 So, we will continue with the discussion where we left in the previous lecture. So, what I was doing just a recoup of whatever has been discussed in earlier lecture, I was actually discussing m by n composite switch. So, there will be m inputs in this case and there will be n outputs. So, these m inputs can come at any point of time and n outputs have to be connected to the outgoing side. So, all subscribers are connected on this side. So, typically this particular calculation will actually be required when you are actually going to compute either for a blocking probability estimates for that and of course, what I will do is I will try to use this to actually give an understanding of what is a call congestion and what is a time congestion. I have to formally define these terms and also convey the meaning of these two because this how the switch blocking probabilities are usually estimated. So, basically what is the amount of call congestion and what is the amount of time congestion in a switch is an important parameter. So, for this the way I actually I had told earlier that for each line a call rate can arrive at lambda calls per unit time. This technically means between when the one call on a time scale when a call gets finished up and when the next call arrives the gap between these two is exponential distributed with mean value of 1 by lambda. And as I have actually mentioned in one of my lectures that exponential distribution as well as a Poisson statistics are kind of two faces of the same coin. So, if I look at the number of events and this events are point events which are happening. So, number of events which are going to happen that probability will be given by in a time t. So, n events is going to happen that probability is given by this and of course, the time gap between them is going to be exponentially distributed with mean value of 1 by lambda that is what actually it means. So, whether you talk in terms of Poisson distribution or you talk in terms of exponential distribution these are technically same things, but when we talk about voice call the things are slightly different because when a call arrives the person actually talks it consumes. So, channel is blocked no new call can arrive at this point of time only when the call gets finished then the next call only can arrive and this inter arrival time basically when the call gets finished the next call arrives that is being characterized by 1 by lambda actually for the circuit switching systems because this is not a point process a call arrival is not a point process it arrives, but then it holds on for some time before the call closes it does not finish is off it cannot be served immediately actually it takes some time. So, usually for this we will have exponential distribution of 1 by mu and for this 1 by lambda and for a small infinitesimally small time delta t that probability that one call will arrive is given by lambda delta t you can put here lambda into delta t risk power on e risk power minus lambda delta t by 1 factorial and you put in limit when delta t goes to 0 this will be actually converging on to this entity. And so this actually means if and we have to define the state of this particular switch. So, state of the switch is how many calls are currently through so if you are in a state 0 you can actually move to a state 1 and this will happen because call can arrive on any one of these lines and that will happen with the probability of because each line is independent. So, that will be happening with a probability of m lambda delta t and similarly when a call is being call is already in process what is the probability that call will be completed actually in time delta t that will be given by whatever be the number of calls multiplied by mu into delta t because mu into delta t is that a call will get completed in a infinitesimally small time delta t if there are m calls. So, the probability with that one call will get completed will be m mu delta t kind of thing. So, in this case there is only one call so you will have mu delta t and then of course, the state can be 2. So, in this case it will be m minus 1 because there is already one call through now second call is arriving. So, blank input ports for m minus 1. So, it will be m minus 1 lambda delta t and this will be 2 mu delta t and so on and then of course, in the n to maximum number of calls I am actually assuming a case where n is less than m. So, the maximum calls which can be done is n. So, when you are in this state you will actually have number blank input ports or input lines will be m minus n. Even if a call arrives it will remain in the same state call cannot be accepted because it cannot be connected to any outgoing port. So, just before this it will be m minus n plus 1 there has to be at least one output port free for coming to this particular state. This state will be called n state will be lambda delta t and when you go out it will be n mu delta t and under steady state operation the probability of being in a state is going to be constant is going to be same for the time to come on an average basis actually. So, which actually means under steady state operation if I make take any close surface like this the rate at which the transition which will be happening to for going to outside the surface and coming into the surface has to be equal that is a balance equation. So, once we create a balance equation I can solve for this. So, in this case for example, it will be m lambda delta t probability that you have to be in a state 0. So, that is the rate at which the transition out of the surface will be happening this has to be equal to p 1 there is only 1 mu delta t. So, that is a rate and delta t of course, cancels. So, in fact the balance equations will not have delta t. So, which actually implies I can actually now need not consider these delta t is here because we are talking about rate equations. So, which implies that p 1 I can now write here p 1 will be nothing, but m lambda by mu by 1 what about p 2. So, for p 2 if I write the equation p 2 2 mu t in fact I am not writing delta t because those will anyway will cancel out and if I am there in p 1. So, this will be m minus 1 into lambda. So, this will become p 2 is equal to m minus 1 by 2 lambda by mu p 1 and I can put a value of p 1 from here this will be m into m minus 1 by 2 into 1 lambda by mu