 start from the place where we left last time and we will solve to get a closed form solution. So, the P L the call loss probability is I think that is where I left probability of being in a state sigma arrival rate probability of call loss conditioned on sigma summation over all sigma actually. So, that is where we started we have identified individual components I think for P of sigma that is what we were doing and we could get to a solution which was I have not proven that it was m c that was probability of being in a state x and correspondingly probability of being in a state y was that is what we had done ok. This I had got from Ankset distribution. So, very quickly just recalling that what we did was being in a state x was 0 to m we are taking case where m is smaller than n actually ok. So, this can only go till n you can only have n you can only have m occupancy not n. So, that was the probability of being in a state x and correspondingly it is true for a state y also you make it I actually that is better. So, it will not confuse ok and what I did I think if you recall quickly I think I did very fast because it was in the end of the class. Now, this is a complete binomial. So, I can always write this thing as 1 plus lambda by nu this power m ok this is coming from complete binomial and then I moved it we just modified this thing nu plus lambda this power m nu this power m ok. And what I did is I said ok let us put lambda plus nu as a the reason for doing this was if I divide by actually if I take nu out this will be n lambda. So, this will be 1 over lambda plus 1 over nu this value will be equal to this. So, on time scale if this is the first call arrives the call remains there call actually goes away call is cleared the next call arrives. So, this duration is approximately 1 over nu and this duration is approximately 1 over lambda. So, this technically gives nothing, but the fractional utilization and we also define this as a load in R lines ok. R line is the name of a gentleman and then since he had contribution to switching to respect him is given as R line. So, now how much R line of loads. So, for one telephone line the fractional utilization cannot be more than 1 if you have 1000 input ports and if you have 0.9 fractional utilization per line. So, it will be 900 R lines load on the whole switch that is a way it will be defined. So, we solve it further actually may because call cannot arrive see when the line is busy the call is on hold no call can arrive during this period and it is not between 2 because it is a special kind of process you take any particular time strength it need not be that call should have been arrived at time and after how much time the call will arrive it will the average time will always be 1 while lambda that is the beauty of this particular distribution. Because call cannot arrive here call will only arrive after this. So, this can should not be included in the arrival rate because that time line is busy occupied call is already there. So, 1 by lambda corresponds to this entity if you converse put all 1 by mu is equal to 0 then will be exactly whatever is the Poisson process which we have learnt. So, that is why it is done this way because telephone it is peculiarly in this fashion because circuit switch scenario. So, now solving I need to just convert everything to a and this will give me this thing actually I think this is a step where I missed out. So, I require and of course, 1 minus of a is 1 minus of lambda plus lambda plus mu. So, I will just do that conversion. So, from here you know what has to be done. So, mu m minus x. So, I have to take out mu plus lambda m minus x. So, here mu minus mu plus lambda x will be left because this sum has to be equal to this mu m minus x I have already taken lambda restore x will be here. So, take this and take this and you will end up in getting this. So, p x and p y both are known to me I have already determined what is going to be lambda x y this also we have determined. We also had determined this is a middle stage switch. So, k is the number of switches in the middle stage with that we had also found this. So, all things are there I have now put it in that summation and then solve for it actually. So, I will do this. So, remember it is a dual summation sorry this has to be. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . That is why I have taken a constant for which I do not know the value. So, I have put in your lambda sigma has been put. Now, let us put p of sigma which is multiplication of these two x and y both are independent. So, I have done this p x p y. Now, conditional probability x factorial y factorial divided by x plus y minus n n factorial. So, numerator part has been done. What I will do is I will try to solve in this equation and I will go slow. So, that because you have to rewrite the equation I can erase the board and do it. This will be I think this cannot be n this has to be k, because the reason for this is this I think the way I have taken, because k is greater than n that is why this gets converted to n c y. Any n maximum connections can take place at any point of time. Input line is m and this is n. So, in that case I have to use m here actually. Look at the expression this should be m. So, I think to be consistent I will just change wherever n is there I will just put this should be again m. I can observe the inconsistency otherwise this is then n it will be fine. So, this is actually n and this we are using n, because I have my derivation with n k actually n where is there anything else wrong? No. So, this I have to change this is fine. So, kindly verify that everything is correct. So, it is as per this notation. Lambda sigma no this cannot be n. Incoming line