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Can Mathematica 9 convert a Boolean representation of an RBD to its symbolic algebraic form?

If I have a simple series Reliability Block Diagram made up of 2 components, c1 and c2

bexpr1=And[c1,c2]

and the corresponding symbolic representation for the system would be RS1 = R1*R2

The system I am interested in is more complex and contains 7 components in a 4 path architecture:

path1=And[c1,c2,c5,c7]

path2=And[c1,c3,c5,c7]

path3=And[c1,c3,c6,c7]

path4=And[c1,c4,c6,c7]

bexpr2=Or[path1,path2,path3,path4]

How can the algebraic form of the reliability function be obtained for bexpr2 using Mathematica?

The solution can be seen on page 5 of the PDF file at the following link

http://www.reliasoft.com/newsletter/2Q2000/index.htm

Thank you.

Daniel,

Thank you for your reply but I think it too is not correct and I' m not sure why. See below for a comparision of the two solution methods. Am I mis-interpreting something here ?

Also, I agree that Eq.(3) is undersimplified. Perhaps the author intentionally left it in that form to help show how this problem can be solved manually (pencil & paper).

equation3 = (r1*r4*r6*r7) + (r1*r3*r6*r7) - (r1*r3*r4*r6*r7) + (r1*r2*r3*r4*
r5*r6*r7) + (r1*r3*r4*r5*r6*r7) - (r1*r2*r3*r4*r5*r6*r7) - (r1*r2*r3*r5*
r6*r7) - (r1*r3*r5*r6*r7) + (r1*r2*r3*r5*r6*r7) - (r1*r2*r4*r5*r6*
r7) - (r1*r3*r4*r5*r6*r7) + (r1*r2*r3*r4*r5*r6*r7) + (r1*r2*r5*r7) + (r1*
r3*r5*r7) - (r1*r2*r3*r5*r7) (* Eq.(3) ReliabilityEdge volume 1, Issue 1 *)

r1 r2 r5 r7 + r1 r3 r5 r7 - r1 r2 r3 r5 r7 + r1 r3 r6 r7 + r1 r4 r6 r7 - r1 r3 r4 r6 r7 - r1 r3 r5 r6 r7 - r1 r2 r4 r5 r6 r7 + r1 r2 r3 r4 r5 r6 r7

Original Problem

path1 = And[c1, c2, c5, c7];
path2 = And[c1, c3, c5, c7];
path3 = And[c1, c3, c6, c7];
path4 = And[c1, c4, c6, c7];
bexpr = Or[path1, path2, path3, path4]

(c1 && c2 && c5 && c7) || (c1 && c3 && c5 && c7) || (c1 && c3 && c6 && c7) || (c1 && c4 && c6 && c7)

Ilian' s Solution (1)

dists = Table[{ToExpression["c" ~~ ToString[i]], 
   BernoulliDistribution[ToExpression["r" ~~ ToString[i]]]}, {i, 7}]

{{c1, BernoulliDistribution[r1]}, {c2, BernoulliDistribution[r2]}, {c3, BernoulliDistribution[r3]}, {c4, BernoulliDistribution[r4]}, {c5, BernoulliDistribution[r5]}, {c6, BernoulliDistribution[r6]}, {c7, BernoulliDistribution[r7]}}

survivalfunction1 = Expand[PDF[ReliabilityDistribution[bexpr, dists], 1]]

r1 r2 r5 r7 + r1 r3 r5 r7 - r1 r2 r3 r5 r7 + r1 r3 r6 r7 + r1 r4 r6 r7 - r1 r3 r4 r6 r7 - r1 r3 r5 r6 r7 - r1 r2 r4 r5 r6 r7 + r1 r2 r3 r4 r5 r6 r7

Simplify[survivalfunction1/
  equation3] (* This show that the results are equivalent *)

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Daniel' s Solution (2)

bexpr2 = BooleanConvert[bexpr, "ANF"]

(c1 && c2 && c5 && c7) [Xor] (c1 && c3 && c5 && c7) [Xor] (c1 && c3 && c6 && c7) [Xor] (c1 && c4 && c6 && c7) [Xor] (c1 && c2 && c3 && c5 && c7) [Xor] (c1 && c3 && c4 && c6 && c7) [Xor] (c1 && c3 && c5 && c6 && c7) [Xor] (c1 && c2 && c4 && c5 && c6 && c7) [Xor] (c1 && c2 && c3 && c4 && c5 && c6 && c7)

survivalfunction2temp = bexpr2 /. {And -> Times, Xor -> Plus}

c1 c2 c5 c7 + c1 c3 c5 c7 + c1 c2 c3 c5 c7 + c1 c3 c6 c7 + c1 c4 c6 c7 + c1 c3 c4 c6 c7 + c1 c3 c5 c6 c7 + c1 c2 c4 c5 c6 c7 + c1 c2 c3 c4 c5 c6 c7

survivalfunction2stringtemp = ToString[survivalfunction2temp]

