 So, last time what we have done was m by n composite switch and a blocking probability estimation for that and we also understood what is time congestion and what is the call congestion. So, we will just move it further, but before that let me introduce something called time switching because this will be an important thing and most of time in most of the commercial systems this is what actually is being used we will always try to use that. So, basically the idea is very simple most of the voice which is transported is basically usually you will be doing a sampling at 8 close samples per second that is and once you do the sampling you will for each sample you will use it you will do analog to distal conversion of that and what you will do you will get 8 bits for each voice sample and that actually means you will get 64 kBit per second stream and usually 64 kBps is a very small bandwidth and if you have a fiber if you have a higher bandwidth link you will not transmit one single voice channel. So, already there has been standards which make a higher bit rate stream out of multiple voice circuits will this is a very common practice. So, I think one of the examples I had done earlier in the previous semester was even carrier system even in T1 both. So, even carrier system is actually is going to have 32 slots and as I have already mentioned in the previous lecture that every framing whatever is being used anywhere is going to be of 125 microsecond duration ok everything is of 125 microsecond duration whether it is SDH frame whether it is a E1 frame T1 frame E3 frame whatever you take only number of voice slots which are transported is going to change ok. So, this actually means I am anyway getting a time multiplex signal which is available. Now, how to do switching with that is the question. So, usually there will be slots now another important thing that these frame boundaries are extremely important because slots location is what it will identify the circuit. So, when you are setting up a path from the source to destination it is a slot number which actually matters ok. So, if you say this is going to be from your phone to some other's phone this is the slot number being given your voice sample will only be going in this particular slot in this particular frame ok. So, now this is something which we are going to use to build up a time switch. Now, one thing is this frame boundaries need to be identified. So, frame delineation mechanism has to be there. So, usually there will be some mechanism of that kind. So, even system usually uses two slots one slot is for identification of the frame boundaries and second slot which is used is for signaling purpose ok. So, usually this will be and similarly there is another format called T1 carrier system. This one is still this not used in India what we use in India is this, but with voice over IP come in even this will not be required in time to come actually. So, T1 carrier system will be using 24 slots in 125 micros this is again 125 microsecond duration frame duration. We do not bother about this you have to push in 32 8 bit 32 octets in 125 microsecond duration. Because every person who is talking over the phone is going to generate one voice sample every 125 microsecond. We have to give him chance again. So, frame has to be over within 125 microsecond you cannot stop. So, usually in this case I think 16th 0 it goes to I think it is starts from 0 the 15th one and the 31st is what is being actually used for this one is used for framing purpose this one is used for signaling purpose that is the way actually it is ok. In this case there is no octet reserved for signaling. Signaling is done in a different fashion here one extra bit is actually going to be used for framing. So, here this one is used for framing the whole octet itself here there is only one bit which is being added ok. Now, this extra paraphernalia and similarly how the signaling will be done is every 6th frame. So, there are many frames you will usually create a super frame of 12 frames and 6th and 12th frame the voice sample will not be of 8 bits it will be of 7 bits ok. So, lower most order bit for every voice channel will be used for signaling. So, this provides a low bandwidth low capacity signaling channel between two end points ok. Now, the problem is the signaling this framing cannot be is not required in switching. Switching is a box is a device. So, first thing which you do is you remove all this extra stuff out of the frame. You generate your internal time frame and when the internal time frame is generated inside the box this will contain purely the markers where the frame boundaries are starting and ending is handled by the device inside handled by the box or the switch. It is either on a separate channel or a marker or a clocking whatever way ok, but it is not part of this stream this is a serial stream this is going over a single wire or single fiber. So, you cannot keep these frame markers cannot be sent separately within a device this can be done within a box this can be done. Similarly, the signaling channel is being already separated out signaling is usually separated out and whatever is a processor of the switch that will take care of that particular information. So, signaling is never required to be switched signaling is between two boxes which are connected. For example, these are two boxes connected signaling is between these two intelligent entities. So, that is either carried by that one bit in every octet every sixth frame in T 1 carrier system or either by this 15th slot I am starting always the slot number from 0. So, 0 to 31 it is a 32 slot system 0 to 23 one extra bit will be added. So, all this is going to be stripped off before you can do switching because this is usually is a confusing point to the students you should be careful in this you have