1 00:00:14,469 --> 00:00:23,980 Okay, so in my previous lecture I was talking about the expansion of crossbars using cross 2 00:00:23,980 --> 00:00:29,050 bar basically building up a larger switches using multiple of them in multistage interconnection 3 00:00:29,050 --> 00:00:35,489 configuration. We figured out that two stages cannot give me strictly non blocking property, 4 00:00:35,489 --> 00:00:42,780 but three stages probably can and I will show yes they can and they this requires much less 5 00:00:42,780 --> 00:00:48,489 cross point complexity compared to using a single crossbar of the same size okay, we 6 00:00:48,489 --> 00:00:53,680 will actually show that. But before we do that we would like to actually 7 00:00:53,680 --> 00:00:58,650 look into how we will estimate the blocking probability of a system. And then of course, 8 00:00:58,650 --> 00:01:04,049 they are two kind of blocking estimations blocking probability estimation that we need 9 00:01:04,049 --> 00:01:10,400 to also learn to differentiate and we need to also learn how these two are virtually 10 00:01:10,400 --> 00:01:11,400 related to each other. 11 00:01:11,400 --> 00:01:12,670 Unable to capture the image because lecturer writes and erases the content immediately. 12 00:01:12,670 --> 00:01:18,870 So I will start with a very simple cross bar which is M/N, so remember I am not taking 13 00:01:18,870 --> 00:01:25,110 my inputs are not same as the outputs okay. So I am actually having M lines which are 14 00:01:25,110 --> 00:01:31,760 coming in and there are n lines which are going out. And they were like the telephone 15 00:01:31,760 --> 00:01:37,350 communication which is happening the calls will come and calls will be them connected 16 00:01:37,350 --> 00:01:40,271 to some outgoing line. So if a call comes on this line this will 17 00:01:40,271 --> 00:01:47,690 be connected and call is true okay. So the rate at which the calls will come here will 18 00:01:47,690 --> 00:01:55,010 be defined using a poison statistics that actually means a point event on a time horizon, 19 00:01:55,010 --> 00:02:03,560 I will take time difference time duration of T how many events are going to happen. 20 00:02:03,560 --> 00:02:10,590 So if I actually assume the calls do arrive and they are instantaneously completed and 21 00:02:10,590 --> 00:02:16,209 instantaneously removed. So even happens that is it, normally whenever 22 00:02:16,209 --> 00:02:21,600 a phone call will come it will be held up for certain time so line will get occupied, 23 00:02:21,600 --> 00:02:26,510 new call cannot come, but if that is not happening at what rate the call will be coming when 24 00:02:26,510 --> 00:02:33,680 call is not going on. So that we will define using a poison statistics which is a commonly 25 00:02:33,680 --> 00:02:43,980 used model. So we define something called ? which is the number of calls per unit time 26 00:02:43,980 --> 00:02:49,970 I will call it call arrival rate okay or arrival rate. 27 00:02:49,970 --> 00:02:57,110 And this also being used this technique is also used for packet switching systems also. 28 00:02:57,110 --> 00:03:05,300 So in time T ?xT calls will be arriving on an average okay. So the probability if this 29 00:03:05,300 --> 00:03:18,790 is a person statistics the probability that N calls will be coming will be arriving in 30 00:03:18,790 --> 00:03:25,459 time T will be given by, now remember when the call is arriving is being instantaneously 31 00:03:25,459 --> 00:03:31,349 served and line is made fries word new calls can come in, because if you occupy the line 32 00:03:31,349 --> 00:03:35,209 new call cannot come in actually then I cannot make this estimate. 