 So, you are actually making the probability of blocking estimate p b and for a least approach which is very simplified thing. The basic assumption was that irrespective of the state probability that input link and output link both will get occupied was all independent of the states. It was just simply p and we approximated that using a balance equation and we got the expression. This will usually work for larger dimensions of the switch. So, whenever the value of a is very large value of in that case we have taken I think m and n as the input and n as the output. So, whenever this is going to be very large this m and n that approximation usually will work, but not for the smaller values, but we can make a still better approximation then this least approach this was because of least. So, today we will do Carnot's approach and in fact, nowadays because computers are available you can actually look you can iterate through all possible states of the switch and actually can numerically compute the precise call blocking probability that is technically feasible. You have to just write the appropriate software for a corresponding architecture do all that computation and come up with probability of blocking for a given load condition. Remember probability of blocking will always be function of it will be always be a function of arrival rate lambda and the ports which are there m and n. So, even in the expression actually which we did there was m there was m and there was a, a was equivalent to lambda arrival rate arrival probability is always function of these three things. So, this can also be done numerically. So, but we are not bothered about the numerical simulation and estimating from there this is a close form thing a classical approach and we will do that. So, we have to assume the probability of blocking in fact, I should call it a probability of loss. So, number of calls which are lost or basically you are trying to make a call, but you find that resources are not available or path cannot be set up and we call the call is lost in that case. It is not that call was already there is something happened and call was lost that is a call drop call losses when you are trying to set up a call, but call cannot get through because the resource was busy in the switch. And call actually means a connection between input and output port of the whole switch is again is a three stage switch. Number of calls which are lost number of calls which were attempted. So, this has to do with the this is call loss probability technically. In earlier case Lee's thing we were only estimating the probability that switch is in the blocked state. So, that was time congestion. So, here Lee's approach gives you time congestion and then there is a third approach which is again very precise very similar to this, but which is used for estimating the time congestion directly, but we would not do that. I will just state the result of the Jacobius approach which is the third one and this gives you the probability of switch being in blocked state for the same switch using very similar model. So, in this case I will assume that sigma be the state, sigma is state of the switch. So, I am not depend going for the state dependent variables state of the switch. So, I can write this thing as probability that switch is in state sigma that is a first condition you have to do always summation with respect to that. The calls will arrive arrival rate also depends on the state of the switch and probability you are in a state sigma under that what is the conditional probability that call will be lost. So, call has arrived and then there is a conditional probability that call will be lost. Sigma is a state of the switch how many calls are through what is the current setup. So, you can always say if the switch there is no connection setup for example, this 2 by 2 switch no connection is set up this is one state. Another possible state is there is only one connection being set up this is this. There can be another state which is this for single connection only, but these two connections are mathematically equivalent can be represented by only one entity there if you set up two connections one possible state is this one possible state is this they are also equivalent you just twist the switch input output thing. How you enumerate or put the numbering this can be converted to this. So, these are the possible five states of a switch. So, in abstract sense I am is identifying the switch state by sigma no call one call two calls if there are three switches 3 by 3 switch there will be more number of states. Of course, this is sigma is over all states this is the general switch you can actually do an exercise that take n by n switch cross bar you have to set up one connection. How many possible variations are there in this case how many states will be there you can take the first to top input it can connect in n possible ways. So, n states next one n states n square states total will be there with single connection no connection only one state with two connections. So, there in n c two possible ways you can take the inputs they can be connected to n factorial ways sorry n c two ways they can be connected to the output. So, based on that you can actually make an estimate of what will be total number of switch sets it will be huge number is not going to be small actually for 0 connections for 1 connection for 2 connection 3 connection. We will use a technique here an abstraction and as I told that for example, these two switch states are equivalent states. So, technically this switch has only 1, 2 and 3 only 3 possible states because these two states can be merged in one state only. So, we are going to also use and limit our search space or what you call state space we will try to limit that to a very narrow range. But that trick we have to always use even if you are doing a computer simulation to identify a switch performance you have to do this otherwise you required a