 So, we will actually go ahead with the calculations which we are actually trying to do last time. So, I thought I actually have to last was the two matrix which have to be used for performance analysis. One of them was throughput and other one was turn around time. So, the way they are defined that throughput is nothing but average number of packets which are being pushed out per unit time that is how throughput is taken. It has nothing to do with the input of course, if you keep on changing your input rate throughput will change and throughput ultimately should saturate throughput ultimately should saturate turn around time will be your average time interval between instance when packet arrives at input of network and when packet leaves at the output of and of course, the maximum throughput will be existing when there are no conflicts. So, for example, at each input you have packets in such a way that there are no conflicts and all of them can be pushed without any contention to the outgoing port. So, that will be the case for maximum throughput and minimum turn around time also will happen when there are no conflicts because you will not be storing anything in the buffer just pass through. So, usually then we use the normalized version of these. So, whatever is the maximum throughput possible you can normalize with that this will ensure your normalized throughput is always less than 1 this for comparison sake it is independent of sizes now. So, that is a normalized value similarly for turn around time also. So, what is the basic unit of time counting you are essentially looking at that. So, you will divide it by minimum turn around time. So, these are the two normalized versions which actually have to be used. So, first of all we will actually estimate the same thing for unbuffered delta not the buffered one, but unbuffered one which we had done earlier. So, this one is pretty simple. So, for a 2 raise power n by 2 raise power n unbuffered delta you can assume p k will be the probability that. So, remember the k subscript. So, for different edges there will be different probabilities. So, I can actually iteratively compute if I know the input probability I can compute what will be the output probability and that will give me a throughput that is basically is the idea here. And from here you can define p n in the similar fashion this is probability that the packet arrive at output length all n stages have been finished. The total n minus 1 0 I am counting 0 1 2 3 n minus 1 are stages the moment I talk about p n that is the output length because there is nothing like a s n stage is not there s 0 to s n minus 1 are there. So, this will be a technically your throughput actually. Now, there are certain assumptions which have to be there. So, at every intermediate switch in general there can be many inputs and as we had done earlier each one of them is going to connect to a set of inputs. So, each one of these inputs will be connecting to set of inputs of the network. So, in earlier case it was only 2 set i x and i y I was talking about they were always disjoint. Here also all the sets which are going to form will be disjoint. Once they are disjoint since all arrivals at the network are independent and they are kind of independent identical process is actually arrival process. So, that independence also holds true here. In fact, it holds true at every intermediate switch. So, this actually means I can assume that p k is going to be same for all links all input links to stage k. So, not only this switch in that stage whatever number of switches are there all inputs in that stage they will be all independent. So, there is no correlation between them and this actually you can keep on extending it means this p n will be same for all output links also. That independence will still be maintained not for output side, but yeah p k is going to be same for all switches in stage k. Each one is an independent set actually and each one of them will be now connected to 2 raise power k inputs. So, since the number of inputs are actually being same to which it can reach and each set is different. Within a switch it is each set will be different. So, this will be showing independence this is actually true for everybody. So, they are going to be same they are independent within a switch. If you take two different switches then they are not independent, because they might be coming from the same input, but since every packet is also coming independently it is ok. The independence comes because of the fact when packet comes with the equal probability it is choosing an output destination. There is no bias with 1 by n probability 1 by 2 raise power n in this case it is going to choose any one of the output outgoing links and each packet in all inputs are going to do the same thing. So, statistically every input port is technically same you just find out probability at one place use it at everywhere unless you have asymmetric situation. It is a symmetric or uniform traffic conditions actually here. And important thing is that it is independent of what kind of delta network you are taking. Whether it is a shuffle net based or it is a inverse bands or whatever you have you can design or it is a omega network or whatever it is. So far it is a delta network the situation is going to be true. In fact it is true for any Banyan network not only this only in the case of Banyan network is that tag the address tag will not be uniform for a given output. Adress tag has to be different at different inputs for same output actually that is only variation. But so far that condition there is exactly one path from any input to any output is maintained and no input links and no output links in the intermediate stages are left open. This is going to be true. Technically it is a demultiplexed tree which you are building all the time and of course this is now throughput I can