 So, I have just redrawn all the 14 states, these were the same which I used last time. So, because we need to keep on referring it, so I have just. . . The formula which I wrote last time, I had not explained that I think, I left some of you might have intuitively figured it out how it happens. So, maybe I can explain how that formula comes. So, I will start with that itself, that was p j tilde that is what you are computing, this was probability that a packet at switch output link in stage j will be passing out in time torque k. So, let me write down the expression and see how that can that we get derived actually. So, this was p j tilde which was there, this is the probability that a packet at switch output link in stage j passed actually forward in time interval and this is a basically a conditional probability. It is conditioned on that the packet has to be there at the output of j th link. So, there has to be a packet first and then it will be transferred on to this thing. So, remember the step one is when this packet goes out and after step one the packet has to jump actually from here to here. So, this one was basically after step one. So, after j plus one if the packet on the outgoing link has gone out that is a step one is over. After that step what is the probability that it will be in which state that will decide what is going to happen in step two here and that in turn will decide what will happen in step one here. So, if you remember the diagram step one when it happens here step two is happening at this point and step three is happening further that is what was usually the case. So, whenever the step I think I had drawn this figure that time two specific cases when I defined step one and two. So, when there are two stages. So, when this packet is going out. So, here it is happening step one the packet is jumping. So, step two is happening here for this packet is step packet is arriving at the input link. So, step three is happening here step three packet arrives in your input buffer step two packet goes from an input to output buffer step one packet goes from your output buffer to onward. Output of j minus one to output of next state whatever is your stage from your output the packet should be moving further that is your step one and once your output link is free then only from your input to output link a switchover will happen that is a step two for you and once your input link is free or available that buffer on that then some packet from the previous stage can come in that is a step three for you. So, basically when this movement will happen will be decide by after step one in which state this particular switches. So, I have to look at all those possible states where there will be a packet here at the input. Now, since actually this the way it will be is I have to look at now basically because this is a conditional probability that first thing the packet has to be there. So, I have to now sum up all the state after step one what are the states all possible states because state probabilities are absolute probabilities these are not conditional ones. So, find out all possible states and sum up all the probabilities combinations by which a packet will be here at this place. So, if you look at for example, state one there will not be any packet here going out I have to look at only those states where this packet will be there and then I have to find out whether packet can move on the other side or not. So, for example, if you look at two there is no packet at the input this cannot be counted three cannot be counted four there is a packet here the question is you might be connected to this link or you may be connected to this link. So, probability that this is in state four you have to multiply by half with only half into being in state probability being the probability of being in state four multiplied by half with that probability you will have a packet here and with remaining half multiplied by the same probability the packet will not be there at least only those situations or those probabilities where the packet is available here. Then I have to now sort out those conditions where the packet can move from this place to this place in step two when the step two will be running here. So, after step one what is the current state that itself governs what is going to happen in step two. So, after step one if you are in this state. So, after step two this packet will be going out. So, that should be there in numerator because of that look at this situation with half half into probability of being in state five you will have a packet half into the same probability you will not have the packet, but when you have the packet this packet cannot move further in step two because there is this is a state after step one. So, there will not be any movement. So, this should not appear in the numerator. So, I am looking into absolute probabilities in the denominator multiplying by the probabilities when of those states where packet is going to move out in a step two in j plus first stage because your condition is this after step one in this stage. So, what is the probability that in step one packet will move out that is what we are saying is p j tilde. So, we have to essentially do this enumeration. So, I think for clarity sake I am let me just write down this expression, but this time building it step by step is last time I did it in hurry I only did it in the end of the lecture. So, today we have time. So, I can do this and remember this is a conditional probability that is why I am doing it. Now, look at one one should not be listed in the denominator two I think should be listed or not no three what about four yes and half of it or one of it half of it. So, it is half and what I should write it is probability after step one and what is a state four this I am talking about j tilde. So, this is j. So, this should be state for j plus one. So, I am going to write j plus one here and then of course, this is in time tau k. So, I have to write k on top of it after time