how many are free? n small n is greater than m. Number of maximum calls can be n even if n is larger you cannot have more than m calls only m m of these a links can be occupied. That is what I derived I think there is same confusion n k thing. In my lecture notes I think it is written as n k on the website. So, that is where the mistake is, but anyway forget that website I have to be consistent whatever I have doing in the class. And you should be capable of modifying all the symbols whenever required, because I also remember it will be n k, but in a sense I have been doing m m. So, I will be doing m m not an issue. So, I have to actually now I cannot take m minus x a m minus y out of this x y summation you have to understand this, but I can take delta very clearly without an issue this delta can be taken out this delta will be cancelled expand these m c x m m c y do the same thing yet is nothing, but a step by step evolution now. So, I can cancel this and I will become minus 1 this will become minus 1 I can write this m factorial as similarly this factorial can be written as I can take these m square out I can do the same thing on the numerator side. So, I will end up in m minus 1 m minus 1 minus 1 this fine n factorial is independent of x and y. So, this can be taken out. So, m square cancels I can remove this now this particular thing can be written as if you look at this expression this is nothing, but m minus 1 c x m minus 1 minus x. So, there is one extra 1 minus a which will be there similarly for this I can also write 1 minus a m minus 1 c of y. So, I can very well replace this whole thing and 1 minus a square can be taken out x and y both are independent here and there was earlier one extra term that will actually become 0. Obviously, which you can verify. So, one term has been now reduced these two are independent. So, it is nothing, but complete binomial and I can replace I can actually simply write this thing as summation over x and this as summation over y that is possible. This thing is nothing, but 1 raise power m minus 1 which is equal to 1. So, this whole denominator is gone perfect we have only 1 minus a square. Now, let us look at the top stuff what we can do I have to use some trick here. m minus 1 factorial square actually can be taken out. So, this is no more required and now I will split into two summations one is x other one is over y I will do it over x later first over y and second over x. So, anything which is having y I will keep it on this side this x factorial cancels with this this is what you have. So, I am just going to rewrite it actually first. So, that is what you are going to have I can now multiply and divide I will do sorry I have to do that I am just splitting that sorry this one I will keep here summation these two terms actually will come here after this summation. So, this is what is the summation for y that is summation for x and I will then multiply and divide by something. So, what is sum of these two that has to be independent of y that is the only condition it should not be function of y. So, one is having plus y other one is having minus y. So, it will actually cancel if I add. So, I can multiply the by that whatever is the sum factorial of that I can put that numerator as well as denominator that is simplest way of doing it. So, it will be m just add these two plus x minus n minus 1 you can put in your numerator same thing I have to put and this I can take out because it is independent of y. So, this one is now this 1 minus a square I need not even keep this line 1 minus a square I can just put it here this thing this summation over y now what are the valid values of y that you have to figure out. So, valid values of y will be they will be ranging from. So, first thing is this actually should become 0 and secondly this should become 0 that is how you will actually get the range. So, y will actually varying 1 extreme it will be x plus y sorry x minus n and minus y you make it 0 n minus x. Now, this is the condition when x plus y is equal to n just when the blocking starts you x plus y is less than n blocking cannot happen this just when the blocking starts this is the first case. So, the y will actually range from this value for rest everything this will go to 0 anyway you cannot have summation over those terms other terms has to be removed even if all other possible values of y is there there is 0 I am only going to sum up only this range when blocking happens only those terms are there. So, I am not estimating when blocking is not possible those cases I am not considering in my term. So, that is why when that condition x plus y is actually great smaller than n happens those cases are not coming in if that situation satisfied for a switch this will give you a value 0 actually. So, we have taken care of earlier case in lease approximating we were not isolating those conditions we are not removing them here in the summation term itself when the blocking is happening I am removing those because those becomes invalid values otherwise I cannot sum it up if I take y is equal to going from 0 to n it is not possible because terms will be invalid in that case I cannot build up a closed form solution. So, while building up closed form solution I am actually removing or excluding those terms this is actually usually is not evident. So, I am explicitly stating this. So, the next range will be coming from here y will be m minus 1 obvious because there is one line which is free for which you are trying to on which the call is coming you are not looking at blocked state you are looking at arrive when a call comes whether it will get through or not get through. So, in worst case it will have m minus 1 states you cannot have higher than this