write a simple statement from here. Now paper actually gives slightly different variation here. I also could not figure out why it is been given that way. But what it is saying is in the whole switching network of K stages when the packet enters and packet goes out that is a period which is min D. So one lot of packet comes in it takes min D to go out and once the min D is over next then the next packet will come in. So there is one packet being injected at each input or if P is the arrival rate or the probability of having a packet on this input side the P fraction of packet on an average are being injected per min D slot. So min D is kind of a slot but actually it is not required in one slot if I am moving from one stage to another stage. So I can use piping when the packet moves here the another packet from input can come in here. So technically my number of packets has to be this particular slot that assumption has not been taken while computing the throughput performance. So throughput actually can be K times technically then what I am writing if this pipelining is also assumed. So pipelining is not assumed in the paper that is one thing which is done there. So throughput will be nothing but P raise power n which is probability that you are going to have a packet at the output link and total number of output links will be 2 raise power n and these many packets will go per min D slot but need not be per min D slot it can be per min D divided by K. So K times higher actually it can be achieved. So that will be the throughput I am just going to state whatever they have told. So this will be the throughput for all unbuffered delta networks in fact for that matter all Bunyan networks. So for that condition is satisfied that no input and output links in the input stage are left open and you have exactly one path from each input to each output and this we have to just estimate what is P n what P n has to be done through a recursion you cannot make a direct estimate once the P at the input link is given. So in this case now every stage you are going to have a switch remember and you are connecting them like this whatever way I am not bothered at currently something must have been done. Now each of this switch can be of A by B size general cross bar or in this case it has been taken as B symmetric case it is not a symmetric case you can solve for a symmetric case also that is also fine where A and B both are not equal. Now number of packets which are going to come are binomely distributed their B inputs as a single unit cross bar remember which is a simple switch. So for a cross bar of actually you can do it for in general for A by B also this paper does for B by B. So the packets which will arrive will be binomely distributed there can be J packets which can be arriving and this arrival can happen with the probability here the probability is P k remember probability that a packet will come will be P k at a input port. So this actually means B C J that is a probability that J packets will come and probability that a packet you take an output link. So this probability is nothing but will be P k plus 1. So probability you will have a packet on the outgoing link that probabilities there has to be at least one packet out of these which is selected here this can happen for all outgoing links with equal probability. So this can be estimated as probability a packet and this is under the condition that given that J packets arrive at inputs. So this value will be nothing but you will simply say that each packet now this is a conditional probability J packets have already arrived I have to multiply by this is a conditional probability remember. So this probability is that out of these J packets which have arrived under that condition none of them is being directed to an outgoing port I can find out that probability and each outgoing port is going to be selected with equal probability actually the 1 by B. So you will have 1 minus 1 over B that is a selection probability see and 1 minus 1 over B that you are not going to select the outgoing port none of the J packets will be going to select an outgoing port is this and 1 minus of this at least 1 is going to select an outgoing port 1 minus of that. So that is a conditional probability that packet will be there given J packets and then of course you can find out what is P k plus 1 from here. So from here P k plus 1 so J can go from 0 to B and of course this is a conditional probability 1 minus 1 minus 1 over B J and I just multiplied by whatever is the condition which was B C J. So this is absolute probability that there will be at least one packet at the outgoing port I have just summed up with all possible J's condition thing I have removed technically. So this actually can be solved and this will turn out to be. So this whole thing now can be written as this one actually I am taking out summation of 1 minus 1 over B J and P k I can combine actually. So this term is a complete binomial so it will become 1 this one is also I can always write A raise power J and B raise power P plus Q raise power B that form I can use again it is a complete binomial and based on that I can get the solution. So it is a iterative thing which we will get actually. So it will be 1 minus so that will be the probability of P k plus 1 in terms of P k at this I think you can also figure out was for input Q's we have done similar thing when the packets were dropped except this P k was converted to something else there and of course you can always take for the maximum loading condition P 0 can be taken as 1 and from there I can find out what is the maximum possible throughput which you can have. So even buffer delta will always have throughput which is lower than this that is the maximum which you can get this is not maximum sorry this is not maximum. So this throughput is P 0 is 1 you can compute and maximum throughput is