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Reliability Exercise

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1. Given the following component reliabilities of a simple system consisting of 3 components, find the reliability of the system (an electrical current will travel from pt. A to B successfully)

A→.996—.990—.86→B (Series System)

2. Given the parallel arrangement of the same components determine the reliability of the system (an electrical current will travel from pt. A to pt. B)
A→ .990 →B

3. Which arrangement (1 or 2 above) would you use to build the system and why would you choose the arrangement you chose?

4. Assume you "tested several Fireston tires and recorded their failure times (as defined by an engineer). After doing a Chi-Square test, you decide that failures are "normally" distributed, with mean failure of 50,000 miles and a standard deviation of 10,000 miles. Determine the following:
a. Reliability of the tires at 40,000 miles
b. Given that a tire has been driven on a test machine 20,000 miles, determine its reliability at 60,000 miles.

5. Assume you have a product with the reliability of .90 at 1000 hours. Your customer wants a reliability of .99. Explain what you can do, if the "best technology was used to produce the components of the system, to achieve the .99 reliability?

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Solution Preview

1. In order for the system to work, all three components must be functional. We know that the probability of all three events occurring is the product of the individual probabilities (since components are independent). Hence P(system reliable) = 0.996 X 0.99 X 0.86 = 0.8480.

2. In the case that they are in parallel, we know that the system will work if at least one of the three is working. In other words:

P(system reliable) = P(at least one is reliable) = 1 - P(none reliable) = 1 - (1-0.996) X (1 - 0.99) X (1-0.86) = 0.9999944. This is true because the systems are independent.

3. Clearly the second arrangement is much better because its reliability is higher than the first arrangement.

4. a) Reliability of the tires at 40,000 miles mean, in technical language, P(X >= 40,000) given ...

Solution Summary

This solution shows how to solve problems related to the reliability of a system.