Solving convolution? Int[0,t] [f(t-x)*F(x)] dx while f(t)=0.01 And F(t)=0.01*t

[Elirazn]

Hello everybody
I tried but still can't solve this integral:
Integral from 0 to t of the expression f(t-x)*F(x)dx
While f(t)=0.01
And F(t)=0.01*t
The result should represent a system reliability (must be smaller than 1). If you can explain to me how you actually solve it, it will be awesome.
Thanks

 

[Dr.Peterson]

I tried but still can't solve this integral:
Integral from 0 to t of the expression f(t-x)*F(x)dx
While f(t)=0.01
And F(t)=0.01*t
The result should represent a system reliability (must be smaller than 1). If you can explain to me how you actually solve it, it will be awesome.

Can you show what you tried, so we can see where you are stuck?

Possibly you are just confused because the functions are so simple. Function f is constant, so f(t-x) is just 0.01. And F(t) = 0.01t, so F(x) = 0.01x. Therefore, the integral is INT{0 to t} 0.01*0.01x dx. That is not hard to integrate.

Again, if you have trouble, show where; and if I have misinterpreted the problem, make sure we understand it.

 

[Elirazn]

Can you show what you tried, so we can see where you are stuck?

Possibly you are just confused because the functions are so simple. Function f is constant, so f(t-x) is just 0.01. And F(t) = 0.01t, so F(x) = 0.01x. Therefore, the integral is INT{0 to t} 0.01*0.01x dx. That is not hard to integrate.

Again, if you have trouble, show where; and if I have misinterpreted the problem, make sure we understand it.

First of all thank you for replying.
I will explain how I got to this expressions.

The aim is finding a system reliability, while we have 2 parallel components (one operative and the other one is a cold backup, so it starts operating only when the first is broken).
We have a formula for finding a reliability in such a case.
Rs(t)=1- integral from 0 to t of F(x)*(t-x) dx
Rs(t) is the system reliability (possible from 0 to 1)
f(t) is the derivative of F(t)
F(t) is the possibility that the component fails before reaching time=t

It was given that a component has a lifespan of 0 to 100 hours with a randomal distribution. Hence f(t) is a probability density function and equals to 0.01.
F(t) is a probability and is the integral of f(t) and equals to 0.01*t.
The expected final result,Rs(t), should get the values from 1 to 0, while it equals to 1 when t=0 and to 0 when t=200.

 

[Elirazn]

Description of the exact problem

First of all thank you for replying me, I appreciate it.
I am confused since it indeed seems to be a simple problem, but I probably miss something

The exact description of the problem:
I am trying to calculate the time dependancy reliability of a system, that comprises of 2 parallel components.
One of the component is the main one that usually operates, while the other one is a backup that starts working only when the first one is broken, and hence the system is categorized as a "2 parallel cold redundancy type".

The formula for finding the system reliability in such a case is:
Rs(t)=1- integral from 0 to t of [F(x)*f(t-x)dx

Rs(t) is the system time dependent reliability
f(t) is the derivative of F(t).
F(t) is a probability that a component fails before reching to time=t.
if we sum it up, f(t) is a PDF (probability density function) and F(t) is the CDF (cumulative distribution function).


I was given that the component has a life that ranges from 0 to 100 hours.
So f(t)=0.01 and F(t)=0.01*t.

I am adding a MATLAB code that represents the resulf we should get and a photo of the expected graph.

 

[Dr.Peterson]

First of all thank you for replying me, I appreciate it.
I am confused since it indeed seems to be a simple problem, but I probably miss something

The exact description of the problem:
I am trying to calculate the time dependancy reliability of a system, that comprises of 2 parallel components.
One of the component is the main one that usually operates, while the other one is a backup that starts working only when the first one is broken, and hence the system is categorized as a "2 parallel cold redundancy type".

