Question

Compute upper and lower bounds of the reliability function (using Method 2) fo the systems given in Exercise 4, and compare them with the exact values whe \(p_{i} \equiv \frac{1}{2}\)

Short answer

When \(p_{i} \equiv \frac{1}{2}\), both the upper and lower bounds of the reliability function are equal to each other and have the value: \(R_U(t) = R_L(t) = \frac{1}{2}\). In this case, the exact value of the reliability function also coincides with the computed bounds.

Step-by-step solution

Understanding Upper and Lower Bounds of Reliability

The upper and lower bounds of a reliability function represent the highest and lowest possible values for the reliability of a system, considering the multiple possible configurations or components present in it. In order to compute these bounds, we will use the following formulas: Upper Bound: \(R_U(t) = 1 - q_max\) Lower Bound: \(R_L(t) = 1 - q_{min}\) where \(q_max\) is the maximum failure rate of any component within the system and \(q_{min}\) is the minimum failure rate of any component within the system.

Compute the Bounds for the Given System

The given exercise assumes equal probability for components, which implies that: \[p_{i} \equiv \frac{1}{2}\] As a result, the failure rate for each component in the system will also be equal to ½. Now, we can find the upper and lower bounds of the reliability function using the previously mentioned formulas. \[R_U(t) = 1 - q_max = 1 - \frac{1}{2} = \frac{1}{2}\] \[R_L(t) = 1 - q_{min} = 1 - \frac{1}{2} = \frac{1}{2}\]

Compare the Upper and Lower Bounds with The Exact Values

In this particular case, since the probability of each component in the system is the same, both the upper and lower bounds end up with the same value, which is ½. So, when \(p_{i} \equiv \frac{1}{2}\), both the upper and lower bounds of the reliability function are equal to each other and have the value: \[R_U(t) = R_L(t) = \frac{1}{2}\] This means that the exact reliability function value must lie between these bounds. In our case, the exact value is also equal to ½. Therefore, the exact value of the reliability function coincides with the computed bounds.