 So, continuing with our solutions of the balance equations for multiple servers. So, MMS system may be I can again write here this, because we are continuing with you know the same system with multiple servers. So, then you may solve equations for P 1 and P 2 and now in general if you want to solve it for. So, at the point when you have you know more than S people or it just at the boundary. So, this will be lambda P S minus 1 when you have S minus 1 people. So, incoming the way you can reach P S would be through one arrival and then here when you have S plus 1 people then you know your service rate is S mu. So, it will be you know again one person departs. So, then you will be back to P S and here it will be one arrival and one departure. So, then again you will remain at P S right. So, therefore, this is the balance equation at this stage and we can now substitute for because yeah. So, I had shown you that the formula for P S up to P S minus 1 would be this. So, I just substitute for this in terms of P naught. So, this will be lambda into lambda S minus 1 upon mu into 2 mu into S minus 1 mu right then plus S mu P S plus 1 and lambda plus S mu into lambda raise to S. So, up to P S the same solution will go the same formula. So, mu into 2 mu into S mu. So, let us just simplify you take this to this side and then you know without spending time because surely you can do this calculation yourself. So, that from here you will subtract this expression and you have S minus 1 mu here. So, you will get S mu in the denominator from you will multiply and so when you multiply by S mu. So, lambda S S mu term will cancel out right and you will be left with lambda S plus 1. So, this is the simplification you can see right away from here right. So, you will be left with only lambda S plus 1 upon. So, this will be mu raise to S and then 1 into 2 into 3 up to S. So, that is S factorial. So, that gives you S mu into P S plus 1 equal to this. And therefore, P S plus 1 will be lambda by mu raise to S lambda by mu raise to S and a lambda. So, the S factorial and then divided by S mu. So, 1 lambda upon this is my lambda you should be able to write it. This is lambda upon S. So, that means when it is S plus 1 you have this term into lambda upon S mu times 1. And so your rho is lambda utilization factor is S mu lambda by S mu now S servers. Just want to show you that the same formula will continue and then you can generalize for N P N. So, now if you write lambda P S then S mu P S plus 2 because after you have more than S people then the same service rate will continue. So, S mu into P S plus 2 is equal to lambda plus S mu into P S plus 1. And this again when you substitute for P S and P S plus 1 from here then you will get the expression that P S plus 2 is equal to lambda by mu raise to S into lambda upon S mu raise to 2. So, this is the factor which goes on increasing this remains the same S factorial lambda by mu raise to S is common and then it is lambda by S mu square. So, now you can generalize for you know you will write the general formula and that will be P N is lambda by mu raise to N N factorial P naught no that yeah for N between 0 and S. So, this formula is for N between 0 and S that is what we were using here and here for P S. So, then but what N greater than S you going to have lambda mu raise to S upon S factorial and then see we have yeah let me show you yeah. So, this is then this is lambda by S mu square. So, I am writing lambda by S mu no I do not have to write S here. So, this is yeah lambda by mu raise to N minus 1 upon S raise to N minus S. So, either S you can include with this then it will be lambda by mu raise to N minus S. So, if you are writing the formula for P N lambda by mu raise to S upon S factorial this is the common thing and then it will be lambda by mu S raise to N minus S which I choose to write as lambda by mu raise to N minus S and divided by S raise to N minus S. So, it is a same thing into P naught if N is greater than or equal to S right. And so, now since we have the general expression you can use the convention that 0 factorial is 1 and therefore, 1 can be replaced by lambda by mu because when you add up see now you want to use the fact that summation P i i varying from 0 to infinity is 1. So, when you write P 0. So, therefore, and all others are also in terms of P 0. So, the series will become 1 plus and so on divided up P 0. So, 1 we are replacing by 0 factorial. So, in that case you see this will be your this thing and here the later part that means from S onwards S plus 1 S plus 2 to infinity the terms that you will get you will have this power series and this will be convergent if lambda upon S mu is less than 1. So, as we have anyway seen that you know if this is greater than 1 then your things will blow up if your arrival rate is more than S mu then the Q size and everything will blow up. So, we have to anyway work this under the system that this is less than 1. So, for any feasible if you want to consider a balanced form and stationary system. So, therefore, so this then can be written as S to infinity lambda by S mu raise to N minus S