. So, really the way we always teach the solution is we always say, let us build up a Markov chain where we have some state probabilities and based on that, then you start building of what we call balance equations and then using a fundamental axiom that all state probabilities sum has to be equal to 1. We actually get all simultaneous equations to solve them, get a state probabilities. That is what usually we do. Here actually, I am not interested in that. I am more interested in what is going to be my throughput performance, what is going to be the outflow, but that actually depends on the states. So, last time I had drawn 14 states of a 2 by 2 switch and we will take two modes. One case will be when in fact, for every switch when the packets are there at the input, there are two ways I can actually understand. Two ways the analysis can be done. One is the total period which is required in this case is what we call that I just t delay actually. So, t delay is one step delay from here to here. This delay usually will consist of t select plus t pass. t select is the time required to identify to which outgoing port these packets have to go and then when actually the transfer happens that requires this much time. So, we will take actually first the case when t pass will be t select will be 0 and then t pass will be 0. There are two cases actually. So, now one important thing I have given you 14 states. I will assume that because now switch will actually consist of buffer delta, consist of many switches in a state and there are many states actually and so on. So, all state probabilities within a stage are going to be same for all switches. All state probabilities for all the switches in one single stage is going to be same that is the assumption. So, I need not bother about all these switches because otherwise you see 14 states I have given and then you have how many switches. So, 14 possibilities raise power number of switches those many number of states will be there and even if it is a very small number say you are going to have 8 by 8 switch. For example, 8 by 8 switch requires 12 such switches. So, 14 raise power 12 is a huge number actually and maintaining those kind of things in a Markov chain or in any finite state machine is going to be complicated. You cannot handle that actually. So, analysis cannot be done in that way. So, first simplification which was done is this. This is usually true because number of ports which are coming in these are independent states and that is true for all switches because what is true for first switch result into for the second and so on iteratively. So, this is all switches technically see the same thing. So, their state probabilities has to be actually same. Now, this is going to be true. So, I am going to take this case this is the one important thing. You have to keep your state diagram handy all states actually. So, I will just still put them here I will not erase them during the lecture because I have to refer again and again back to these states. So, I think I can redraw them. So, this is one. Now, before I go onward with the iterative method of computing the state probabilities that is going to be pretty simple. For example, if you want to find out for a cube for a single cube I think I can just explain how it will be done for that. For that actually usually I can find out a very simple solution because my balance equations can be formed. So, with this if I actually being given you a 0 transition probabilities 1 transition probabilities and so on. For example, this is an infinite cube. So, usually actually you can always run an algorithm on this. So, you can start with a cube when it is empty and now we talk about the state probabilities at various time instance that is very important. While when we did a queuing analysis of a simple cube I was not using time instance for a state probabilities. So, at t is equal to 0. So, I will always define p 0 as the state at time t. So, this is k. So, k will be 0 in the beginning. So, your initial boundary condition can be that there is no q that is a probability and then you will have transition rates. So, you can have to define transition probabilities. So, lambda is the rate for going into one direction and mu is not going into that mu is coming back. So, you will define this as lambda and the not going to go there is 1 minus lambda actually. Important thing is transition is happening at every time instant it is a differential. So, I am now making a very different kind of q here at on a time scale at t is equal to 0 packet will come t is equal to 1 something will happen t is equal to 2 this will keep on happening it is not as such a Markovian thing. So, lambda can I can replace this thing by actually p the probability p is a discrete time system now. So, with probability p the event packet comes 1 minus p packet does not come with packet again the same thing is going to happen at every time instant. So, from here actually you can find out what is going to be your p 1. So, from here you can actually find out whatever was the value p 1 at instant 1. In fact, you can build up an equation based on this. So, for 1 I will show then I can write it for general I will use a very similar procedure there. This one will be that you have to be in this state in the previous time instant then there is a transition probability which was I call it something else I think let us call it lambda lambda is not a rivalry it is a probability you have to understand it. So, lambda is probability, but this is a computational procedure you cannot have a closed form solution here. So, with this you will be able to come here second thing which you can get is p of 2 from here also you can come and this probability is defined as mu, mu is the probability that packet will go out of the packet actually out within