33 00:03:35,209 --> 00:03:48,730 So ?TN E-?T/N!, so this will be the distribution okay. So if you look for the probability of 34 00:03:48,730 --> 00:03:53,629 no call being arriving here probability of one call probability of two calls arriving 35 00:03:53,629 --> 00:03:59,120 in time T and so on. The total sum of these are all mutually exclusive events so when 36 00:03:59,120 --> 00:04:03,879 one call is arriving within time period T two calls will not be arriving when two is 37 00:04:03,879 --> 00:04:08,220 arriving one will not be happening. So these are all mutually exclusive events 38 00:04:08,220 --> 00:04:12,720 when I sum up all the probabilities this should be equal to 1 which actually implies that 39 00:04:12,720 --> 00:04:24,220 summation of ?TN E-?T/N! this should be equal to 1 which 40 00:04:24,220 --> 00:04:42,820 is obvious because, so n goes from 0 to 8 and this is nothing, but a series expansion 41 00:04:42,820 --> 00:04:50,370 for exponential. So this will be written as E-?T which it turnout to be equal to 1. So 42 00:04:50,370 --> 00:04:56,520 this is a probability distribution and we call it Poisson distribution actually okay. 43 00:04:56,520 --> 00:05:02,960 So that is what is happening when the line when actually in real life when a call will 44 00:05:02,960 --> 00:05:07,940 come people will start talking so there will be some time after which call will get cleared 45 00:05:07,940 --> 00:05:25,080 okay. So I define that the call duration is exponentially distributed that is my assumption 46 00:05:25,080 --> 00:05:35,430 is exponentially distributed and the mean value mean duration okay for 47 00:05:35,430 --> 00:05:48,110 every call is 1/ Á okay this also comes actually can be derived from poisonous statistics. 48 00:05:48,110 --> 00:06:04,570 So before I move further I need to clarify this so I had define ?Tn e- ?t / n! that is 49 00:06:04,570 --> 00:06:14,620 a arrival statistics but if I want to measure to Britain between two consecutive arrivals 50 00:06:14,620 --> 00:06:21,940 what is the time duration let me find out the statistics for this time okay so I want 51 00:06:21,940 --> 00:06:29,090 we can figure out a PDF but for finding out a PDF, PDF is probability density function 52 00:06:29,090 --> 00:06:42,820 you can always take probability distribution function and do that derivative of this 53 00:06:42,820 --> 00:06:49,080 and if the variable is T it has to be / T it is derivative as to with respect to T. 54 00:06:49,080 --> 00:06:58,650 So I think this is very similar that your x is less than X that gives you the your PDF 55 00:06:58,650 --> 00:07:06,069 probability distribution function you take the density you will define something like 56 00:07:06,069 --> 00:07:15,380 probability density function, so the variable X is going to be in range x + x2x + dx within 57 00:07:15,380 --> 00:07:20,771 this range the probability that x will be lying within this is always given by px x 58 00:07:20,771 --> 00:07:28,389 dx okay that is what the density actually means and when it is a probability distribution 59 00:07:28,389 --> 00:07:35,370 function this defines probability that x will be having a value smaller than this X that 60 00:07:35,370 --> 00:07:39,800 is a PDF okay. And this is incremental technically that is 61 00:07:39,800 --> 00:07:46,050 why you have to take derivative of this to get this okay when I am actually I can use 62 00:07:46,050 --> 00:07:54,569 this same mechanism to estimate the exponential distribution for the time so Poisson distribution 63 00:07:54,569 --> 00:08:01,980 is related to exponential distribution in that sense the time between two consecutive 64 00:08:01,980 --> 00:08:09,780 arrivals actually is exponentially distributed so this can be estimated by if the time is 65 00:08:09,780 --> 00:08:18,300 t < T the inter arrival time is smaller than this is this will happen this probability 66 00:08:18,300 --> 00:08:37,580 is nothing but that probability that you have more than one or more one arrivals in time 67 00:08:37,580 --> 00:08:46,570 T. Okay and of course this actually means 1 - there 68 00:08:46,570 --> 00:08:53,079 is no arrival in time T okay so this I can use this same expression Poisson thing 1- 69 00:08:53,079 --> 00:09:03,949 no arrival when n is 0 this will be e -?t and this is nothing but a distribution function 70 00:09:03,949 --> 00:09:11,249 see here so I need to just take the derivative to get the density is the probability that 71 00:09:11,249 --> 00:09:21,519 the value of x is lying going to lie between x to nx + dx divided by this differential 72 00:09:21,519 --> 00:09:28,329 so this is the probability and so this px x will be given I will be giving you the density 73 00:09:28,329 --> 00:09:30,839 function. So let us take the derivative of this will 74 00:09:30,839 --> 00:09:47,369 give me now remember this the random variable here is T so this time so I can actually replace 75 00:09:47,369 --> 00:09:54,749 it by a t also and if I take this if this is a PDF the area under this should be equal 76 00:09:54,749 --> 00:10:04,290 to 1 which is actually true if I take ? e - ?t dt now this t cannot be negative this 77 00:10:04,290 --> 00:10:10,449 can only be positive so it can take value from 0 to 8 this will be equal to 1 so this 78 00:10:10,449 --> 00:10:19,529 actually can be proven now this time exponential this is the exponential distribution. 