 So, we will continue from where we left yesterday. In the previous lecture, what we did was, I was talking about basically how to build up a larger dimensional switch or larger size switch using a smaller size basic switch elements. And in that sense, what we want is the preservation of the certain property which we call as strictly non-blocking property. And therefore, the question was whether I can reduce my cross point complexity in some way or not. First thing was these are creating larger size by using a smaller thing and further reducing the cross point complexity. So, what we did was, I took a case where we were trying to build up using two stages and I took some switch with certain basically kind of a small n as the input ports and then we build up the interconnect by using, if I am going to use n by n, I said that one can go to, one link can go to each one of the switch in the output stage or in the second stage, but this was leading to a blocking situation. So, if certain inputs are free and as well as output is free, I cannot set a path between them in certain situations. So, in order to resolve that, what we did is, we actually start using n by, this n by n square kind of switches were actually were used instead of n by n and in that case, this there were actually n wires which are going out and this actually means I need to put n wires to each one of these. So, all these n can be connected to any one of these n without any blocking to happen. So, far this number is going to be smaller than n, then it will not be possible because whatever is left over and there is one which is free here cannot be connected, only part of the connections can be made and of course, we figured out that the complexity still remains because if you have only there are k such elements. So, you require you have switch size of n k by n k which usually will require, if I build up a single cross bar will require n square k square. In this case, I will have a switch of n by n square which becomes n cube complexity and I have actually two k such switches. So, I will be requiring O 2 k n cube kind of thing and then of course, if I actually take from here, you can actually clearly see that if I am actually making kind of k will become equal to n in that case both will be O n raise power 4 or since your capital n is nothing but technically equal to n square. So, this is nothing but complexity of O n square only. So, that is what we did last time and then I thought we can go to three stage interconnection and I defined a closed network in that case and I also define what is going to be strictly non-blocking structure, what is rearrangeably non-blocking, what is wide sense non-blocking and what is going to be blocking structure. And then of course, I define the closed network which was you will have some inputs. So, for example, n input ports and they are k such switch any switching elements. So, total number of incoming ports is n into k and we were connecting in the middle stage many such switches and we have to find out what will be the dimension. Now, each switch should connect to each one of those switches. So, I will be actually having something like this. So, I might end up in choosing some number here, some middle stage. So, in fact, the way I defined it was it was this was not n, this was m 1, m 1, this side was n 1, this is also n 1. So, all switches where which were here was m 1 by n 1 switches and I was using k 1 of them. Then I was using the middle stage k 2 switches which was of m 2 by n 2 size and then there was a third stage configuration third stage switches where I will be using m 3 by n 3 and there were k 3 such switches. Condition is that only free incoming ports are available only in the first stage and only free outgoing ports are available only in the last stage which is the third one in this case or output stage. None of the incoming ports and outgoing ports in the intermediate stage should be left free they have to be connected. So, in that case the number of because I need to connect everybody here. So, this number which actually means that m 2 has to be equal to k 1 because k 1 actually lines will be coming and connecting on to this and similarly here the number of inputs everywhere will be m 2 will be equal to k 1 and similarly now n 1 has to be equal to whatever be the number of switches in this case this will be equal to k 2 and this also implies because same structure will be applied here. So, you will have n 2 will be equal to k 3 and m 3 will be equal to k 2. So, you can see this condition. So, m 3 will be always equal to n 1 is equal to k 2 this condition will always be satisfied and I said this actually switch can become strictly non-blocking and I took a very simple example this is this will be later on proved as a class network. This is actually the class network we will use a class theorem to find out what will be the condition for strictly non-blocking nature of this particular switch. So, if m 1 minus 1 lines are already busy occupied here the same is true at this place that m 1 minus 1 lines are occupied here there is only one line which is left in which I am interested there is only one line which is left in which I am interested I want to connect these two. In worst case scenario now these out of these lines which I am putting here m 1 minus 1 are already occupied they might be using certain middle stress switches for connecting to some incoming ports. From this side in worst case scenario these can be using a different set of switches there is no overlap between them. If I had one more extra this can always be used to set up the path between this free input and output link and switch will be always strictly non-blocking switch it do not need to disturb any existing connections which actually implies that I need to have if I am going to take a symmetric case where by m 