 So, now the last topic that I want to consider in this course is applications to reliability theory. That means again the concepts that we have learnt during the course of probability theory, I want to show you some applications. Of course, everything may not be very new, but still we want to put them all together. So, that you have a good feeling about and of course, see the reliability theory is a very, it is become very important now with since the systems very complex systems, you know coming in and then besides that you know you have lot of dependence on the systems functioning. Otherwise, if something fails then lot of things are connected with it also fail and you know there is a chaos. So, this is a very growing area and we see that the applications are really interesting and very meaningful. So, you can model situations here also through the tools that we have learnt during the course. Now, so first of all let us first understand what we mean by reliability of a system. Now, just for example, consider steel beam under load and so you have whole structure you know resting on a beam and there is a heavy load, then you have a fuse inserted into a circuit, so the source of stress. So, you know there are systems and then there are some source of stress. So, here and for example, an aeroplane wing, the planes fly at high speed, there is lot of friction and so there is some sort of stress on the aeroplane wing also, then electronic device you have, this is when it is functioning again that is called that is a sort of a stress and therefore, so one can go on and writing a number, the list can be very very big you know anything that you use as a tool and then so when you are using it, there is some stress, some kind of stress on the tool and so things will change or things will happen. So, that means, so you have a source of stress, you have a system that means a component or a whole system and then you talk of state of failure. So, what can happen the beam can crack and similarly, a fuse can burn out or the aeroplane wing can buckle because of pressure or stress and an electronic device can just fail. You suddenly find that your PC is not working or your you know electric kettle is not working or whatever it is right. So, therefore, you know you can write down your own these things, you can take a component or a system, you can talk about what kind of stress that system is facing and then you can also define the way the system will stop functioning. So, essentially what we are saying is that we have this. So, now here we can and suppose we I define therefore, so we need to talk about time to failure or life length and it will be a random variable and why because you see if you take identical components under identical stress may fail at different and unpredictable times, you really have no idea sometimes a thing just stops working. So, some may fail early, some may fail at later stages and so on. So, therefore, you can see that whatever the systems we have talked here and talked about and many other, you see that you cannot really for sure say that this component or this system will work for so long. So, there is lot of unpredictability if they are not. So, these things can not be predicted very well and therefore, because the you never know how the stress works on a particular component or a system and also the manner of failure also varies as I told you because the beam will crack, the fuse will burn out and so on. So, the manner of failure you can say fuse working one moment will fail the next moment, you suddenly you find that thing is not working, you are having good output the radio was functioning very well broadcasting and suddenly it goes away goes out. So, the beam you know steadily over a long period it becomes weaker and weaker and then it cracks. So, here again and many other situations you can talk about. So, therefore, it is reasonable or it is appropriate to construct a probabilistic model and then treat the lifetime of the system or the component as a random variable. So, this seems to be very appropriate and now we will what we will do is we will talk about different models that we can use for predicting or for the lifetime of a component or a system and we will be using lot of tools that we have learned so far. .So, first of all let us define because now with the understanding that what we mean by system failing and the manner of failure can be very different and the unpredictability of these components or systems failing under different situations. So, we can now define a reliability of a component or a system at time t say by r t. So, we will define so r t will be the will define the reliability of a system and we will say r t is equal to probability t greater than t. So, at time t we want to know the reliability of a system then this is equal to probability t greater than t where t is the life length of the component. So, suppose that we are talking about a component then t is the life length and we want to know the probability that t will be greater than t and r t is defined as the reliability function. So, this is the reliability function and so about different values of t we can get the value of this function which will tell us the probability that the system is functioning at that time. So, essentially when we say that r t when we are computing this way this probability. So, if t is the life length that means here it is saying that the life length is more than t. So, that means at time t when you are computing this the system is functioning. So, the probability is not 0 well I am not saying that I am saying that the probability that t is greater than t that means the life length is greater than the. So, the system is still that is how you will interpret this because this is the life length. So, the life time of the component is greater than t. So, at time t that means it is functioning the system is functioning. So, this is what now for a particular item suppose r t 1 is 0.95. So, at time t 1 you compute this and it turns out to be equal to 0.95. This means that