2. RESULTS

Suppose that two components in a system are structurally permutation equivalent (i.e.,

. It is easy to see that they possess equal Birnbaum structural importance measures; however, their Birnbaum reliability importance measures I_B(i;p) and I_B(j;p) may not be the same when p_i ≠ p_j. A k-out-of-n system is the only system in which all components are symmetric. Let h_k(p) denote the reliability function of a k-out-of-n system. Boland and Proschan [3] obtained that if p_i ≤ (k − 1)/(n − 1) for each i, then the reliability function h_k(p) is Schur-concave and, hence, I_B(i;p) ≥ I_B(j;p) [hArr ] p_i ≤ p_j; however, if p_i ≥ (k − 1)/(n − 1) for each i, then h_k(p) is Schur-convex and, hence, I_B(i;p) ≥ I_B(j;p) [hArr ] p_i ≥ p_j (1 ≤ i, j ≤ n) (see [3] for the definitions of Schur-concavity and Schur-convexity).

For the case that

, the following result is presented in [10], which provides a criterion to compare I_B(i;p) and I_B(j;p) for a general system, not necessarily a k-out-of-n system. Note that in the following theorem, the strict inequality “>” in Theorem 2 is replaced by “≥.”

Theorem 3 (Meng [10]): Suppose that

and ∂²h(p)/∂p_i∂p_j ≥ 0 (≤ 0) for all p. Then, I_B(i;p) ≥ I_B(j;p) for all p satisfying p_i ≤ (≥) p_j.

A real-valued function f defined on Rⁿ is called L-superadditive (-subadditive) if f satisfies the following condition:

where x ∨ (∧) y is the vector of componentwise maxima (minima). It is known that if f has second partial derivatives, then f is L-superadditive (-subadditive) if and only if ∂²f (x)/∂x_i∂x_j ≥ 0 (≤ 0) for all x ∈ Rⁿ,1 ≤ i,j ≤ n (see Marshall and Olkin [6]). Boland and Proschan [3] established Schur-concavity (-convexity) of a symmetric reliability function on a region of component reliabilities and obtained results for comparing the Birnbaum reliability importance of components in k-out-of-n systems. In Theorem 3, by restricting the L-superadditivity (-subadditivity) property on two components, such comparison results are obtained for more general coherent systems, with less restrictions on component reliabilities. In Theorem 4, we observe that the condition stated in Theorem 3 is also necessary, and this property will be utilized later in this article to derive further results.

Theorem 4: Suppose that

. Then, ∂²h(p)/∂p_i∂p_j ≥ 0 (≤ 0) for all p [hArr ] I_B(i;p) ≥ I_B(j;p) for all p satisfying p_i ≤ (≥) p_j.

Proof: (⇒) This part has been shown in [10].

(⇐) By expressing I_B(i;p) (similarly for I_B(j;p)) as

we obtain that

Thus, when

and hence

The right-hand side condition in Theorem 4 then implies that ∂²h(p)/∂p_i∂p_j ≥ 0 (≤ 0) for all p. █

We are now ready to present one of the main results of this article: We replace the restricted L-superadditivity (-subadditivity) property stated in Theorem 4 by equivalent conditions in terms of minimal cut (path) sets, which is much more convenient for engineers to use in practice. First, some notations are needed. Denote by C(i)(P(i)) the collection of minimal cut (path) sets in which each minimal cut (path) set contains i. Also denote by C(ij)(P(ij)) the collection of minimal cut (path) sets in which each minimal cut (path) set contains both i and j, and by C(ij)(P(ij)) the collection of minimal cut (path) sets in which each minimal cut (path) set contains i but not j. It is known that if

, then C_k ∈ C(ij) ⇒ {j} ∪ C_k − {i} ∈ C(ji) (see [7]). Thus, C(ij) = {Ø} ⇒ C(ji) = {Ø}. The condition that

can be divided into three cases: (i) C(i) = C(j), (ii) P(i) = P(j), and (iii) neither (i) nor (ii) holds. Note that in this case

, (i) is equivalent to C(ij) = {Ø} and (ii) is equivalent to C(ij) = {Ø}.