1 is equal to n 3. So, you will find that I need k 2 should be equal to or greater than or equal to 2 into m 1 minus 1 plus 1. So, that is the condition which will come from this thing and this is nothing but 2 into m 1 minus 1 this should be the condition and k 2 will then actually decide what will be value of n 1 and m 3. So, strictly non-blocking switch will look something like this it will have m 1 which is going to come in and you will have these ports which is 2 m 1 minus 1 and this will be going to various switches here m 1 ports there are total k 1 such switches. So, again this dimension will be 2 m 1 minus 1 total number of switches will be 2 m 1 minus 1 the number of ports here will be k 1 and since I am actually taking m 1 is equal to n 3. So, I will call it again m 1 this will be also m 1 and since you will have in this case again these lines have to be 2 into m 1 minus 1. So, number of middle stress switches will be these many and these of course, because I want a symmetric thing. So, I will be also taking this thing as a k 1 k 3 will be equal to k 1 that is a symmetric case and this will be the condition which will be satisfied. So, m 2 will be equal to n 2 in this scenario this switch will also everything actually symmetric in this case and this will strictly non-blocking switch. Now, what will be cross point complexity for this particular switch? So, if I want to find out cross point complexity number of cross points I can very well estimate that for this particular switch I will be requiring m 1 multiplied by 2 m 1 minus 1 number of cross points and number of switches which are here is k 1 and exactly similar stage also exist in the third one. So, I have to multiply it by 2 and then what I will have is how many middle stress switches will be required is 2 into m 1 minus 1. What is the size of the number of inputs which are coming into this? This will be nothing, but k 1 number of inputs which will be coming in and k 1 will be going out. So, this is what will be the number of cross points. So, we can try to optimize this we have to first of all find out what is going to be the value of because the total number size is actually now n m 1 into k 1 is what is your is a constant and I need to find out what are the optimal values for these. So, I will write everything down in form of m 1 and we will try to optimize with m 1 and then find out what should be value of m 1 for optimality basically the minimum size of the number of cross points delta c. So, for this let me now first of all repeat whatever I have written. So, these are total number of cross points. So, I will now replace all k 1 by n by m 1. So, that I have only one variable sitting in there. So, n I can take a derivative. So, k 1 will be n by m 1. So, this what we will get we can also actually assume that let m 1 be much larger than 1. So, that I can now make an approximation to m 1 minus 1 can be approximated as 2 m 1 this actually will simplify my life. So, approximately the cross points will be now varying as 2 m 1 into 2 m 1 into n by m 1 plus and now solving this gets reduced I will have 4 m 1 n plus 2 n square by m 1. Now, taking derivative delta c over delta m 1 for the optimal value of m 1 and then we will put that m 1 value and find out the total number of cross points approximately and what is going to be their complexity. I can now actually keep this particular assumption true and then use this expression for finding out the number of cross points. So, once I take the derivative this will turn out to be 4 n plus 2 n square minus 1 over m 1 square and I can make it equal to 0 and this will give me in fact, I can now reduce this will be 2 n I can take out. So, this will be 1 I will let me keep it as it is. So, this actually implies either n is equal to 0. So, which is the condition I am not looking for. So, I think this argument if this is 0 this is what will give me the answer. So, this has to be 2 plus. So, 2 will be equal to minus n by m 1 square. So, m 1 will be nothing, but n by 2 root. So, this will be the answer. So, I can now put this thing in the to find out the number of cross points in this expression and this will turn out to be 4 root of n by 2 into n plus 2 n square by root of n by 2. So, this will turn out to be 2 root 2 n 3 by 2 plus again from here 2 root 2 n is power 3 by 2. So, now the complexity is now of the order of 3 by 2. So, we have now able to go from n square to this. This is lower complexity than what we had in a single cross bar. So, I am able to reduce my number of cross points. Of course, only thing is that now I have introduced complexity in terms of the routing algorithm which is required because you have to search for a path and then you have to set up a path, but path can always be set up. So, this is a pretty efficient system. Now before actually moving forward, I need to understand now this kind of switch requires large resources. Now, usually network operations or whenever we do a practical design of a switch, we will never build up a actually a strictly non-blocking switch because you do not require it. If I can introduce a very small amount of blocking probability say 5 percent or 10 percent and have much less hardware probably I will be able to provide an acceptable service to the users. So, it is always a good idea that or I make a very small size strictly non-blocking switch and then I actually use what we call expenders and compressors which basically is a special kind of a switch where large number of incoming ports are there on one side and smaller on this side with the assumption that all users will not be active at any point of time. So, chances you will be blocked is very small, but because of which my amount of hardware reduction is much drastic which will help in reducing the call cost. So, I will appreciate that. So, usually you will put a compressor like