large tremendously large amount of computing capacity and you will be doing redundant computation. So, with that of course, if you try doing the simulation you will understand that. So, let us come I will just come back to this state, but let me write down the expression first. So, I think notion of a state is clear and state transitions happen whenever a call goes off whenever a call arrives in and whenever reorganization happens there can be multiple state transitions which are possible whenever rearrangements are done. So, this is the call loss calls which are lost this is equal to whatever was written in the numerator this is a number of calls which would have arrived total number of calls which would have been attempted. So, this is your blocking probability a call loss probability. So, we have to find do this now what is sigma question is this sigma has to be interpreted in some way I have to find out some state variable. So, that it is finite actually and we can easily solve it a small sigma right not capital sigma this I will always call this as summation this is sigma I will never call that thing as a sigma actually that will be summation. So, now let us do this small sigma is not this abstract thing you cannot identify by for example, this figure you cannot give a number, but this is a possible state this can be identified by some variable some notation some abstract notation. So, it is a set of all these possible states and each set you can call it a for example, you can call it this b you can call it c there is no number. So, sigma is that abstract entity which belongs to set of all states in this case you cannot do there also Markov chains are also similar concept when you write the state variables that is a abstract thing of course, when I did the queuing a queue thing I always say let the state being represented by number of elements which are there in the queue. So, for with my convenience I transformed that abstract notation of a state to the number of nodes which are there sitting inside it. So, I was able to identify state by the number of packets which are queued up in the queue that was my convenience, but in general state space notation in Markov chain or any place does not bother about a numeric value. So, all states form a set whatever state variable you are taking that is member of that particular set. In fact, for that matter even numbers are nothing, but they are also members of a set the whole mathematics actually is about sets. So, 0 1 2 3 whatever it is it says going to be member of set which goes from 0 to 9 there only 0 to 9 is 10 possible members of a set which represents a digit and every number is every digit is coming from the set combining multiple digits defining rules you are able to build up the whole mathematics after that. And only what you we do is we start making a b c d also as members of the set you start doing algebra on that we feel awkward because we do not count our oranges by a b c d we count oranges 0 1 2 3 or any counting which we do. So, that is a perception, but number of packets inside this or number of incoming ports outgoing ports which are getting occupied here also I will do the same thing, but it is not going to be state variable one single state variable there will be two variables which are independent of each other combination of those will give me the state definition, but we have to make certain assumptions before that before I can define this thing. So, again I am taking lot of middle stage switches are there I am only taking one input and one output because if I take some other one this will also be connected to everybody. So, between this it is independent it is no conflict when I am trying to set up connection between these or I take this particular pair. So, I can take only one input one particular case and one particular case here and based on that I can make a judgment this should be true for any other input switch or any other output switch also. So, I need not bother about the whole design I will only bother out one single input stage switch one single output stage and all middle stage switches because they do not have any conflict they can be independently analyzed I can take this pair and analyze. So, we are now going to take a very important approximation, approximation is from each input port to each output port the connection is made with uniform probability, uniformity of traffic actually extremely important thing. Once this is there the probability each one of these links will get occupied I am also again assuming to be same. So, when you want to for example, set up a connection between a free link here and free link here you have to find out first of all a free path here then a free path here has to be figured out. Now, there are ways and means of doing this you can start doing search from 1 2 3 4 5 you get a free link find out there whether there is a free link available not available go next one 2 3. So, you can you are now counting in an order till you find a free link and you set up the path next connection request come again you will go in the same order I am not assuming that it is not an ordered setup what it will do is it will randomly pick up just randomly pick up whatever is a free link and based on that it will choose it is not in order that is very important we call it random route hunting approximation no otherwise I cannot build up my state variable model. So, I want to simplify my calculations. So, at least whatever I am going to build you can only what we call it will give me some kind of an idea some kind of an bound actually on probability of call loss probability. Because at least this much I can achieve it can be only worse than this it cannot be better than this it can be better depending, but this random route hunting gives me a bound essentially. So, we are estimating that. So, under random route hunting these and these are independently being made busy what we do is we assume the x links are busy here