approximately 95 percent of such items used under similar conditions will be functioning at time t 1. That is what we mean. So, there are 95 percent of the items will still be functioning will be functioning at time t 1. So, this is what we mean by probability being 0.95. So, probability that the life length is more than point is more than t 1. So, this here this you will write as equal to probability that t is greater than or equal to t 1 for t greater than t 1. So, that is equal to 0.95. So, this probability is 0.95 that means 95 percent of such items under similar conditions will be functioning at time t 1 and we continue with this. So, you see we saw that if f is the pdf of t the random variable t the probability density function. So, in that case we can write the reliability function r t as t to infinity because remember this is probability t greater than t. So, therefore, this will be the integral of f s from t to infinity. So, because we were saying that this definition tells us that the reliability gives you the probability that the system is functioning in the interval 0 t. That is it is not sorry the way to say it is that the reliability tells you that the system is functioning is has not failed in the interval 0 t that is the better way to put it right because the life time is greater than t that means the system has not failed in this interval right. So, it can fail anywhere in the interval t to infinity that is what reliability is right probability t greater than t. So, that will be written as t to infinity integral of f s from t to infinity which if capital F as we denote the capital F as the cdf of cumulative density function of t then this is 1 minus f t right. So, this is what we can express the reliability function as. Now, we will associate another function with reliability and which is with the random variable t which is the failure rate z and this is also called the hazard function. Now, if you recall while talking about exponential distribution I had introduced this hazard function and of course, we did not talk much about it only I simply I showed you that if the pdf of a random variable t is exponential negative exponential then the hazard function will be a constant. And so, we will see many more interesting kinds of hazard functions and but anyway we had come across this at that time, but now I am using this. So, that is why I said that some of the things which I talk about here may already have been discussed, but we are putting them all together. So, that it becomes a complete unit. So, now failure rate we are defining as z t is equal to f t upon r t. So, the pdf divided by the reliability function which you can write as f t upon 1 minus f t right from here defined and of course, this is defined for f t less than 1 because remember f t is your f t is probability t less than or equal to t. So, if f t is equal to 1 that means and so this will say that the life time is less than or equal to t. So, if this is 1 then this that means it is a certain event f t equal to 1. So, f t equal to 1 implies that by time t the system has definitely failed, it is a certain event right. So, therefore, this has meaning only when the system is still functioning and therefore, this is a reasonable condition to impose that this is defined for f t less than 1 because f t equal to 1 would mean that the system has or the component has failed. So, therefore, so this is your failure rate and why is this, why are we calling it as a failure rate? This is the definition, but now let us understand why this does covers the failure rate of the system. So, consider the conditional probability t capital T lying between small t and t plus delta t given that capital T is greater than t. So, since I am asking for see the failure rate has to be after time t. So, here you have to consider the inequality strict inequality, I cannot allow equality here because my conditional event is that t must be greater than t. So, that is why you cannot write less than or equal to here. So, this is what we have to consider, I mean what I am saying is that the event 2 should be described properly. So, then the probability this less than or equal to t lying between. So, this is strict inequality and that is less than or equal to t plus delta t. So, then by our rule because this and this when you take the intersection actually means just this event because capital T is greater than t already satisfied which is here. So, therefore, this conditional probability can be written as probability capital T between small t and t plus delta t divided by probability t greater than t. And this integral form if I take f to be the pdf then this would be my failure law then this would be t to t plus delta t f s d s divided by r t. Now, for small delta t you see I can take this to be by the mean value theorem or here you do not need small delta t. So, by the mean value theorem of integral calculus this integral can be written as delta t the length of the interval of integration into. So, there exists a value point real number psi between the interval t and t plus delta t such that this integral can be written as delta t into f psi. So, this is the well known mean value theorem of integral calculus divided by r t. Now, for small delta t then I can say that this is approximately z t into because your failure rate is f psi f t upon 1 minus f t. So, then I can because my if delta t is small then this interval is very small. So, I can treat this as the value at t and therefore, this will become f t because if this interval is small then your psi is close to t and so approximately I can say this is equal to z t into delta t because your z t is f t upon r t and so this is what your and therefore, z t represents the proportion of items that will fail between t and t plus delta t. That means, during the time span delta t so this is the proportion of items and that is why we call it the failure rate. So, among those items which have not failed till time t remember because we are computing this. So, this is p t greater than t. So, all items