Theorem 5: Suppose that

. Then, the following hold:

Proof: Case (i). (⇒) Recall the minimax representation of the binary structure φ and that h(p) = Pr{min_1≤k≤c max_i∈Ck X_i = 1} (see [1]). Since the assumption is equivalent to i ∈ C_k [hArr ] j ∈ C_k, it is easy to see that the structure allows for a modular decomposition with a modular set, z = max{x_i,x_j} and that h(p) = p_z h(1_z,p) + (1 − p_z)h(0_z,p), where the reliability of the module p_z = Pr{Z = 1} = p_i + p_j − p_i p_j. Hence,

The conclusion then follows from Theorem 4.

(⇐) Suppose that the right-hand side of (i) holds. Then, from Theorem 4, ∂²h(p)/∂p_i∂p_j ≤ 0 holds for all p. Note that

Thus, the assumption implies that

Now suppose that C(i) ≠ C(j) (i.e., C(ij) ≠ {Ø}). We will derive a contradiction. Let C_k ∈ C(ij); then, {j} ∪ C_k − {i} is also a minimal cut set. Hence, φ(0_i,1_j,0^C_k−i,1) = φ(1_i,0_j,0^C_k−i,1) = 0 and φ(1_i,1_j,0^C_k−i,1) = 1, where 0^A means that x_i = 0 for all i ∈ A. It is seen that Eq. (8) fails to hold when x = (0^C_k−i,1).

Case (ii). (⇒) The assumption is equivalent to i ∈ P_r [hArr ] j ∈ P_r. Recalling the maxmin representation h(p) = Pr{max_1≤r≤p min_i∈Pr x_i = 1}, we can let z = min{x_i,x_j} be a modular set with reliability, p_z = p_i p_j. It is then easily obtained that

The conclusion follows from Theorem 4.

(⇐) In this case, by Theorem 4, ∂²h(p)/∂p_i∂p_j ≥ 0 for all p. Suppose that C(ij) ≠ {Ø} and let C_k ∈ C(ij). Then, φ(0_i,1_j,0^C_k−i−j,1) = φ(1_i,0_j,0^C_k−i−j,1) = 1 and φ(0_i,0_j,0^C_k−i−j,1) = 0. Hence, φ(1_i,1_j,x) − φ(1_i,0_j,x) − φ(0_i,1_j,x) + φ(0_i,0_j,x) < 0, when x = (0^C_k−i−j,1). Thus, from Eq. (7), we see that the condition ∂²h(p)/∂p_i∂p_j ≥ 0 for all p fails to hold. █

Example 1: Let φ(x) = x₁ ∧ (x₂ ∨ x₃), x₁,x₂,x₃ ∈ {0,1}. In this example, there are two minimal cut sets: {1} and {2,3}. Clearly,

and Theorem 5(i) applies to the two nodes. Hence, I_B(2;p) ≥ I_B(3;p) if p₂ ≥ p₃.

Example 2: Let φ(x) = x₁ ∨ (x₂ ∧ x₃), x₁,x₂,x₃ ∈ {0,1}. In this example, the minimal cut sets are {1,2} and {1,3}; Theorem 5(ii) applies to nodes 2 and 3. Hence, I_B(2;p) ≥ I_B(3;p) if p₂ ≤ p₃.

Next, we consider the case that nodes i and j are not symmetric but are ordered by their criticality. First, the following theorem is analogous to Theorem 4, but inequality is replaced by strict inequality.

Theorem 6: Suppose that

. Then, ∂²h(p)/∂p_i∂p_j ≥ 0 (≤ 0) for all p [hArr ] I_B(i;p) > I_B(j;p) for all p satisfying p_i ≤ p_j (≥ p_j).

Proof: (⇒) This part has been shown in [10].

(⇐) To show that ∂²h(p)/∂p_i∂p_j ≥ 0 for all p, we will prove that φ(1_i,1_j,x) − φ(1_i,0_j,x) − φ(0_i,1_j,x) + φ(0_i,0_j,x) ≥ 0 holds for all x. Suppose that our claim is not true. Then, there exists an x* such that φ(1_i,0_j,x*) = φ(0_i,1_j,x*) = 1 and φ(0_i,0_j, x*) = 0. Choose a probability vector p* such that Pr{(·_i,·_j,X) = (·_i,·_j,x*)} → 1. Then, when p = p* in Eq. (5), h(1_i,0_j,p) − h(0_i,1_j,p) → 0 and ∂²h(p)/∂p_i∂p_j → −1. Hence, I_B(i;p) < I_B(j;p) holds for some p_i < p_j, which contradicts our assumption. (The case ∂²h(p)/∂p_i∂p_j ≤ 0 can be treated in a similar manner; the details are omitted.) █

Now, similar to the symmetry case, we divide the ordering

into three cases: (i) P(j) ⊂ P(i), (ii) C(j) ⊂ C(i), and (iii) neither (i) nor (ii) holds, where “⊂” denotes strict containment relation.