this then is a strictly non-blocking switch of n by n. So, you are using some capital M which is going to be larger than n. So, either you will be using this kind of configuration for a switch. So, where this will be strictly non-blocking switch or you will itself will design a m by m blocking switch with less number of cross points because every cross point require a management. So, in fact cross point need not be actually physically made in this fashion. I can build up a strictly non-blocking switch even using time slot interchanger which I will be covering sometime later in the lecture. And we need to estimate what is going to be the blocking probability if I build up a three stage structure because I need to have a blocking probability estimation if I am going to design a blocking switch. Now that requires some understanding of Q-ing theory. So, I am just actually digressing from here and moving to the what we call the probability or what we call state probability estimation in case of a switch. But before this let me do a very simple Q because this need to be understood and I will use a similar procedure now for what we call simple m by n. I will call this as a composite switch m by n composite switch and we will try to find out how you will estimate the blocking probability for this thing. So, so far the number of calls are going to be less than n which are passing through this switch this will remain in non-blocking state. The moment you get a call which is going to be a number of calls will become equal to n after that any call comes in cannot be made through it is then going to be in blocking state. We need to estimate that blocking probability and for that we need to know understand the Q-ing theory basic fundamentals of that. So, in this case it is very simple the way we actually model it. Let me go to a simple Q and then I will come. So, what happens Q is always represented in this fashion. So, there can be tasks which can come into the queue and they can be served. So, what we do is the how the task arrival task will be arriving randomly actually there is no deterministic thing how you can actually model this particular task arrival process. So, what we do is we assume it to be a Poisson distributed phenomena. So, what do we mean by Poisson that if you take a time interval. So, you are starting from say time 0 to capital time t. So, how many number of packets will be arriving in this time interval t. So, how we will what is the probability that k packets will be arriving. So, packet arrival is an instantaneous process in one single instant a packet come. So, two packets can come very close to each other, but since it is a point process two packets cannot come at the same instant there will be infinitesimally small gap between them and we define a parameter called lambda here which is the arrival rate. So, this average number of packets which are going to arrive per unit of time. So, that is I think something which can be done. So, in capital time t on an average lambda t packet will be coming. Now, the probability that k packets will be there in time t. So, probability for this will be defined as lambda t raised to power k e raised to power minus lambda t divided by k factorial. This is a probability distribution and of course, one can verify that if I do summation of this if I actually do sum of this over k I should actually end up in getting equal to 1 which of course is obvious because e raised to power minus lambda t can be taken out and this summation as all of us do know over k this summation is nothing, but e raised to power lambda t. So, e raised to power minus lambda t and lambda t will give you 1 always. So, this actually is a probability distribution and we call it a Poisson statistics and of course, we can look at the same particular distribution in another way that what is the chance that a packet arrives and the next packet arrives what is the distribution of inter arrival time the time gap between them. If I want to find out the distribution of this what should be the distribution this actually can be derived from here we call it exponential time distribution and the derivation is again very simple and elementary. What I can do is I can now take let a packet arrive what I can find out at least packet is not arriving in this time period I can find out this probability. So, if I know at least one packet arrive this is not that packet does not arrive in time t this is means that probability that arrival time is greater than or equal to whatever is the t prime which I am taking. So, remember this is a cumulative distribution function once I take derivative I will get the pdf of this probability that arrival time is greater than this actually means that 1 minus probability at least one packet arrives in this time distribution or probability that no packet arrives in this particular time distribution. So, that probability I can get from that Poisson statistics. So, this will be 1 minus e raise power minus the so probability no packet arrives in time t prime that is what it means and this will be I just use the Poisson statistics. So, lambda t prime will be 0 e raise power minus lambda t prime divided by 0 factorial it will be e raise power minus lambda t prime. So, that is the probability or alternatively we can say is that the probability that inter arrival time in fact cdf is always defined that probability of x being less than x that is what is cdf and then we take the derivative here I have taken it on the higher side that t is greater than t prime I should actually take where t is less than t prime. So, this implies that at least one packet should arrive probability that at least one packet should arrive I should have derived it this way at least one packet arrives in time t prime which is nothing 1 minus no packet arrives. So, it will be this thing. So, this is