79 00:10:19,529 --> 00:10:21,709 Unable to capture image because lecturer writes and erases the content immediately 80 00:10:21,709 --> 00:10:26,869 So I can represent that how many calls will be served or how many calls will be finished 81 00:10:26,869 --> 00:10:33,959 essentially is a service rate the Á number of call swill be serviced per unit time by 82 00:10:33,959 --> 00:10:37,519 the line that essentially service rate will be decided by the call duration average call 83 00:10:37,519 --> 00:10:46,480 duration so that is also essentially being assumed to be exponentially distributed okay 84 00:10:46,480 --> 00:10:52,230 so with this actually now I can analysis this switch and try to estimate what is the blocking 85 00:10:52,230 --> 00:10:58,509 probability now before that when the blocking is going to happen. 86 00:10:58,509 --> 00:11:08,180 If my number of outgoing links n is greater than M will there be a blocking whenever a 87 00:11:08,180 --> 00:11:14,230 call will arrive call can always be connected to an outgoing link because it is available 88 00:11:14,230 --> 00:11:18,790 and when you are talking anyway the call cannot come at your line only on the free lines the 89 00:11:18,790 --> 00:11:24,519 call can come but you will always find an outgoing link so there is no blocking here 90 00:11:24,519 --> 00:11:32,749 so blocking will always happen if n is going to be < than M so when m is > than or equal 91 00:11:32,749 --> 00:11:37,429 to M then blocking will not happen blocking will happen only in this case. 92 00:11:37,429 --> 00:11:43,699 So what is the blocking probability so I have to find out that is all what is the probability 93 00:11:43,699 --> 00:11:50,009 that all the lines here are going to be occupied we need to estimate that how to do this so 94 00:11:50,009 --> 00:11:58,220 we can actually use here Marko chains the concept of that so the idea is that I will 95 00:11:58,220 --> 00:12:03,319 identify the state of this switch if only one line is busy switches in one particular 96 00:12:03,319 --> 00:12:08,329 state this line can be anybody it does not matter so for one line is busy switch in state1 97 00:12:08,329 --> 00:12:16,290 when no line is occupied switch is in state 0 and so on I can actually then been switch 98 00:12:16,290 --> 00:12:21,759 can be in state n when it is in a state n m is larger than n. 99 00:12:21,759 --> 00:12:26,399 So there are some free lines call can still come in but they cannot go through so switch 100 00:12:26,399 --> 00:12:34,179 is in blocked States so probability that this switches in state n will give you the probability 101 00:12:34,179 --> 00:12:39,910 of the switch being in blocked state okay this estimate or this number will be actually 102 00:12:39,910 --> 00:12:45,499 useful to us at some later point of time then we will actually start doing estimation for 103 00:12:45,499 --> 00:12:49,899 a three-stage interconnection so when you are switch so what I will do. 104 00:12:49,899 --> 00:12:50,899 Unable to capture image because lecturer writes and erases the content immediately 105 00:12:50,899 --> 00:12:57,359 Is I will represent these small round circles will represent States so I am I can actually 106 00:12:57,359 --> 00:13:04,149 be instate 0 I can be in state 1 I can been straight to and so on I can be in some state 107 00:13:04,149 --> 00:13:12,029 i and I can be in some state n in fact truly speaking if this one is line is occupied. 108 00:13:12,029 --> 00:13:13,029 Unable to capture image because lecturer writes and erases the content immediately 109 00:13:13,029 --> 00:13:19,359 All over all free this is one particular state next line is occupied all others are free 110 00:13:19,359 --> 00:13:23,800 is another state but they are equivalent states because there is only one outgoing line which 111 00:13:23,800 --> 00:13:30,739 is occupied so I am actually not I have actually have reduced by state space by merging all 112 00:13:30,739 --> 00:13:35,329 the equivalent states so there are only ten non-equivalent states which are possible in 113 00:13:35,329 --> 00:13:44,069 the system so when you are in state 0 there is no call I need to find out at what rate 114 00:13:44,069 --> 00:13:54,189 I will be transiting to state 1 so with ? number of calls per unit time the calls are arriving. 