it is not possible. So, this is the range over which this summation will be done then only I will get a closed form solution otherwise I cannot get a closed form solution. So, because the moment I take the summation because this is nothing, but a binomial now I can now convert it into a binomial. I can I have to put something this a raise to power y is there x I can take inside now I can call it a x plus y there is minus n I can put just a minute this is minus 1. So, I have only one left here I will also even remove this and this will be converting into minus 1 I will put I require minus n here I will put a raise to power n here. So, still everything is in balance now this is value matches to this and m minus of this or basically this value comes here this is now complete binomial this is the range which is also the valid range when the blocking happens. I am removing all the terms. So, I can do the summation and I will end up in getting. So, range concept is very important because usually unless you under appreciate this you will not be able to figure out that why you are getting the correct probability of call blocking correct you are getting because I am not taking this summation other terms which corresponds to the cases wave blocking does not happen those have been removed. So, this now you can solve. So, y variable is not there on this side. So, what will be the value a plus 1 minus a raise to power m plus x minus n minus 1 that should be the value p plus q raise to power n in binomial this is nothing but 1 raise to power something this whole thing is unity. Once it is unity I can simply remove this excellent you have got this one step closer to further solution. Now, if I add m minus x minus 1 this m plus x minus n minus 1 right if I add these two again I will get a range of x by keeping this as 0 and then this as 0. So, some of these two is how much 2 of m x will cancel minus 2 minus n. So, I have to just take the factorial of this n multiply and divide this any I can go out. So, I am not bothered I am only bothered about now this particular piece again you have to look at the range. So, what terms you are excluding what all values possible values of x you are excluding is important here. So, now one possible range will be m minus x minus 1 is equal to 0 you put. So, x will be m minus 1 which is valid because that is the maximum which you can get, but what is the minimum value of x? x will be n minus 1 n plus 1 that is a minimum value of x which is there. I remember in earlier case it was based on y was ranging y range was depending on x actually when you are computing range for y it was depending on x. So, whatever but x has to be minimum chosen as we any one of them can be least actually depending on what is the value of n. If n plus 1 is on this side remember this these many number of ports and these ports physical into these number of ports are double the this n plus 1 is greater than or equal to double the value of m then x will be this value will be this usually it will be lower actually this will only be higher if it is more than double the value of n. Only m minus 1 calls are there worst case scenario you are right this will be because if you actually go out of this range again the blocking will not be possible blocking will not happen. Now, why this is happening if n minus m minus 1 is this if my x is less than that on the other side m minus 1 can be occupied and I am having value less than this there is always possibility whatever combination will take it will be always non blocking only if this difference is fully covered remember I have done a case in this case can blocking happen blocking cannot happen blocking has to happen only if whatever is the left over is covered by this minimum minimum value has to be this and that is what I am writing n minus m minus 1 n minus m minus 1 is also to substitute that is this value. So, this value minimum has to be equal to this and maximum it can go to m minus 1 anyway. So, this value has to be lower because x cannot be higher than m minus 1. So, this is a lower range lower value and this is higher and the case which I am taking when n minus m minus 1 becomes equal to m minus 1 that is a strictly non blocking switch in that case probability of call loss will be 0 that will immediately if the summation all terms will go out actually in that case you will not be including any term in that case. So, again it means I am only taking the values in this range then it will be a complete binomial otherwise it will not be. So, I can now solve it. So, I have m minus 1 factorial square n factorial 2 m minus n minus 2 and this is nothing but a plus 1 minus a whole restored to m minus 2 minus n complete binomial a cancels with this it is 1. So, your no I have made a mistake what mistake I have made I have done something, but it is incorrect you have to have these terms which corresponds to these I do not have term for a which corresponds to this here. So, I cannot simply write a plus 1 minus a and then say equal to this no that is incorrect this a raise power n actually should come out this is independent of x is not very. So, I have to take it out I have to put some other term here. So, I will put 1 minus x minus n minus 1 that is a way of doing it actually. So, once I do it it is 1 plus 1 minus a raise power n because v at minus n was used here earlier in the previous step in the summation where we did summation over y we have moved it inside a raise power minus n was required. So, we created a raise power n here a raise power x actually was moved inside there. So, that you can have a raise power x plus y minus n. So, this is what is your the closed home expression for probability of call loss. So, I will write it is m minus 