which have not failed till time t then the proportion of those which continue to work in the interval t t plus delta t so that is represented by z t. So, this is you know interpretation of what I defined here z t as f t upon 1 minus f t. So, this was the conditional probability so life time is more than t and therefore, this is so therefore, z t is the rate of failure so that makes sense. Now, we could we could define z t given the pdf of t. So, the given the pdf of t that means, the f determined z t z uniquely the failure rate the converse is also true. That means, if you are given the failure rate then you can determine your t uniquely and determine your pdf f uniquely and once you know f then you know the cumulative density function also. So, let us see that so the theorem is that if t the time to failure is a continuous random variable with pdf f's and if f 0 is 0 where f is the cdf of t sorry where capital F is the cdf of t then f can be expressed in terms of the failure rate z t as follows. So, then f t can be written as z t into e raise to minus integral 0 to t of z as z s. So, that means, once I know z then through this function I can compute my pdf. So, if suppose you are you somehow know the failure rate you know empirically or somewhere then you can sort of compute the pdf also of t and of course, this makes sense what are we saying that f 0 is 0 implies that r 0 is what 1. So, that means, certainly the function this is we are not going to when the time is 0 then your system is not going to fail it will take some requires some time you start the function the system working or the component is working then only after little lapse of time there is a possibility that the system may fail or something. So, the reliability will be 1 at 0 time and so this is not you know this is a reasonable assumption that m 0 is 0 because you are talking of reliability and the reliability comes only when the system starts functioning. So, some time has to lapse before you can say that the component has failed. So, with this condition now let us start computing f from given z from the given failure rate. So, let us see r t is equal to 1 minus f t differentiate both sides that will give you r prime t as so the derivative of f t is the cumulative density form is the pdf. So, this is minus f t. So, r prime t is minus f t and therefore, your z t which is defined as f t upon r t. So, for f t you are going to write r prime t. So, this is minus r prime t upon r t. So, f t is minus r prime t and therefore, if you integrate this from 0 to t this is of the kind 0 to t r prime s because that is what you have here. So, this is 0 to t z s d s and this will be. So, now you have this integrand of the kind where you have numerator as the derivative of the denominator and so immediately you know that this is the integral of this is minus l n minus sign is here. So, l n of r s log of r s from 0 to t and now here you see this that means this will be. So, if you write it out this will be minus l n r t plus l n r 0. This is what this integral will be, but then since f 0 is 0 r 0 is 1 and log of l n of 1 is 0. So, therefore, there is no contribution from here and you have minus l n r t. So, since f 0 is 0 therefore, r 0 is 1 and so from here z t is this. So, I should write therefore, r t I wrote down this way and this is just integrated 0 to t. So, r t is. So, this is l n r and this is l n r t. So, you have the equation that z t. So, what have we obtained? Now, let that be there. So, the long equation has become long that is why. So, what we obtained is 0 to t. So, z s d s is minus l n r t. So, from here we are saying that this implies that r t is so l n. So, e raise to remember when we write l n it means to the base e. So, this will be e raise to minus 0 to t z s d s and now you have it from there see from this equation our original definition because you want to compute. So, I should have finished it here. So, once you have this now your z t is f t upon r t. So, therefore, f t is r t into z t and r t we have just computed as this. So, therefore, f t is z t into e raise to minus integral 0 to t z s d s. This is what our result was we wanted to show this. And so, given your failure rate z I can compute the pdf of the life length random variable t uniquely. So, there is an interesting relationship between the reliability function r t and the mean time of failure. So, expected t would be the mean time to failure. And so, I want to show you that. So, this is the theorem it says that. So, the t is the random variable and r is the corresponding reliability function. So, I want to show you that e t the expected mean time expected value of t is nothing but the integral 0 to infinity r t d t. So, in other words if you have this then you can integrate this function and also compute e t. There is nothing much great about it we use the same concept that we have. So, remember your definition of r t is the probability t greater than t. So, integral 0 to infinity r t d t will be integral 0 to infinity probability t greater than t d t. Now, this was also I think I gave you this as an exercise in one of the earlier exercises when we were defining expected value of a random variable. But let me now just spend time and show you why this will be true. So, t greater than t if you are integrating now this I can write as integral t to infinity of f s d s probably t greater than t this is what we wrote down just now. So, it is t to infinity f s d s and then 0 to infinity. Now, let us integrate by parts. So, I will treat you know this one as a first function and this is a second function. So, integral of the first function would be simply t and so we want to have this t into integral t infinity f s d s from 0 to infinity plus why will it be plus because when you differentiate this. So, integral of the first into the derivative of the first second and the integral of the whole. So, when you want to differentiate this you see the lower limit is a function of t. So, it will be as it is when you do the you know the first integral the integral of the