 So, I will continue discussing the exponential law, it is very important one and anyway there are so many aspects that there may be some repetition also, but does not matter. So, we said that for example, I have been giving the example of a fuse, which does not show the wearing house effect, wearing out effect and similarly, a jewel bearing, you know you have them in watches. So, there is no bearing out effect on such components and they are good as good as new while they are functioning. So, it does not matter how long they have been functioning, they are as good as new and if a fuse has not burnt out, it is like a new one, a jewel bearing does not wear out. So, this is again and again I am trying to give you examples of components for which and therefore, for such components the exponential law is appropriate and is supported by empirical evidence. That means, when you have collected data for such components, how long it takes for them to fail and so on. So, then and then fitting the curve to the data, it turns out that exponential law is an appropriate one for such components. So, another way of saying that there is no wearing out effect is as follows, consider the conditional probability that capital T, the lifetime lies between T and T plus delta T given that capital T is greater than T. Now, the intersection of these two events is simply this because here also T is greater than T and it is less than T plus delta T. So, therefore, the intersection of these two events is this. So, then the conditional probability can be written as probability of the intersection of the two events divided by probability that T is greater than T. Now, this you can see immediately is the probability T less than or equal to T plus delta T minus probability T less than or equal to T. So, this event and since this is 1 minus e raise to minus alpha T plus delta T minus of 1 minus e raise to minus alpha T. So, the 1 1 cancels out and you will be left with e raise to minus alpha T minus e raise to minus alpha of T plus delta T divided by this probability which is e raise to minus alpha T. This of course, I am showing you for the exponential law. So, therefore, this is 1 minus e raise to minus alpha delta T which is nothing but the probability that T lies between 0 and delta T. So, this tells you that this probability is independent. So, this conditional probability is independent of T depends only on delta T that is depends only on the length of the interval that you are considering that means you want to consider. So, that means here from T to T plus delta T. So, the length of the interval is delta T. So, this probability this conditional probability is independent of T and depends only on delta T. So, therefore, this is another way of saying that it is memory less or the varying out effect is not there. That means it is not it does not matter for how long the system has already been working, but now when you want to look at the probability that it will be working in the interval T to T that means it fails before T plus delta T then that is dependent only on delta T. So, this is the whole idea and so we have seen that there are many situations where this is a very appropriate law. Now, see this one to make a note here that even though this probability will all if you would say T less than or equal to capital T less than or equal to T plus delta T this will also come out to be the same as this one. But since we are considering the conditional probability that T is greater than capital T. So, therefore, we have to consider this event and not this event. So, sometimes inadvertently may be one can I may have written it like this, but the when you are looking at this conditional probability then it has to be T strictly less than capital T and here it is less than or equal to T plus delta T. So, that would be the right way to write the event even though because of the continuous case the two probabilities may come out to be the same important part. So, thus as long as the component is working it is as good as new. Now, in this equation suppose I expand the right hand side so expansion for e raise to minus alpha delta T right. So, all of you should be familiar that much calculus everybody has done. So, you can write down the expansion for e raise to minus alpha delta T which will be 1 minus alpha delta T plus sorry I am sure that means 2 factorial and so on powers of delta T. And then when you open up the brackets one cancels out and then we have you have alpha delta T plus higher power terms containing powers of terms like delta T square delta T cube and so on fine. Now, when delta T is small delta T is small we can just ignore these terms and therefore, for small delta T probability of failure in time delta T is proportional to alpha delta T. So, this is what again just reiterating what we have been saying. So, now in fact we have given it a better expression from here this is more you can immediately conclude quite a few things from here. So, that means in a small interval no matter where that time interval is length delta T is the probability of a failure in that time period is proportional to time period itself delta T. Now, this is if you again see I am saying the same thing again which I said. So, this is nothing but your Poisson process this is this is the one of the basic assumptions of a Poisson process. In