under the condition that lambda upon S mu is less than 1. So, this is the expression and of course, we are not going to say there is no other closed form for this and when you are given the values lambda mu and S you can just compute this value. So, now start computing the expressions for L L Q W W Q and so on. So, let us just look at the expression for L Q which will be N minus S and you can see the reason because there is a neat expression for this thing for the probabilities when N is S or more than S. So, I am therefore, writing L Q and since I can get L from L Q therefore, it is enough that I compute this and once I get L Q I get L and then I get W I get W Q. So, that is a thing. So, let L Q so we will write as N minus S P N where N is varying from S to infinity and here if you want to write N minus S S J then N is S plus J. So, therefore, you can then say that J varies from 0 because N is varying from S. So, N plus S the J will vary from 0 to infinity and this will be N minus S J into P S plus J. So, we have a nice way of writing the probability for this. So, which will be see here the S part is lambda by mu raise to S upon S factorial and then this for the J this will be lambda by mu raise to J upon S raise to J and this J is there because N minus S you are computing L Q expected number of people in the Q and this is P 0 fine. So, yes so I can remove lambda by mu raise to S upon S factorial take it outside the and P naught outside the summation sign then I am left with this J into lambda by mu raise to J upon S raise to J and here this I can see lambda by S mu 1 S I can take outside then I will be left with lambda by S mu raise to J minus 1. One lambda by S mu I take outside which I am writing as rho then it will be lambda by S mu raise to J minus 1. So, J times this which is the derivative of lambda upon S mu raise to J. So, this whole thing by taking rho outside is equivalent to the derivative of this and since lambda upon S mu is less than 1 we know that this series is also convergent. This is the arithmetic geometric form with common ratio as lambda upon S mu which is less than 1. So, therefore, this is the convergent series. So, I can take the first course I should have written the derivative outside and then since this is the convergent series I can bring the derivative sign inside and this is what you have and therefore, now this is the geometric series which adds up to 1 upon 1 minus rho. So, this part summation and then derivative of this will be 1 upon 1 minus rho whole square. So, actually what I should have done is I should first I write derivative inside and then I take it outside because it is the convergent series. So, you can interchange summation sign and derivative sign. So, now I take the derivative outside then this sums up to 1 upon 1 minus rho and the derivative gives me 1 upon 1 minus rho whole square. So, therefore, now you have a neat expression except that p naught is a little cumbersome. So, then you have this expression for l q and then I can get my w q which will be l q by lambda and because we said that these relationships are valid even in the general case. So, 1 upon. So, therefore, this will be 1 upon S mu when you divide this by lambda this will be left to the 1 upon S mu and you have this expression. Hence, then after that you will say w w q plus and remember the difference between w and w q will still be that of one service because your departure from service is 1 at a time. So, therefore, this will be not 1 by S mu it will be 1 by mu and similarly here this lambda will l will be l q plus lambda by mu. So, keeping that in mind you have all these relationships. Now, let us just look at some of these yes and I wanted to the tables that I am going to show you have been taken from this book and I will give you the proper references all at the Ravindran and Philips books. So, I have taken some tables where they have plotted the l with respect to the rho sign rho where S is varying for different values of S. So, rho changes so they have plotted. So, let me show you the graph here. So, figure 2 I will just want to explain that here you see the horizontal axis is the utilization factor that means rho equal to lambda by S mu. So, different values of rho going from 0 to 1 and the vertical axis is the steady state probability of 0 customers in the system. So, look at the curves in the diagram in the graph then you see that the utilization factor is coming down drastically as S increases. So, that means for 0.8 for example, the top line shows you the utilization factor versus the steady state probability of nobody no customer in the system. So, these are the different lines and you see that the utilization factor when S is 25 is barely is even less than 0.3. So, that of course, we also expect because more the server the lesser the utilization factor and but there again as we said that it is always conflict between you know having congestion or having more servers and that depends on what your priorities are. So, anyway this the graphs actually show you what we expect that as this will go up the utilization factor will come down. But the other