# Homework Help: Reliability of a system (with simple "complex" configuration)

1. Sep 9, 2014

### kalizazzz623

1. The problem statement, all variables and given/known data

Question: A system consists of 7 equal reliable components having the following configuration with each component has a reliability of 0.9. Find the reliability of the system.

2. Relevant equations
I was taught (and only been taught) with the calculations to obtain reliability of a system with its components arranged in either in series or in parallel. However, I have real trouble in solving these kinds of complex configuration.
For series: Rs = R1R2...Rn
For parallel: Rs = 1-(1-R1)(1-R2)...(1-Rn)

3. The attempt at a solution
I have search through the internet and realized that this can be solved with the solution used for bridge structure. I tried to mimic the way the solution from the resource (Reliability Engineering: Theory and Practice, by Alessandro Birolini) found (but with only 5 component instead of 7) and come to this:

The first picture showing the reliability of the system with the center component not fail, and the second picture is when the center component has fail. (component number order: 1 to 6, from top left, clockwise; center = 7)
Then, Rs=R7(R1+R6-R1R6)(R2+R5-R2R5)(R3+R4-R3R4) + (1-R7)(R1R2R3+R4R5R6-R1R2R3R4R5R6)
This solution doesn't lead to the answer (0.9575). I tried using Binomial method by considering all sort of possibility but still no help (duh, obviously), so can anyone enlighten me?

2. Sep 9, 2014

### Ray Vickson

Label the components as
$$\begin{array}{ccc} A & B & C \\ &G& \\ D & E & F \end{array}$$
Imagine we want to send a signal from left to right; the component reliability is the probability that a signal gets through that component. So, to send a signal from left to right, at least one path from left to right must be "open". There are four paths: P_1 =in-ABC-out, P_2 = in-DEF_out, P_3 =in_ABGEF-out and P_4 = in-DEGBC-out. If we let $E_i$ be the event that path $P_i$ is open, can you figure out $\Pr(E_i)$ for each $i$? For $i \neq j$, can you see how to get $\Pr(E_i \cap E_j)$, etc? You need to find
$$\text{system reliability} = \Pr \{ \text{at least one of the events }E_1, E_2, E_3, E_4 \text{ occur} \}$$

Last edited: Sep 9, 2014
3. Sep 10, 2014

### kalizazzz623

P(E1)=0.9x0.9x0.9 =0.729
P(E2)=0.9x0.9x0.9 =0.729
P(E3)=0.9x0.9x0.9x0.9x0.9 =0.59049
P(E4)=0.9x0.9x0.9x0.9x0.9 =0.59049

For $\Pr(E_i \cap E_j)$, thus this means I have to do for $\Pr(E_1 \cap E_2)$, $\Pr(E_1 \cap E_3)$, $\Pr(E_1 \cap E_4)$ and the remaining? And by the equation $\Pr(E_i \cap E_j)$, should I use $\Pr(E_i \cup E_j)$x$\Pr(E_j)$?

And for the final system reliability, is $Pr \{ \text{at least one of the events }E_1, E_2, E_3, E_4 \text{ occur} \} = \Pr(E_1 \cup E_2 \cup E_3 \cup E_4)$?

4. Sep 10, 2014

### Ray Vickson

I strongly recommend that you use a symbol $r$ instead of an explicit 0.9, at least until the very last. This is because by using a symbolic constant you can keep separate and keep straight the different effects. So $\Pr(E_1) = \Pr(E_2) = r^3$, $\Pr(E_3) = \Pr(E_4) = r^5$, etc.

The general inclusion-exclusion says that
$$\text{answer} = R = \Pr \{ E_1 \cup E_2 \cup E_3 \cup E_4 \} = S_1 - S_2 + S_3 - S_4, \\ \text{where}\\ S_1 = \sum_i \Pr(E_i)\\ S_2 = \sum_{i < j} \Pr(E_i E_j) \\ S_3 = \sum_{i < j < k} \Pr(E_i E_j E_k)\\ S_4 = P(E_1 E_2 E_3 E_4)$$
Here I have used the simpler notation $AB$ or $ABC$ instead of $A \cap B$ or $A \cap B \cap C$, etc.

For $E_1 E_2$ all components except G must work; for $E_1 E_3$ all components except D must work. Look at all the other $E_i E_j, i < j,$ in the same way, and do the same for $E_i E_j E_k, i < j < k$ as well as for $E_1 E_2 E_3 E_4$. In this way you can obtain $S_1, S_2, S_3, S_4$ in terms of $r$. And yes, it is a bit messy and lengthy, but welcome to the world of reliability computation.

5. Sep 10, 2014

### Ray Vickson

I strongly recommend that you use a symbol $r$ instead of an explicit 0.9, at least until the very last. This is because by using a symbolic constant you can keep separate and keep straight the different effects. (Besides which, this allows you to vary $r$ and see how the answer changes, because rarely would you know an exact value of $r$ accurate to infinitely many decimal places.)

So $\Pr(E_1) = \Pr(E_2) = r^3$, $\Pr(E_3) = \Pr(E_4) = r^5$, etc.

The general inclusion-exclusion says that
$$\text{answer} = R = \Pr \{ E_1 \cup E_2 \cup E_3 \cup E_4 \} = S_1 - S_2 + S_3 - S_4, \\ \text{where}\\ S_1 = \sum_i \Pr(E_i)\\ S_2 = \sum_{i < j} \Pr(E_i E_j) \\ S_3 = \sum_{i < j < k} \Pr(E_i E_j E_k)\\ S_4 = P(E_1 E_2 E_3 E_4)$$
Here I have used the simpler notation $AB$ or $ABC$ instead of $A \cap B$ or $A \cap B \cap C$, etc.

For $E_1 E_2$ all components except G must work; for $E_1 E_3$ all components except D must work. Look at all the other $E_i E_j, i < j,$ in the same way, and do the same for $E_i E_j E_k, i < j < k$ as well as for $E_1 E_2 E_3 E_4$. In this way you can obtain $S_1, S_2, S_3, S_4$ in terms of $r$. And yes, it is a bit messy and lengthy, but welcome to the world of reliability computation.

6. Sep 10, 2014

### kalizazzz623

Sorry for being bad at elementary probability, but does $E_1 E_2$ equals to $r^6$? And for $E_1 E_2 E_3$, is it equals to $r^7$? Thanks in advance.

7. Sep 11, 2014

### Ray Vickson

You tell me.

8. Sep 11, 2014

### kalizazzz623

Well I'm not really sure if I'm doing it wrong or correctly, cause it doesn't yield the answer.
My solution:
$$S_1 = \sum_i \Pr(E_i) \ = 2r^3+2r^5\\ S_2 = \sum_{i < j} \Pr(E_i E_j) \ = 5r^6+r^7\\ S_3 = \sum_{i < j < k} \Pr(E_i E_j E_k)\ =4r^7\\ S_4 = P(E_1 E_2 E_3 E_4) =r^7\\$$
then
$$Pr \{ E_1 \cup E_2 \cup E_3 \cup E_4 \} = 2r^3+2r^5+2r^7-5r^6 = 0.9383 (≠0.9575)$$
It seems some how my approach is wrong, but I can't think of other ways.

9. Sep 11, 2014

### Ray Vickson

Your answer is correct (although I would round it off to 0.9384). The 0.9575 figure is wrong. In fact, using your (correct) expression for $R(r)$ you can solve the equation $R(r) = 0.9575$ numerically, to get $r = 0.91797978$.

Last edited: Sep 11, 2014