1 factorial a raise power n 2 minus a 2 m minus 1 minus n n factorial 2 m minus 1 minus n and remember this is nothing, but call blocking probability and we have estimated that call blocking probability will be function of n minus 1 if n minus 1 is replaced by n this will end up in this is a switch being blocked this is a time congestion this is a call congestion this we had already computed earlier for a composite switch. So, there is an alternative proof because this is a call blocking probability you can also do when the switch will be in the blocking state that estimation that is known as Jacobius approach for call congestion estimation in the same three stage network and he will actually get the result where p of this is p of l p of b will be this is m Jacobius got this expression and can you observe these two and these two the same relationship still holds same relationship still holds. So, this there is an alternative approach through Jacobius name of the gentleman who did this alternative derivation. So, that gives you the call blocking probability. So, now next we have to go into the all kind of theorems. So, that we can formally figure out that why a switch is strictly non blocking or why a switch is the rearrangeably non blocking and why a switch is wide sense non blocking and can we reduce their cross point complexity further I want to minimize on cross point complexity as far as possible. So, for what we have figured out is O n 3 by 2 for strictly non blocking switch can I make still better than that. Yes, we can do still better than that we can get O n log 2 n log 2 n power 2 point 5 8, but how this 2 point 5 8 comes is interesting this can be derived it is not a heuristic number we will do that derivation. But before that when I go through now formal proofs and formal theorems regarding this you have to understand the concept of Paul's matrix. So, again it is a name of gentleman in switching theory and this is a very handy tool in giving all kind of formal proofs theorems lemmas everything comes from here. So, I have sufficient time actually to explain what is this Paul's matrix what is the meaning of this 2 by 2 it is simple to do, but if multi states number of ports can be arbitrary any kind of thing and then you have to do a formal proof you require this. So, you will have 3 stage switches and I have lot of middle stage ones important thing is that if I take any other input and output pair switch the path between these switches is not blocked by the paths which are corrected between these 2 switches that is extremely important. I can set a path between this and this independently of this and this. If a path between this and this is being set up that may lead to blocking of a path being set up between these 2. So, at least either input or output switch has to be common for the blocking to happen or for the interference in setting up of the switch. So, in fact, I need not look into this very generic I think you should be able to appreciate that and I will give now some values say a and b this can be numbers numbers are also nothing, but symbols to represent their form from a set, but we have built up our own arithmetic on that or an algebra by which we can define the complete number system. So, it does not matter for me it can be any arbitrary symbols that is good enough because the algebra here is pretty simple. I can write this as the middle is f g h i j whatever you want to put and what we will do is we can always create a matrix that is why it is known as Paul's matrix. And if for example, this between a and b if I want to set up a connection. So, all the rows corresponds to the input stage switches. So, these are input stage switches and what should be these columns output is obviously see remember the signs is very symmetric in nature. So, if you understand symmetry and duty of it is going to be very very simple anything which is ugly probably it is incorrect that is a thumb rule and you will have here output stage switches. So, I will just explain something and then leave it for you and we will go ahead with this particular thing later on. If a switch a is here switch b is here. So, number of switches number of columns number of switches number of columns a and b I can set up path through many of them. So, if I have set up a path through f there is already a path happening. So, I will put a entry f in this cell which is common, but the cell need not have one entry it can have multiple entries also. If I am using g h i j also to set up path between these two I will put those entries also f g h i j, but f and g has already been used for a and b, but a I am using to connect to some i to some others a some z. So, for the z column in that cell I will put this i ok. So, that is what is known as pulse matrix only thing now you have to understand is I am actually giving it as an assignment you try to think about it. You can even look into the nodes that is your choice, but I will appreciate if you try to think on your own and come with the conditions and later on verify with the nodes or anyway I will tell that ok. What are the conditions of validity of this particular pulse matrix? If I give you a pulse matrix can you check whether it is a valid pulse matrix or not. My conditions like how many times entry can come in a column can it repeat? Will I have only f can come only once in the row or f can come only once in the column ok. Can f come at 2 or 3 places how many entries can be there at most? So, try to think about all those conditions you have actually understood quite a bit already only thing you have to just convert it into now this and amazingly you need not visualize the switch after this you can just work with this pulse matrix blindly after that ok.