fact, what you can say now here is that suppose you have a electronic device and you have lot of components which have the same which follow the same exponential failure law and they have the identical distribution that is the same the same parameter lambda let us say and then you are and of course, the components behave independently. So, that condition also for a Poisson process is satisfied that is the you know arrivals are independent and so here the components will behave independently. So, their failures will also be independent of each other. So, that assumption plus the assumption that within in a small interval the probability of a failure is proportional to alpha delta T. So, then in that case with those two assumptions you can then say that if you are considering let us say time period 0 comma T then the number of failures within this time interval will follow a Poisson process. So, you can see that the arrival and the inter arrival times that we will talk in detail later on. So, the inter arrival times and the arrival pattern. So, inter arrival times would be exponential and the arrival patterns would be Poisson and so on. So, we will continue with that discussion. Now, because again we want to say this thing now that since the for the exponential failure law the failure behavior of an item depends only on the length of the time period being considered and not on its past history. So, therefore and for the non exponential just like we have so far considered the normal failure law. So, for non exponential failure laws the past does matter it does matter when you are under you know stress then the varying out effect is there and so it depends on how long the stress has been and so on. So, for non exponential failure laws the past does matter. So, therefore it is important to understand that when you talk of the time T like time you know life length. So, for the exponential failure law T will denote the time in service up to failure because it does not matter when you start counting it is if the component is functioning when you start counting the time from then. So, up to failure T will denote the time it does not matter when you start counting it right, but for other for non exponential components T will denote the total life length up to failure. So, you started functioning from whatever time you you you count your time from that and so capital T will in that case the normal random the random variable will be will denote the total life length up to failure. So, total means that whenever you start service or whenever you start using the item then you start counting your time from here whereas, here it does not matter you start counting from any point and then up to failure. So, the life length will denote that. So, T will denote that time period. So, it is important to understand these two differences between you know how you count the time for exponential failure law and for a non exponential failure law. Let us look at the various graphs connected with the exponential failure law. So, this is a familiar one right at t equal to 0 this will be 1 it will be equal to alpha right and then it will go down this way right as t goes to infinity it will go to 0 and then correspondingly you see z your failure law rate of failure z t is alpha right which is a constant right and so it continues to be alpha value alpha for all values of t this is your t axis this is your z t axis then f t would be of course. So, f t would be what 1 minus e raise to minus alpha t right sorry this function right. So, 1 minus e raise to minus alpha that t equal to 0 this will be 0 and then as t goes to infinity this goes to 0. So, the function finally goes up to 1 and then your r t the failure function sorry the reliability function yes the reliability function r t which is 1 minus f t and therefore, this is equal to e raise to minus alpha t and so here again at t equal to 0 the value will be 1 and then it will go down as t goes to infinity. So, the reliability goes down as t goes to infinity. So, this is the graph for the pdf of an exponential failure law. So, at t equal to 0 it is equal to alpha and then it goes down to infinity as t goes to infinity then we know that the failure rate z t or the hazard function this is the constant right and the constant values alpha then cumulative density function which is f t is equal to 1 minus e raise to minus alpha t would be again at t equal to 0 it will be 0 because this will be 1. So, 1 minus 1 is 0 and then it goes up to 1 the reliability function would be e raise to minus alpha t. So, at t equal to 0 this will be 1 and then again it goes down to infinity. So, these are the four graphs that the four functions that you relate with the exponential law and you have picture of the all four of them. Now, let us look at this example if the parameter alpha is given and reliability is also given that means r t is specified we can find t. So, that means given alpha that means you have specified the failure law and then you are asking for a certain level of reliability. So, you want to know what would be the time required for the equipment or the component to operate to achieve that reliability. So, the number of hours of operation required for the specified reliability level I should say. Now, let alpha be 0.01 and reliability is 0.9. So, you want this to be obtained