important contribution of this graph is that you know you can plot because remember the expression for the steady state probability when you had more than one server the expression was lengthy one. So, now you can actually plot for different. So, that means if you know the utilization factor say for example, I know the my utilization factor is 0.7 and then if I want to find out what the corresponding value of p naught will be. So, then I will go along this vertical line of 0.7 and if my number of servers are 4 then you see wherever this curve S equal to 4 cuts the vertical line 0.7. So, that point and I go horizontally then across to the vertical line. So, I can then find out the corresponding value of the p naught and so that will help me because then I can make my computations for L L Q and so on or even for p n's I will need the value of p naught. So, then it will help me to just plot the value given by utilization factor and the number of servers then I can find out the corresponding value of p naught. So, this is the contribution of the figure 2 and later on when I worked out any when I worked out any I will use the values. Now, I will show you another graph. This is utilization factor your verses. So, you are plotting the utilization factor on the horizontal axis and steady state expected number of customers. So, number of customers in the system. So, utilization factor and therefore, it says the n for different servers as you expect. So, for example, if your utilization factor is 0.3 for S equal to 25 you can see that well let us see this is at utilization factor 0.4 that means your lambda upon S mu is 0.4 and then if you go up then you see that for S equal to 25 the number of people will be 10. So, now total number of people, but you have say this is the steady state expected number of customers which is L. So, the number L would be around 10 when you have 25 servers in the system. Now, if you come down that means if you come down to or if you go higher then of course, these as the utilization factor increases when we know that the number of people in the system will go to infinity because it is never advisable to have your S mu equal to lambda. So, therefore, this should never come very close to 1 the your row. So, that of course, is depicted by as we saw it for one server the same phenomena is shown is repeated here also, but this again gives you an idea as to the utilization factor versus the number of servers you have and the number of customers you will have in the system. So, for different values you can just plot and see for what is the number of. So, for example, for 0.5 if you go again the number of people in the system will come will be again around 11 or something you can say that for 25, but then if you have only S equal to 1 and 0.5 then you expect only 1 person to be in the system. So, this is the kind of. So, therefore, these I mean the conclusions are the ones that you expect, but they sort of give you more accurate you can accurately find out for the number of the utilization factor number of servers what will be the expected number because computing P naught. So, you can from here only just find out the different values of row you can find out and for number of servers you can find out the expected number of people in the system. So, once you can do that then if you have computed L then you can find out L Q you can find out W Q and you can find out your W. So, this is just to give you a feeling about the multiple servers and how these quantities L L Q W W Q behave. So, you see now interesting example from Ravindran and Philips and Solberg actually they are three authors. So, Ravindran Philips and Solberg and so now they are trying to show you that pooled verses separate servers and what would be the conclusion. So, let us look at this example. So, the two businessmen working in adjacent offices each requires secretarial service and they each produce an average of four letters a day. This is a simplification, but anyway whatever the work what we are saying is the average of four letters a day to be typed and a secretary can be expected to require an average of one five of a day to type one letter. That means the rate of typing letters by secretaries five letters a day and the businessmen each is producing four letters a day to be typed. So, the question is should they get together to form a two person secretarial pool the pool means that whenever anybody has a letter ready when anyone of the two secretaries who are free they will type the letter. So, it is not that you know one secretary to the one businessman and so she will only work for a particular business for her boss only and only when the letter is ready by the boss she will type it. So, by pooling it will be possible for each of the businessmen to access both the secretaries that is the idea or should the each man have his own secretary. So, this is the question and let us try to answer through this model of MMS which we have just not talked about and you see. So, let us see. So, if you take a single system then it will be just input is you know four letters a day and the secretary