that slot is it require same slot it is technically is not arrived at it is a probability it is a conditional probability technically. So, if you are in this state chances that will go here is this, but it also turns out to be same as rate. So, this method is not used while doing for a simple mm 1 infinity q. So, I have modified the q somehow. So, you get one equation through iteration remember my time constants are different here and to begin with my initial condition is that this is 1 rest everything has to be 0 because of this rest everything has to be 0 and q is also in state 0. So, I will be able to get p 1 1 from here I can in fact solve for all these in general. So, I can get p of say m in state k as p of m minus 1 lambda k minus 1 plus p of m minus 1 m plus 1 actually k minus 1 into mu. So, now you actually can you have to just run a computer program start with this initial condition and as your time moves you are computing every time p 1 k p 2 k p 3 k p 4 k you are just iterating and p 0 k will also get adjusted correspondingly you have to understand p 0 k will also be changing will be a function of p 0 k minus 1 1 minus lambda. So, what I am doing is we are adjusting continuously we are reducing it by some amount and increasing all other state probabilities and after some time in the computational procedure itself it is always bound to converge you will find there is a steady state condition which will operate and steady state condition is when the flow rate almost becomes constant out of a system. So, when for all of them you will find out that it happens that p of any value any m and say k this almost dot lambda lambda is actually constant here departure rate. So, the flow rates here statistically has to become constant they will not be varying or when the flow rate becomes gets balanced that also you can figure out. So, in this case condition is pretty simple here probability of 0 are we not transferring the probability from state 1 to 0 yeah that also need to be there you are right that also has to be there you are right I agree k minus 1 that also you have to make that. So, when this condition or the balance condition actually you see start seeing your computation things have converged take those state probabilities and you have got your results. So, with various values of lambda and mu actually you can find out that thing it will turn out to be almost very same what we did for the simple queue. Now, this same procedure we are going to invoke here because here we cannot get the close form solution this Markov chain was very simple could have been solved, but that one we cannot do, but just to give an example I have just simplified this thing into this particular this is a computational procedure, but we this comes in the category of still analysis it is not a simulation because we are not simulating anything we are just computing and it is a iterative computation till the computation is stabilized or converges. So, we are technically solving balance equations and balance equations are when the flow rates from in to out and out to in both are same. So, you are just finding out when the flow rate converges and everything is going to keep on it will actually balance remember it is like a first order equation it is not a second order. So, there is no question of oscillations here and if you carefully observe it is a first order actually equation it is always trying to move to a stable condition. So, convergence is guaranteed actually this is fine this one was a simple q now I am going to come to complex one. So, state transition probabilities are complicated I think one important thing is you have to understand how state transition probabilities are estimated and how the how you actually understand the whole abstraction that is the important thing. So, far the things were I think little simpler now coming to this first case which we will take is when t pass is 0. So, there are two cases when I can take t pass is equal to 0 or I can take t select is equal to 0 this is all condition now what is the meaning of this actually what is the implication of this. So, I think once you understand this then it will be much more clear in the paper it has been given later I am doing it early actually I somehow feel this should be done. So, I have to take the cascades of the switch let us take this case. So, all switches are empty there is no nothing in the buffer and they are having exactly one buffer at every input port does not matter you can actually then do the simulation even for two buffers or infinite buffers whatever you feel. So, if you have two packets coming in now you have to look at the time instants. So, this is at one part of a time instant it has come then the t delay will elapse after this and then the next state will come. So, how the state transition will happen and this case I am taking as t pass is equal to 0. So, this t delay will be nothing but equal to t select this is the time required for identifying to which outgoing port the packet has to go, but packets move almost instantaneously from input to output from any buffer to any buffer. So, once the decision will take this much time you know there is a packet at the end of this t delay what will happen one packet will be almost instantaneously just before my time period finishes just before that 0 minus t minus actually. So, if it is 0 instant then just 1 minus at that instant the packet would have moved here your situation will be something like this if both are I am assuming contending for the same port you will come in this situation. Another t delay will happen you will get a situation like this now. So, in this t delay they will figure out it has to go to which port this has to go to this port earlier also tried, but there was a contention. So, it was buffered this also will be identified