and your parameter is 0.01. So, if alpha is 0.01 then your mean will be what? So, for exponential distribution the mean is so expected t would be 1 by alpha which is 1 upon 0.01 which is 100 hours. So, if you are talking if your unit of time is hours then this is 100 hours. So, now so therefore, the number of hours that are required is given by to achieve this reliability level. Then you are saying that e raise to minus 0.01 t remember this is the reliability function. So, e raise to minus 0.01 t should be equal to 0.9. So, you want that value of t which will satisfy this equation. So, you take log of both sides and this will then give you minus 0.01 t is equal to ln of 0.9 and remember ln of number less than 1 is negative. So, therefore, this is so there is a minus sign here and so if you now divide by 0.01 then you get the t is equal to 100 into ln of 0.9 which comes out to be if you look up the values of ln 0.9 then multiplied by 100 that gives you 10.54 hours. So, therefore, the way you can interpret this is that out of 100 components all are working simultaneously and 90 what we are saying is and if they all operate for 10.54 hours then our expectation is that 90 of them will not fail. So, after 10.54 hours 90 of them will be working. So, this is what we mean by the reliability level and so on. So, therefore, you know so essentially it is question of given what and then what you can compute. So, sometimes you may be given the time then you can compute the reliability level. So, if you are given t then you will be able to determine this number and if you are given this reliability level then you can determine the time or if you are given time and reliability then you can determine the parameter. You can uniquely determine the exponential failure law because you only need the parameter alpha to determine the failure law exponential failure law. Let us look at another example. So, here you are given the cos c in terms of mu. So, c is 3 mu square be the cost of producing an item where mu is the mean time to failure. So, again we are talking of a exponential distribution exponential failure law and mu is the mean. So, therefore, if mu is the mean then remember the distribution the failure law would be minus 1 by mu t. This is what you have. Now, cos is 3 times mu square. So, which means that if mu is small then the cost is small. If mu is large then your cost would be accordingly large which makes sense. So, may be because if your mean time to failure is small then you expect that the cost is also small and if the mean time to failure is big is large number. Then that means you expect the component not to fail very early or has a long life time and in that case the cost of producing that item would be also high. So, it is reasonable to assume the cost in this way. Now, suppose rupees D are earned for every hour the item is in service. So, you earn a profit of D rupees per hour when the item is functional. Now, you want. So, therefore, profit per item is given by the profit would be D into t if the life time is t hours then D into t minus the cost 3 mu square and t is the number of service hours. So, find the value of mu for which the profit is maximized. So, expected profit. So, certainly because this is a random variable. So, we will maximize the expected profit. So, the expected profit is this because D of e t this is the I mean this is not a random variable this would be some fixed number. So, then D of e t minus 3 mu square and e t is mu. So, therefore, D mu minus 3 mu square. So, exactly what I was saying that. So, this is. So, therefore, to maximize this expected profit I would differentiate this respect to mu. It is a function of mu and put it to 0. So, that gives me D minus 6 mu equal to 0 which implies that mu is D by 6. And of course, only critical value and if but still you need to verify the D square by D mu square is of the function is minus 6 which is less than 0. So, therefore, the critical point is the point of maximum. So, the value that we get of mu here is value which maximizes the expected profit. So, therefore, mu equal to D by 6 is maximizing the profit and the maximum profit is D square by 12 rupees. So, just trying to give you a feeling for the failure law exponential failure law and the kind of problems that can be discussed and that arise corresponding to this. So, again the level is at very basic. The level I have kept is very basic because this is just trying to give you a glimpse of how these probability tools that we have learnt can be used for answering so many questions about day to day operations of systems, service systems and so on. So, this is the whole idea. Otherwise, reliability theory has become very complex and in fact, the next failure law that we will discuss is a complex one and again we will just try to understand the basics of the viable failure law. So, let us now I will talk about the viable failure law and you can see that now the degree of complexity is going up and we have not so far discussed this distribution probability law also. So, let us look at it and the idea here is to because the constant failure rate was only applicable to a special kind of components which were not let us say for which there was no wearing out effect. So, that was a special situation and we have seen that there are so many almost electronic devices or