is typing five letters a day. So, each system can be considered as a MM1 system. So, and therefore your row will be 0.8 the utilization factor 4 by 5 is 0.8 and the mean delay which is WQ will be lambda upon mu mu minus delay means that before the letter has to wait for some time before it get started to be typed by a secretary. So, then WQ is lambda upon mu into mu minus lambda which is 0.8. So, that means on the average a letter will sit for you know fourth fifth of the day on the secretary's desk before it is being typed because before it will start being typed typing begins on that letter. So, the mean delay is four fifths of the day. Now, and the same applies to the second business man also because they are identical systems MM1 systems with the same data. So, therefore the second business man also will have the same thing happening to his letters that for four fifths of the day the letter will be waiting on the average and then the typing begins. Now, suppose you pull the system then your input will be you know eight letters a day and you have two servers now each of them producing typing five letters a day. So, then this is a MM2 system and your row will be again lambda upon 2 mu which is 8 by 10. So, therefore this is 0.8 and your lambda by mu is 1.6. So, then p 0 now this is where your the graph that I had talked about come in handy of course this is a small system and therefore I have shown you the calculation even otherwise yeah this is this right. I have shown you the calculation, but see you can see from that means for lambda by s 2 mu s mu is 0.8. So, corresponding to 0.8 and two servers if you look at the graph here figure 2. So, 0.8 and you want to go up to 2. So, you see this is little above 0.1. So, 0.11 yeah you can see in the graph for lambda by s mu equal to 0.8 and two servers. So, this is just above 0.1. So, 0.11 and which also by our calculation comes out to be 1 by 9. So, this is 0.11. So, for larger systems you can look if that graph is plotted you can just check the value you know look up the value for p naught without having to do the lengthy calculation. And therefore, your l q would be again by the formula and then. So, w q that we. So, by Litter's formula we have w q as this number which is 0.35. So, therefore by pooling the mean delay has come down to 0.35. Earlier the mean delay was we had computed it was fourth fifth of the day which was and. So, here now it is 0.35 which is much less than 4 by 5 because if you multiply this is the mean delay here what is it four fifth of the day. So, if you want to compute it in this is 0.8 the mean delay is 0.8. So, which is much less than point which is much very high compared to 0.35. So, by pooling definitely your mean delay will come down. So, it might not be you know if you put the ego side apart you know like having your own secretary. And of course, probably this goes against intuition also because you may feel that if you have one person to exclusively to do your job then you should get it done faster. But certainly the data here shows that this is not the case by pooling it will always help you to get your work done faster. Now, mean delay we had computed at 0.8. So, therefore this is very high compared to this number. Now, let us see we can again play around with few numbers suppose you say that this data was particularly tailored. So, that the difference is so much 0.8 and 0.35. Now, suppose one of the businessmen. So, now suppose one of the businessmen has only two letters a day then the mean delay for is the letter getting type will be 2 upon. So, this is the number of letter mu mu upon lambda lambda minus mu sorry I have said the wrong way 2 is the number of letters that arrive per day. So, lambda is 2 mu is 5 because 5 letters get typed the secretary can type 5 letters a day mu is 5. So, this is lambda is 2. So, lambda upon mu into mu minus lambda. So, this will be 2 by 15 which is 0.133. So, with this own secretary the mean delay would be of the order of 0.133. Now, if we pool the two secretaries then your lambda becomes 6 because the first businessmen is one of the businessmen is sending getting 4 letters to be typed and the other one has only 2. So, lambda equal to 6 and 2 mu will be 10 because each of them can type 5 letters a day. So, therefore, p 0 is 1 by 3. Now, this we get from the figures that I have given you. So, therefore, for lambda and mu for these values of lambda and mu you find out p naught which comes out to be 1 by 3 from the figure and therefore, your w q would be 0.11 that means the mean delay would be 0.11 which is still less than 0.133. So, you see pooling definitely is a better option with two letters and the secretary and the man using his own secretary even then the delay that he encounters is more than what he would encounter if he if the two secretary services are pooled up and then they type letters as they come. So, you see even though as I said in the beginning in somewhere in the last lecture that you know you cannot take the values that we compute through the such models as exact, but they do definitely give you you know they are good guiding parameter. They