Question

4. Reliability of Systems - Take n components to have failure times Ti, T2, ..., Tn If we construct a complex system out of these distribution of the failure time T of the entire svstem in terms of the distributions of Ti, T2, ..., Tn. There are two basic networks. In a series hookup, the system fails as soon as any one of the components fails. Hence T - min(T1, T2, ...,Tn). In a parallel hookup the system is operational as long as at least one component is working. So here T-max(Ti, T2, ...,Tn) a. If 10 components all have probability .99 of lasting at least 100 hours and they are hooked in series, what is the probability that the system will not fail in 100 hours? components, the problem is to get the b. What if they are hooked in parallel? c. With the component failure times Ti,T2,, Tn exponentially distributed with parameters λι, λ2, , λη show that in a series hookup the system failure time exponentially distributed with parameter λ1 λ2+ + λη d. If all components have the same expected component time is the expected component time until failure is e. What is the expected time until failure in parallel hookup if all components have the same parameter ! parameter α, then in a series hookup the

Answer 1

a.

Given, P(Ti $\ge$ 100) = 0.99 for i = 1,2,...,10

Probability that the system will not fail in 100 hrs. = P(T $\ge$ 100) = P(min(T1, T2,..., T10) $\ge$ 100)

= P(T1 $\ge$ 100, T2 $\ge$ 100, ......, T10 $\ge$ 100)

= P(T1 $\ge$ 100) * P(T2 $\ge$ 100) * .... * P(T10 $\ge$ 100)

= 0.99 * 0.99 * ... * 0.99

= 0.99¹⁰

= 0.9043821

b.

Given, P(Ti $\ge$ 100) = 0.99 for i = 1,2,...,10

=> P(Ti $\le$ 100) = 1 - 0.99 = 0.01

Probability that the system will not fail in 100 hrs. = P(T $\ge$ 100) = P(max(T1, T2,..., T10) $\ge$ 100)

= 1 - P(max(T1, T2,..., T10) $\le$ 100)

= 1 - P(T1 $\le$ 100, T2 $\le$ 100, ......, T10 $\le$ 100)

= 1 - P(T1 $\le$ 100) * P(T2 $\le$ 100) * .... * P(T10 $\le$ 100)

= 1 - 0.01 * 0.01 * ... * 0.01

= 1 - 0.01¹⁰

= 1

c.

Given, Ti ~ Exponential($\lambda_i$) so, P(Ti $\ge$ t) = exp(-$\lambda_i t$) for i = 1,2,...,n

Probability that the system will not fail in t hrs. = P(T $\ge$ t) = P(min(T1, T2,..., Tn) $\ge$ t)

= P(T1 $\ge$ t, T2 $\ge$ t, ......, Tn $\ge$ t)

= P(T1 $\ge$ t) * P(T2 $\ge$ t) * .... * P(n $\ge$ t)

= exp(-$\lambda_1 t$) * exp(-$\lambda_2 t$) * ... * exp(-$\lambda_n t$)

=  exp(-$(\lambda_1+\lambda_2+...+\lambda_n) t$)

Thus, system failure time is exponentially distributed with parameter $\lambda_1+\lambda_2+...+\lambda_n$ .

d.

Given, $\alpha = 1 / \lambda_i$ for all i = 1, 2, .., n

or, $\lambda_i = 1 / \alpha$

Since, system failure time is exponentially distributed with parameter $\lambda_1+\lambda_2+...+\lambda_n$ .

Expected component time until failure = $1/(\lambda_1+\lambda_2+...+\lambda_n)$

$= 1/(1/\alpha+1/\alpha+...+1/\alpha) = 1 / (n /\alpha) = \alpha / n$