first function into the second function minus it is a minus sign. But since this is the limit is a function of t the lower limit is a function of t. So, there will be another minus sign. So, and the derivative of this will be 1. So, then f t so plus. So, therefore the sign will become plus and this will be 0 to infinity t f t d t. So, if you apply integration by parts to this treating this as the first function and this whole thing as the second function then you can write this. Now, you can immediately see that the limit at the 0 point. So, this whole thing has to be integrated has to be computed between 0 and infinity. So, at 0 this is 0 and this will be 0 to infinity f s d s which is equal to 1 because f is the p d f and the variable t varies from 0 to infinity. So, this integral is equal to 1. So, this is 0 and now here this is little complicated, but we can see that the limit here s t goes to infinity. So, you see here s t goes to infinity this part becomes 0 and this is going to infinity, but it can be shown that the product here the limit of this product will go to 0. So, once that happens then that integral reduces to simply 0 to infinity r t d t which is I mean it reduces to simply this which is t f t d t from 0 to infinity and this is your p t and so you have this relationship. So, either way you can use it I mean if you know this then you can say that this integral is equal to this or if you know r then you can by integrating this you can compute the expected value of the random variable t. So, now it is we want to study various failure laws that means we want to find out the different p d f s which will be suitable for different situations where we want to study and of course we will keep it very simple here not going to complicated results, but just look at various distribution some of the distributions and we will show why they are appropriate for modeling certain situations for reliability. So, after defining your reliability function and your failure rate we are going to show how important tools they are to you know study these failure models that we will be discussing. So, these are the two basic tools that we need and continuously we will be making use of them. Now, the questions that arise the first question of course then we will ask what underlying failure laws are reasonable to assume that is what should be the appropriate form of the p d f of t you want to know and of course people have you know computed data I mean collected data and then try to fit these various probability laws which are normal exponential and so on. So, that is what we want to say that some of the tested p d f s we are going to look at and tested by tested we mean that you know you try to you have the data you know system is going on and then you compile the failure times and so on of the components and then you try to fit curve and these the ones that we discussed now have been very well tested and you know found suitable for the some particular data that we are going to talk about right. So, when there is a varying effect on the component just as we said that you know beam under heavy load. So, slowly gradually there is a varying effect and then it breaks down. So, for such situations where the varying effect is the one which is the cause of the failure then normal failure law is considered to be appropriate. Consider in the sense that again it has been tested and by fitting the data by fitting you know having the particular kind of data which is you know which is for components failing under the varying effect and then you know finding out that yes normal failure law or the p d f normal p d f seems to give the quite accurate results. So, now therefore, we want to and there are many situations where the varying effect is the prominent reason for the failure of the or the breakdown of the component or the system. Now, when you look at the normal law. So, because the normal law is such that you know if mean is the mean expected value. So, mu is yeah. So, I am looking at the normal law where which is mu sigma square. So, mean is mu expected value is mu and variance is sigma square then this as the bell shape the normal. So, we have already studied I mean in this course I do not have to spend time on describing the normal curve to you. And we know that if you take the area between mu minus 2 sigma and mu plus 2 sigma then area under this these 2 is 0.9572. And if you go up to 3 sigma mu minus 3 sigma mu plus 3 sigma then of course, very little areas left out it is almost I think 0.99 something the area within these 2 limits. And you also see that for t equal to mu this point is the mean as well as the mode that means the maximum failure will occur at around the time t equal to mu. But, since our variable t is takes only non negative values therefore, we will consider the normal distribution that is only the portion from 0 to infinity and not from minus infinity to infinity. So, this normal failure law implies that most of the failures occur around t equal to mu which is the expected value and number of failures decrease as t minus mu decreases in absolute value on either side. So, the number of failures will decrease and the probability goes down and so for example, if you take normal failures law means that 95.72 percent failures take place for t satisfying this mod t minus mu less than mu. So, that means, if your t is in this area then so that will be from 0 to infinity only this portion. So, we do not have to worry about the yeah. So, whatever the mean the expected value the maximum failures will occur around this time. Now, if you look at r t the reliability function then the reliability function yeah. So, r t is equal to 1 minus probability capital T less than or equal to small t right because this is r t is equal to probability t greater than t which can be written as this. And now here you can write this probability for in terms of the standardized normal variate which we have been doing all along in this course. So, this would be equal to 1 minus probability t minus mu divided by the standard