to strip all these off before you can actually build up a time switch. So, you will get actually a raw frame raw frame will be like this there will be nothing, but voice samples voice circuits. So, these are for example, I am showing eight of them. So, these are channel numbers there is no separate frame markers these are internally being maintained by the device or the switch box. So, again the next frame will come one will repeat here and so on the eighth one. So, this is what will be the raw thing which will be coming into the switch box and how the switching will happen because the location of a slot inside the frame itself identifies the source as well as destination. So, this corresponds to some source one which will depend on the interconnection of the previous switch this also identifies to which output port it is going to go. If my this box can somehow swap these positions it can read this octet and put this octet here read this octet and put it here swapping can be done or moving the thing I will be able to achieve switching and that is what the time switch actually does. So, I can give an example I am just taking a four slot a simple system and I have to now this is the input and then there is output thing I am showing two frames. Now, the problem is if these are happening at the same instant I want to do switching between 1 and 3. So, the input which is coming from port number 3 has to go to port 1 I certainly cannot move it here that is not possible because this has already passed through there at the same time is what can happen is three can be moved to any slots which is later not earlier. Now, this is actually one subtle thing another two ways you can handle the switching in this case again and this is I think implementation dependent does not matter actually honestly speaking, but one important thing that all these actual values may not remain together in the outgoing frame that is very important. So, one very good example is say I want to do a switching I let me just put a switching pattern say 2 to 4 say 3 to 2 and 4 to 1 I have just taken something arbitrary. So, these are the again I am taking all unidirectional and not bidirectional switching. So, that is a way input to output macking inside a switch I am only worried about the switch part and switch is a generic device I can use this thing even for packet switching if I can change the circuits or these maps every packet duration I can even create a packet switch also. So, that is why I am doing it in a generic fashion, but voice you are right if one has to connect to 3 3 always has to connect to 1 usually it should be, but it need not be. If for example, I can give you an example if it is a voice circuit it is a radio transmission over a telephone there used to be that same this kind of service earlier days. Radio over a radio over a telephone used to be there. So, in that case there is no reverse channel is not required and it usually used to be a multicast. So, one is being copied to multiple slots kind of thing if it is a time switch or if it is a space switch I will do duplication. Remember in that space switches I have not discussed about that actually. So, if it is a cross bar this certainly can be done. If it is not a cross bar then this cannot be done if 1 to 1 map is there, but remember I have when we were discussing switching I have always told it is a strictly non-blocking switch and it is nothing but equivalent to a cross bar. So, if this input has to go to all 4 outputs this can always be done this can always be done that is my assumption. And if it is a true cross bar this can always be done this depends on how you build up the implement the switch mostly time switching time and space. Routers work in slightly different way. They are sorted structures they actually use they have a time and time is sorted into small things and switch will have cross or bar state or may be this interconnection per map I call it input to output map. So, for one slot duration there is one I O map depending on whatever packets are coming they will move to the output depending on that I O map. And next slot again it will change and this has to be dynamically computed depending on the packets which are going to be there in the head of the queue. And if your speed here is very fast. So, number of packets which are coming per slot is 1 outgoing packet number of packets going per port per slot is again 1, but incoming I can actually do the transfers at much faster rate which actually means I can actually read 2 packets for same output port and transmit them simultaneously that is also possible. So, that will lead to what we call the blocking will not be there in that case and that is known as speed of factor. So, if it is running at end times higher speed even if all packets are for same output in port in one single slot I can read all of them and then put it at the outgoing thing, but outgoing port cannot transmit it remember. So, it has to use something called buffer. So, it will put on the queue, but only one packet per slot is still will be going out when this buffer is full packet will be dropped you cannot help it actually. So, technically it is similar thing switching is the core internally. Routing is identifying what we call the paths. So, who is the next destination to whom I should forward the packet. Switching is actually creating the path forwarding is by looking at routing table finding out which outgoing port it has to go. So, these are three different processes which run in every router or every switch. Well if you call them switch or router, but it contains all the three things actually embedded. So, in this case once you do this map I can just I will actually do the exact implementation of this particular time switch, but as of now you assume we are going to implement it through delay lines, some kind of