you know behave that way and so the exponential law is appropriate for them. So, now if you want to modify this constant rate then the viable failure law was thought of and this is alpha beta t raise to beta minus 1. So, now you have introduced power of t of the time and of course, two parameters and there can be more than two also. So, we are talking of a two parameter viable failure law and so here alpha and beta are positive and t is greater than 0 as usual because it is the lifetime variable. So, therefore, has to be non-negative. Now, you can look at I have drawn various pictures of z t when for the various values of beta and different values of alpha and beta and so maybe we will just look at it. Let me just first compute your f t. So, remember we said that we can compute f t uniquely given the failure rate function z t. So, then this is z t e raise to minus integral of 0 to t of z s d s integral 0 to t of z s d s right. This is our formula for computing f t given z t. So, then here if you substitute for z t this is alpha beta into t raise to beta minus 1 e raise to minus integral 0 to t alpha beta s raise to beta minus 1 d s right. So, if you just look at this integral let us just compute. So, this will be s raise to beta minus 1 d s from 0 to t and the integral here is 1 by beta s raise to beta 0 to t. So, therefore, this is 1 by beta t raise to beta. So, if I make that substitution here I get my f t as this. So, this is the p d f connected with the variable failure law and the failure law is specified there. So, the way it looks it is little complicated, but just see when you put alpha equal to 1 and beta equal to 1 alpha equal to 1 beta equal to 1 and this is 1 and this will be e raise to minus t. So, e raise to minus t would be your yeah. So, this would be your p d f right exponential with parameter 1 and that is the one I have drawn here right. Then if you look at the value alpha equal to 1 and beta equal to 2. So, therefore, in this case the beta equal to 1 this will be exponential and the failure rate will be constant which will be 1. So, beta of course, here I have drawn it only for beta equal to 1. So, alpha is as it is this one is of course, you put alpha equal to 1 also. So, anyway this is your function for the failure rate. Then when you put alpha equal to 1 and beta equal to 2 then I have drawn this one. This is the 1 for beta equal to 2 of course, not very accurate figures. So, you can always Google search and then you can find nice pictures very accurately drawn graphs. So, beta equal to 2 now you look at this thing here. This is alpha 1 beta equal to 2. So, this is twice t that means it is a linear function of t. So, therefore, you see for different values of beta things are changing. So, alpha equal to 1 beta equal to 1 you got constant alpha equal to 1 beta equal to 2 you get and in fact, any value of alpha alpha does not have to be 1. Then in that case it will be a linear function of t if beta is 2. So, here of course, I have drawn it for alpha equal to 1 and beta equal to 2 the diagram. Then and the corresponding pdf will be 2 t e raise to minus t square. Then for beta equal to 3 for example, I have drawn I have drawn the picture here also for beta equal to 3. So, then this starts taking a bit more bell shape and beta equal to 3 alpha equal to 1 will be 3 times t square. So, that would be that means that t is a quadratic function of t and correspondingly your this thing will be. So, here when beta is 3 then of course, this is t square and then e raise to something right. So, quadratic into a financial function. Now, and for beta equal to 5 for example, it will become more steep like this and then as beta goes to infinity you see you can show that it will simply be a spike. Now, just a spike because beta is going to infinity then this will simply just spike into this thing which you call as a delta function. So, at the point so, here it will become a spike as beta goes to infinity. So, the thing becomes narrower and narrower as your beta goes up. So, here beta is a shape parameter therefore, you see beta is a shape parameter and 1 by alpha is the scale parameter. So, I use scale the whole thing right that means when you are drawing the thing. So, failure rate is proportional to power of power beta minus 1 of t. So, therefore, this is the generalization that we have made to the constant failure rate and so, this is the failure rate is now proportional to t raise to beta minus 1. So, you can see that yeah and then this also gives you good feeling. Now, for example, if your beta is less than 1 then this will become negative. So, t will be in the denominator and so, alpha beta upon t raise to 1 minus beta and that will be for as t goes to infinity. So, the denominator will go to 0 and therefore, this will let me so, alpha beta upon t raise to 1 minus beta. So, as t goes to 0 this goes to infinity the denominator t goes to 0 this goes to infinity and so, it should be the other way. I want to show that for beta less than 1 for beta less than 1 this is negative. So, when I take it here it will be positive this power is positive and so, as yes as t goes to 0 this goes to infinity because this goes to 0. So, this goes to infinity as t goes to 0 right yeah and therefore, it is this way and then as t goes to infinity this goes to 0 and so, failure rate decreases with