and y links are busy here we really call them a links and b links. So, number of a links which are busy are x number of y links which are busy are y it is not question of one path or two path. Now, if you are careful enough in observation while removing these I have taken I have actually merged lot of other states into the states corresponding to only these two switches already one merger has been done otherwise there could be a large R1 number of switches here again same number of switches output state switches. So, all other inputs also will be connecting to these. So, actually number of states will be huge, but I have merged to all these possible states into states which are only because of this one input and one output and middle stage switches right. But for analysis purpose because I say because these are independent whether I do analyze for analysis for this analyze for these two pair analyze for this or this pair they are all independent. So, I am only analyzing this technically means for all other pair combinations of input and output switches I have all of merged all of them into one single scenario. So, already one merger has taken place random route hunting again further simplifies I can now only represent by x and y if to have been ordered situation the things would have been far different. So, I can now represent my sigma as nothing but x comma y. So, x and y pair I cannot add x plus y. So, it is not a single state variable it is a two dimensional state variable. So, when I am doing summation over sigma this will be actually two summations one over x and one over y and now blocking probability is dependent on my values of x and y I will explain how that happens. So, I have to now estimate what is p sigma what is lambda sigma what is conditional loss probability for certain states loss will not happen I have to take care of that while in case of Lee's approximation we were not bothered about state there were states where there was no blocking, but we are just making an approximation there. So, my state will be sigma is equal to x y you have to find out p sigma lambda sigma and p l given sigma three entities. So, we will take them one by one and solve we will try to understand how to build up a model for this. So, let us take I think the first thing which I should do is which is nothing but probability of call getting lost or probability of switch being in blocked state because call arrives at that time switches in the blocked state then only the loss will happen. So, is a conditional probability that switches in blocked state and in a given state and within that state I am looking in multiplying it by arrival probability. So, it becomes call loss actually in that case, but this is a blocking switch being in blocked states probability for that conditional one. So, this is a interesting thing I am taking a very simple example of say four middle stage switches these links can be occupied at any point of time. So, let me take x is equal to 2. So, x can take value from 0 to 4 there only four middle stage switches no rearrangements are permitted remember. So, x is equal to 2 and y is equal to 2 I have taken and if this is a scenario these two links are being used these two are free remember, but these two are used because this switch this particular path is being used to set up may be somewhere else actually like this. So, that is why I am this is because of the independence of these a links and b links you cannot set up a path in this case your switches in blocked state you can have a scenario when you can set up a path through the bottom link. This arrangements can happen this combination of two occupied ones happens independently out of all possible variations this happens independently actually of a links for b. So, what I have to do is I have to only enumerate all possible combinations and find out the cases when the switches in blocked state. So, number of enumerated states because now I am looking at this state even for x is equal to t 2 and y is equal to 2 I told you x is equal to 2 y is equal to 2 is 2 comma 2 that is one state, but that actually is not one state. There are so many possible states now again this is a question of merger I am representing this state by 2 2. So, all states for which x is equal to 2 y is equal to 2 are taken care of by this Jacobius approximation the same gentleman. So, is there another merger of states which happen. So, I have to find out all possible combinations which can exist and for which the blocking happens and for and total number of possible combinations. So, is very simple I will assume that there are m and n I have been taken a m and n this fine. So, there are n links here m links here. So, take one particular combination some particular value of x fixed. So, this pattern is fixed now. Now, this pattern can vary independently in how many possible ways you can create two busy links 4 C 2 there only these two guys which are there which are actually occupied or in fact there these are two actually now free ones. In how many possible ways these two free ones will coincide with these busy ones 2 C 2 that is only possible way out of these two. Now, these free has to coincide with this in how many possible ways you can do this 2 C 2 only ways and total number of possible ways with which these free links can be unaged here is 4 C 2 again does not matter why this a busy links or free links. So, there are total number of ways in which these two free links are coinciding with this divided by this value will give you the call loss probability approximately if everything can become busy or free with equal probability. I am still assuming the same thing equal probability that is a uniform loading condition what we did for least approximation. So, this divided by 4 C 2 will give you the loss probability given 2 2 I can take up an another example actually here to exemplify this whole thing. So, this time I let me take 6. So, x is equal to 4 and y is equal to maybe I