what is consistent definition with the cumulative distribution function. So, this is what is the cdf and once you take the derivative of the cdf you will get the pdf. So, once you do that take the derivative. So, pdf of t prime distribution will be e raise power minus lambda t will remain as it is minus lambda will come out on this side this is minus. So, this is what it will be at t is equal to 0 this value will be lambda and then it will fall down and we call it as the exponential distribution function. So, that is the inter arrival time which is going to be there between the two. So, coming back to this q. So, this arrival process is characterized by lambda we also define an equivalent process which is for departure which is technically nothing, but again if there are always packets number of packets which will be departing if packets are continuously there in the q that will also follow a Poisson statistical process, but we cannot do it because if there is no packet in the q that time the packets cannot go out. So, the outgoing process cannot be represented by Poisson statistics because that is dependent on the arrival process. So, what we do is we define a complete packet processing time. So, if a packet is there in the q. So, packet the moment you start serving it how much time does it take to serve the one full packet or for its transmission. So, that is essentially now we assume it to be exponentially distributed and we define this distribution as this. So, average value of the distribution time or the time average time which a packet takes for moving out is 1 over mu actually in this case and usually what will happen is if the packet the process if the packet if this q is going to be empty for at least some fraction of time if you observe for a longer duration then the q is stable. So, only when suddenly large packets will come the it will the q will fill up, but if the all the time packets are coming and you are not serving at faster rate it means that mu is going to be a smaller than lambda then there is a problem actually if mu is smaller than lambda then q will keep on building to infinity it will be unstable thing we will actually prove it that it is what happens. So, this is also technically is a Poisson process, but only if the packets are continuously available because packets will not be there some once in a while we cannot use this kind of thing the number of packets going out is equal to this this is only going to be true if packets are available all the time. So, technically it is also being derived from Poisson statistics. So, what you do is now we have to now build up what we call Markov chain for this and we are going to assume a steady state condition. So, we define the state of this q is being defined by number of packets which are there in the q. So, the first state will be the state 0 when there is no packet. So, if I take a very small time. So, now the time of observation is extremely important if you are in state if I am to actually having a time interval delta t and I will actually make this delta t goes to arbitrarily 0 value, but ultimately this will cancel out in the balance equations. So, the probability that a packet will arrive is defined by delta t by n factorial and if delta t limit going to 0 then of course, for value of 1 for n is equal to 1 the probability will be lambda into delta t for higher order higher probabilities this value will because now this argument will become 0 actually. So, this will be 1 and for higher values the probabilities will become extremely small it will become lambda into delta t square by 2 factorial and so on. So, this values can be neglected they are very very small. So, I need not consider them. So, at any point of time when delta t is infinitesimally small only one packet can arrive and if that happens I will be coming to state 1. So, you will never be going from 0 to 2 that is not possible in this case. So, similarly a packet goes out that probability will be mu into delta t and you will come back to state 0. So, this happens at any point of time when you are observing. So, similarly from 1 to 2 you will get lambda into delta t that will be the probability and this probability will be mu into delta t and so on. Since, I am actually taking infinite size q. So, this can keep on happening we in fact call it a Markovian arrival Markovian departure single server there is only one server and infinite buffer size mm 1 infinity q we call it. Now, under a steady state condition the probability that you will be in certain state will be constant which actually means if probability that I will find this system in this state 1 or state 2 are going to be constant. So, the probable if I make any close surface if I make any close surface the chances that you will exit out of this surface and you will enter into the surface has to be exactly same that is a balance condition. So, this balance condition if it is satisfied will give me the steady state of I can actually now use this particular thing to identify the state probabilities. I will exactly use the same thing later on to make an estimate for my m by n composite switch. So, in this case if the p 0 is the probability steady state probability of being in a state 0 and I make a surface something like this which actually means lambda into delta t this is a transition probability then I am going out of the this surface this particular surface and what is the probability of coming into the surface that I am in probability of being in state 1 into mu into delta t and I can see actually sorry I have. So, this delta t cancels out so delta t is actually a material. So, p 0 and p 1 they are related by ratios of lambda and mu that is what is more important. Similarly, if I actually look at this particular thing I can then same thing I can do it p 1 into lambda delta t is equal to p 2 into