time. Now, failure rate decreases with time that means, see essentially defective items fail early and the failure rate decreases over time as the defective items have been weeded out. So, all the defective items have been weeded out from the population. So, that case the failure rate will decrease with time right. So, all defective items fail early and therefore, as time progresses your failure rate will decrease. So, this is the situation that gets modeled when you are by putting beta equal to less than 1. So, all values of beta between 0 and this thing here all right. So, you take this then this is the situation if you suppose to model and beta equal to 1 we have already discussed thoroughly. Now, here of course, since the failure rate is constant. So, as time goes on the failure rate does not change. So, this is because random external events are causing the failure that could be one of the reasons. So, random external for example, if a fuse will blow out if the high current comes suddenly in the line right. So, therefore, that is the external event and many others can be explained high wind and so on. For other high tension wires you can snap and so on. So, therefore, beta equal to 1 because the failure rate is constant it is understood that external events would cause the failure could be the reason for the failure. And for beta greater than 1 as we have seen failure rate is increasing with time and therefore, this models the situation where aging process has a role to play in the failure of the system. And that is parts are likely to fail as time goes on and this is when the stress part. So, you see this suddenly captures it is a more complex and it will more comprehensive failure law which captures more than one situation and you know you can play around by manipulating the value of beta and alpha and try to get accurate results. So, this is the whole idea and we will continue with the discussion on web distributions. So, let us make this the computations about the expected value and the variance of a viable distribution. So, E t the theorem says that E t is alpha is to minus 1 by beta gamma of 1 by beta plus 1. So, we all know the gamma function and then V t the variance would be alpha is to minus 2 by beta gamma of 2 beta plus 1 minus gamma of 1 beta plus 1 whole square which is we are using the formula that variance is E t square minus expected value of t square minus expected t whole square. Now, so just apply the because we have already computed the f t the p d f for t when t has a viable failure law then this is t into alpha beta t raise to beta minus 1 raise to minus alpha into t beta d t. So, yes so now you can see that this t beta minus 1 and t beta. So, this prompts you to make the substitution that t beta is y. So, we will do this. So, again you are all familiar with this part of the calculus you can do this integration. So, t raise to beta is y that will make beta t raise to beta minus 1 d t is equal to d y the limits will not change they will remain from 0 to infinity since beta is positive. So, for t equal to infinity y will also be infinity and t 0 y is 0. So, now therefore, this thing whole thing gets replaced by d y. So, you have a beta t beta minus 1 and d t. So, this we will replace by d y then you are left with a t and an alpha and then e raise to minus alpha y because t raise to beta is y. So, therefore, this is what you have alpha e raise to minus alpha y and y raise to 1 by beta and there is a yeah. So, y raise to 1 by beta because you have a t here. So, t will be y raise to 1 by beta. So, therefore, this is what you have now you see this looks familiar because you can now relate this with the gamma function gamma p d f. So, here alpha e raise to minus alpha y then you have to have alpha y here as the variable. So, alpha y raise to 1 by beta now y 1 by beta is there. So, alpha 1 by beta I am adding here. So, therefore, I will divide by alpha 1 by beta and then I need a gamma 1 by beta plus 1 for this integral d y to be 1 because this is the p d f of a gamma distribution with parameters alpha and 1 by beta. So, then I am left with this and this finally. So, therefore, the expected value of the random variable t where t is has a variable failure law is given by gamma 1 by beta plus 1 into alpha raise to minus 1 by beta and then the second part should be straight forward. So, you have this. So, therefore, I need to compute expected value of t square and so it will be t square alpha beta t raise to beta minus 1. So, now you see you are again beta t raise to beta minus 1 and e raise to t raise to beta would become this thing yeah I am sorry I mean beta t raise to beta minus 1 d t that will be d y from here right and then you have t square. So, that will be y raise to I mean I have not written the integral here. So, anyway this will this will reduce this will be equal to 0 to I have I written t this is 0 to infinity yeah this is 0 to infinity. So, y raise to 2 by beta and then you have an alpha and then you have e raise to minus alpha y d y right. So, here again I will do the same trick that I did here. So, alpha e raise to minus alpha y is there then this you need to write this 0 to infinity alpha y raise to 2 by beta alpha e raise to minus alpha y is exactly the same thing. And then you will divide by alpha raise to 2 by beta then you need a 2 by beta plus 1 right