I have been asked to solve some questions in the statistical program R.

First, I want to find the expected life length (by numeric integration) of a network with 3 parallel components (T1,T2,T3) and 1 serial component (T4).

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The hazard functions of the components are: $\\$

$\lambda_{T1}(t) = \lambda_{T2}(t) = \lambda_{T3}(t) =\frac{1}{2*\sqrt(t)}$, and $\lambda_{T4} = \frac{t^{11/10}}{50}$

From my understanding, to be able to calculate the expected life length, I want to find the survival function of the network which leads me to

$E[T] = \int_{0}^{inf}S(t) dt$.

I know that the hazard function is

$ \lambda_{T}(t)$ = $-\frac{\partial}{\partial{t}}log S_{T}(t) $

and thus

$S_{T}(t) = exp(- \int_{0}^{t}\lambda_{T}(t) dt.$ (Survival function = Cumulative hazard function)

My first problem is that I have no idea how to find the survival function S(t) when I´m dealing with network components?


I am guessing that since the problem is not supposed to be solved explicitly, the solution is to estimate the expected life length via simulation. The life length of the system is $$T = \min((\max(T_1, T_2, T_3), T_4),$$ if you generate many monte carlo realisations $T^*_b$ of this variable, the estimate of the expected life length is $$\frac{1}{B}\sum_{b=1}^B T^*_b.$$

If you want to do the numerical integration of the survival function by a simpson type approach to find the expectation instead of monte carlo, the survival function of the system is

\begin{align*} S(t) &= 1 - P(T \leq t)\\ &= 1 - P(\min(\max(T_1, T_2, T_2), T_4) \leq t)\\ &= 1 - (1 - P(\min(\max(T_1, T_2, T_2), T_4) > t))\\ &= P(\min(\max(T_1, T_2, T_2), T_4) > t)\\ &= P(\max(T_1, T_2, T_2) > t)P(T_4 > t)\\ &= (1 - P(\max(T_1, T_2, T_2) \leq t))(1 - P(T_4 \leq t))\\ &= (1 - P(T_1 \leq t)P(T_2 \leq t)P(T_3 \leq t))(1 - P(T_4 \leq t))\\ &= (1 - F_1(t)F_2(t)F_3(t))(1 - F_4(t))\\ &= (1 - F_1(t)^3)(1 - F_4(t)). \end{align*} where the last equality is due to that $F_1(t)=F_2(t)=F_3(t)$. Also, $F_1(t)= 1 - \exp(-t^{\frac{1}{2}}), $ and $F_4(t) = 1 - \exp(-\frac{1}{105}t^{21/10}).$

  • $\begingroup$ Wow, thanks for the quick reply Simon. $\endgroup$
    – Ole G S
    Oct 1 '19 at 17:26
  • $\begingroup$ So basically all I need to do is to Integrate $(1-F_{1}(t)..*F_{3}(t)*(1-F_{4}(t))$ to get my expected life length? How does the hazard function look like for the whole interval then? I`ll get confused when there exist hazard functions of several components. U got any idea or tips regarding this matter? $\endgroup$
    – Ole G S
    Oct 1 '19 at 17:37
  • $\begingroup$ The hazard function of the lifetime of the system is $$ \lambda_T(t) = -\frac{d}{dt}\log(S(t)) = \frac{3F_1(t)^2\frac{dF_1(t)}{dt}}{1 - F_1(t)^3} + \frac{\frac{dF_4(t)}{dt}}{1 - F_4(t)},$$ since F_1(t)=F_2(t)=F_3(t). $\endgroup$ Oct 1 '19 at 17:51
  • $\begingroup$ And yes, just integrate the survival $$S(t) = (1 - F_1(t)F_2(t)F_3(t))(1 - F_4(t)) = (1 - F_1(t)^3)(1 - F_4(t)),$$ from $0$ to $\infty.$ $\endgroup$ Oct 1 '19 at 17:53
  • $\begingroup$ You sure it suppose to be $S(t) = (1-(F_{1}(t)^{3})(1-F_{4}(t)))$ and not $S(t) = (1-(F_{1}(t)^{3})*(F_{4}(t))$ I am suspicious because $F_{4}$ is serial connected to $F_{1,2,3}.$ $\endgroup$
    – Ole G S
    Oct 2 '19 at 11:27

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