deviation. So, this is less than or equal to t minus mu by sigma right. And so this becomes 1 minus phi t minus mu by sigma which is equal to 0.9. Well that I was just considering the value. So, this is the function functional form for your reliability function. And if you plot it here on the t axis then it will be something like this. So, at 0 t equal to 0 your probability t less than or equal to 0 is 0 right because t has takes only non-negative values. So, this probability is 0 at small t equal to 0. So, for this is equal to 1. So, at 0 your value of r t is 1 and when you take r mu then this will be t less than or equal to mu. Now, since t is normally distributed we know that area this area that means this whole area is equal to 0.5 right half the area is on this side and half the area it is a symmetric curve. So, therefore this will be 1 minus 0.5 which is 0.5. So, therefore for mu equal to so this is at mu t equal to mu the value of r t is equal to 0.5. So, this is the kind of curve and then it goes to as t goes to infinity this goes to 0. So, the reliability function decreases with time. Now, so I was saying that see what you can say from here is this is your reliability function. If you want a high value for the reliability function then obviously. So, that means you want this whole thing to be high which that is what I was saying here. So, suppose this is equal to 0.9 and this implies that this must be small and if for this to be small you can again tell by the graph that t must be away from mu this is the whole idea right. Because as we said that maximum failures will occur around t equal to mu and as you get away from mu the values become smaller. So, if you want this whole thing if you want high reliability then your value of t must be removed from mu right. So, for example, if this is 0.9 then this implies that your phi of t minus mu by sigma should be 0.1. This you bring here then 1 minus 0.9 would be 0.1 and so that will make it make t minus mu by sigma equal to minus. So, you look up the tables right again I do not have to spend time on this the standard norm because this is now standardized normal variate. So, you look up the numbers the table among the tables corresponding to 0.1 area you look up. So, it will be somewhere here for the standard normal thing. So, it will be somewhere here point because you want only area 0.1 up from up to this point. So, it will be very small right. So, this is minus 1.2 and so t comes out to be minus 1.2 times sigma plus mu. So, that much removed from mu your value of t is if you want reliability of the order of 0.9. So, take another example now here suppose the failure rate the time life length life time of any component is normally distributed with mean mu and variance 100. So, the standard deviation is 10. Suppose this is this and you are told that the reliability for the 100 hours is 0.99. So, r of 100 is 0.99 you have to find the value of mu that is the expected value of t you have to find. So, since we know the functional form for r t that means again 0.99 is 1 minus phi of. So, we standardized the normal this variate t which is 100 minus mu upon 10 standard deviation right. So, this is phi 100 minus mu by 10. So, that gives you that phi of 100 minus mu by 10 is 0.01. So, again we look up the standard tables corresponding to the area 0.01 this value the value of z is equal to minus 2.33. So, the tables so the see the smaller this becomes the further away you go from the mean value that is what we are saying. So, this implies that mu is see here you will multiply this with this and then. So, it will become 23.3 and mu comes to this side this gets added to 100. So, this will be then 123.3 hours. So, the mu that means the mean is here and so you see this is removed from the mean this is 123.3 hours. So, at 100 hours if you are asking for reliability 0.99 then your mu is this. So, this is the whole idea that you if you have if you if your data or your experience with the system that you are working with is that the normal distribution is appropriate for studying the failure rates then. So, here of course you should also look at what your z t will be. So, if you want to compute which may not be a very. So, for example, your z t we said is f t upon r t. So, which will be 1 upon root 2 pi sigma or let me just so this will be equal to your f t is 1 upon root 2 pi sigma e raise to minus 1 by 2 sigma square x minus mu whole square divided by 1 minus phi of t minus mu by sigma. So, something like this this is your failure rate. So, if you want to compute z you will get this complicated expression, but it seems that here your reliability function is the one which is more useful and has a simple form. Because all you have to do is to for a fixed value for a given value of t mu and sigma if given then you just have to look up the normal tables and compute the reliability of the system at any given time t. Failure law I said to be applicable we want this because obviously, the time to failure cannot be less than 0 with only when this apparatus or the instrument starts functioning then you talk about its failure time and so on. So, time to failure. So, therefore, since this is less than or equal to 0 and normally our the normal distribution extends from minus infinity to infinity, but what we are saying here is that the most of the curve should lie to the right of 0. So, that means here yes the curve should be like this. So, very small area is to the left of 0. So, most of it lies to the right of 0 then the computations would be fine. So, this is very important. So, that means this probability t less than or equal to 0 should be essentially 0 negligible very small that is what we want to say. Now, another way to handle the situation could be that I consider the truncated normal distribution. That means you truncate when you may have your normal distribution this and then you truncated this portion and so that means then you would consider, but then since it has to be a pdf. So, then you will have to so that is what I am trying to say that you