and you will multiply by 2 raise to beta by gamma of 2 beta plus 1 2 by beta plus 1 and so this whole thing will be and there is a d y right. So, then you will be left with gamma of 2 by beta plus 1 into alpha raise to minus 2 by beta. So, this will be expected value of t square right and therefore, the variance will be this minus the expected t whole square. So, you take out alpha raise to minus 2 by beta common and then you are left with it. So, this is just for your now and the here again we could make use of the gamma distribution computations to compute the expected value and the variance. Again, so we have seen that variable distribution represents an appropriate model for the failure law whenever the system is composed of a number of components right and the failure is primarily due to the most severe flaw among a large number of flaws in the system. So, this is what is happening. So, you have lot of components and lot of parts in the device that you are using and each one of them has a flaw, but then it will be governed by the most severe flaw among all the components. This is what the you know you can say that this is the representation of this is the situation, but which a viable distribution will represent and your alpha beta and of course, you have seen that you know by changing the values of beta you can either have a increasing failure rate or a constant failure rate or a decreasing failure rate. So, this we have already seen right and yeah. So, here you know I am just being able to give you you know short glimpse into the what this distribution can do and one needs to really work at lot of examples to understand the implications of the importance of this distribution. Now, let us just look at this example. So, each of the six tubes of a radio set has a life length in years. So, our time unit is a year which may be considered a random variable. So, the lifetime of a radio tube in a radio set number of years should be considered as a random variable. Suppose, these tubes function independently of each other. So, that is important they are working independently of each other. What is the probability that no tube will have to be replaced during the first two months of service. So, we will have to translate this to years because our unit of time is years. Now, the pdf of the failure time to failure is given by this 50 into t e raise to minus 25 t square. So, immediately you can recognize that this is a variable distribution and since t has power 1. So, your beta is 2 this is beta is equal to 2 because t raise to beta minus 1. So, therefore, if beta is 2 then alpha is and of course, from here alpha is 25. So, this is alpha beta alpha beta into t raise to beta minus 1 e raise to minus 25 t square because this is beta. So, this is a variable distribution. Now, of course, we have not made this computation of the for the for the viable distribution, but certainly you can do it or maybe you can use numerical method. So, essentially yeah it is a viable distribution with these are the parameters. Then, you since the tubes are functioning independent of each other and you do not want and of course, let me say that your t is 1 by this is 2 months. So, 1 by 6 which I have written here. So, t is 1 by 6 you want. So, yeah so we we want their 6 tubes they are working independently of each other. We do not want any one of them to fail. So, therefore, in the first 2 months. So, the probability of let us say the first tube not failing in the first 2 months is t 1 greater than or equal to 1 by 6. So, the time unit is 1 by 6 and then since all of them are independent of each other we do not want any of them to fail. So, then this would be probability t 1 greater than or equal to 1 by 6 raise to 6. So, this we will you know this is not a difficult integral again you know because see you have seen my computations here for e t and various t. So, you can just use those you can use the gamma distribution computations to do the computations here and then you can find out this probability raise it to 6. So, the answer is approximately 0.5 raise to 6. So, this integral you can handle now since I have given you the method of computing e t. So, exactly it just boils down to that for different values of t and so on. So, now this does not exhaust the failure laws I have only as I told you and I have been repeating it that we are only considering very basic failure laws here and my of course aim was to since we have discussed the probability theory and so the various tools you have learnt about. So, I just thought that I would like to show you the various applications also of these tools that we have learnt in the course. So, that has been the whole idea theme across the course that you learnt the theory and then you learnt to use it also and so marco processes discrete marco processes then we will we have talked about continuous marco processes and in the process special cases which are poison and exponential distributions and then birth and death processes and finally applications to reliability theory and here also the basic concepts have been given to you, but as I said it is a very growing large growing area very important and lot of applications of course probability theory and there are many more people have come up with failure laws which probably supplement the theory that has been discussed here.