can take it say 3 is fine I am just taking any arbitrary value. So, I have free is up and remember if I whatever is true for these 4 busy and these 2 free links there are many other combinations this ratio will still remain same irrespective of whatever I take here. So, I did not count the number of ways in which this arrangement can be done I am only worried about the outgoing side or you can do it reverse way around you fix this one and make an estimate based on this you will still get back the same result both ways. So, in how many ways these 3 busy links or 3 free links which are there these 3 free links can be arranged in this case sorry no no their total 6 possibilities and these 3 have to be arranged 6 C 3. So, 6 C 3 is the total number of possibilities and these 3 free links are can be arranged over these 4 busy links in how many ways 4 C 3. So, this should be the call loss probability which one you have to tell me no there can be block why not there is a possibility in this case the case which I have shown it is not blocking let us draw the case like this. There are only 3 free links now can you set up the connection. So, there are certain combinations for which blocking happen and if you observe this actually means your x plus y has to be greater than or equal to n then only you will have blocking. If your x plus y is less than n then you cannot have blocking it is not possible because even in worst case offset scenario you will have certain free links. So, you your total number of what is the value of n is 6 here it is 4 here is 3 you make it 2 sorry make it actually say 1 or make it 3 and 2. So, 3 will get occupied 2 will get occupied you take whatever combinations there is always one free link available one free path which is available call can be always set up. So, 3 2 case you take for example, I am just modifying this now there are only 2 busy links worst case scenario 2 are here there is no overlap there is still one free path available blocking is not possible in this case. So, blocking this is the condition in fact you would not be able to estimate p l given x y unless this condition is met I will we will come to this our factorial will not get a give a solution. So, now generalizing this whole thing. So, I will not be using now numbers this was an example. So, let us put this as x and this is y. So, in how many possible ways you fix up one particular combinations for x connections I have already free for y how many n minus why those many links are free big type links I have actually free through which the connection can be set up in how many possible ways you can arrange these n c n minus y and in how many possible ways these free links will be overlapping with the busy links on the input side or a type links. So, that will be x c n minus y same thing which we did this x c n minus y this n c n minus y same logic you expand on this. So, this is the numerator and let me expand now on the denominator part this is what it will be this cancels and you note and observe this particular factorial this only valid when x plus y is greater than or equal to n for anything is smaller you cannot define this you do not have factorial for negative values this is only for positive values you define and 0 factorial is always 1 that is a condition and then you are totally define the factorial. So, that condition is has to be met whatever was written here. So, this clear just a minute you are right because when I do x minus n plus y. So, it has to be yeah thanks for correcting it has to be this way. So, everybody agrees with this. So, we have already got this thing next we need to get what is the probability of being in a state sigma n minus 1 y is the free links. So, these are the how these free links which are there in how many possible ways they will overlap with the busy links on the a side. So, a links I have taken the busy ones which is number is x. So, x c n minus y possible ways the overlaps will be here and what are the total possible combinations n c n minus y. So, that ratio will give this because I am using random root hunting remember if it is not random root hunting this will not be correct expression you have to change that. So, how you are setting up root is important. So, you cannot simply blindly use this formulation if the root hunting is different. If the call is being set up through a different hunting process hunting is you will keep on observing keep on finding out when the free path is available here you do it randomly. So, with the random root hunting this is ok that is why the equal probabilities have to come into picture. So, I can just do the enumeration and ratios of that can be used to estimate the probability which is cannot be done otherwise. So, next is p sigma you will remember this I have written it at the corner. So, I will keep it here. So, I have to now find out probability of being in a state sigma. So, one of the important thing is that in this switch the value of x here and value of y both are independently driven they are not dependent on each other because some other people will be setting up the connections and the y might get y links are busy only because of other nodes and these links are busy because this guy is trying to set up to any one of the outputs. So, x and y can be independent. So, I can safely assume that for each line I am getting lambda x is the arrival rate I am going to use whatever we did for composite switch same thing and of course, if there are x lines which are already occupied. So, lambda x into n minus x. So, because each line I am looking at now the arrival probability. So, I am not worried about this actually cancels out whatever it is the value. So, actually what is occupancy probability of a link I am bothered about that m does not matter because again I can do balancing and then find out what is the probability of occupancy of each line here. So, call arrival rate is proportional to lambda x into n minus x for this particular links. On