Template:Parameter estimation fielded rga

Parameter Estimation
Suppose that the number of systems under study is $$K$$  and the  $${{q}^{th}}$$  system is observed continuously from time  $${{S}_{q}}$$  to time  $${{T}_{q}}$$,  $$q=1,2,\ldots ,K$$. During the period $$[{{S}_{q}},{{T}_{q}}]$$, let  $${{N}_{q}}$$  be the number of failures experienced by the  $${{q}^{th}}$$  system and let  $${{X}_{i,q}}$$  be the age of this system at the  $${{i}^{th}}$$  occurrence of failure,  $$i=1,2,\ldots ,{{N}_{q}}$$. It is also possible that the times $${{S}_{q}}$$  and  $${{T}_{q}}$$  may be observed failure times for the  $${{q}^{th}}$$  system. If $${{X}_{{{N}_{q}},q}}={{T}_{q}}$$  then the data on the  $${{q}^{th}}$$  system is said to be failure terminated and  $${{T}_{q}}$$  is a random variable with  $${{N}_{q}}$$  fixed. If $${{X}_{{{N}_{q}},q}}<{{T}_{q}}$$  then the data on the  $${{q}^{th}}$$  system is said to be time terminated with  $${{N}_{q}}$$  a random variable. The maximum likelihood estimates of $$\lambda $$  and  $$\beta $$  are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).


 * $$\begin{align}

& \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} \end{align}$$

where $$0\ln 0$$  is defined to be 0. In general, these equations cannot be solved explicitly for $$\widehat{\lambda }$$  and  $$\widehat{\beta },$$  but must be solved by iterative procedures. Once $$\widehat{\lambda }$$  and  $$\widehat{\beta }$$  have been estimated, the maximum likelihood estimate of the intensity function is given by:


 * $$\widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}}$$

If $${{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0$$  and  $${{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}}$$   $$\,(q=1,2,\ldots ,K)$$  then the maximum likelihood estimates  $$\widehat{\lambda }$$  and  $$\widehat{\beta }$$  are in closed form.


 * $$\begin{align}

& \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{\mathop{\sum }}}\,\ln (\tfrac{T})} \end{align}$$

The following examples illustrate these estimation procedures.

Example 1
For the data in Table 13.1, the starting time for each system is equal to $$0$$  and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates $$\widehat{\lambda }$$  and  $$\widehat{\beta }$$.

Solution Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of $$\hat{\beta }$$  and  $$\hat{\lambda }$$  are then calculated as follows:


 * $$\begin{align}

& \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{\mathop{\sum }}}\,\ln (\tfrac{T})} \\ & = & 0.45300 \end{align}$$


 * $$\begin{align}

& \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\ & = & 0.36224 \end{align}$$



The system failure intensity function is then estimated by:


 * $$\widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t>0$$

Figure wpp intensity is a plot of $$\widehat{u}(t)$$  over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.

Economical Life Model
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if $$\beta >1$$. It does not make sense to implement an overhaul policy if $$\beta <1$$  since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul? Denote $${{C}_{1}}$$  as the average repair cost (unscheduled),  $${{C}_{2}}$$  as the replacement or overhaul cost and  $${{C}_{3}}$$  as the average cost of scheduled maintenance. Scheduled maintenance is performed for every $$S$$  miles or time interval. In addition, let $${{N}_{1}}$$  be the number of failures in  $$[0,t]$$  and let  $${{N}_{2}}$$  be the number of replacements in  $$[0,t]$$. Suppose that replacement or overhaul occurs at times $$T$$,  $$2T$$ ,  $$3T$$. The problem is to select the optimum overhaul time $$T={{T}_{0}}$$  so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since $$\beta >1$$, the average system cost is minimized when the system is overhauled (or replaced) at time  $${{T}_{0}}$$  such that the instantaneous maintenance cost equals the average system cost. The total system cost between overhaul or replacement is:


 * $$TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S}$$

So the average system cost is:


 * $$C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T}$$

The instantaneous maintenance cost at time $$T$$  is equal to:


 * $$IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{S}$$

The following equation holds at optimum overhaul time $${{T}_{0}}$$ :


 * $$\begin{align}

& {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\ & = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{S}} \end{align}$$


 * Therefore:


 * $${{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }}$$

When there is no scheduled maintenance, Eqn. (ecolm) becomes:


 * $${{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}$$

The optimum overhaul time, $${{T}_{0}}$$, is the same as Eqn. (optimt), so for periodic maintenance scheduled every $$S$$  miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.