know truncated normal distribution. So, if you consider the truncated this normal distribution truncated to the left of t 0 then the pdf would be this, but that is not depicting the actual situation because by this what we are doing is since I want to call this a pdf. So, this integral and of course, this is 0 for x less than 0. So, this is from 0 to infinity. So, what we are saying is that this will be from 0 to infinity this will integrate to 1, but this is not what I want because I want the normal failure law, but it should be such that the most of the curve lies to the right of 0 of t equal to 0. So, that is the meaning and therefore, this of course, this will complicate your computations also, but besides that it will not give you the desired results. So, therefore, the truncated normal distribution is not to be considered. It is simply that keeping this in mind we have to make sure that you know most of the curve lies to the right of 0. So, therefore, when we compute the probabilities they would be approximately alright for and that as I said that the normal failure law is for the aging where the aging is prominent. . So, now we will study we look at the other failure laws which have again been tested for certain different situations and have proved to show good results. So, this will be exponential distribution and available distribution and some others. So, let us look at the exponential failure law. Now, of course, obvious way to define the exponential failure law would be by defining the pdf. So, we say that the pdf of t is alpha e raise to minus alpha t where t is positive takes positive values and then alpha is also some positive constant and then we can obtain r t and z t, but that does not really have that dramatic effect as when you say that the law we say that the failure rate that is z t is a constant is equal to alpha. So, here I had just said that alpha is some constant, but actually when your law is exponential law failure law is exponential then negative exponential then your failure rate or your failure rate would be a constant is a positive constant is equal to alpha actually. So, alpha is your failure rate. So, this is constant and immediate consequence of this is because remember we said that we can given z we can uniquely determine the pdf and also given the pdf we can uniquely determine the function z the failure rate. So, here you see the definition was that f t would be z t e raise to minus integral 0 to t of z s f s. So, here z is a constant. So, therefore, this is alpha into d s integral this leads to t. So, t alpha and here z t is alpha again. So, this is alpha e raise to minus alpha t for all t non negative. So, therefore, this is your so immediately you compute the pdf of t and the converse is also immediate because if you are given that f t is alpha e raise to minus alpha t then this we have already done this computation, but anyway let us just go through it again. So, z t is f t upon 1 minus f t which is r t and so f t is alpha e raise to minus alpha t and you know that 1 minus f t is e raise to minus alpha t. This is probability greater than capital t greater than small t and so it will be e raise to minus alpha t and so it comes down to alpha. So, this is the constant. So, your z t the failure rate is constant. So, immediately we can write down this theorem that and so if I remember correctly we had not done this part that means given the failure rate when I had talked about the exponential when we would we had just introduced the exponential distribution. I had defined the hazard function or the failure rate, but I did not at that time discuss this part that given z t you can obtain your p d f uniquely. So, now we can immediately write down this theorem because I have shown you both ways that is and so for constant failure rate the only p d f that can be there is your negative exponential and if the p d f is negative exponential p d f of the life time random variable t then it has to be the failure rate has to be a constant. So, immediately we have the theorem that let t the time to failure be a continuous random variable assuming all non-negative values then t has an exponential distribution if and only if it has constant failure rate. So, this is now neat way to present the whole thing that is they can be no other p d f which satisfies the condition that the corresponding failure rate is a constant. So, if a constant failure rate would always be exponential p d f and constant failure rate implies that it is time independent. So, no matter that is the failure does not change with time the failure rate does not change with time. So, no matter therefore, this will be appropriate for situations where there is no bearing effect. So, that means no matter how long the component has been working it does not matter it has no bearing on the failure of the component it will only be some external. So, maybe I will come out with in the next lecture I will try to show you some more you know ways of describing the situations where these different failure laws can be appropriate. So, essentially now try to think of these situations where the failure is not because of the bearing effect at least not because of some kind of stress or load, but it is. So, therefore, as we try to say that if you take a fuse then fuse can be working fine for a long time and then suddenly it fails. So, it is not because may be because of high current or something high current has come and then the fuse burns out otherwise it may continue for a long time. So, therefore, such situations would be very appropriately modeled by the exponential failure law and one can you know now that you read about this you can yourself you know try to think of components or systems where the failures are not because of the bearing effect or this kind of some kind of load or stress, but it is a different kind of failure. And therefore, this can be modeled by the exponential law and you can see that how effectively it will give you the parameters that you require for you know predicting things about the model.