step k tk. Now, with five what should I write half agree. Now, go to six half of being in probability of six. Now, moving over to seven again half eight has to be full probability of being in a state eight what about nine all are full. So, I am going to write all of them. So, only these many states have to be included in the denominator. Now, let us look at the numerator when the packet will move out I am looking at p a tilde conditioned on that the packet happens. So, this states will actually give you essentially one over this. Technical gives nothing, but probability that this is actually all possible states which. So, this technically gives nothing, but probability that your packet is here one over that. Input of j plus one. Yes, right. Sir, step one in j th state After computing step one in j th state j plus first I am looking into the state of j plus first block or switch in j plus first stage after step one. See, then only I will know what was the state after this what will happen step two. So, this is situation after step one. Now, I am going to look into what can happen possibly after step two. So, if you look at step one that state one nothing is going to happen that state is not of our interest there is no packet. So, it cannot contribute to the probability that this packet will be there. See technically what happens is you will have always p conditional the packet has to be there and there is a conditional probability that packet will pass. So, absolute probability when you multiply by that when packet probability that packet will be there this conditional probability is coming from here actually. So, this is probability that packet is going to be there. The above one gives the absolute probability. So, absolute divided by this thing this is like probability that packet will pass divided by probability that packet is going to be there. So, this is absolute probability you have to divide by the probability that packet is there that will give you condition. So, this is the probability that a packet is present at the output of j th stage that is why I have written that thing denominator because what I am going to write in numerator is nothing but absolute probability of packet being going out, but I look I want to have a conditional probability instead of that. So, look at state one cannot come here because there is no packet can move out. So, it cannot contribute packet two there is no packet. So, packet cannot move out it will not be there state three no state four packet will move to the second step the output of j plus first. So, j plus first will be executing when it goes through second step this movement will happen and then the step one will be happening in j th stage. So, here step two will happen see it will be like this. So, the step one has already finished here. So, now when the step two will be executed this will be moving on this side and then that time there is step one will be running here that is what this p j tilde is about after step one packet moves out. So, after step one what was the state will decide what will happen in a step two. So, that is why I am going to write in numerator the state probabilities after step one. So, four will contribute and how many times with half probability only it will happen because I may be connected to this link or this link if you are connected to this link there is no packet going out when you connected here then only packet is going out. . . . . . .So, I am all equivalent states are being represented see no one state can be mapped on to another state by any kind of rotation flip or whatever it is that has been ensured. So, you will have half of p after one four j plus one k. Now, look at step state five even if after step one in this particular stage you are in a step state five a state two when we is going to be executed a step two is going to be executed packet cannot move because one packet will be present here. So, this cannot contribute to p j tilde. So, I should not put it in the numerator even when packet is present actually at the output port. So, five will not contribute in the numerator look at six packet will move out after step two in j plus one and packet will be emptying the outgoing buffer after step one in j th stage. So, that has to be included the six one and that will again happen with half probability because I can be connected to either of the two links input links look at seven no packet cannot move out. So, it should not be listed in the numerator state eight only one will move out and you can be connected to any one of the input links with equal probability 50 percent chance it will move out remember this one is eighth one is listed with full here it is not half in the denominator, but in numerator it will become half because only one packet can go out not both. So, I will be listing ninth both of them will go out with 100 percent thing. So, it is there is no half here in this case 10th nothing will go out. So, it should not be listed 11th only one will go out. So, it will be half not 10th it says to be 11 sorry 10th nothing can go out 12th again only one can go 13th nothing can go 14th nothing can go. So, those will not be included in the numerator. So, this gives you p j tilde. So, I think now this is more clear in the second iteration last time it may not have been and most important thing at the end the last link last stage. Since, remember now what the case which we are looking is when t select is 0 and sorry t passes t passes 0 that actually means when two packets are there in the last stage and they are both can go out. This buffer will always be empty because packet transmission takes 0 time is instantaneous even if both have to be switched out they both can be read instantaneously and taken out of the buffer. So, what will happen is this is n minus first stage or n minus n minus 2 because we have 0 to n minus 1 sorry. So, it means p j p n minus 2 is 1.0 that is a boundary condition. So, when we will iteratively do the solution this will be the boundary condition at in