Lifetime Evaluation vs. Measurement. Part 3.

Sometimes shifting your perspective
is more powerful than being smart.

—Astro Teller

Guest post by Oleg Ivanov

A common approach for “no failure” testing is the use of the well-known expression

$$ (1) \quad 1-CL={{R}^{n}}$$

where CL is a confidence level, R is a required reliability, n is a sample size. Its parent is a Binomial distribution with zero failures. This expression is like a poor girl:

When she had done her work, she used to go to the chimney corner, and sit down there in the cinders and ashes, which caused her to be called Cinderwench. Only the younger sister, who was not so rude and uncivil as the older one, called her Cinderella.

1 – CL makes sense a consumer risk β. It is a probability that the product passes the tests, but doesn’t meet operation requirements. There is some probability that the product passes the tests on the right side, but a probability that the product doesn’t meet operation requirements is absent. We correct these mismatches.

In the first, we don’t know the reliability r the tested product. We know only the bounds of a reliability uncertainty are from 0 to 1. And exactly, that the reliability r is not equal to required reliability R.

In the second, we add the probability that product doesn’t meet the operation requirements. In this case, there is a requirement of the product reliability . So we must transform (1) into

$$ (2) \quad \beta ={{r}^{n}}\centerdot \left\{ \begin{array}{l}1,\text{ }r<R\\0,\text{ }r\ge R\end{array} \right.$$

…she struck it with her wand, and the pumpkin was instantly turned into a fine coach, gilded all over with gold.

We don’t know a value of probability r and so we cannot use the expression (2). The “worst case” method is the best way to decide the problem of uncertainty, so we find the reliability r which gives us the maximum risk β:

$$ (3) \quad  \beta =\underset{0\le r\le 1}{\mathop{\max }}\,\left( {{r}^{n}}\centerdot \left\{ \begin{array}{l}1,\text{ }r<R\\0,\text{ }r\ge R\end{array} \right. \right)$$

A solving of (3) gives us expression (1), but now we know how and for which conditions it is received.

As each mouse went out, she gave it a little tap with her wand, and the mouse was that moment turned into a fine horse, which altogether made a very fine set of six horses of a beautiful mouse-colored dapple-gray.

How to evaluate system reliability from tests of the components, when there are no failures? The probability that all components pass the tests is , where m is an amount of components, is a reliability of component i, is an amount of tests of component i.

Instead of expression (3) we have expression

$$ (4) \quad \beta =\underset{\begin{array}{c}0\le {{r}_{1}}\le 1\\0\le {{r}_{2}}\le 1\\\cdots \\0\le {{r}_{m}}\le 1\end{array}}{\mathop{\max }}\,\left( r_{1}^{{{n}_{1}}}\centerdot r_{2}^{{{n}_{2}}}\centerdot \cdots \centerdot r_{m}^{{{n}_{m}}}\centerdot \left\{ \begin{array}{l}1,\text{ }{{r}_{1}}\centerdot {{r}_{2}}\centerdot \cdots \centerdot {{r}_{m}}<R\\0,\text{ }{{r}_{1}}\centerdot {{r}_{2}}\centerdot \cdots \centerdot {{r}_{m}}\ge R\end{array} \right. \right)$$

In the case when the amount of tests of the component is identical () we can substitute and receive the same expression (3) and its solution (1). So the evaluation of the system reliability from tests of the components is the same as for tests of a system as a whole. It meets to a conclusion obtained earlier in Part 2.

The fairy made choice of one of the three which had the largest beard, and, having touched him with her wand, he was turned into a fat, jolly coach- man, who had the smartest whiskers eyes ever beheld.

In the case when the amount of tests of the component is different the solution of (4) is more difficult. Decide this. You will receive an unexpected and paradoxical solution.

She had no sooner done so but her godmother turned them into six footmen, who skipped up immediately behind the coach, with their uniforms all bedaubed with gold and silver…

In this approach, we can set up and consider clear reliability goals. The expression (3) consider the requirement of the product reliability only. If the request of absence of failures for N products will be established, instead (3) we can consider expression

$$ (5) \quad \beta =\underset{0\le r\le 1}{\mathop{\max }}\,\left( {{r}^{n}}\centerdot \left( 1-{{r}^{N}} \right) \right)$$

Here makes sense of the probability of one or more failures in the N products operation.

Her godmother only just touched her with her wand, and, at the same instant, her clothes were turned into cloth of gold and silver, all beset with jewels. This done, she gave her a pair of glass slippers, the prettiest in the whole world.

As a method of control, the reliability test has no only the consumer risk β but and a supplier risk α:

$$ (6) \quad \alpha =\underset{0\le r\le 1}{\mathop{\max }}\,\left( \left( 1-{{r}^{n}} \right)\centerdot {{r}^{N}} \right)$$

This approach allows us to consider it separately without consumer risk or together with consumer risk in an aggregate. It does a solution more interesting.

But after midnight, if she stayed one moment longer, the coach would be a pumpkin again, her horses mice, her coachman a rat, her footmen lizards, and her clothes become just as they were before…

…and we’ll return to the expression.

$$ (7) \quad 1-CL={{R}^{n}}$$