delay. So, this can be done by delaying 1 by 1 can be moved easily to 3 remember 2 can be moved to 4, but whenever this destination port number is smaller this cannot be done. So, 3 cannot go to 2 here. So, 3 has to go to this thing and of course 4 has to go to 1. So, 4 will go to 1 and similarly this 1 will come here. A third slot and 2 will come here at the fourth slot. Now this frame contains 2 octets from the current frame itself incoming frame and 2 octet from the earlier frame. So, this is one technique you do not worry, but you try to minimize on the delay of transmission ok. Second possibility could have been you will get exact one slot delay, one slot plus higher something. It will not be minimizing the delays actually if you want to maintain all these 4 together all the time. Then this 1 and 2 will be shifted here and remember the delay line length requirement for this kind of system will be what is the maximum. In this case I have already done 3 is going to 2 that is the maximum exactly this much slots 4 slot delay will be required 4 minus 1 actually 3 slot delay will be required in worst case ok. But in case other scheme where I am not going to do it this way the 1 has to go to 4 sorry 1 has to go to 3 in this case 2 has to go to 4, the 3 has to go to 2, 4 has to go to 1. Worst case scenario would have been if 1 would have gone to the 4 actually. So, you require 2 into n minus 1 that much maximum delay I have given this map I am trying to implement this map. So, this pattern will keep on changing if you change your map actually ok. So, but this will require higher amount of delay usually this is not preferred 2 n minus 1 is 2 into 4 minus 1 7. So, you require 1 2 3 4 5 6 7 if 1 is going to go to 4 you require 7 yeah that is that is the advantage of that scheme. But you are not putting the things together voice if it is a circuit switch system does not matter and if it is a packet switch system you will always be happy. A packet switching system does not have this time switching concept you just do the scheduling of packets and packets are just inserted whole packet belongs to one source n is going to go to one destination you transmitted faster rates the packet duration goes down here it does not matter 1 slot 1 octet is going to come every 125 micro second ok. So, you do lot of aggregation this way ok. No usually it will never be done it will never be done this cannot go to you cannot have any octet from here will not go to here that is not possible that is usually will never be done. And I think there is no case where it will be required 1 will come to 4. Then any. The 2 the 2 wants to go where for example, you give me the position for example, I change 1 to 4 that is the map you give me any other map actually say. So, in this case 1 will come to 4 2 will come to 3 3 will come to 2 4 will come to 1 that is it does not matter it will never be going to the another one within this you should be able to finish it. So, there are 2 possibilities actually. So, this is what basically is the time switching concept is. And the way I think it has been told at lot of places is by using multiplexer and demultiplexer. So, if you have. So, this is the equivalent and I put a time switching between which will do the job of mapping and I have 1 2 3 and 4 what I have created. So, I am still using a time switch, but using a MUX and DMUX rotary MUX and DMUX they do it synchronously actually remember. It always go to 1 2 3 4 that also goes to 1 prime 2 prime 3 prime 4 prime. So, if I do this I am able to create a space switch remember there is nothing, but technically a space switch. So, I am not worried about this input is sampled only every 125 micro second if it would have been analog. So, there is a ADC and then there is 1 sample generated to every 125 micro second that always goes into the first slot of the frame. And that can be switched to any outgoing port and there is a DAC you will put it will become nothing, but a crossbar. You put a DAC here distal to analog converter here you put ADC this is equivalent of a crossbar, but you can do time switching because you are using a distal format. So, far we have whatever we have done till the earlier lecture we are not bothered it could have been very well analog telephone, but only with distal format you can do time switching. So, so far this is fine, but we have also done a class network. Now can I expand on to the class network with this. So, can I use time switch there my question is this. So, let us draw the class network and then. So, as I told the class network is get total R 1 switches and you have M 1 incoming ports you have N 1 outgoing ports and you have here M 2 N 2 outgoing ports and then there is a third stage that is what I had done earlier. We also had the condition that N 1 is equal to R 2 is equal to M 3 R 1 is equal to M 2 this is 1 and N 2 is equal to R 3 that is what we had done. So, that is why this guy can connect to each one of the switches if there is some middle stage 1 it will connect to here. So, every switch is connected to everybody every switch in the next stage by exactly one link and we of course, have done the case I also proved what is strictly non blocking switch in what what condition if it is symmetric. Symmetricity actually means that M 1 is equal to N 3 and R 1 is equal to R 3 under that condition you have a variable R 2 you change R 2 N 1 and N 3 also will change that was the case which we took and intuitively I gave a proof that R 2 has to be greater than or equal to 2 into M minus 1 for the switch to be strictly non blocking. So, formally there is actually is minimum of something it has to be greater than or equal to that we will come to that thing later on. Now, my question is this purely is a space switch how to put in a time switch here can I do something with this time switch because time switch