any stage q j will be defined as p j tilde. If p j is probability of packet going out q j is it will not go out it will remain there. Only one boundary condition that is yeah I am now coming to the other formulation also. So, all of them will be used continuously in tandem then I will give the process the method. So, I have to give all this then I have to give transition probabilities and then I have to give the method which is has to be executed. This all is done in a computer program it is not a simulation it is a computation. But it is a iterative computation it will converge at some point of time and when it converges that time you take out all the results and find out the throughput. So, that is the procedure there is no close form solution. You can write 1.0. You can write 1 not an issue probability cannot be. So, somehow we always write it as 1.0. This probability both t pass and t select are present in this probability. T select t pass is 0 t select is complete what we call t delay. Sir if t pass is 0 then 8th state those two packets can be sent outside. Depends on in which state you are in which a stage you are. If there is a last one see for example, this states cannot exist where the outgoing ports outgoing packets are there these states cannot exist in the last stage because outgoing will be immediately emptied out. This will and in n minus 2 stage this is going to be there for some time because where this is selecting these packets have to be held here. When the t select is finished then only they will be removed. So, you can actually have these all the states are possible in n minus 2. But in n minus 1 only this state is possible this state is possible and these two states are possible. No other thing will be existing. So, 1, 4, 8 and 9 that those are the only possible states for the switches in n minus 1 stage. From there we will do I will actually tell you how this backward calculation is done. Sir, we have to do this computation process. It is a iterative computation process that is the only way. Yes, it will be computation and simulation. Simulation what you will do is you will actually create you will actually generate packets. You will let the packet flow through. See it will be like kind of you will have a clock a universal clock at first clock there are no packets then the clock goes from 0 to 1. You will generate packets at the input and then what happens in that clock you will do find out the whether conflict is not there or not there you will actually move the packets. You will maintain a data structure which will maintain the complete state of the switch and now this state will be evolving on the basic principles and you will keep on moving. Ultimately at the output you will observe how many packets have passed successfully. Now you have to observe for multiple clock instants. There is nothing like what we call a time after which you find out there is a convergence. There is nothing like convergence there and in simulation there is another very tricky thing. Here I will just look for stabilization. When convergence happens I will just take the results in this computation. In simulation there are no packets so it is not a steady state. So initially there will be a transient. So I will discard that actually all that observations and after that I will observe per unit time per unit slot how many packets are successfully going out. I will keep on making observation. I will observe for certain number of time slots. It is like you implement actual switch but not in actual reality but in computer program. So you are simulating the algorithm itself not computational algorithm but operational algorithm by which they are being or they are operating. So whatever is that patrinette that essentially you will be implementing there or equivalent of that but the paper does not actually paper do have simulation results which are matched with the computational result. These are very precise. It is like I ask you what is the how you will compute the value of pi. So one very simple way how you will do it through simulation. Can anybody tell how you use simulation to generate the value of pi? We keep on drawing circles and keep on taking the ratios. No that is a different thing by simulation. Can anybody think? We have circle of radius r and then r arc length. That is fine that you know all by mathematics. You do not need to remember you only need to remember that the area of a circle is pi r square and area of a square is r square and you always know that if I make a square I can always create a something like this. So what is the area? 4 r square for the square. Circle has area of pi r square. So what is the ratio? Pi by 4. Pi by 4. Suppose circle to this thing. Write a very simple program with uniform probability I do a coin toss and I put a point. So I choose a random number generation and the points will be just randomly plotted. You put 1 million points. How many fraction of points will be there inside the circle and how many will be total points which are plotted inside this? So find out the fraction which lies inside the circle and that fraction most likely will be nothing but pi by 4 and once you know that fraction of points which are there inside the circle multiplied by 4 that will give you value of pi approximately. So this is a pi measurement through simulation. This is not computation. So I think now you understand what is the difference between simulation and a computation. Otherwise you write a series thing and do it or do whatever calculations draw circle make length or whatever it is that is a computational process. But pi there is a closed home solution no it does not exist. So closest approximation is 22 by 7 which is we all know that is approximation. But best is draw a circle measure the length find out the radius length take the ratio and that is a value of pi. Larger the circle better will be the estimate. So that is a computational procedure. So I am not still simulating you have to understand this is still a computational procedure. My