115 00:13:54,189 --> 00:14:01,589 What is the probability that a call will arrived in any one of these lines so there is only 116 00:14:01,589 --> 00:14:10,290 one line which I will consider probability that one call will be coming in sometime elemental 117 00:14:10,290 --> 00:14:15,799 time ??T I am going to take it is very small time and as Weill solve we will figure out 118 00:14:15,799 --> 00:14:20,669 this ??T is immaterial because it cancels out when we will try to estimate the state 119 00:14:20,669 --> 00:14:27,410 probabilities so I can use the poison statistics so this will give me ? into ??T1 is found 120 00:14:27,410 --> 00:14:38,050 one in fact I can estimate the probability for to caller i will also in time ??T but 121 00:14:38,050 --> 00:14:42,050 ??T I will take so small that probability of having two calls. 122 00:14:42,050 --> 00:14:51,029 Will be still smaller actually so and then of course I will have 1 factorial n in limit 123 00:14:51,029 --> 00:15:00,019 when this ??T goes to 0 this will turn out to be nothing but ? into ??T so that is a 124 00:15:00,019 --> 00:15:06,350 probability or that is a instantaneous probability in small time ??T that one call will be arriving 125 00:15:06,350 --> 00:15:14,540 on one line, there I am such lines so the probability that one call will arrive will 126 00:15:14,540 --> 00:15:22,339 be given by m x ? ??T that is the rate at which the calls will be arriving, so that 127 00:15:22,339 --> 00:15:28,259 is a transition rate for going from state zero to state 1, okay. Now when you are in 128 00:15:28,259 --> 00:15:33,619 state 1 there is only one call which is going through. 129 00:15:33,619 --> 00:15:38,610 Remember it is like I told you that exponential distribution and Poisson distribution are 130 00:15:38,610 --> 00:15:44,629 related to each other, so it is like there was only one call and this call is going to 131 00:15:44,629 --> 00:15:53,179 be serviced and there is only one server and the service rate is Á so in time ??T Á x 132 00:15:53,179 --> 00:16:03,019 ??T en e- Á ?? T by those many calls will be served so probability that n calls will 133 00:16:03,019 --> 00:16:11,429 be served is Á ?? T en e- Á ?? T/n! since there is only one server only one call can 134 00:16:11,429 --> 00:16:23,419 be served so in time ?? T when limit when ?? T goes to 0. 135 00:16:23,419 --> 00:16:30,009 n service to be given so when I am looking for n = 1 if this becomes Á x ?? T that is 136 00:16:30,009 --> 00:16:36,470 a service rate. If the two lines which are occupied in time ?? T it will be 2 Á ?? T 137 00:16:36,470 --> 00:16:47,059 that will be the service rate. So I can actually come from one back using Á x ?? T when you 138 00:16:47,059 --> 00:16:55,939 are in state 1well line is occupied from this side. The free incoming lines are M - 1 so 139 00:16:55,939 --> 00:17:07,350 the rate h way at which the calls can arrive is. M - 1 x ? ?? T and where you are in state 140 00:17:07,350 --> 00:17:11,980 2 there are two calls which are there. With a small time ?? T the probability of 141 00:17:11,980 --> 00:17:16,670 chances this basically is the probability transition probability, okay. In small time 142 00:17:16,670 --> 00:17:21,480 ?? T remember is a probability expression which have been approximated as ? ?? T there 143 00:17:21,480 --> 00:17:32,840 is a probability expression, so chances are that it will be 2 x Á x ?? T and now important 144 00:17:32,840 --> 00:17:39,070 thing is that if I want to if this is a system which is in a steady state condition. 