square p 0. So, that is what will be the term and so on. So, p of i will be in this case will be m into m minus 1 m minus 2 and so on m minus i plus i i minus 1 2 1 and of course, clearly this is nothing, but m committed to real i lambda by mu risk power i into p 0. So, that is how the state probabilities for this particular Markov chain actually can be computed and then we also know the rule that sum of all state probabilities because all of them are mutually exclusive and system has to be in one of those states this is a complete definition. So, summation of all state probabilities p of i for i is equal to 0 to n has to be equal to 1. Remember we have done it for m m 1 infinity q also in the one of the previous lectures. So, I can actually use this exam and then of course, do the summation this will give me summation of p 0 plus m c 1 lambda by mu risk power 1 p 0 m c 2 lambda by mu risk power 2 p 0 and so on last term will be m c n of course, lambda by mu risk power n p 0. So, of course, I can actually put here m c 0 which is 1 lambda by mu risk power 0. So, this gives me a complete arithmetic progression sorry this is geometric progression because the binomial actually series and once I do the summation of this, but the problem is I cannot solve it because I need to get the terms from 0 to till m I am actually having only term till n. So, I have to just keep it as it is that is a sum which is going to be there and if I do this. So, I can write this thing as m c i lambda by mu risk power i i goes from 0 to n where n is less than m p 0 is equal to 1 and from here I can get an estimate of what is p 0. So, p 0 will turn out to be 1 over summation of i 0 goes to from 1 to n m c i lambda by mu risk power i and then of course, I can actually further solve it I will get p i as lambda by mu risk power i m c i whatever was the p 0 value which is now this is the probability distribution and we call it a ang set probability distribution for a composite switch. So, this gives the blocking basically the probability that a switch will be in state i. So, now if I put instead of i I put n which is the when the switch will be in the blocking state. So, once you are in state n any new call which arrives the call cannot be completed because all the outgoing ports will be busy. So, probability of blocking is can be identified as p b is equal to m c n lambda by mu risk power n lambda by mu risk power n k is equal to 1 to n m k m c k is lambda by mu k. So, that is the blocking probability which we have now when I am talking about a switch it is very important there two things even if the switch is in this state for which I have defined the probability and no call arrives for example, you are operating a switch does goes into this state, but when it is in this state no new call comes actually. So, in fact none of your calls will actually see blocking. So, there is a difference. So, switch just remains in blocking state does not mean that a call will get blocked the call will only get blocked if a call arrives. So, we have to now define something called call congestion and a time congestion. So, now I will explain this thing like this that there was a switch and all calls the switch is in the blocking state because now if no if in some new call arrives it do not get through. So, the probability your switch is in the blocking state is what is known as time congestion actually. So, that is time congestion. So, clearly what you are seeing here is nothing, but the time congestion it is not the call congestion actually and if you want to actually make a measurement of a switch you just keep on observing the switch. This probability will give you nothing if you observe the switch for some time t it is a fraction of time fraction of t for which switch was in blocking state. It does not talk about whether calls were blocked or not blocked it is talking about only switch being in a certain state. So, this since it is in terms of fraction of time we call it time congestion it is a time for which switch was in blocking state. I can now change the definition and probably some of you may not be interested in this parameter. People can ask what I would like to know is if I make 1 million calls how many calls were not switch was not able to complete switch was not able to connect them actually. So, the fraction of calls which saw the switch was in blocked state now that actually depends on the call rate when you are making the call and that fraction of calls which will see the switch in blocked state is what is known as call congestion. So, this is fraction of calls seeing switch in blocked state is what is known as call congestion. So, we would like to now actually find out a relation between these two. So, how that actually will be done. So, what we can say is let the switch was in blocked state that probability is P b and when you are in this blocked state this basically is call congestion what is the probability condition on this that you are in state n a new call arrives that probability this will give what we call in absolute sense a call loss rate number of calls which will be lost on an average basis if you may 1 million calls how many of them will be lost is not conditioned actually it is not conditional probability, but it is absolute sense probability that a call arrives and it will be lost it is not that what is a probability that a call will be lost conditioned on that a call arrives remember that two different thing if a call arrives now call has arrived is already sure now what is a conditional probability that it will be lost. So, call arrival is already there if call is not lost this probability of call loss is going to be 0 probability if the call is going to be lost surely it is 1, but when I am talking about this kind of thing probability that a call arrives and call is lost that will never be equal to 1 that will be less than 1 actually usually. So, this should be that probability call arrives and probability arrives and it is lost. So, I should write this that this is a probability that a call arrives. So, call arrival probability is also included and it is lost both things this should be equal to that what is a probability that a call will arrive p of a and then it will see that switch was in blocked state. So, we call it p of l that is basically is probability that a call arrives is a conditional probability