students do have a tendency of mixing of the two things. So usually computation simulation is nothing but whatever happens in real life from those observations you make an estimate that real life experiment is actually simulated in a computer program is conducting experiment in computer program through a computer program and then from there making observation. So usually random number generation will be part in partial of any simulation process. I am not doing any random number generation here. So it cannot be simulation and there is no statistical estimation. So this is a computational procedure. So paper actually does give both results and they actually match pretty well. So this is seems to be a logically right way of getting the solution. This is nothing but a iterative method of getting to a solution of a Markov chain which I explained earlier in the last class. Where I give it for a Markov arrival Markovian departure and single server queue with infinite buffers. So how we can get a solution to that. So it will essentially convergence will happen and that time you get all state probabilities. So for all complicated Markov chains I think this is a very good method and here what the nice thing which has been done is how many switches are there in a stage those are immaterial. How many stages are there those what matters. So all state probabilities within a given stage are all for every switch it is going to be same. That is assumption and we have built up the relation between state probabilities of various stages. That is what I am doing actually now. So next probabilities which we need to compute is probability that a packet will I have to essentially erase this. See because remember state transitions how it is happening. When I am looking at jth stage when a packet is going out state transition happens then from there from my input link to output link there is a packet movement that is a step 2. Step 3 when the packet will come from the previous stage. So I have to look at the state transitions. So this first transition after step 1 in this thing happens as per this relation p a tilde it depends on that. So if the packet moves out or does not move out we will decide what will be the state transition matrix here. So if what was a state at t is equal to 0 or just before the first step and what happens after the first step. First step is this moving out when this has a step 2. Then once we know what happens if the step 2 is run. So if I know the state after step 1 whether packet is there at the input what is situation at my output that will decide. And once if I am available what is the chances the packet will come from the previous stage that will decide for the step 3. So this is for that situation time interval talk. So again I have to write down all the state probabilities. This will also be called pj tilde this also will be called pj. Now it is not pj tilde it is pj bar sorry. So now what is happening is these are situations. So I am now looking at the states here and step 3 is step 2 is running here. And what happens in the step 2 will decide what will happen after step 3 here. And what will happen in a step 2 will be decide by what was the state after step 1. So now let us look at it. I am trying to find out what is the chance that a packet will come from this input to here. So after step 1 if this j minus first stage in a state 1 can a packet come here. First thing I have to find out I am actually I am having an empty state here there is no packet. I have to look at only all those possible states after step 1 here. Then only I can figure out what will happen in step 2. But first you have to check whether packet is there or not. Packet should not be here. So a step 1 a state 1 no packet at the outgoing port. So it is free. So I have to list that actually. But step 1 the packet will be sent out from the output port. Which one? After step 1 for step 2 output port will be always green also because in step 1 whatever the existing packet is already sent out from the output. No that can only be sent out it is possible if you look at earlier case. Sometimes packet cannot be moved out again after step 1. If there is a packet in the next buffer you cannot just simply push you have to just keep on waiting you have to just keep on waiting. So after step 1 what are those states where this port is free? I have to look at those. That has to go in the denominator. Numerator will be those probabilities of those states where the packet will move from input to output of j minus 1. So look at this. So we will not consider anything which is not available. I mean packet has to be there there is a probability of that also in the input stage shall we not. I am only looking at the relations as of now. After step 1 what is the state here? That will decide what will happen in step 2. Because when the step 2 will be running here step 3 will be happening here. I am looking at the probability after step 3. Probability that the packet will arrive in the input buffer. This can only happen this will be decide by what happens after step 1 here. What is the state? That will govern the computation of this probability. So after step 1 if there is a state 1 j minus 1 this is important. Both of the outgoing links are free and I can connect this is the remember earlier case was like this. We are sitting here in jth stage. So you are looking at j plus 1 at that time. We are looking the backward direction. So with probability 1 I can be both of these links can be connected to me is not half both with equal probability. So both times it will be empty if after step 1 this as if after step 1 this is in state 1. So I will write this state 2 half because there is a half chance because this is after step 1 you are in state 2. So you can be here or you can be here packet can be any one of the 2. So with half probability you will actually may get it empty or may not get it empty. And if it is not empty the packet cannot come if it has to be empty first. So I will write now half of p 1 2 j minus 1 k