and just before the next time slots just minus just slightly earlier the packets will be moved instantaneously. So, you will actually have this packet coming on to this position and this packet coming on to this position. So, you will have this situation. So, that is when your t passes 0 drawing with a different color when t select is 0 then what will happen. So, it actually means when this particular period from here to here that slot which is starting within 0 time I am able to figure out which packet has to be going to the outgoing port then it takes one full slot for the packet to be transferred to the outgoing buffer. So, after this situation this is what is going to be there perfect. So, this is what will be the situation another time slot delay. So, in the beginning of this particular slot they will know that this packet has to go here and this packet has to go here, but the problem is this packet cannot move because this buffer is not empty because in the beginning of the slot itself you have identified where the packet has to go, but when you start transferring this packet will take full one slot for transmission to the outgoing buffer and unless this buffer is empty this cannot move. So, this will still remain stuck there. So, you will end up with a situation here where this packet will be here and this packet will be here. So, that is the difference between the two. So, analysis when t select is equal to 0 is simpler actually. So, I will do a complicated version first when t passes 0. So, final state will be there will be no packet in the middle stage? The one with the red ones you look at maybe I can draw it separately if you wish. So, colors you have to match. So, red one corresponds to red one, white one corresponds to white one. So, now for when t passes 0 you have to understand now if there are packets which are queued up like this in a string all of them packet will move in the same slot to the outgoing port. Decision will take the same time one slot period then there is instantaneous movement. So, all of them will move in a chain. Now, solving this is slightly tricky. So, what happens is each time slot or each event is being now identified into three phases. So, first phase we call when the packet actually goes out when you are this is the outgoing buffer when the packet from your outgoing buffer or the input of the next stage input buffer of the next stage is pushed out that is a phase one when the phase one is happening. So, this is a step one actually we do not call it phase one this is a step one here a step two will be when the packet from here to here will come. See, I have broken things into steps. So, that when this goes out I can immediately move the packet in the same slot here which is not possible when t select is 0 for t pass for simulation or iteration I need to do this. So, that is a step two and step three is that packet comes to your input buffer. So, you have to understand when step one is running here which step is running here is step three when step two is running here no this is not a step three this is step two actually packet is going from input to output of a buffer this is a step two the same thing is happening here for this. So, while at the same time the step three is happening for this person these are for t pass t select is equal to 0 you do not require this definition this break up is not required per slot well if you carefully think I think this is difficult to explain, but this is the way it happens this is the only way it can happen actually. More complicated is t select is equal to 0. More complicated t pass is equal to 0. But then I have to split it when I am doing iteration see it is a problem of how you implement the computation their computation is simple every one step one time slot one step, but here every time slot is broken into three steps and this actually does allow all the chain to move in one go. So, when a time slot starts you will say step one here then step two then step three. So, everywhere you will do the step one step one will happen first then step two then step three for whole switch. So, usually what will happen is this is the packet at the outgoing buffer step one this packet will go out instantaneously here when the step one happens here then you will apply a step one here and at this point step two is going to happen here is a chaining which is happening. So, when one is happening here two is going to happen here and then when one is going to happen here then two will happen here and three will happen here and then of course, two will happen here three will happen here last three will happen here and well full chain of event. So, one complete slot will be simulated one two three all three has happened for this there is actually nothing to do in with one or two one has happened in the previous time slot. So, currents and time slot I can implement this way, otherwise it is not feasible. So, now coming to the formal definitions, which are required here. Never done t pass is equal to 0, we require minimum 3 states 2 states. 