Theorem 7: Suppose that

. Then, the following hold:

Proof: Case (i). (⇐) Suppose that there is a minimal path set P_r such that j ∈ P_r and i ∉ P_r. Then, since

, {i} ∪ P_r − {j} is a path set (not necessarily minimal). Hence, φ(0_i,1_j,1^P_r−j,0) = 1, φ(1_i,0_j,1^P_r−j,0) = 1, and φ(0_i,0_j,1^P_r−j,0) = 0. We then choose a probability vector such that Pr{(·_i,·_j,X) = (·_i,·_j,1^P_r−j,0)} is sufficiently close to one. Then, from Eqs. (5) and (7), I_B(i;p) < I_B(j;p) for some p_i < p_j, which contradicts our assumption. Thus, P(j) ⊆ P(i) holds. Clearly, since

, there is a minimal path set P_r such that i ∈ P_r and j ∉ P_r; hence, P(j) ⊂ P(i).

(⇒) Suppose that P(j) ⊂ P(i). It suffices to show that, by Theorem 6, ∂²h(p)/∂p_i∂p_j ≥ 0 for all p. Let A_ij be the event that at least one minimal path set, containing both i and j, is functioning; let A_i be the event that at least one minimal path set, containing i but not j, is functioning; and let A be the event that at least one minimal path set, containing neither i nor j, is functioning. Further, let B_ij be the event that there exists a minimal path set, containing both i and j and in which the components other than i and j are functioning. Similarly, let B_i be the event that there exists a minimal path set, containing i but not j in which the components other than i are functioning. Then,

Thus,

Case (ii). (⇐) Suppose that there is a minimal cut set C_k such that j ∈ C_k and i ∉ C_k. Then, since

{i} ∪ C_k − {j} is a cut set (not necessarily minimal). Hence, φ(0_i,1_j,0^C_k−j,1) = 0, φ(1_i,0_j,0^C_k−j,1) = 0, and φ(1_i,1_j,0^C_k−j,1) = 1. Choose a probability vector such that Pr{(·_i,·_j,X) = (·_i,·_j,0^C_k−j,1)} is sufficiently large. Then, by employing similar arguments to that in (i), a contradiction is derived.

(⇒) Suppose that C(j) ⊂ C(i). We then let A_ij be the event at which at least one minimal cut set, containing both i and j, is working, where a minimal cut set working means that all of its components have failed. Define A_i, A, B_ij, and B_i similar to that in case (i), except that minimal path sets are replaced by minimal cut sets. Then,

Thus,

The conclusion then follows from Theorem 6. █

Example 1 (continued): Consider nodes 1 and 2 in this example. Clearly, P(2) ⊂ P(1) holds. Hence, I_B(1;p) > I_B(2;p) for all p₁ ≤ p₂.

Example 2 (continued): Consider nodes 1 and 2 in this example. Clearly, C(2) ⊂ C(1) holds. Hence, I_B(1;p) > I_B(2;p) for all p₁ ≥ p₂.

Example 3: Consider the bridge structure shown in Figure 1. The minimal path sets of the system are P₁ = {1,4}, P₂ = {2,5}, P₃ = {1,3,5}, and P₄ = {2,3,4}; the minimal cut sets are C₁ = {1,2}, C₂ = {4,5}, C₃ = {1,3,5}, and C₄ = {2,3,4}. Consider nodes 1 and 3. Clearly,

holds, but neither P(3) ⊂ P(1) nor C(3) ⊂ C(1). Thus, the third case of the ordering

holds between nodes 1 and 3. In this example,

Bridge structure for Example 3.

Thus, ∂²h(p)/∂p₁∂p₃ > 0 if p₅ → 1 and p₂,p₄ → 0, whereas ∂²h(p)/∂p_i∂p_j < 0 if p₅ → 0 and p₂,p₄ → 1.