mu delta t. You can do whatever surface you can will be able to make if they are say k states possible states you will be able to build up k minus 1 equations and for k state probabilities you want to estimate it require k equations the last equation will be the first axiom of probability that sum of all mutually exclusive events the probability of those has to be equal to 1. So, we will use that axiom as the last equation and then we will to get all state probabilities for a steady state system. So, again delta t goes out same condition holds true. So, I can keep on extending this so which actually means I can write p 1 as lambda by mu p 0 p 2 I can write as lambda by mu p 1 which is lambda by mu square p 0 n turn p raise power n will be lambda by mu raise power n p 0 and n will go till infinity actually. So, now almost all equations I have all probabilities I have now being represented in terms of p 0, but what is p 0. So, if I do p of i summation over all possible i's this should be equal to 1 which implies that p 0 plus p 1 plus p 2 plus p 3 and so on till infinity this should be equal to 1 and I can write this as p 0 plus lambda by mu p 0 plus lambda by mu square p 0 and so on lambda by mu raise power n p n and so on. So, this should be also equal to 1. So, this is nothing, but I got a series and this is a geometric progression and I can very well actually solve it I can call lambda by mu as rho and this value will be using a series things. So, series says that once I want to solve this particular sum so series will be nothing, but a b raise power total number of terms. So, that what should be the sum. So, I am just going to use that same formula and I will end up in getting is because a in this case is 1 rho raise power infinity minus 1 rho minus 1 and interestingly what I will find out rho raise power infinity can be 0 only if rho is less than 1. So, p 0 will be rho minus 1 divided by rho raise power infinity minus 1 and if rho is less than 1 then only I will get a solution which will be 1 minus rho some positive value and a probability of course, has to be always less than 1 which will always be guaranteed if rho is going to be less than 1. If rho is 1 p 0 will be 0 and of course, rho is 1 p 0 is 0, but 1 raise power infinity is also going to be undefined in that case. So, this becomes undefined so rho has to be less than equal to 1 for a stable situation. So, once I know p 0 I can now actually find out all state probabilities with that. So, I have now got my p 0 which is 1 minus rho p 1 which is rho into 1 minus rho p 2 rho square 1 minus rho and so on and henceforth I can find out what is the average Q length of this packet. So, average Q length of course, I need not I need only the state probability thing because I will be using similar expression similar procedure actually now to for m by n composite switch. So, average Q length in this case because this will be required so I call it l bar this will be nothing but summation of that number of packets which were there in the Q i and what is your probability of being in a state i. So, this is for all i's and this will be your l bar. So, this can be estimated again as i rho raise power i 1 minus rho and i goes from 0 to infinity and of course, you can solve this and you should get a value of rho 1 minus rho I am not solving it I am just writing down the result directly the easier way to solve it actually take the derivative take 1 minus rho out take the derivative and then of course, replace that is the best way of doing it. So, I can I can try doing that actually it is not. So, let me do it slightly. So, 1 minus rho comes out straight i is equal to 1 to infinity for i is equal to 0 this does not make sense I will have rho i 1 minus rho I can take also a rho out and of course, now this thing can be always written as and summation and derivation both are linear operators. So, I can always exchange them. So, I will end up in getting d over d rho summation of rho raise power i i is equal to 1 to infinity. Now, this summation is again nothing but a geometric progression with the first value is going to be rho in this case. So, rho rho raise power infinity minus 1 rho minus 1. So, rho raise power infinity minus 1 this thing actually will turn out to nothing but derivative of rho 1 minus rho that is what you will get remember whenever you are taking 1 over 1 minus rho derivative will be minus 1 over 1 minus rho square and minus rho again. So, it will become positive actually that is how it will be and I can solve it 1 minus rho plus rho this cancels with this I will end up in rho over 1 minus rho. So, that is what was the result which I wrote here. So, this gives you the average q length and you can actually now plot rho into 1 minus rho is equal to l bar and that is when rho is equal to 1 and rho is equal to 0 this goes from 0 and then this explodes and goes to infinity. So, q becomes unbounded it will not be bounded in length and it will be unstable at rho is equal to 1. So, we will use the same now technique essentially these are basics of I think everybody in Queen Curie does this thing first. Now, we need to do it for m by n composite switch. Now, here what is happening is a call actually arrives. So, call request arrives and those arrival will happen now you have to understand if whatever as a free ports only call can arrive on those once a call arrives that line is now busy. So, if you look at a telephone line if a call arrives call will now remain hold a line is now occupied till the call is over the person is now talking for this period and when call is over again line is free then only the new call can arrive. So, in this period no arrival is possible in the line actually. So, arrival is only possible for the free lines. So, but for a free line when a line is free what is the arrival rate. So, I can take some lambda. So, if