A system is maintained according to Policy 3 over an infinite time span. It has the same lifetime distribution and minimal repair cost parameter as in exercise 7.20. As with exercise 7.20, let \(c_{r}=2000\).

(1) Determine the optimum integer \(n=n *\), and the corresponding maintenance cost rate \(K_{3}(n *)\).

(2) Compare \(K_{3}(n *)\) to \(K_{1}(\tau *)\) (exercise 7.20 ) and try to intuitively explain the result.

Data from Exercise 7.20

The lifetime \(L\) of a system has a Weibull-distribution with distribution function

\[F(t)=P(L \leq t)=1-e^{-0.1 t^{3}}, t \geq 0\]

(1) Determine its failure rate \(\lambda(t)\) and its integrated failure rate \(\Lambda(t)\).

(2) The system is maintained according to Policy 1 over an infinite time span. The cost of a minimal repair is \(c_{m}=40\) [\$], and the cost of a preventive replacement is \(c_{p}=2000[\$]\).

Determine the cost-optimum replacement interval \(\tau *\) and the corresponding minimal maintenance cost rate \(K_{1}(\tau *)\).