3, t pass is equal to 0, 3 steps per slot. And 3 stages also. Any number of stages, there can be a state. Minimum 3, minimum 3 stages. Not necessary, even with 2 also you can. So, it is like a state machine. One full slot is gets done, when all 3 steps are executed. So, what happens is, second step. Second step packet is coming from your input to your output. First step packet from your output has to go. Output has to go outside. From your output the packet is moving further, going out, passed from your input to output then packet has to come. Then from a packet has to come to your input. These 3 things will give a slot for a switch. So, which is my input set in this case out of 1? Any particular switch. In particular. Which one? In step 2 and 3. Then what is it? This is the, I am talking about this switch. When for this switch, this step 1 is happening. This packet is moving out. For this switch, step 2 is happening at that time. No, this is also for in step 2 and 3. This is step 2 is for this. Here the packet is going from input to output. But then for this switch, the packet is coming to the input. So, there is a step 3 for this and step 2 for this. And in the last one packet is coming to my input. So, the previous switch, you are having a step 2 happening. So, it is always 1, 2, 3 and it actually propagates backward. That is what is happening. So, when full 1, 2, 3 set propagates from input to output to input, one slot has finished. But that is the way we assume. T delay is 1 delay. T delay is delay here, remember. And every after T delay, packet moves further. So, 1 T delay requires 3 instances, 3 steps. And selection time is full T delay in this case. But packet passing is instantaneous. So, we define now what we call our probabilities before we come to the states. So, notation is very simple. So, at what happens at time when slot starts, you are at state, you are at step 0. Then step 1 will happen, then step 2 will happen, step 3 will happen, step 3 will be nothing but a step 0 for k plus first slot. So, step 3 for kth slot and step 0 for k plus 1 are same. So, now we will actually define at time is 10 T k. Now k can go from 0, 1, 2 and so on because iterative calculation, these are stages in the switch. So, I am not worried about how many input ports are there, 2 raise power n by 2 raise power n, but number of stages matter because I have made an assumption which is a correct assumption for almost all delta networks that in a stage all switches are equivalent. The state probabilities are going to be same, so far they are in the same stage. So, under that assumption and we will define now a probability we call it p 0. This is after step 0 actually, in time slot k, k starts from, so that is a slot and I think do you understand what is the meaning of this, close and open intervals. So, in this case T k is included, T k plus 1 is not included, that is what it means, it is open on this side and close on this side. So, this particular probability is probability that your switch in jth stage, this is this, is in state, this is not a stage, this is state i, this one. So, notation you have to remember, do not change it, it has to be consistent actually at time T k. So, if this is the T k instant, when the time starts this is that value. Now, this whole interval is known as T k, T k is this interval, this edge is not included, this is included. So, in T k I am assuming that states 1, 2, 3 are there. There states are there, steps are there, steps 1, 2, 3 are happening. I am going to write down those probabilities, p 0 then p 1, p 2 and p 3, p 3 will be nothing but become p 0 for the next time step, next t delay. Time t k, the switch is in state i, switch is switch in jth stage at state i, there are many stages. So, for multiple stages we have to solve and is giving the formula is an actual calculation has to be done through a computer, p 1. So, this number identifies the step after which you are looking at the probability, state, stage, time instant, time slot. This is pretty common in most of the switching analysis, you will use lot of subscripts, superscripts with all kind of notations. So, this is a probability that your switch in stage j is in state i in time interval. I am not writing time instant now, time interval tk after step 1, there is no step 0, third step is step 0 for the next time. Which one? This after 0th step, just before the first step, then first step has taken, second has happened, third has happened, third is nothing but the 0th of the next one, next time plot. So, I will change there, k will be changed. Three steps, but counting 0 and 2. I am counting 1, 2, 3. 0 is same as 3. 0 is same as 3, but time instant changes, k will become k plus 1. See why I will step, I will not count 0, 1, 2, 3. These are objects which are counted or ports which are counted 0, 1, 2. I am being consistent with the paper, because ultimately I know this is difficult to communicate in the class. You will anyway have to depend on the paper. And if my notations are not same, there is going to be a problem in that case in understanding. So, I have to be exactly same what is given in the paper. The certain nuances which have not been explained. For example, t pass and t select difference has not been explicitly told. The procedure of computation is not explicitly stated, because the author assumes that what happens in most of the papers we do not give details sometimes. We assume there is a kind of a common knowledge and person can figure out on the other side, but I know the class cannot figure out as of now. Unless you are a PhD student with a good amount of experience, it is going to be difficult to figure it out. So, I could do that, that does not mean that everybody else can do. So, I have to be explicit and at least you have an option to fall back. You can always ask me if there is a problem. I never had that option. With this paper is pretty old. I do not know where I had these guys, but I think this was the first one which gave for buffered data the analysis part. And it matches with the simulation that was good thing. And this has taught me a different method of solving Markov chains iterative computation, which is again usually not taught at most of the places. So, switch is in, switch in stage j is in state i in time interval tau k after step 1 and 2 both has happened. Similarly, you will have P 3 k and this has to be equal to P 3 i j k plus 1. Rest everything follow exactly except in the last you will write step 1, 2 and 3. Rest everything is same. Next time it will be tau k plus 1. Yes, next will be tau k plus 1. So, this let me