Question

Let \(t_{i}\) denote the time of failure of the \(i\) th component; let \(\tau_{\phi}(t)\) denote the time to failure of the system \(\phi\) as a function of the vector \(\mathrm{t}=\left(t_{1}, \ldots, t_{n}\right) .\) Show that $$ \max _{1 \leqslant j \leqslant s} \min _{i \in A_{j}} t_{i}=\tau_{\phi}(\mathbf{t})=\min _{1 \leqslant j \leqslant k} \max _{i \in C_{i}} t_{i} $$ where \(C_{1}, \ldots, C_{k}\) are the minimal cut sets, and \(A_{1}, \ldots, A_{s}\) the minimal path sets.

Short answer

In summary, the time to failure of the system, denoted by \(\tau_{\phi} (\mathbf{t})\), can be represented in two different ways: using minimal path sets and using minimal cut sets. For minimal path sets, the time to failure is the maximum of the minimum times of failure for all components in the sets, given by \(\max_{1 \leqslant j \leqslant s} \min_{i \in A_j} t_i\). For minimal cut sets, the time to failure is the minimum of the maximum times of failure for all components in the sets, given by \(\min_{1 \leqslant j \leqslant k} \max_{i \in C_j} t_i\). By analyzing both expressions, we have shown that they are equal and correctly represent the time to failure of the system.

Step-by-step solution

Understand the definitions

To start, it's important to understand what minimal path sets and minimal cut sets represent in terms of time to failure. - For a minimal path set, when all components in the set are functioning, the system remains operational. Consequently, the system fails when the first component in a minimal path set fails. Therefore, the time to failure of the system under a minimal path set is the minimum of the times of failure of all the components in the set. - For a minimal cut set, the system fails when all components in the set fail. Consequently, the time to failure of the system under a minimal cut set is the maximum of the times of failure of all its components. Now that we have understood the relationships, we can move forward to proving the given identity.

Proving the identity using minimal path sets

First of all, we will deal with the left-hand side of the given identity. From the definition of minimal path sets, the system will fail when any of the minimal path sets fail. This can be represented as: $$ \tau_{\phi} (\mathbf{t}) = \max_{1 \leqslant j \leqslant s} \min_{i \in A_j} t_i $$ Here, \(\min_{i \in A_j} t_i\) represents the failure of the system under minimal path set \(A_j\), and the maximum of all these values is the time to failure of the system.

Proving the identity using minimal cut sets

Next, we will deal with the right-hand side of the given identity. From the definition of minimal cut sets, the system will continue to function until all the components of any minimal cut set fail. This can be represented as: $$ \tau_{\phi} (\mathbf{t}) = \min_{1 \leqslant j \leqslant k} \max_{i \in C_j} t_i $$ Here, \(\max_{i \in C_j} t_i\) represents the time up to which the system continues to function under minimal cut set \(C_j\), and the minimum of all these values is the time to failure of the system.

Combining both expressions

Now, we've found two ways to express the time to failure of the system, one through minimal path sets and the other through minimal cut sets. The given identity is: $$ \max_{1 \leqslant j \leqslant s} \min_{i \in A_j} t_i = \tau_{\phi} (\mathbf{t}) = \min_{1 \leqslant j \leqslant k} \max_{i \in C_j} t_i $$ Through steps 2 and 3, where we analyzed minimal path sets and minimal cut sets separately, we have successfully shown that the expressions on both sides of the identity are equivalent to the time to failure of the system. Thus, we can conclude that the identity is correct.