all outgoing packets will have a without conflict this is a maximum throughput you divide by this actually. So normalized throughput your you will get is nothing but P of n which you have to get from here but this has to be solved through recursion. So P 1 will be 1 minus 1 minus 1 over b because P 0 is 1. So this is going to be smaller value actually every time the value will keep on reducing decaying actually. So every step you will do it after k stages find out what is the value of P n and that is your throughput that is a normalized throughput the whole P n you are dividing by this value actual throughput which I defined was P n into 2 to the power n by but maximum which you can achieve is when all outgoing ports will actually have one packet every min D slot and if you allow pipelining you multiplied by k. Min D s contains k sub slots is sub slot you go from one input stage from the input of a stage to output of this stage for every stage. Sir you have to sum up I mean recursively means P you assume P k plus 1 and then further you have to go or this P k 1 P k plus 1 will be. You know how to get P k plus 1 from P k. Yes. So if you know P 0 is being 1 under maximum loading condition what is P 1 you can find out please. So P 1 will be. You need to sum up all these after this. Why you need to sum up you are getting absolute value sum up is only when P 0 P 1 P 2 P 3 all values are there you do not know their values but you know their relations. So then use last one when you will say sum of all probabilities has to be equal to 1 is the last equation which makes your number of equations become equal to number of variables to solve each one of them. Here you are getting the values actually directly. This is no Markov chain actually it is not a Markov chain. You will put 1 over 1 minus B this is the P 1 with value B P 2 you will can find out now putting P 1 find out P 3 P 4 so on iteratively is a recursive relation which you can get at best. That is what I get. There are n inputs you can use exponential or even Poisson. I am not worried about the input statistics input statistics the packet is always there. So in a input statistics cannot be exponential or Poisson remember there is a reason for that I am talking about a discrete even discrete time system is a slots. So packet can only start here and it has to close by one slot. So only statistics which you can use is either a packet can be there or cannot be there. So this is nothing but P you use the probability. So on an average how many packets per unit slot you will have it is P. P is the probability then and that is the arrival rate per line. So most of the discrete time system that is what you are using going to use for packet switching systems. Packets are not coming at any arbitrary time interval if that is permitted then that is a different situation. In our case it is not it is a fixed length packets coming one packet can come in one time slot. You cannot have two coming in actually remember those systems actually can be done but usually the way we do it in that case is if you have a switch for example and there is one outgoing link with buffer the many inputs which are coming in they are not time synchronized. All these packets which are coming will be put in one single queue. Now since they are not time synchronized I cannot get in this case technically what I am assuming because all slots are synchronized they are fixed length. In one time slot I can get either 0 packet I can get 1, I can get 2, I can get 3 that is a binomial thing. But if I do not I am not permitting synchronization and all packet are of fixed length then packet can arrive at any point of time on time scale they cannot then all of them need not come simultaneously. It is not 0, 1, 2, 3 in a slot. So packets will come sometimes here, sometimes here, sometimes here then I talk about inter arrival time which is exponential distributed or alternatively a Poisson distribution that how many number of packets will come in per minute time because these are all independent they because of that independence I can take that assumption not at a source. Source usually actually have a correlated what we call arrival process but as you go into the core when the packets will get dispersed and packets are coming from various different sources then only Poisson's statistics can be assumed. Actually sir had I made some assumption like number of packet tends to infinity and probability of each packet arriving means very close to 0 then I put an approximate equivalent Poisson distribution. In that case you could have turned out the limiting case of binomial distribution. Fine but then what is happening is you are assuming this time slot to be. Very small. That is what you are technically doing. So continuous time variable is nothing but a limiting case of this when packet can arrive not at discrete intervals but at any time interval at any time instant. A real life always works this way it is not this but if there are many incoming lines you cannot control what is happening in the behind you can always use it as a Poisson thing. So when you look at this kind of queuing system then you will say that the arrival and statistics here is Poisson but that does not matter we will always be doing the Poisson this thing binomial only. Poisson will not give the length of the packets it will give the only. Poisson will give only the number of packets which are going to come per unit time. The length of the packet will tell you what will be the departure process. If all packets are of fixed length it is a deterministic departure process. So you know that what is going to be it is a fixed length. If you know packets are only of length l and l 1 and l 2 they are only two deterministic departure time. So either it will take time t 1 or t 2 with some statistical distribution and if it is exponentially distributed that is what we assume most