p 3 p 3 should not be there because then my input buffer link itself is not input buffer is not free it is already occupied. So no packet can come so it cannot contribute 4 yes it is there and 4 means both. So p 1 4 there is no half in this case state number 5 half state number 6 again half state number 7 no state 8 full state 9 full state 10 is half state 11 is half state 10 will be half right 11 also will be half 12 also will be half no 13 14 cannot be half half they cannot exist we got everything right here the upper ones we have to look state 1 packet cannot come in packet cannot go actually in step 2 there is no transition possible. So this cannot contribute packet state 2 cannot happen state 3 is anyway not possible state 4 yes if the switch at j minus 1 is in state 4 after step 1 in a step 2 packet will transition. So this will contribute to p j bar so that has to be listed but this will be happening with half probability because your stay your switch can be connected to any one of the outgoing. So it will be half of p 1 4 j minus 1 k p 5 no there is no possibility if you are in that state packet will not come in packet 6 yeah state 6 this is going to happen with half probability then after 6 7 nothing 8 you will get with half probability a packet while in the denominator it must be actually full probability because both outgoing links are free 9th yeah it will be full 10 nothing will be there no I mean 11 yes packet can come it will be half 12 also will be half but both cannot go simultaneously only one of them will be there is got absolute probability then the question is you can be either connected to this or you can be connected to this with half probability no see j switch is connecting to j minus first. So you should have connected with half here you will get a packet if you are connected you would not get a packet. So that is why this is listed now with this thing this is remember what is happening is this is transition after step 1 this tells what is going to happen in step 2 essentially. So I am looking into arrival probability here so what is going to happen in step 2 I am looking at that now for this the boundary condition will be p 0 bar means the first switch the input of that I am looking at a maximum loading condition if this is free then the probability that arrival will happen is 1 that is a maximum loading condition again sorry I can write 1 and of course q j bar 1 minus p j bar that condition will be there. Sir in this zoom denominator 1 by 2 p 5 state how will be denominator in denominator. 1 this one look at the fifth one see you can be connected here then your switch was state that buffer was not empty. So that should not be counted with half probability you will be connected and you will have a buffer which is empty. If it is connected like that it will be. Conditioned on that in j th stage my switch input is free there is no packet there on that condition what is the chance that packet will arrive in step 3. Actually in state 5 sir step 1 has not happened in j minus 1 in state 5 sir step 1 has not taken place in j minus 1 stage. Does not matter I am looking at after first stage whether what is the chance of being in this state if you are in this state you are this is free but no packet will be transiting this is not there in numerator this is only in denominator. So I am looking at what will be probability that a packet will be arriving if my input is free I am estimating that. So this contributes in the denominator only 50 percent when the that input link will be free for that it is contributing to that actually it is not there in numerator. Now let us come to the state transition tables these two things essentially we will be handling try to build everything here this figure is still has to remain there. So I call this transition matrix as p j in the j th stage this is after step 1 going from initial stage is m and you want to go to initial state is m and next state is i after step 1. So I make the probability is going to 1 2 3 these are the states only 14 are required. So if you are in state 1 what will happen after step 1 there is no packet at all outgoing thing. So there is no question of it is going to be certain probability p j. So with probability 1 you will remain in state 1 only. So we will keep this as 1 so I am going to put only the values which are existing rest everything is 0. I am not filling up all the cells here all other cells will contain 0 except the way where I have written the values. Now look into state 2 state 2 will remain now the probability that packet which is sitting there goes out is p j tilde. So if that packet goes out you can come back to state 1 if that packet remains there you remain in the same state. So it actually means this will be p j tilde no this should be a p j tilde you will come from 2 to 1 you will remain in 2 with q j tilde rest everything is 0 not clear. So you have a switch and if it is in state 1 this is a state 1 and what is step 1 packet from at your outgoing port goes out and if you have a packet here this will go out with probability p j tilde and it will remain there with probability q j tilde which is 1 minus of p j tilde in general stage because if it is a n minus second stage packet will certainly go out there is not an issue after step 1. And if there is another one so this will be going independently with p j tilde I am going to talk about 1 packet condition if the packet is there if there are 2 packets both of them will go out what is the probability p j tilde square p j tilde square only one of them goes and one of them remains multiplied by 2 there are 2 possible combinations both of them remains there p j tilde square and all these 3 events will lead to certain state transformation. So there is no packet there is no packet here with certainty with 100 percent probability or 1 probability it will remain in state 1 only that is what I have written there. So this is your m and this is your m to i so m to i transition this is initial state this is next state after step 1 in