1. ## Reliability Question

I have a circuit diagram of 5 switches.
Reliability of switch A and B are 99.5%
while C,D and E are 98%.

What is the reliability of the circuit working?

I understand how to do a straight circuit multiplying the percents such as A and B being .995 * .995= .99025 or 99.025% but seeing how this circuit spreads out I am having a hard time finding the equations I must do and put together in order to find the percent if reliability. I do understand with all these switches the percent of reliability must be extremely high and close to 100%.
Anything will help at this point. Thanks!

2. ## Re: Reliability Question

Originally Posted by Launcher
I have a circuit diagram of 5 switches.
Reliability of switch A and B are 99.5%
while C,D and E are 98%.

What is the reliability of the circuit working?

I understand how to do a straight circuit multiplying the percents such as A and B being .995 * .995= .99025 or 99.025% but seeing how this circuit spreads out I am having a hard time finding the equations I must do and put together in order to find the percent if reliability. I do understand with all these switches the percent of reliability must be extremely high and close to 100%.
Anything will help at this point. Thanks!
The probability of a parallel circuit working is the same as 1 - P(all branches not working). Hence, the probability of E/D working is: 1-(1-0.98)(1-0.98) = 0.9996. Now C/D/E is parallel to A/B, and since you know how to calculate the probability components in series, apply the parallel reasoning as above to compute the total prob (I get 99%).

3. ## Re: Reliability Question

Thanks for that.
With what you gave me I did this for all 3 combinations and multiplied them together giving me
.9995670
or 99.9% does that sound correct?

4. ## Re: Reliability Question

Originally Posted by Launcher
Thanks for that.
With what you gave me I did this for all 3 combinations and multiplied them together giving me
.9995670
or 99.9% does that sound correct?
I punched the numbers into excel and used the wrong cell in the final calculation! Here's the corrected version. I get 99.98%:

Here's what I got:

P(A & B) = (0.995)(0.995) = 0.990025
P(D||E)=1-(1-0.98)(1-0.98) = 0.9996
P(C&D||E)= 0.98(0.9996) = 0.979608
P(A & B || C & D|| E) = 1-(1-0.990025)(1-0.979608)=0.9998