of time then you will define this thing as 1 over mu will be the average processing time or number of packets which can be processed per unit time will be mu is a departure rate but we are not looking at these cases here. Important thing is that we are going to use a iterative thing to build up the outgoing probability. So you will be able to get P n from here and for the throughput performance. Sir what is the iterative acceleration for in unbuffered case we are having throughput higher than in the buffered case. Buffered case you can see the every time the probability is going down. So I am actually assuming if there are j packets for example I am taking these four ports I get all four packets all four wants to go to this outgoing port only one of them will be going. So remember when I am looking at the probability I am actually telling even if one or more than one if they are trying to go to an outgoing port one will go that is how I am estimating P k plus 1. So it means remaining are being dropped while in case of a buffered delta case this is not going to happen and of course we are using a buffered delta model we are using a back propagation of the signal. So at the input itself at least we will get the what is the maximum throughput which can be achieved under maximum loading condition. In this case you put maximal loading the throughput is limited to P n when P 0 is going to be 1. So you will find this as a maximum throughput cross bar I need not do this iterative calculation cross bar and this are only different in what this is a calculation for a cross bar technically B by B cross bar but I am using recursive measure of that. So every time probability is falling down because of recursion in case of a single cross bar same thing will now become your P 1 is 1 minus 1 minus whatever was P 0 2 raise power n 2 raise power n that will be your throughput performance. So P 0 you can make it 1 if it is a full loading condition. So this is what will be your value and of course when n becomes very large 2 raise power n becomes very large this can be approximated as so P 0 you can put to 1 1 minus e raise power minus 1 which is 0.632. So that is a maximum throughput which you can achieve with a cross bar single cross bar the whole switch is implemented using a single cross bar or I am using it implementing by multiple cross bars by creating delta network. Yeah there is one single cross bar being used for the whole switch. So unperformed delta we know that how we will compute for this one we know how we are going to compute but this has to be done computationally there is no closed form expression which can be plotted but these are exact computational results which will be there. First is unperformed delta the throughput is P n P n normalized throughput will be P n. So similarly you can also normalize this also if you wish you have technically normalized this when you say P 1. So maximum is whatever is to this power n divided by min d. So that you have normalized P 1 is your throughput here is that is a P n. So P n certainly will be less than P 1 or more. So what we want to say is for all look across to a delta network the performance under those scenarios performance is almost same. In throughput performance will be same for all delta networks under those conditions what is mentioned under all blocking because there is no blocking technically the switches are all symmetric in one single stage and all inputs are independent. If the independence is not there then there is a performance will be different then you have to assign the input ports such a way they are two correlated inputs which are always going to pump the packets make sure there is two correlated packets are always going at different switches in the different stages. They should not never come to the same stage because they will not they should not try to pump the packets simultaneously. Is it cross bar implementation? This one. This is a single cross bar implementation of the 2 raise power n by 2 raise power n switch. That one I am using b by b switches to create 2 raise power n by 2 raise power n switch delta. So there is one doubt in this a cross b if we have the throughput will be same or. Formula will become different that is the only thing formula will be different in that case. You can compute not an issue you can compute for a by b I should have done that because that is done in the earlier paper I had actually is have not done the analysis at that time because that was anyway required to be done here but this paper only talks about a symmetric case. So for a by b it will be so what is the probability if you have a probability p k known here what is a probability that you will get at least one packet on this side. So each packet is being uniformly distributed over this thing. So number of packets which will be j you can actually compute that. So p k by b that is a packet will going to come to certain outgoing port none of the packets are going to come to this outgoing port 1 minus of this and all possible j scenarios actually no all this will be a sorry this will be a all inputs this one only tells that at this particular input I will have a packet with probability p k with p k by b this packet will be directed to this outgoing port 1 minus of that this packet which is going to come here will not be directed here this should be true for all a. So I have to multiplied by a is not b a b b here that is the only difference which will come and 1 minus of this there will be at least one packet going to which is going to come out here. So your p k 1 this will be equal to p k plus 1. So this recursive relation will change in this case. No unbuffered there is no buffering here buffered delta is tricky to handle. Remember in the switch total outgoing ports are 2 raise to the power n best case can be when every min D slot one packet is going out if one packet is going out