has a peculiar problem this interchange of slots will be done because you are writing into the memory and reading from the memory and memory or RAM as a access time limitation. So, you have to write 2 N words and you have to read the 2 N words back sorry N words you will be writing and you will be reading and this has to be done in 125 microsecond. So, if your access time of a memory is T access. So, 125 microsecond and the and read access and N write access 2 N. So, this is the access time which is available for each word for read write. So, this has to be always greater than or equal to T access which actually implies your N has to be less than equal to 125 microsecond by 2 into T access. So, you require very fast memories if you require a higher N. So, this value if you keep as less as possible. So, that you can have higher and higher N. So, N is also a limited in any time switch you cannot have N very large N is limited so how to solve this particular problem. So, I cannot create a very large dimensional time switch I have to do something and we have to do something here. So, any guess any suggestion from you people go ahead. One frame is coming with 4 slots right 4 sampled times now where you would like to put the time switch in this stage. That is a one possible configuration, but that is tricky to do, but a simpler version is I can use time switch here first and third stage, but I will not be requiring R 2 switches there that is the important thing. So, let me implement using time space and time combination we call it first stage at time, second is a stage, third is time switch is known as TST time space time this also becomes a class network nothing else. The whole mathematics which is represented which is actually is needed for this switch is also valid for this configuration. So, let me do it I will draw it here. So, let me take a time switch and these are the incoming frames I will define this thing as M 1 slots. I am not worried about signaling and framing. So, that we will take care. So, for me it is only pure M 1 voices voice slots which are coming in and I can now, but the problem is I have told you that it whatever is the number of incoming slots has to be the outgoing slot. I have not told that actually I have only used that thing in the example, but that does not mean it is always going to be true. You technically can have N 1 slots going out. Now, that is a complicated thing to handle because whatever scheme I have told for swapping by providing appropriate delays those cannot be maintained in this case. The whole octet has to be written whole frame has to be written first in the memory then only it can be read out at a different rate because remember it is still the frame duration is 125 microseconds that does not change. So, in reality I will again do this particular implementation of M by N time switch later on. We actually use two RAMs. So, when the first frame is written in first frame only writing will be done. Reading will be done the from the others RAM actually and that will be done with a different clocking rate that you can read faster. Only problem is this if this M 1 is smaller than N 1 only M 1 slots will be occupied remaining will be free that is possible switches are implemented that way. And if this N 1 is smaller than M 1 some of the slots have to be dropped those connections cannot be made. It means those voice slots here will be busy because what happens when a frame is generated the switch which generates the frame or the mugs will maintain the flags that which all actually slots are currently being occupied or busy or which are free. So, whichever is a smaller value of these only those many slots or those many voice circuits can be set up and that will be maintained as only those many will be maintained as free status sorry busy status remaining will be kept as free status. Now, this is exactly equal to this only thing M 1 was in space here N 1 was in space there it is in time that is the only difference. So, I can actually now put in the same thing multiple of them maybe I can put them how many R 1. Now, the only problem which comes is these switches. So, how this switch is connected you take the first input here first input from this first input from every switch here first input from every switch here. So, maybe common sense will tell I can actually do a demux here create physically 1 to N 1 and then I can have N switch connecting 1 from here similarly there is a mugs in this fashion this is not required actually I can implement it this way and I require again you will again have a time switch perfectly equivalent configuration both only thing the switching I have implemented in is time here and I have converted from time to space using this demux separating out all voice slots. So, first outgoing port first outgoing port physically is here I have separated out first is here. So, all first are being put together in one switch this switch is equivalent to this. So, this is going to 1 to R 2 this is clear so far or is there any confusion I can repeat they are this one this one will also have R 2 inputs you are right 1 to R 2 inputs. So, remember N 1 is equal to R 2 that condition will always be satisfied N 1 is equal to R 2 is equal to m 3 that condition always has to be satisfied it is a cross network. So, what I have done is or you look at this particular thing this is equivalent to what? Remember I have drawn a figure earlier if I can draw m time multiplexer with m inputs I can generate m slots I did the time switching I am doing demux this is nothing, but this switch. So, I have replaced this thing by a time switch actually is a equivalence. Now, only one important thing which you will notice during the period when the time slot number 1 is there which one of the switches in action now are all the switches in action all the time you tell