145 00:17:39,070 --> 00:17:45,230 So the probability of being in certain state will not change it will remain it is a constant 146 00:17:45,230 --> 00:17:49,909 actually and something can always remain constant if I take any closed surface in this is known 147 00:17:49,909 --> 00:17:57,960 as Markov chain this is called Markov chain and this is being 148 00:17:57,960 --> 00:18:05,909 used extensively for queuing theory for analyzing the Q Q's performance, so this is also technically 149 00:18:05,909 --> 00:18:11,759 a queue, so if I take any closer surface like this. 150 00:18:11,759 --> 00:18:15,399 So the rate at which you are going to move out of the surface and rate at which you are 151 00:18:15,399 --> 00:18:22,830 coming into the surface has to be equal if the switch is in steady state, which implies 152 00:18:22,830 --> 00:18:29,480 that probability that you are going to be instate 0 is p0 and that is a probability 153 00:18:29,480 --> 00:18:32,480 of transitioning going from 0 to 1 so that should be a transition rate now, should be 154 00:18:32,480 --> 00:18:45,480 this and this should become p1 Á??T. Now interestingly this ??T gets cancelled because 155 00:18:45,480 --> 00:18:55,360 it is common on both sides you actually make any closed surface here and build up a balanced 156 00:18:55,360 --> 00:19:00,520 equation this is known as balance equation. So the rate at which you will go out of the 157 00:19:00,520 --> 00:19:06,700 surface and read rate at which you will come into the surface has to be equal, okay. Because 158 00:19:06,700 --> 00:19:14,139 this system is in steady state under that condition ?? T will always cancel out, okay. 159 00:19:14,139 --> 00:19:19,601 So ultimately we normally in when we build Markov chains we do not write this ?? T because 160 00:19:19,601 --> 00:19:27,970 of this condition, when I put ?? T it is a probability, probability of the transition 161 00:19:27,970 --> 00:19:31,450 taking place. But when I am NOT putting ?? T this value 162 00:19:31,450 --> 00:19:37,270 can actually be more than one, remember probability cannot take a value more than one that is 163 00:19:37,270 --> 00:19:42,340 why ?? T was multiplied ideal it should be probability is a transition probability going 164 00:19:42,340 --> 00:19:47,130 from one state to another state what is a chance it will happen, so that is what we 165 00:19:47,130 --> 00:19:55,600 do in finite state machines, okay. So ??T now actually can be removed. So ultimately 166 00:19:55,600 --> 00:20:01,789 what how this whole chain will look like, so chain will now look like if you go from 167 00:20:01,789 --> 00:20:44,889 here M ľ 2? 3 Á. So I have written these are basically now 168 00:20:44,889 --> 00:20:51,669 become transition rates not the transition probabilities actually okay so the probability 169 00:20:51,669 --> 00:20:58,270 are proportional to this so proportionality constant d T is be removed out so balance 170 00:20:58,270 --> 00:21:03,210 equations it does not matter actually. So now I can actually use a balanced equation 171 00:21:03,210 --> 00:21:09,200 to solve it my idea is that I want to figure out what is the probability of being in state 172 00:21:09,200 --> 00:21:18,009 I, I want to estimate that okay so once I know the probability of being in state I, 173 00:21:18,009 --> 00:21:23,299 I can find out the probability of being in state n which is the blocking probability 174 00:21:23,299 --> 00:21:30,309 of the system so let us write down the balanced equation I can write down in many ways okay 175 00:21:30,309 --> 00:21:39,350 so one of the possible ways is this. So I can actually take a closed surface around 176 00:21:39,350 --> 00:21:45,210 every state and then write down the balanced equation so balanced equation around 0 will 177 00:21:45,210 --> 00:21:56,450 be p0 the outgoing rate is m? this should be equal to incoming rate which is p1 Á okay 178 00:21:56,450 --> 00:22:13,230 so which gives me p1 as I can solve it m ? by Á p0 I can look at the second surface and 179 00:22:13,230 --> 00:22:22,960 write down my equation this will turn out to be p0 m ? that is incoming rate one incoming 180 00:22:22,960 --> 00:22:45,580 rate is from this side to ÁP2 the outgoing rate is Á p 1 + p1 M - 1 ? okay. 181 00:22:45,580 --> 00:22:51,330 So I just need to have a relation between p2 and p 0that is what I want I want to represent 182 00:22:51,330 --> 00:22:57,610 all the state probabilities in terms of p0 okay once I have that I can use some fundamental 183 00:22:57,610 --> 00:23:03,840 axiom of probability to figure out what is p0 and henceforth I can actually estimate 184 00:23:03,840 --> 00:23:12,960 all state probabilities so I need to put out p 1so let me put the value of P1 there so 185 00:23:12,960 --> 00:23:52,100 that I can get p 2so this p0 also I need to move on that side so this will turn out to 186 00:23:52,100 --> 00:24:16,190 be m ? p0 this term comes on the site will become p0 m ?. 