that when call arrives the call will be lost this is what we call call congestion this is what we call call congestion this is time congestion we need to find out the relation between these two. So, from here we clearly because we have estimated this p b we have to estimate p l I will do it for composite switch. So, p l will be given by p b now if the call arrival probability when which I have written in the bottom is a call arrival probability when switch is in state n when call arrival rate is independent of the switch state switch is state does not govern it this value and this value both will cancel these both will cancel and call loss probability will be equal to or call congestion will be equal to time congestion both of them will be same this usually will happen in very large gigantic switch and you are operating below a small level actually very large number of things are there. So, it is like if you take m and n and m is very very large than n then this will be very closely call congestion and time congestion will be very closely relate actually become equal to each other. There will be minor difference because in absolute sense in real sense these two probabilities cannot be same, but you can approximate them we now need to make an statement for this we need to find out how we will find out p n a first we need to find out this now this will this is call arrival probability when switch is in blocked state. So, that will happen what is the arrival rate total arrival rate because of the free incoming ports when the switch was in state n over time delta t. So, this will be nothing, but you have to find out this arrival rate is m minus n because that is a free ports into lambda into delta t and I am talking about a arrival chance within a infinitesimally small time delta t. So, that is why I am not taking the full Poisson expression because remaining thing actually goes to 0. Now, p of arrival call arrival because now I have to do what we call as statistical averaging over all possible states for all arrival probabilities. So, I can write this thing as lambda t into delta t where lambda t will be nothing, but it will be an average value probability you are in state k is this and the arrival probability in that state is this one. So, I will do the averaging sorry this will be n because there are only n states. So, that will be your lambda t. So, from where I can get p a. So, this also can be done now if you put it of course, I need to put these two in these this particular equation and I will get p of l lambda n lambda t n p of b. So, that is the term which we will get from here. So, p and a will be nothing, but lambda n delta t p a will be lambda t delta t delta t will cancel. So, again we are talking about rates and lambda n I can write as m minus n into lambda into p b this I can write as summation I will let me use the write c m subscript p k lambda k p b and this solving this further. Now, you can actually clearly observe one important thing I am doing the averaging of arrival rate. So, lambda when you are in 0 state will be very high because you will be it will be m lambda actually in that case as I keep on reducing keep on actually increasing my state value it will be monotonically decreasing function in the end you will have m minus n into lambda. So, average value which you will be computing will always be higher than this particular value which actually means this whole fraction will always be less than 1 which implies that your call congestion p l will always be greater than or equal to p b the time congestion. So, this will be usually holding true unless your arrival process itself is a state dependent then only it will be equal otherwise it will be always less than condition. So, now let us solve this for this particular thing. So, p of l will be given by m minus n lambda p b is. So, I can write now p k as m minus k just a minute m c k sorry lambda k is m minus k lambda and I have to put a value of p k. So, simplifying it now p k will be given by an n set probability distribution m c k just you look what we had derived this is how you will get your m c k. So, I have to write exactly the same expression. So, I am using subscript here is 0 there is a different summation that is what it should be. So, it is exactly from here I have just taken it and we have got this expression and of course, ultimately p b here. Now solving it further and probability of blocking also we have the relation already identified by this particular thing and I have to use this expression. So, let me use this also. So, I am just rewriting everything here on the fresh sheet now p b is going to be m c n lambda my mu risk power n divided by summation actually variable running variable here. So, now this particular entity and this particular entity are same and they are independent of k. So, I can simply cancel them. So, this will give me nothing but p l or the call condition. So, you will end up with m minus n lambda this I can write as m m minus n factorial n factorial lambda by mu n into lambda m minus k m factorial k factorial m minus k factorial there is a lambda lambda by mu risk power k. So, this lambda of course, can cancel with this one without any problem and I will end up in getting. So, this m minus n will cancel with this m minus 1 I will get minus 1 here this will be removed m minus k this will become m minus 1. So, I can take this m out and it will become m minus 1 factorial m minus n minus 1 factorial n lambda by mu risk power n. So, similarly m will be taken out here and then of course, m will cancel I will get m minus 1 c n summation over sorry this summation is over n actually. Now, interestingly this whole expression this particular expression is very similar to the expression given here this particular one except this m is now being replaced by m minus 1. So, which implies that p of l for a value m is nothing, but equal to p of b for m minus 1. So, that is the relationship which will come between call congestion probability and time congestion. So, once you understand this we need to now see we actually have to use it for finding out what we call in a three stage interconnection network how to estimate the blocking probability. So, the same concept will actually be used there. So, in the next lecture we will be actually doing a call congestion estimate for a three stage class network.