every min D slot. So throughput performance is 2 raise to the power n packets per min D slot that is a maximum throughput. So whatever throughput actually compute divide by this to normalize it. So that is what is being done. So you divide this by 2 raise to the power n by min D you will get this. So you will get p k p n will be the throughput I have written somewhere here. So this will be p n will be the normalized throughput. So I just find out what is the value probability at the outgoing port per port throughput I am estimating technically. Per port maximum throughput is 1. So I am just estimating that p n. Sir in the previous this is we took 1 minus 1 by b therefore j. So that is j is the count of number of packets which are coming at the input. I could have directly calculated also by read it explicitly for j packets come with conditioned on j. So since a is the number of input line should not be a p k sir and the it should be raised to a p k because a p k will give the number of packets. Number of I am not clear. I will write it as p k. You are telling this has to be p k by b raise to the power a p k. Yes at least it is raised to that exponential where it is number of packets which are there. Let me compute with j packets if you want the other way around this is a direct calculation. Suppose j packets are going to come here what is the probability for that a c j when j packets have arrived what is the probability that at least one packet will be coming to an outgoing port. So this is a condition so that probability is 1 minus 1 over b j total j packets sum it over j. j can go from 0 to a remember solve it you will get exactly same thing that p k will not come in the exponent. Just a way to understand actually. So both are fine no questions so far. So next is now we have to go to the actual analysis of buffer delta system. Now buffer delta system I have to now look into what we call state transitions of a switch. So before I move on to that particular thing I have to define the switch state. Now remember they are buffers which are involved here. So they are all possible combinations are there. So we will define 14 states of the switches here and that I will use in a building up the recursive relation. Again this is done through a recursive relation which is computationally computed. There is no simulation it is a actual calculation which has to be done. So it is a computational procedure. So there is no close form expression as such but computationally you will get the exact values. So the model goes like this most of the states I am looking into 2 by 2 thing. So one state is when the buffer on the outgoing side which is nothing but the input buffer of the next stage. That is free and this is free that is one particular possible state. So this is known as state 1. So I am keeping all the inputs free and then trying to put packets on the output side buffer. So one possible states I am just numerating the states now when you will have only one packet. Now you can say this packet can be here or this packet can be here but both are equivalent states. So both are going to be all equivalent states are merged together the way we had done earlier. So equivalence is I can rotate, swap, do whatever it is but input and output cannot be swapped the way we did in the earlier case in case of that white sense passed on blocking system. So this is a state 2. Third one will be now when I will have 2 packets which will be there this is state 3. Whenever the packets are there at the input buffer they will be directed to some outgoing port. There was no direction of the packet was shown because there were no packets at the input buffer. So in this case this is what will be the direction. Now remember even if this direction is not this I put it like this way. This is equivalent to the earlier states because I can swap these 2 ports. So it does not matter. So I will represent this thing by this because these numbers will be used remember these state numbers. I will be mentioning in the subscripts later on for the in that computational model. Then you will have a state. So I am looking at one input and then one in output to output kind of combinations all possibilities. So one is this state 5. Now this input can be directed to this one that is one possibility this can be directed to a free port that is a sixth state. Now whether you direct to top or bottom it is all the same. There is only one input. Basically this cannot go because this is already filled up. So this is a state 7. Then you have all cases when one input was there in the switch are taken care of not 2 inputs which are there. So if there are free ports one possibility is both of them will conflict that is a state 8. They do not conflict that is a state 9. So whether it is a bar state or cross state does not matter. There is no conflict when 2 inputs are there both will be state 9. State 7 will be conflicting. Conflict can only happen if you have more than one input. But it cannot go on to the outgoing buffer because the next buffer is occupied for some reason. So no outputs I have taken care of one. One this could be one case this could be one case and then you will have a state number 13 and a state number 14. Cross and bar states in this case are exactly same. They will all merge into state 13. So whether the conflict happens because both want to go up or both who want to go down both are same states here. They are total 14 states which we have to define for a switch. So all switches will always be existing in one of these 14 states in a buffer delta system. So now here I actually close the today's lecture and will start now looking into probabilities and steps which are required for doing computational stuff of buffer delta. So that we will do on Thursday. You have to actually refer to the paper because I am trying to do it as slowly as possible but still there will be confusions.