me that only first switch will be operating only first switch is operating all other are silent because their output will second output will come only during the second period I am not doing any time expansion in the demux if I simply just split them I have just used that pointer thing. So, 1 by 1 by 1 is just splitting out whatever is coming in the frame here. So, only first switch will be operating at this time when the second slot time will come second one will be operating then R 2 will be operating then again 1 will operate then 2 will operate and so on round robin fashion they are actually all operating 1 by 1 I can do something smart actually I can use this itself gives a clue that I can replace this whole thing by nothing, but one single space switch and this is space switches of which how much dimensions R 1 by because the R 1 inputs are there M 2 is equal to R 1 and N 2 is equal to R 3. So, this will be R 1 by R 3 switch technically. So, there is instead of having R 2 switches I have only one switch only problem is when the slot number 1 will come it has to have some interconnection pattern it will act like switch number 1 or middle stage when slot 2 will come my interconnection should change. So, I should be able to build up a control system by which this switch configuration will keep on changing with every slot. So, it will be able to operate as R 2 switches because they are happening in different time slots. So, it is technically possible and this is known as time multiplex switch because of this. This switch is a single switch, but it is acting as R 2 switches and it is in happening in time multiplex mode. So, it is a time multiplex space switch actually. So, it is not multiplexing of signals it is a multiplexing of a space switches which is happening in time. So, it changes every 125 microsecond and 2 times it will change sorry n 1 times it will change. So, n 1 is the number of slots which are here. So, it will take n 1 different configuration every 125 microsecond and this is nothing but equivalent configuration of Kloss network, but this is a time space time configuration. Now, we did the strictly non blocking thing last time. Now, can you give me the condition using the same clues what will be condition here for strictly non blocking switch? How you will find it out here? You should be able to do it actually use same principle that how many these ports will be busy in worst case there is one free port here one free port here they have to be connected. So, how many maximum ports here will be busy how many maximum ports will be busy here. So, how many slots can be in worst case will be busy here m minus 1 take a symmetric case again the way last time we did. So, here it is m n this is again n and here it is again m. So, in worst case here m minus 1 will get occupied because one is there one slot is there to which need to be connected to one slot here. So, only m minus 1 slots are occupied. So, only m minus 1 will be occupied here and this switch is operating as in every slot some switch. In worst case scenario if you look at this particular frame and this frame the occupied slots may not have any overlap they will be like this if you take these two slots this is another one. So, you have m minus 1 being occupied here worst case scenario from here onward again m minus 1 occupied there is no overlap of time is very similar condition and if I can have I have these extra stuff available extra slots available I can always use that to set up the connection. So, condition is 2 into m minus 1 plus 1 and that condition was a number of switches here. Here switches only one it is in time multiplex mode. So, the condition is now on the slots here is condition is on the slots and this condition will turn out to be same 2 into m minus 1 greater than or equal to is m by n switches in the first stage and n by m switches in the third stage and whatever number of switching elements you use here that is a cross bar size in the middle stage, but it should be able to switch very fast. So, we will actually do this exercise of actually building up a circuit or a circuit schematic by which we will figure out how this switching actually happens right you are right this should be this you are right. So, now actually I can also do because I have about 8 minutes. So, blocking probability estimate now this is going to be valid for this case as well as for the earlier class network both. So, this approach because Carnot's approach will take time. So, let me do what is called Lee's approach and we already know when the probability of blocking goes to 0 and you will be able to find out that under this condition Lee's approximation does not give you the blocking probability equal to 0. So, the case here is incoming port it is connected to middle stage switches 1 link 2 on switch. So, this is my free outgoing port free incoming port probability of blocking is when both are free I will not be able to set up the connection. And we assume that this switch has n incoming port I am just let me keep it m only because this for consistency we are using m by n and these are n outgoing ports. There are other switches of similar kind those does not matter for us actually I can only take this it will also give me the same blocking probability. So, when the switch will I will not be able to set up the connection I can now do some approximations one very simple is I can assume that this link is going to be busy I can estimate a probability of that let this probability be p is a probability with which the link will be occupied or busy. Now, remember these two links are independent I can even I can make that assumption actually that is not true, but I am making that this assumption, but you may say because there is only one path this whenever this is busy they should be busy know that is not true there are other