Reliability of a system (with simple "complex" configuration)

[kalizazzz623]

Homework Statement

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.

Homework 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: R_s = R₁R₂...R_n
For parallel: R_s = 1-(1-R₁)(1-R₂)...(1-R_n)

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, R_s=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?

 

[Ray Vickson]

Homework Statement

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.
View attachment 72912

Homework 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: R_s = R₁R₂...R_n
For parallel: R_s = 1-(1-R₁)(1-R₂)...(1-R_n)

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:
View attachment 72913
View attachment 72914
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, R_s=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?

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
$$\text{system reliability} = \Pr \{ \text{at least one of the events }E_1, E_2, E_3, E_4 \text{ occur} \}$$

 

[kalizazzz623]

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
$$\text{system reliability} = \Pr \{ \text{at least one of the events }E_1, E_2, E_3, E_4 \text{ occur} \}$$

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 $Pr \{ \text{at least one of the events }E_1, E_2, E_3, E_4 \text{ occur} \} = \Pr(E_1 \cup E_2 \cup E_3 \cup E_4)$?

 

[Ray Vickson]

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 $Pr \{ \text{at least one of the events }E_1, E_2, E_3, E_4 \text{ occur} \} = \Pr(E_1 \cup E_2 \cup E_3 \cup E_4)$?

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.

 

[Ray Vickson]

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 $Pr \{ \text{at least one of the events }E_1, E_2, E_3, E_4 \text{ occur} \} = \Pr(E_1 \cup E_2 \cup E_3 \cup E_4)$?

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.

 

[kalizazzz623]

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.

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.

 

[Ray Vickson]

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.

You tell me.

 

[kalizazzz623]

You tell me.

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.

 

[Ray Vickson]

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.

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 $r = 0.91797978$.