write it. I think this is better because switch in stage j is in state i in time interval after step 1, 2 and 3. And we will define one more probability actually. Now come the trickiest part. Now what I am defining is actually will be the probabilities which are used for computing what we call transition probabilities at every state. And these will also keep on changing with time as time evolves and they will stabilize after sometime in the Markov chain. So, first probability is P j tilde. So, I am going to use three of them. I think one has to be careful. I will be using this, I will be using this. These three have all different meanings. So, first I will define the topmost one. This is the probability that a packet at switch output stage j, this j actually identifies this is passed in time interval tau k. So, this actually is indeed a function of k. I am not explicitly mentioning it, but all these P j tilde, P j bar and P j will be function of k. I will not be writing them except paper also has not written it. So, I am keeping it as it is. Ideally I should have actually put a k, k superscript there for doing this. So, this actually means this will be in terms of now the states. Whatever those 14 states are there and I have already defined how the probability will be defined after every step. And we know what these probabilities actually mean. What happens in step 1, what happens in step 2 and what happens in step 3? So, from here if there are two stages, if this is the jth stage, what is the probability that a packet will go out? I am trying to estimate that. Now, usually what will be the function? This will depend on at the start of the time slot. What was the state of this switch? And then the step 1 will be happening for this and step 1 for this and step 2 will be happening for this. And there is a switch previous to this that also you have to take into account because their index j actually will change when I am going to estimate for P j. I will be using in this case j plus 1 actually. And that decision will happen is what happens after step 1 here because step 2 outcome will be decide by what has happened in the step 1 here. So, that is why the expression actually will contain all j plus 1's terms, but what is going to happen after step 1. So, let us see what is going to happen. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . It is a conditional probability that packet is sitting here, what are the chances it will move here? You have to identify basically the states for this, what is the earlier state and after step one what is happening here? That will decide the probability of transition for this. See, whatever happens here will have an impact here, if packet moves here then only state will change here, packet does not move, state does not change here. So, this is a transition, this actually will be used for computing transition probability here, this is outgoing packet probability, but that will be a function of what happens in step one here after step one, if the packet goes out or does not go out, packet does not move out from here then nothing will happen, this packet cannot move. So, essentially what I have to do is I have to now list, this thing can be written as, so this will be nothing but probability of this such that as t, as to essentially identify states when this is possible and remember this is a conditional probability. So, I have to identify all the states when the packet is there at the input of j plus first stage. This will become more clear when I will give the example now, build up all that, it will give the examples how this actually works out and of course for this, the boundary condition before that is this. At every stage you will identify q of j by this, these two are why these two in the last stage if the switch is there, there is one buffer here packets are instantaneously removed, there is no packet queued up here, there is no buffer required at the output, if these two packets are there in the last stage and they both have to go to any port, even if they have go to the same port, in the there is only time required is t delay which is for selection. Selection is done when both of them will be instantaneously put here and they will be instantaneously taken out. So, probability that this particular packet will be which is there in, this is n minus first stage remember, this is n minus 2. So, probability that the packet at the output of n minus 2 will be passed out, has to be always equal to 1, if the packet exist condition on that, so that is the initial condition. So, this is known as boundary condition of the solution, you have to start with certain boundary condition always. Initial condition will be when all switches will be in this state, state 1, no packet in the system, when you start your iteration. So, boundary condition t tilde n minus 2 is equal to 1.0, there is no output condition at t n minus 1. Not possible because packet is taken out instantaneously, t select is equal to 0, when t select is equal to something is equal to 0, then this will be problem will be there, you cannot take, then it has to p n minus 1 tilde is equal to 1, it is not p n minus 2. t passes equal to 0, t select is what we have gone through, t select is equal to t tilde. So, in full slot both of them will be selecting the one output, but if there is output for it does not matter because time t is 0 for reading out the packet. So, both of them can be read out instantaneously and they can be read out from the output port. Sir here since. Very closely. 