switches also. So, if I set up a connection this way I can occupy this one, but this link is free that is possible. So, input links and output links that is the way we define are independent of each other and they get occupied with probability p both of them. So, this also gets occupied with probability of p it is a symmetric condition. So, this is m here this is n here and when you will not be able to set up the path if you can find out even one switch and any one of these for which incoming and out way input and output links both are available you will be able to set up the connection this is strictly known blocking this element and this element is also strictly known blocking. So, probability of blocking now this two pair is not free what is the probability of that 1 minus that both of them are free that is a probability that will happen either this is busy or this is busy or both are busy you have the sum of these three possibilities. So, I will do 1 minus p that the link will be free both the links will be free because they are independent. So, I can simply multiply the probabilities 1 minus of this is a probability that the path will be through that switch will be busy. So, 1 minus of that and how many such links are there n such bridges are there n such branches. So, this is a probability of blocking estimate, but how you will get the p. So, we take here simple example n usually is larger than m if it is smaller than the blocking will happen at this itself. So, blocking is not happening because of switch it is always happening because of the interconnection network. So, I will assume that a is the arrival rate or arrival probability a link incoming link will be busy that occupancy probabilities a. So, a into m should be equal to p into n the balance condition always statistically because whenever this link will be busy one link has to be busy here and since the n is larger. So, probability a link will be busy will be lower actually. So, n multiplied by the p should be equal to m multiplied by m. So, I can get p from here as a into m by n. So, because n is larger m is a smaller. So, a is multiplied by a fraction. So, p will be smaller actually. So, I can write that thing here. So, this was one of the crudest approximations which was invented earlier. Now, you put value of n is equal to 2 m minus 1 do you get a 0 m by n sorry you put n is equal to 2 m minus 1 you cannot get a 0, but we know we get we actually have no blocking at that point p b should be 0. So, this estimation is an approximation this does not give the actual estimation and reason is that there are actually the reasons are that your arrival rate here is dependent on the state. I can define the state as number of incoming links is busy and number of outgoing links is busy probability of blocking dynamically changes actually with that depending it is a state dependent. So, if I define this sigma as a state technically probability of blocking is function of a state sigma. So, I have to average it out over all states. So, here it was all state independent thing. So, tomorrow's class what I will do is because we are making the recording. So, quiz will be done separately this time it will not be in tomorrow's class. So, most likely tomorrow night you have to sit on your computer. So, we will do a trial run and after that I think quiz will be done live online itself. So, there will be a time window it will be open for I think 20 minutes within that you have to write down the answer on the computer ok. So, we will try that thing out because I cannot afford to waste time in taking the quiz otherwise recording will be of shorter duration all recording has to be of 50 minute duration. So, tomorrow we will have a regular class. So, quiz most likely will be happening I think tomorrow night I will be doing the I will be doing trial and then. So, that you also know how it is done you will get some mock result and mock marks also in that. So, day after tomorrow we will actually have the regular quiz depending on the state is a result. Marks you will get automatically at the same time it will be automatic grading by the machine objective. But then the problem is certainly made in such a way that you have to do something before you can give the answer it will be with negative marks ok. So, you can get minus also. So, your mark sheet will can contain negative marks. So, that is the only thing they can be more they have to be made more because you should not get time to do the cut phase from all them one more percent you all might sit down on the same laptop. So, from one single laptop two people cannot give the answer I have to ensure that ok. So, that way our system will also get tested. So, tomorrow we will do the trial around I think it will be 10 midnight 10 evening it will be fine. So, all of you can be in your halls and do it from there itself, but you have to arrange for a machine that is the only thing ok. So, and everybody has the connectivity and machine login the same recipe login will work. So, only thing it opens at that particular time it actually closes also at dot time. So, if even if you are one second late it will not accept the answer after that. So, that is why I will give multiple questions if you miss you miss only one question not one not actually there is no it will not be one question may be say 5 questions. Most likely it there should not be that is why there is a rear cell. Now, if it is true for that time most likely it will be working. We have tested the system and if there is a problem we have to understand and resolve that issue and we make it more and more steady. We should be able to figure out that there is actually is a problem you cannot simply say there is a problem. And if it is going to be a problem I will take I will just call you for 15 minutes probably on Saturday and have the quiz that time.