The formula for finding the system reliability in such a case is:
Rs(t)=1- integral from 0 to t of [F(x)*f(t-x)dx

Rs(t) is the system time dependent reliability
f(t) is the derivative of F(t).
F(t) is a probability that a component fails before reching to time=t.
if we sum it up, f(t) is a PDF (probability density function) and F(t) is the CDF (cumulative distribution function).


I was given that the component has a life that ranges from 0 to 100 hours.
So f(t)=0.01 and F(t)=0.01*t.

I am adding a MATLAB code that represents the resulf we should get and a photo of the expected graph.

Thanks for the details, but I still can't say what you are missing, because you have shown no work.

I see what I was missing in my previous attempt to solve it, which did not yield a graph like what you show: You haven't explicitly stated that f(t) = {0.01 for 0<=t<=100, 0 otherwise}; and similarly for F(t) -- that is, they have a restricted domain, and in the integral will have to be treated as piecewise-defined functions, since t is allowed to go up to 200, not just to 100. With this modification of my work, I get exactly what you show, which itself will be defined piecewise.

But you may have made a mistake before reaching the point where this matters. Please show me what you have tried.

 

[Elirazn]

Thanks for the details, but I still can't say what you are missing, because you have shown no work.

I see what I was missing in my previous attempt to solve it, which did not yield a graph like what you show: You haven't explicitly stated that f(t) = {0.01 for 0<=t<=100, 0 otherwise}; and similarly for F(t) -- that is, they have a restricted domain, and in the integral will have to be treated as piecewise-defined functions, since t is allowed to go up to 200, not just to 100. With this modification of my work, I get exactly what you show, which itself will be defined piecewise.

But you may have made a mistake before reaching the point where this matters. Please show me what you have tried.

Hey again...
After thinking about it over and over again I also concluded that the problem is probably the fact that f(argument) is defined as long as its argument is smaller than 100. So I wanted to split the integral to several ranges, so the use of the function will be realistic.
I am reminding what function I have:
Rs(t)=1- integral from 0 to t of [F(x)*f(t-x)dx
The time the first main component fails is defined as (t-x) and the operating time of the backup component is t, so it will actually work from (t-x) to t, and hence we substitute X in its argument (it only works part of the system lifespan).
But now comes the major problem -
I know that the argument of f is defined differently when reaching to 100. in other words as long as (t-x)<100 the function equals to 0.01 and when (t-x)>=100 the function (which is actually a probability distribution function) becomes 0.
But please pay attention that the range of the integral is defined in terms of t (and not t-x).
So I am really confused and have no idea how to manipulate the integral so I will have the same variable in the boundaries/range (instead of t) and in the argument of f. I thought of variable exchange but got no result.
please, if you have an analytical solution - send me a detailed way, so I will be able to understand exactly ehat to do.
Thanks!

 

[Dr.Peterson]

Hey again...
After thinking about it over and over again I also concluded that the problem is probably the fact that f(argument) is defined as long as its argument is smaller than 100. So I wanted to split the integral to several ranges, so the use of the function will be realistic.
I am reminding what function I have:
Rs(t)=1- integral from 0 to t of [F(x)*f(t-x)dx
The time the first main component fails is defined as (t-x) and the operating time of the backup component is t, so it will actually work from (t-x) to t, and hence we substitute X in its argument (it only works part of the system lifespan).
But now comes the major problem -
I know that the argument of f is defined differently when reaching to 100. in other words as long as (t-x)<100 the function equals to 0.01 and when (t-x)>=100 the function (which is actually a probability distribution function) becomes 0.
But please pay attention that the range of the integral is defined in terms of t (and not t-x).
So I am really confused and have no idea how to manipulate the integral so I will have the same variable in the boundaries/range (instead of t) and in the argument of f. I thought of variable exchange but got no result.
please, if you have an analytical solution - send me a detailed way, so I will be able to understand exactly ehat to do.
Thanks!

I'm not sure I follow your thinking. You are right that you need to split the integral; but I don't think it's necessary to analyze the relationship of the integral to the application. I suspect you are confusing the roles of t and x. As far as we are concerned in the integral, t is a constant; only x is varying. You don't want to change the variable in the limits of the integral; x can't be there, because after integration, there is no x.