jth stage. So this is a transition probability in jth stage after step 1 for going from mth state to ith state. So you can actually write going from state m to state i in jth stage after step 1 and this is nothing but the probability and this is what is this transition you build up the matrix you find out from each state to other state what is a transition probability. So 1 to 1 is obvious actually if you are in state 2 you will have only one packet sitting here if this packet goes out what is the probability of going out p j tilde. So if it goes out in which state you will be state 1. So I have written here p j tilde from 2 to 1 this is the probability that you will be transition and 1 minus of that packet will not go out you will remain in the same state if you remain in the same state that is what I have mentioned here and remember rest all these blanks will be 0 you cannot transition from state 2 to state 4 for example that is not possible look at a state 3 now which contains 2. So with probability as all of you have told p j tilde square all both the packets will go out you will come to step state 1. So you write here p j tilde square n f of course only one packet goes out and one remains you will remain in state 2. So that is 2 p j tilde and q j tilde of course this is very small. So maybe I can write it each event is happening independently. There are 4 possible cases there are 2 possibilities here and there are 2 possibilities here with equal probability. So one case is p j happens other says p j happens here q j happens here p j happens here p j happens here q j happens and here q j happens here q j happens each possibility is 1 by 2 1 by 2. So these 2 are same they always lead to the same state so that is why 2. Actually this is a tabular description sir of Markov chain tabular description. It is a transition probabilities is still Markov chain has 2 things the state probabilities and then transition probabilities Markov chain has these boxes. So there is a probability of being here and this is the transition probabilities. But you can do the same thing with. That I am not doing tabular sir this condition that does. No because I am not actually making any Markov chain. Yeah technically yes I am doing the same thing you are right. Technically I am doing the same thing. We will also have a transition matrix for that bar state p j bar similarly which you just derived now. A transition matrix. Yes sir. It will come this after step 1 I have to draw t j 2 also t j 3 also there are 3 matrices which will be there they have to be maintained as data structures during your computation. Simulation is far simpler actually remember if you want to write a program for simulation that is going to be simpler than this. But the good thing is that you do not have to remember all the possible switches which can exist in a single stage. So only states you have to take care and you have to take care of stages. State to state actually transition will be there and of course q j you will both of them will be remaining there so 3 2 3 q j tilde square. So 4th will be similarly if you are in 4th one then what will happen? Step 2 may happen. Step 2 may happen. After step 1. Step 1 is packet from your outgoing port goes out. There is no packet at outgoing port. So nothing will happen after step 1 only step 2 can make a transition in this. Step 1 cannot make any transition in this state. So this will remain in the same state this cannot do anything. So 4 will remain in 4 with probability 1 5 with probability p j this can be converted to 4 with q j it will remain there that is it. So with 5 can be converted to 4 p j thing and it remains 5 with q j tilde. This is sparse matrix actually. 6th same way after step 1 packet can go out or packet can go out. So if the packet goes out this becomes nothing but 4 otherwise it remains 6. So it is if it goes out it will remain p j tilde 6 it will remain q j tilde. Of course 7th state this can be converted to 4 p j square 7th if it is there p j tilde square and if one of them goes either 5 or 6. 5 or 6 with equal probability. So you have to write p j q j tilde p j q j tilde this has been remember the complete row when you sum up has to be equal to 1 sum across all the rows has to be equal to 1. And of course then q j if you have it will remain 7 8 what will happen 8 will remain 8 not an issue 9 will remain 9 10 will become either 8 or 10 will remain 10. So 10 will remain 10 will become 8 with probability p j tilde and 10 will remain 10 with q j tilde 11 11 will either remain 11 or will become 9. It will be 8 10 row 10 row 10 row it should be 8 column p j tilde. P also has to go to 8 column. Same you can do with 11 11 goes to either 9 or 11 either become 9 or remain same 12 will either remain 12 or will become 8 13 there will be 4 entries for this. So either it will become 8 with p j square and p j q j either it will be 12 or 10 10 and 12 p j q j and of course 13 none of them goes 13 cannot go to 9 same directions the same 4. See direction you have to always take care 13 cannot go to 9 13 only can go to 8 can go to 10 or can go to 12 or remain to 13 14 14 can go to 9 11 14 can go to 9 11 may go to 9 where is 9 yes 11 it is 2 p j q j tilde that is a complete transition matrix. Similarly, you can draw the transition matrix for after step 2 now important thing if you have this so after 0 step you know the probabilities this basically is the beginning of the slot. You can multiply it by now transition probabilities after t 1 j m i sum over all probabilities I think now you can appreciate if I know the state probabilities in state j after 0 I can compute from there my state probabilities after step 1. And these are the transition probabilities which will come and transition probabilities now depending on so this here where I am actually finishing next we will look into the next state transition diagram which is after step 2 and then step 3 and so on and then actually the method of calculation. So, that will you will covering in the tomorrow lecture morning.