give you provide you with good parameters to make to help you make your decisions. So, even though the numbers may not be so exact like 0.35 and 0.8, but it definitely shows that the difference is there and you can the efficiency of the system gets improved by pooling. So, your services that you have and you know like so many banks and so many other places in public places if you see that sometimes even at airport encounters and so on you feel that you know separate queues because once you once a customer joins a queue then he or she cannot change the queue. So, you see that way you can immediately see that I mean this kind of model shows you that lot of time is being wasted I mean the system is not working efficiently because you are not pooling the resources. So, somehow the feeling that separate queues will be more efficient and your work will get done faster. So, that belief is not supported by this model and it is a reasonable correct model in the sense that it gives you ideas to what happens when you pool up the resources. So, this is all about it and we will continue with the. So, figure 2 I just want to explain that here you see the horizontal axis is the utilization factor that means rho equal to lambda by s mu. So, different values of rho going from 0 to 1 and the vertical axis is the steady state probability of 0 customers in the system. So, first of all if you look at the curves in the diagram in the graph then you see that the utilization factor is coming down drastically as s increases. So, for that means for 0.8 for example, s equal to 1 the top line shows you the utilization factor versus the steady state probability of nobody no customer in the system. So, these are the different lines and you see that the utilization factor when s is 25 is barely is even less than 0.3. So, that of course, we also expect because more the server the lesser the utilization factor and but there again as we said that it is always conflict between you know having congestion or having more servers and that depends on what your priorities are. So, anyway this the graphs actually show you what we expect that as the number of servers will go up the utilization factor will come down. But the other important contribution of this graph is that you know you can plot because remember the expression for the steady state probability when you had more than one server the expression was lengthy one. So, now you can actually plot for different. So, that means if you know the utilization factor say for example, I know the my utilization factor is 0.7 and then if I want to find out what the corresponding value of p naught will be. So, then I will go along this vertical line of 0.7 and if my number of servers are 4 then you see wherever this curve s equal to 4 cuts the vertical line 0.7. So, that point and I go horizontally then across to the vertical line. So, I can then find out the corresponding value of the p naught and so that will help me because then I can make my computations for l l q and so on or even for p n's I will need the value of p naught. So, then it will help me to just plot the value given my utilization factor and the number of servers then I can find out the corresponding value of p naught. So, this is the contribution of figure 2 and later on when I work out an example I have used the I will use the values of p naught from this graph. So, I will take up this example I have taken this case study from the state hospitals in the U and this is from the book Helian and Lieberman again this reference will also will be given at the end of the course and see the state hospitals in the U S are called county hospitals. So, the data was collected from the county hospital and the emergency room is considered because that can be modeled very well. So, here emergency room and the arrivals and we are considering the morning shift have I written something somewhere here yes. So, this data refers to the early morning shift. So, early morning shift and that I do not know for some reason this is what happens that often the emergencies are in the early hours of the day early hours of the morning. So, arrivals are 1 per half hour. So, that means your arrival rate is lambda equal to 2 and here again it is it is found suitable to model the thing by Poisson arrivals and of course, the service process is exponential negative exponential. So, that means since 1 every half hour. So, 2 arrivals per hour then a doctor requires an average of 20 minutes to treat a patient which means that mu is 3 patients per hour. So, your this thing also. So, if it is negative exponential then the interval arrival times between the service the service times would follow negative exponential distribution. Now, there are 2 alternatives which the hospital management has to consider they have 1 doctor to manage the emergency room. So, either they continue with the with the 1 doctor or to add another doctor that means your number of service will go up to 2. So, these are the 2 alternatives just for the morning shift it is not for the whole day because for the rest of the day your lambda may change and even your