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.

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 *)

1

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 *)

1

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

9

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.