0. Is 0, even in case. Since I am only considering here packets at the input, that is why I am taking only step 1. Packets at the input, I am bothered about the j plus first stage, what is happening here? See if what I am in some state, I have some packet queued up, what happens to my outgoing packet that will decide what will be my next state, whether packet can move or not move. I am looking into transition probabilities now, whatever is my current state here, this can only get modified if my packet moves and that depends on what is the state probability after step 1 here. So, that is why this has been done this way. If there is contingency in that case, they will not be able to do both the packets in case there is contingency. Only at the last port, only at the last stage, not here. See what happens for example, if you put packets here, if these two packet does not go out, both of these want to come here. So, you have consumed the full period and finding out they have to come here, but this buffer is not free, buffer is not free they will remain there. Remember the patterned model of the switch, there is a backward flow of the signal which is there, packet can only move if the buffer is empty in the next, at the input of the next switch. Here you do not consider packet dropping here. There is no packets are not dropped, they are all buffered, it is a come full loading condition, remember the initially assumption is there. There is a maximal loading condition, I am only estimating what is the maximum achievable throughput, but I am actually using a trick, I can actually modify the input probabilities, but input probability initial conditions or boundary condition I am taking such that it is a maximal loading condition. I will write actually more, this is not the only expression, I have to write p j bar also. So, now this will be, I have to first of all write down all the states when the packets are there at the input. See then only they can go out, if the packet is not there at this port, they cannot go out, I am worrying about here. So, I have to look into all the states in j plus first stage where the packet is there at the input. So, first one, can I put one, I cannot, there is no packet at the input. So, 1, 2, 3 cannot be used, 4 can be used, there is only one packet. So, the next stage switch is in 4, there is a chance I might be connected to the one where there is no input buffer is not occupied or I might be connected to the top one where the buffer is occupied. Both of these can be connected to my switch in j th stage with equal probability. So, I will use half of this p 1, 4, j plus 1, I have to write all that time is 10 k, half is because I can be equally likely connected to any one of those, if the switch in the next stage in this state. So, my possibility is half 50 percent of time I will be connected to the one where input buffer is occupied 50 percent time, buffer is not occupied. Even if I flip these input ports, this I could take on top, this on bottom, it is still the same state remember plus what is the next one? You have 5 also similar situation, but remember in 5, if the next stage in that state the packet will not move, packet cannot go because the next buffer is not empty after step 1, this is after step 1 actually. Step 1 is already taken place in j plus 1, outgoing packet has not moved out. So, my packet also cannot move in. So, 4 will be sitting in the numerator, but 5 will not be sitting in the numerator, it is only in the denominator for this conditional probability. So, I will write P 1 5. So, I am intentionally writing it first thing here then I will only take some of them on the above part, all of them will not go. P 1 same as with the 6th, there is only one input port, I cannot correct it to any one of these, but 6 packet will go out. So, it will go in the numerator also. Similarly, next one will be half P 1 7 j plus 1. I think you have to take a large copy, if you take a A 4 sheet print of the paper that tilde and bar are not clearly written on the paper actually. So, if there is a confusion then you ask me, because there tilde will be actually is visible as a bar. So, in the paper. Sir, in the 6 y, half factor will come, because now the 2 states are not. But my stage, j th stage is going to connect to j plus 1 either here or here, with 50 probability I can be connected here, 50 probability I can be connected here. So, when I am connected here, input buffer is not occupied. So, 8 it should not come, state 8. Step 8 will be half, no step 8 will not be half, it will not be half, it will be full. See, this is the probability that packet is there at the input port. So, here only 50 percent of time packet will be there at my outgoing port or the input port of the next stage. There it will be sure that packet will be there at the port. So, P 7 and then of course, then after that I do not be putting half here, it will be now as you have correctly identified. P 8, similarly you will have P 1 9, P 1 10 will be there. Yes, it will be there. I think now almost all of them will be there. 11, P 1 12, P 1 13, P 1 14. So, I think we are now going to close. So, only I will put the numerator part remaining we will do in the next lecture. So, numerator part will be, you can quickly put. In case of 4, yes, packet is going to go out. So, I am going to write it positively here. When packet can move, half of P 1, 6 j plus 1 packet can move in case of 6 also. It is free, 5 packet cannot move actually go out. So, that should not count to my probability by definition. So, half of P 1 8, P 1 9, I think both the packets can move out. So, it is 100 percent probability, half of P 1 11, half of P 1 of 12. So, this gives you the probability of P j tilde to you in terms of the state probabilities after step 1 in the next stage. So, I will use this P j tilde to essentially compute the. So, we will start from one edge and then compute all the way in the forward and backward direction. Two ways we will do the computation actually. So, I think here I will close today and then we will move forward. We will find out what is P j bar and so on in the next lecture.