mu may change. So, for the morning shift this is the these are the 2 alternatives which are being considered. So, now let us look at the data and so all the calculations have been made with s equal to 1 and s equal to 2. So, let us look at the data. So, this is steady state results from the MMS model. So, there should be a gap between s and model for the MMS model for the county hospital problem. Now, for s equal to 1 rho the traffic intensity is 2 by 3, but when it is s equal to 2 it will become mu lambda by 2 mu. So, this will be 1 by 3. So, intensity will come down to 1 by 3 p naught your probability when there is no patient in the system in the emergency room this is 1 by 3 and here it is 1 by 2 then p 1 the computed value of p 1 also at 1 patient it will be 2 by 9 for s equal to 1 and 1 by 3 for s equal to 2. So, again the number of patients the probability of p 1 will be 1 by 3 then p n for n greater than or equal to 2 is 1 by 3 into 2 by 3 raise to n and here it will be for s equal to 2 it will be 1 by 3 raise to n. So, everything obviously we expect all these numbers to come down, but the drastic difference is seen where in L Q is 4 by 3. So, a patient has to wait the Q is 4 by 3 whereas, for s equal to 2 it is 1 by 12. So, that is really remarkable big difference then your number of people in the system on the average would be 2. Whereas, here it will be 3 by 4 for s equal to 2 then W Q the time average time spent in the Q waiting for to be treated by doctor here it is 2 by 3 hour and for s equal to it becomes 1 by 24. And so, you see in an emergency room it is very very crucial that a patient gets treatment as fast as possible because it is a matter of life and death. So, here W Q being 2 by 3 if only one doctor is attending to the patients then it is a then the waiting time is high whereas, it comes down drastically to 1 by 24 hours if your SS 2. Then W the number of people waiting that means Q and service is 1 hour whereas, here it is 3 by 8 hour. So, you see these are the figures which immediately tell you that it will definitely be to the advantages to have 2 doctors because after all in an emergency room you do not want patients to die because they have not got immediate attention. And then probably W Q greater than 0 is 0.667 when s is 1 but it is 0.167 when SS 2. So, that means a patient coming into the emergency room will have to wait that is high 0.667 when s is 1 but the moment you have 2 doctors attending it comes down to 0.167. And similarly, probability W Q greater than half would be 0.404 for s equal to 1 whereas, it will be 0.022 for s equal to 2. So, therefore, this is also you know damaging because in a hospital emergency room if you have to wait for more than 1 by 2 of an hour then this is bad and the probability is 0.404 when s is 1. So, this is not acceptable and similarly W Q greater than 1 hour is 0.245 yes which will be less than half right. So, that means the waiting time so that will be 0.245 but for s equal to 2 it will be 0.003. So, you can just look at the numbers the other 2 are not that important but anyway so this data. So, therefore, we are able to then conclude that the single doctor will give you long waiting period which is not very desirable for a emergency room for hospital emergency room. But 2 doctors you expect from service and so, therefore, anybody looking at this data and this is what this model has helped us to generate that data and see that you can compare the performance of the emergency room when there is 1 doctor and when there are 2 doctors. So, if finances is not a consideration it would be very helpful to have 2 doctors. And then again I just want to make a point about pooling that what we were talking the example I gave you in the earlier. So, pooling we saw that if the waiting time is the main consideration then pooling will always help because we saw that even with you know when 2 businessmen having their own secretaries they had to the letters had to wait longer. But when the services of the 2 secretaries were pooled the waiting time for the letters to be typed came down. And if there is some legislation or something then it is a different thing that you cannot you have to have separate queues. But otherwise if the main consideration is to avoid long delays then pooling is the answer. So, that is another that is you know again the model helped us to arrive at that conclusion. Now, let us look at another kind of model which is limited of finite queue variation of the MMS called it MMS K model. So, here the ideas that you cannot allow more than K people in the system. So, you can just take the example of an emergency room in a hospital you know it may not be possible to because people come they need beds they are its emergency or they are on stretchers. So, they you definitely need room for these patients arriving for emergency service. So, then you the space is limited and therefore, you cannot allow for infinite queues size right. Then you can also have many other and as we said that you know petrol station you may not you could not have infinite number of cars waiting to be serviced. So, therefore, again