there n free lines my arrival rate will be n into lambda. So, lambda is the arrival rate on single thing now once somebody is talking usually what I am assuming is the person when he talks the duration of his talk is now exponentially distributed which actually means this call completion takes on an average 1 over mu time. So, what I will call is a call departure rate is mu for every line. So, if there are k lines which are occupied which are busy which are which are being talking. So, k here also going to be occupied connected to these. So, k into mu times will be the departure rate in this case. So, k into mu into delta t is a probability of coming from state k to k minus 1. Once you got to k minus 1 it is k minus 1 into mu into delta t and similarly arrival rate is going to be whatever is the number of free lines. So, number of free free lines is l. So, it will be l into lambda into delta t now that will be variation here and the state of this m by n composite switch it is m by n cross bar technically and only n connections can be put on the outgoing ports. So, the state of this particular switch is being governed by number of lines which are occupied here at the outgoing port. You cannot have more than n lines that is not possible m because m has been taken as larger than n. So, it is not possible to have more than n line getting occupied and after that even if a call arrives it has to be dropped it cannot be taken up. So, it is a blocking system and it you do not have the way we had in a queue you could have infinite packet being queued up here that is not possible you can only have till n and after that even if call comes you cannot hold it actually it is gone. So, you are in the blocking state. So, your queue will have only capacity of n here. So, that is how m by n composites which can be modeled with this particular thing. So, let me now build up a Markov chain for this entity. So, your m by n composite switch is going to be. So, there is no call there. So, the arrival which can happen will happen with the rate m lambda delta t. Remember delta t again we is going to cancel when I am going to have a balance equations to be built for a steady state situations. So, I need not I can actually remove delta t in fact earlier time the delta t cancels out from both sides I need not keep it. So, in fact when I have built the Markov chain I could have removed this delta t without any problem kept now that is what actually is being shown in most of the places. So, I need not keep I just keep the rates here it is proportional to rates and so on. And of course, if one call arrives you will be in state one and then how you will come back to 0 delta t. So, I am now going to remove delta t because of the reasons which I have mentioned I need not keep it. Then if you are in state one. So, one call is coming in one call is through. So, the only now m minus 1 lines are available on which the call can come and here n minus 1 lines are available. So, if the next call comes this will come with a rate m minus 1 into lambda that will be arrival rate and from a state 2 you can transit back to state 1 with 2 mu now. So, remember this instantaneous probabilities are fixed and the only one call can be completed at any point of time. So, if 2 2 0 will always happen from 2 2 1 and 1 2 0 that is the rate is always going to happen because delta t is in limit is going to 0. So, from 2 I can come to now 3 and this will be m minus 2 into lambda 3 mu perfect I can keep on doing it. But for how long ultimately remember I have the maximum size n minus 1 calls can be set up and when the nth call is set up you have had only one line free. So, it has to be lambda and you can go to with n mu and after this even if arrival happens you will come back to the same state does not matter after this. So, with this transition probability this actually loop backs m minus n lambda arrival you come back to the same thing. You do not change the state. So, this here actually now this should not be lambda, but this should be m minus n minus 1 lambda actually sorry I need to correct this and you can actually now note that the sum of these 2 if you make m minus 3 forget lambda n mu in this case m minus 3 plus 4 this will be m plus 1 m minus 2 plus 3 m plus 1. m minus n I think it should be n plus 1 m minus n plus 1 plus n m plus 1. So, that is how you can verify this will be the Markov chain for a composite switch and of course, now I can write down the balance equations and then solve them and then get the probabilities. So, let me do that. So, I am going to build up a first of all this surface p 0 into m of lambda should be equal to p of 1 into mu. So, which implies p of 1 is m lambda by mu p 0 that is what you will get. If I make the second surface like this or you can make a surface like this also does not matter ultimately you will be actually now canceling all the terms and trying to get all the state probabilities in terms of p 0 and sum of all probabilities will be always equal to 1 that is the way it is going to evolve. So, p of 2 now in terms of p 1 I can write m minus 1 lambda sorry it has to be p 1 p 2 into 2 mu. So, that is a transition rate at this surface which will give me p 2 is equal to m minus 1 lambda divide by 2 mu into p 1 m into m minus 1 2 into 1 lambda by mu square p 0 keep on doing it. So, p 3 similarly will be m into m minus 1 into m minus 2 3 into 2 into 1 lambda by mu cube p 0 and of course, you can clearly observe what is this is combinatorial m c 3 lambda by mu 3 p 0 and in general I will have p n is equal to m c small n n has to be actually smaller than n lambda by mu raise power n p 0. So, now at this point actually let me stop and we will now take up this thing further and try to understand how to compute call congestion probability and time congestion I will also define those two terms in the next lecture.