187 00:24:16,190 --> 00:24:28,770 So this cancels with this and this Á will become Á2 this will be 2 so ultimately you 188 00:24:28,770 --> 00:24:41,389 will have 189 00:24:41,389 --> 00:24:47,070 alternatively I could actually have taken a surface like this so incoming rate has to 190 00:24:47,070 --> 00:24:52,690 be equal to outgoing rate and you would have got this equation directly okay so this would 191 00:24:52,690 --> 00:25:00,510 have become a relation between p2and p1 and p1 I could have replaced from the previous 192 00:25:00,510 --> 00:25:08,950 one to get this particular equation. So doing it this way for this surface when 193 00:25:08,950 --> 00:25:20,800 I am taking to enclosing this P 1 M - 1 ? should be equal to p2 2Á you so which becomes p2 194 00:25:20,800 --> 00:25:39,629 is equal to M - 1 / 2 ? by Á 1 which is I can actually keep on doing it so ultimately 195 00:25:39,629 --> 00:25:56,509 I can find out what is the probability of being in state I so this will be 196 00:25:56,509 --> 00:26:16,020 probability of being in state I will be given by MM - 1 M - I + 1 okay. 197 00:26:16,020 --> 00:26:31,090 Remember their inner brackets okay and of course I* I ľ 1* 1 and a hair it will be 198 00:26:31,090 --> 00:27:11,149 ? by Á this power I p 0 now whatever what is this expression they think right so this 199 00:27:11,149 --> 00:27:22,740 particular expression is nothing but a combinatorial it is MC IM ! m- I! I effect Orion the lower 200 00:27:22,740 --> 00:27:32,379 part is I !l and this part is what gives you the numerator here so I can write P I s M 201 00:27:32,379 --> 00:27:45,080 commute oriole I ? / Á I into p 0 so for every state the probability now can be represented 202 00:27:45,080 --> 00:27:50,530 in terms of p0 .So now let us solve for what the p0 will look like. 203 00:27:50,530 --> 00:27:52,780 Unable to capture the image because lecture writes and erases the content immediately 204 00:27:52,780 --> 00:27:59,970 So to find out p 0 remember all the states which always has to be in one of the states 205 00:27:59,970 --> 00:28:05,429 so sum of all state probabilities has to be equal to 1 so technically sum of all mutually 206 00:28:05,429 --> 00:28:11,649 exclusive events all possible of them and the probability of those events been happening 207 00:28:11,649 --> 00:28:15,610 when you sum up all the things together that should be equal to 1 that is a fundamental 208 00:28:15,610 --> 00:28:28,190 axiom of probability so I will use that so p0 + p 1 + p 2 and so on + P I and PN should 209 00:28:28,190 --> 00:28:53,530 be = 1 so which I can right now p 0 as 1 + MC 1 ? by Á 1this should be equal to 1 which 210 00:28:53,530 --> 00:29:00,420 gives me nothing but p 0 is this is not a closed form solution if n would have been 211 00:29:00,420 --> 00:29:05,750 equal to M then I could have got a closed form solution here I cannot okay. 212 00:29:05,750 --> 00:29:28,139 So p0 into ? i going from 0 to n MC n ? / Á i is equal to 1 so p 0 is 1 divided by ? i 213 00:29:28,139 --> 00:29:42,139 = 0 to n okay so you can now find out what is going to be probability of being in state 214 00:29:42,139 --> 00:30:08,029 I mci in fact I should write a different index here so I can put j because I have used I 215 00:30:08,029 --> 00:30:13,990 here so I have to use this so this is the probability now what is the probability of 216 00:30:13,990 --> 00:30:20,799 being in blocked state so what is the blocked state when the switch is instate n and when 217 00:30:20,799 --> 00:30:33,580 I do that probability of switch being in blocked state we call it also as a PB okay. 218 00:30:33,580 --> 00:30:38,769 Probability of being in blocked state that is the way I will be representing this so 219 00:30:38,769 --> 00:30:58,210 this will be MC n ? / Á that is power n ? i goes from 0 to n but that is a blocking probability 220 00:30:58,210 --> 00:31:07,779 for am by n composites which we call it m by n composites which 221 00:31:07,779 --> 00:31:57,399 and this also is known as AND set distribution okay.