the other side on this particular side y I can again take as if the call is coming from the output side actually it is not it is coming from these other switches and causing this to be busy, but this is symmetric thing. The switch is symmetric. So, I can very well assume that this should be the arrival call arrival probability from the other side. So, sorry this is not the probability p sigma this is lambda not p sigma, but this is lambda sigma which I am estimating. So, this can be represented as I will come to this probability later on lambda x lambda y n minus x n minus y. So, these two I have used here and I can write this thing as delta n minus x n minus y. So, this is the second element which we will be using now coming to probability of being in sigma state. So, take a composite switch m by n and so p of being in sigma is nothing, but p of x comma y. I have already mentioned that these are independent. So, I can always write x and y you can be independently controlling their probabilities of being in those states. So, what is the value of p of x? Remember that composite switch thing we made a Markov chain and there was a state variable thing which was done and we got the state probabilities. So, we will use that and that value will be we will use the angst set probability distribution. So, probability of you are going to be in state x can be written as now when n is less than m in that case probability that you are in state n was represented by this was for a switch of m by capital N. This was the angst set probability distribution. This is the condition which is satisfied. This is the probability being in state small n where n is number of outgoing links which are busy. That is the estimate which was there. Now, when n is greater than or equal to m in that case summation will be this m k will not be required this can be replaced by sorry this capital N can be replaced by m. When n is greater than or equal to m you can only have go till m state. You cannot go to m plus first state because all the incoming links will be busy. So, when this condition is satisfied you will have p of n. So, I will write with c actually summation of k goes from 0 to m your state cannot be higher than this in Markov chain. So, this will be now m c k lambda by mu k. Now, this is nothing but a complete binomial thing whole range is there now whatever I am using the factorial I am also having those many terms. So, I can convert it to a closed form solution. So, closed form solution will come from. So, I think all of you are aware of m c i summation of p raise power i q n minus i where i goes from 0 to m. This is nothing but p plus q raise power m sorry this m minus i right. So, I will just use simply the same thing here and you will have p of n will be given by m c n lambda by mu raise power n divided by now what is the other variable this p raise power k. So, I can safely write this thing as 1 m minus k. So, this will be 1 plus lambda plus mu lambda by mu raise power m and I can further modify it we will define load actually for the telephone switches. So, I can write this as m c n mu raise power m lambda plus mu raise power m I have actually solved this particular thing expanded basically mu will come here mu by mu and expanding I will get this. So, this can be now rewritten as oh sorry this n right sorry sorry. So, we will define some factor called lambda and why I have done this because on an average a telephone line gets occupied for or any circuits line for 1 by mu duration. This is a duration once the call is finished the time average time after which the new call will arrive this 1 by lambda is that time. So, on time scale the first call arrived this was the time till the circuit actually was set up. So, this is 1 over mu. So, I will call it call arrival then there is a blank period and new call arrives this is 1 over lambda. So, this is nothing, but the fraction of time for which a line will be occupied on an average or we also call it occupancy probability and I will define this thing by a. This a is same factor which we used in the balance condition for doing the least approximation. So, there it was I defined that a into m is equal to whatever is the lambda p n into p and then from here we got our p is equal to a n to m by n. So, this was used in least approximation this a is actually same a because load cannot be identified in case of telephone circuits because in packet switching system we assume that packets are atomic entity they come at an instant. So, how many of them are coming? So, lambda and mu we decide based on that basis and for the mu that departure rate is such that we actually assume that packet is going as 1 instant, but 2 packets cannot go consecutively there is a gap. Gap is because of the packet length of course, but it is not holding up the circuit. In this case I cannot do it because there is nothing else can arrive till the time you are busy. In packet switching system in the input of the queue a packet arrives as 1 single atomic entity at the next instant at t if you get a packet at t plus also you can get a packet. In circuit switching you cannot get a call till the call itself is over. So, here loading condition is different we call it a. A usually will be represented in terms of Erlang. Erlang is fractional utilization of a line. So, only for one line if it is 0.95 you say 0.95 Erlang load on one single line. If a switch is of say 1000 incoming ports every line fractional utilization is 0.95. So, 0.95 into 1000 950 Erlangs will be the input load. So, we define it that way. So, expanding it further we will have the same thing can be expanded into this form. You can verify this. So, now you can represent this by a and this by a and I will get an expression. So, we will actually in the next class we will go move ahead with this with this a thing and we will now put all the three things together into a summation and solve it to get the call loss probability. So, that will be the our next step.