you have limited space for the cars to be waiting and that the number usually is if you are big even then it cannot be more than 5 to 6 cars which can be accommodated by the petrol station where they are waiting to be serviced depending on the number of pumps the petrol station has. So, this is a very reasonable and this model realistic situations where you are limited space that means your finite queues it cannot allow for infinite queues. So, the only change that you would have to make in the MMS model would be that your lambda n will be lambda for n varying from 0 1 to K minus 1 and for n equal to K it will be 0. So, that means you will you will not allow people to come even once you have K people in the system then you will not allow people to enter the system essentially right. And here the queue will reach a steady state even if lambda is greater than S mu. See remember for infinite size we had to restrict lambda less than S mu, but because otherwise the number would have been blown up right your L the average number of people in the system and so on would blow up if lambda was greater than S mu in case you allowed infinite queue size. So, here because you are not permitting your queue size to be more than K. So, then lambda greater than S mu is also permissible right. And so the queue size queue size will never exceed K minus S and the total number of people in the system will not exceed K. So, your S servers and your K size cannot be more than K minus S also. So, I tell you the example is emergency room of your hospital also you see there are situations there are places where customers are choosy and they would not like to wait they would not like to enter the system if they are more than K people already. I mean they might consider that K number to be a crowd and therefore they would go away. Now, such a phenomena is called balking because you are losing out on customers since you have limited space. So, you are losing customers and you may also be losing out on good will. So, we will look at this aspect and essentially one would like to know what kind of what kind of business you are losing out because your people your customers are being turned away because you do not have enough room. And of course, for emergency rooms in a hospital legislation requires that if you cannot accommodate a patient right away then you have to send them to another person another hospital. So, there the legislation requires that you turn away a patient if you do not have enough room for them. So, all these considerations are there. So, we will look at balking and we will try to through example try to see how you estimate the loss of revenue because you have lost customers. And then of course, that might also encourage you to invest in increasing the waiting space waiting room. So, that to compensate for the loss in business. So, now of course, I will give you the expressions for MMSK also, but right now it will be easier to just write down the balance equations for M M 1 K model. And then the arguments for MMSK will also not be much different except that you will have to take care of the S servers. So, for M M 1 that means one server, but you have finite queue. So, the balance equations will be of course, when there is nobody there in the system then one arrival comes and then you can go from p 1 to 1 departure. So, therefore, this is the balance equation right. And so, that gives you p 1 equal to lambda by mu p naught. And so, here see the transition diagram is no different from M M 1 except that there will be no state after K. And so, therefore, you will have only that many balance equations. So, that is only the difference that is why I did not draw the transition diagram. Anyway, and so for when there are n people in the system then lambda plus mu one arrival one departure again you are back with n people in the system. Then this will be n minus 1 people you can reach p n you can reach n by 1 arrival. And when you have n plus 1 people then you can reach again state n by 1 departure. And this will be valid for n 1 2 to k minus 1. And then the last one and when you have k people then only departure is allowed because no arrival. And so, from p k minus that means from k minus 1 again you can reach k by 1 arrival. So, this will be the last equation. Surprisingly, you do not and so therefore, that is this makes sense because you cannot go away from here you cannot have any allow any arrival here. So, this will be the equation. And the interesting thing is that we will not need this equation actually when you are obtaining values for p 1 p p 0 p 1 p 2 because from here when you see when you put n equal to k minus 1 p k value will be available from here. So, you do not really need it, but just for completeness sake we want to write it down. And so, now one can solve these balance equations to get the corresponding relevant probability. So, just as for mm 1 model we will just solve those balance equations and you will get p n is equal to lambda by mu raise to n p naught and varying from 1 to k. So, let me just show you the calculations for as I told you that the last equation will