What I did was this: First, if t < 100, we just do the integral, and get a quadratic in t.

Now, if t > 100, then F changes behavior at 100, so we need a separate integral from 100 to t. Similarly, f(t-x) changes behavior when t-x>100, i.e. when x<t-100, so we need a total of three intervals over which we integrate: 0 to t-100, t-100 to 100, and 100 to t. Determine what the integrand is in each interval, and go to it.

 

[Elirazn]

still no right result :-/

I'm not sure I follow your thinking. You are right that you need to split the integral; but I don't think it's necessary to analyze the relationship of the integral to the application. I suspect you are confusing the roles of t and x. As far as we are concerned in the integral, t is a constant; only x is varying. You don't want to change the variable in the limits of the integral; x can't be there, because after integration, there is no x.

What I did was this: First, if t < 100, we just do the integral, and get a quadratic in t.

Now, if t > 100, then F changes behavior at 100, so we need a separate integral from 100 to t. Similarly, f(t-x) changes behavior when t-x>100, i.e. when x<t-100, so we need a total of three intervals over which we integrate: 0 to t-100, t-100 to 100, and 100 to t. Determine what the integrand is in each interval, and go to it.

I am attaching a photo of my solution, which is still isn't right.
Please help me to get to the rigtht analytical solution.

 

[Dr.Peterson]

I am attaching a photo of my solution, which is still isn't right.
Please help me to get to the rigtht analytical solution.

First, I hope you are aware that what you show here is only for the case when t>100, as I stated before. For 0<t<100, you just have a single integral, and the result matches the first half of the graph you showed.

Now, in the case where 100<t<200, your integrals are almost correct, but in the middle one, you used F(t) instead of F(x); your t should be x. Somehow, though, your integral turns out to be correct!

You got 0.00005 t^2 - 0.02t + 2, which is (completing the square) 0.00005(t^2 - 400t + 40000) = 0.00005(t - 200)^2, which is the second half of the graph you showed. It reaches 0 at t=200, and is 0.5 at t=100.

 

[Elirazn]

Thanks, finally got it!!!

First, I hope you are aware that what you show here is only for the case when t>100, as I stated before. For 0<t<100, you just have a single integral, and the result matches the first half of the graph you showed.

Now, in the case where 100<t<200, your integrals are almost correct, but in the middle one, you used F(t) instead of F(x); your t should be x. Somehow, though, your integral turns out to be correct!

You got 0.00005 t^2 - 0.02t + 2, which is (completing the square) 0.00005(t^2 - 400t + 40000) = 0.00005(t - 200)^2, which is the second half of the graph you showed. It reaches 0 at t=200, and is 0.5 at t=100.

thank you very much!
after your detailed and patient repeated explanation I got it.
you were right that the main problem was that I got confused between x and t.
I am attaching the graph I finally got - combining numeric and exact solution.

thanks again!

 

[HallsofIvy]

Hello everybody
I tried but still can't solve this integral:
Integral from 0 to t of the expression f(t-x)*F(x)dx
While f(t)=0.01
And F(t)=0.01*t
The result should represent a system reliability (must be smaller than 1). If you can explain to me how you actually solve it, it will be awesome.
Thanks

If f(t)= 0.01, a constant, then f(t- x) is that same constant, 0.01. So this integral is just \(\displaystyle \int_0^t (0.01)(0.01x) dx= 0.0001\int_0^t xdx\).

 

[Dr.Peterson]

If f(t)= 0.01, a constant, then f(t- x) is that same constant, 0.01. So this integral is just \(\displaystyle \int_0^t (0.01)(0.01x) dx= 0.0001\int_0^t xdx\).

Yes, this is what he eventually worked out for the case 0<=t<=100, and also for the middle of three integrals (with different limits) in the harder case. The x, not the constant, apparently was the difficulty.