"c1 c2 c5 c7 + c1 c3 c5 c7 + c1 c2 c3 c5 c7 + c1 c3 c6 c7 + c1 c4 c6 c7 + c1 \ c3 c4 c6 c7 + c1 c3 c5 c6 c7 + c1 c2 c4 c5 c6 c7 + c1 c2 c3 c4 c5 c6 c7"

survivalfunction2string = 
 StringReplace[survivalfunction2stringtemp, "c" -> "r"]

"r1 r2 r5 r7 + r1 r3 r5 r7 + r1 r2 r3 r5 r7 + r1 r3 r6 r7 + r1 r4 r6 r7 + r1 \ r3 r4 r6 r7 + r1 r3 r5 r6 r7 + r1 r2 r4 r5 r6 r7 + r1 r2 r3 r4 r5 r6 r7"

survivalfunction2 = ToExpression[survivalfunction2string]

r1 r2 r5 r7 + r1 r3 r5 r7 + r1 r2 r3 r5 r7 + r1 r3 r6 r7 + r1 r4 r6 r7 + r1 r3 r4 r6 r7 + r1 r3 r5 r6 r7 + r1 r2 r4 r5 r6 r7 + r1 r2 r3 r4 r5 r6 r7

Simplify[survivalfunction2/
  equation3](* This show that the results are not equivalent *)

(r4 r6 + r2 (1 + r3) r5 (1 + r4 r6) + r3 (r5 + r6 + r4 r6 + r5 r6))/( r4 r6 + r2 (-1 + r3) r5 (-1 + r4 r6) + r3 (r5 + r6 - r4 r6 - r5 r6))

Let' s do a quick numerical sanity check.

r1 = r2 = r3 = r4 = r5 = r6 = r7 = 1;

survivalfunction1  (* This is the correct numerical value *)

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survivalfunction2  (* This is numerically not possible, must be less than or 
equal to one *)

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Upon examination of survivalfunction2 I notice there are no negative signs in the algebra . . . not sure why. Also, I apologize here if everything is not formatted with all proper formatting, I'm still unfamiliar with much at StackExchange.

Clay,

Thank you for the reply. I believe there may be a problem with your code as it produces a result that differs from Eq.(3) in the referenced PDF file.

I tested the following code on the system architecture that I posted (associated with Eq.(3) ) and other simpler systems and it appears to accurately produce the algebra for the reliability function.

dists = Table[{ToExpression["c" ~~ ToString[i]], BernoulliDistribution[ToExpression["r" ~~ ToString[i]]]}, {i, 7}];

Expand[PDF[ReliabilityDistribution[bexpr2, dists], 1]]

The above code was produced by Ilian Gachevski in his posting at Wolfram Community. Thanks again for your inputs.

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  • 2
    $\begingroup$ It might be beneficial to point out cross-postings (which are fine, like in this case to WRI community) to avoid duplicated efforts and to help later visitors in search of a solution. $\endgroup$
    – Yves Klett
    Commented Aug 24, 2013 at 6:38
  • $\begingroup$ Possibly my use of Xor is not what is wanted for this problem. Maybe it should be a boolean counting function that gives 1 is exactly one clause is true, else 0. $\endgroup$
    – Daniel Lichtblau
    Commented Aug 28, 2013 at 14:22
  • 1
    $\begingroup$ Hopefully version 10 (which I understand is in the works) will have this transform as a more native feature. An integrated block diagram editor that can pictorially display the boolean logic would also be helpful for this kind of analysis. $\endgroup$
    – Steve
    Commented Aug 28, 2013 at 21:59
  • $\begingroup$ Malte, thank you for your solution. I think I'll stay with Illian's approach since it's being documented should provide stability into the future. $\endgroup$
    – Steve
    Commented Sep 6, 2013 at 14:04

3 Answers 3

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The most direct way, but undocumented and therefore subject to change in future versions, is to use a Library function:

In: Reliability`Library`StructureFunction[bexpr2]
Out: c1 c3 ((1 - c5) (-1 + c6) - (1 - c2 c5) (-1 + c4 c6)) c7 + c1 (1 + (1 - c2 c5) (-1 + c4 c6)) c7

Use Expand if you would like the simpler structure showing the cutsets in a clearer way:

In: %//Expand
Out: c1 c2 c5 c7 + c1 c3 c5 c7 - c1 c2 c3 c5 c7 + c1 c3 c6 c7 +
    c1 c4 c6 c7 - c1 c3 c4 c6 c7 - c1 c3 c5 c6 c7 - c1 c2 c4 c5 c6 c7 +
    c1 c2 c3 c4 c5 c6 c7

Note: I am one of the developers originally developing reliability functionality in Mathematica

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Possibly you want to convert to "algebraic normal form" and, from there, to a boolean algebra expression in terms of plus and times.

path1 = And[c1, c2, c5, c7];
path2 = And[c1, c3, c5, c7];
path3 = And[c1, c3, c6, c7];
path4 = And[c1, c4, c6, c7];
bexpr2 = BooleanConvert[Or[path1, path2, path3, path4], "ANF"]

(* Out[25]= (c1 && c2 && c5 && c7) \[Xor] (c1 && c3 && c5 && 
   c7) \[Xor] (c1 && c3 && c6 && c7) \[Xor] (c1 && c4 && c6 && 
   c7) \[Xor] (c1 && c2 && c3 && c5 && c7) \[Xor] (c1 && c3 && c4 && 
   c6 && c7) \[Xor] (c1 && c3 && c5 && c6 && c7) \[Xor] (c1 && c2 && 
   c4 && c5 && c6 && c7) \[Xor] (c1 && c2 && c3 && c4 && c5 && c6 && 
   c7) *)

bexpr3 = bexpr2 /. {And -> Times, Xor -> Plus}

(* Out[26]= c1 c2 c5 c7 + c1 c3 c5 c7 + c1 c2 c3 c5 c7 + c1 c3 c6 c7 + 
 c1 c4 c6 c7 + c1 c3 c4 c6 c7 + c1 c3 c5 c6 c7 + c1 c2 c4 c5 c6 c7 + 
 c1 c2 c3 c4 c5 c6 c7 *)

By the way, that equation 3 in the PDF is probably undersimplified. You will notice the same terms appearing more than once.

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I have not found an inherent translation. For these problems I create the boolean functions:

rOr[list__] := 1 - (Times @@ (1 - # & /@ {list}))
rAnd[list__] := Times[list]

Once you enter the logic you want (using your example)

path1 = And[c1, c2, c5, c7];
path2 = And[c1, c3, c5, c7];
path3 = And[c1, c3, c6, c7];
path4 = And[c1, c4, c6, c7];
bexpr2 = Or[path1, path2, path3, path4]

Which will give the logical expression of the cutsets:

 Out[35]= (c1 && c2 && c5 && c7) || (c1 && c3 && c5 && c7) || (c1 && c3 && c6 && c7) || (c1 && c4 && c6 && c7)

At this point substitute the functions And and Or for the boolean functions. You can expand the expression and define values for the variables at this point.

In[36]:= bexpr2 /. {And -> rAnd, Or -> rOr}
Out[36]= 1 - (1 - c1 c2 c5 c7) (1 - c1 c3 c5 c7) (1 - 
c1 c3 c6 c7) (1 - c1 c4 c6 c7)

In[38]:= bexpr2 /. {And -> rAnd, Or -> rOr} // Expand
Out[38]= c1 c2 c5 c7 + c1 c3 c5 c7 + c1 c3 c6 c7 + c1 c4 c6 c7 -  
c1^2 c2 c3 c5^2 c7^2 - c1^2 c2 c3 c5 c6 c7^2 - c1^2 c3^2 c5 c6 c7^2 -
c1^2 c2 c4 c5 c6 c7^2 - c1^2 c3 c4 c5 c6 c7^2 - 
c1^2 c3 c4 c6^2 c7^2 + c1^3 c2 c3^2 c5^2 c6 c7^3 + 
c1^3 c2 c3 c4 c5^2 c6 c7^3 + c1^3 c2 c3 c4 c5 c6^2 c7^3 + 
c1^3 c3^2 c4 c5 c6^2 c7^3 - c1^4 c2 c3^2 c4 c5^2 c6^2 c7^4

Hope this helps.

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