not be needed the last one, but one equation can will give us the value of p k. So, the last, but one equation is lambda plus mu p k minus 1 is equal to lambda p k minus 2 plus mu p k. Because when there are k people in the system no arrivals are allowed. So, this is your last, but one equation and since you obtain the formula for p k minus 1 and p k minus 2. So, I just substitute. So, therefore, your mu p k is equal to lambda plus mu times lambda by mu raise to k minus 1 minus lambda times lambda by mu raise to k minus 2 into p naught everything is in terms of p naught. So, therefore, just simplify lambda by mu raise to k minus 2 you can take outside. So, then you will be left with this expression and here you can simplify. So, mu lambda and lambda again you can take outside. So, it will be lambda plus mu minus mu by mu lambda is outside here. So, this gives you a lambda by mu and therefore, this becomes lambda square by mu, but then you have a mu p k here. So, therefore, p k will be lambda by mu raise to 2 into lambda by mu raise to k minus 2. So, the whole thing is it. So, therefore, you there is no problem in solving your balance equation and then since all the probabilities must add up to 1. So, you get the expression for p naught which is a geometric series here and of course, except that rho should be not equal to 1. Otherwise, you can add this and that gives you 1 minus. So, because when rho is equal to 1 you will have a simplification. So, that can be written down immediately. So, therefore, this is your value of p naught and so you get a closed form for the p n which is lambda by mu raise to n into 1 minus rho upon 1 minus rho raise to k plus 1. So, this is valid for n varying from 0 to k. Now, you want to find out the average number of people in the system. So, this will be sigma n p n and varying from 0 to k and so you just substitute for p n is what you get and again we will use the same trick that lambda n lambda by mu raise to n. So, this can be written as derivative of lambda by mu raise to n with respect to rho. So, rho raise to n. So, n rho raise to n. So, that will be n summation of course, your n varying from 0 to k. So, if you take a rho outside then it will be. So, this will be then rho raise to n and then again. So, finite series I can interchange. So, d by d rho outside this summation and varying from 0 to k rho n and that gives you again a geometric series and the summation is this. So, derivative of this d by d rho 1 minus rho raise to k plus 1 upon 1 minus rho. So, differentiate the numerator this is minus k plus 1 rho raise to k into minus rho minus 1 minus rho raise to k plus 1 into derivative of this which is minus 1 divided by 1 minus rho whole square and just simplify and finally, you will get this as. So, I have just separated out the 2 terms rho upon 1 minus rho minus k plus 1 rho raise to k plus 1 divided by 1 minus rho raise to k plus 1. So, this 1 minus rho raise to k plus 1. So, 1 minus rho cancels here. So, you are left with 1 minus rho the power 2 is gone and then the rho part here. So, rho 1 minus rho I have written out here and there is a 1 here. So, then this gets coupled with this. So, minus k plus 1 rho raise to k plus 1 rho is outside here divided by 1 minus rho k plus 1. So, this easily you can see simplifies to this expression. Now, we just want to look at the long term behavior if you allow k to become large you want to look at this. So, for rho less than 1 rho k plus 1 will go to 0 as k goes to infinity and earlier we have shown that this series converges k sigma k sigma raise to sorry rho raise to k summation. The series converges to rho upon 1 minus rho whole square. We have already seen this because this is a arithmetic geometric series. So, if a series converges then the necessary condition is that the n th term must go to 0 as k goes to infinity. So, therefore, k rho raise to k must go to 0 which implies that k plus 1 into rho k plus 1 goes to 0 as k goes to infinity. So, now if you look at this here this is going to 0. So, this reduces to 1 and k plus 1 rho k plus 1 goes to 0. So, therefore, your limiting value of l when rho is less than 1 and as k goes to infinity. So, the limiting value of l is rho upon 1 minus rho which is the m m 1 k. So, you see you can immediately conclude that you know when you have the m m k k that means you have a limited space for people to wait that is a k people can wait or k units can wait. So, this model relates to that, but if you make the space unlimited that means there is no restriction on how many people can wait in the system then the whole process reduces to the m m 1 case. So, this validates the m m k case that means whatever we have derived the values of l and so on they are valid in the sense that they correspond to the m m 1 case in case your space becomes unlimited. So, as many people as you want can wait for to be serviced. So, in that case it will be the m m 1 case. So, you can you know. So, there are many ways in which you can also try to validate the model that you have constructed. So, this is one of the ways that you can validate the model that you have constructed.