An automotive component has been designed to withstand certa

An automotive component has been designed to withstand certain stresses. It is known from past experience that because of variations in loading the stress on the component is normally distributed with a mean of 35000 kPa and a standard deviation of 3000 kPa. The strength of the component is also random because of variations in the material characteristics and the dimensional tolerances. It has been found that the strength is normally distributed with a mean of 40000 KPa and a standard deviation of 4000 kPa. Let x denote the stress on a given component and let y denote the strength of that component. The component does not fail if x is less than y.  The Reliability may be expressed as P[X<Y].

Use @Risk to estimate the reliability. One approach is to set up a cell to yield random values of X and another cell to yield random values of Y. Then compute X-Y in a third cell and make it an output cell.  From the simulated distribution of X-Y determine the probability that X-Y is less than 0.

A second approach is to compute a binary variable having the value 1 if X<Y and 0 otherwise.  The average value of the binary variable will be an estimate of the reliability.  Whichever method you choose use 1000 iterations. This problem may be solved analytically but for other distributions than the normal it may not be.   The simulation will be valid for any choices of the distributions of stress and strength. The theoretical value for the present case is 84.1%. If you do it correctly your answer should be close to this.

How do you set this up in @Risk? Trying to learn how to do this

Solution

By using R

t=runif(1000,0,1)
x=rnorm(1000,35000,9000000)
y=rnorm(1000,40000,16000000)
Z=x-y
n=Z<0
P=sum(Z>0)

The probability that X-Y is less than 0 is

P(Z>0)=P/1000

The theoretical value for the present case is 84.1% that is the value of Z is less than 0 is 841 numbers.