# Optimal Maintenance Schedule for a Wind Power Turbine with Aging Components
## 9 Proofs of Propositions  
**Proof of Proposition 1. ** We will need the following result from the renewal theory; see for example [11]. Let  
\[(X_{i},R_{i}),\quad i=1,2,\ldots\]  
be independent and identically distributed pairs of possibly dependent random variables: \(X_{i}\) values are positive inter-arrival times and \(R_{i}\) values are associated rewards. The cumulative reward process is defined as  
\[W(u)=R_{1}+\ldots+R_{N(u)},\]  
where \(N(u)\) is the number of renewal events up to time \(u\), that is, \(N(u)=k\) if  
\[X_{1}+\ldots+X_{k}\leq u<X_{1}+\ldots+X_{k+1}.\]  
According to the renewal-reward theorem, the per unit of time reward  
\[\tfrac{W(u)}{u}\rightarrow\tfrac{\mathrm{E}(R)}{\mathrm{E}(X)},\quad u\to\infty\]  
almost surely converges. Recall that the full maintenance cost of the gearbox due to a CM is \(g\), and a similar cost for a PM is \((ma+h)\), where \(a\) is the gearbox age at the replacement. Thus, if we plan our next PM for a new gearbox at time \(t\) after each renewal event, then the corresponding renewal-reward process has inter-arrival time \(X=X(t)\) with  
\[X(t)=L\wedge t=L\cdot 1_{\{L\leq t\}}+t\cdot 1_{\{L\geq t+1\}} \tag{10}\]  
and the reward \(R=R(t)\) with  
\[R(t)=g1_{\{L\leq t\}}+(tm+h)1_{\{L\geq t+1\}}. \tag{11}\]  
Since  
\[\mathrm{E}(X) =\mathrm{E}(L\cdot 1_{\{L\leq t\}})+t\mathrm{P}(L\geq t+1)=\sum_{k= 1}^{t}(e^{-\theta(k-1)^{\beta}}-e^{-\theta t^{\beta}})k+e^{-\theta t^{\beta}}t,\] \[\mathrm{E}(R) =g\mathrm{P}(L\leq t)+(tm+h)\mathrm{P}(L\geq t+1)=(1-e^{-\theta t ^{\beta}})g+e^{-\theta t^{\beta}}(tm+h),\]the renewal-reward theorem implies that the time-average maintenance cost is computed as the following function of the planning time \(t\)  
\[q_{t}=\frac{\mathrm{E}(R)}{\mathrm{E}(X)}=\frac{(1-e^{-\theta t^{\beta}})g+e^{- \theta t^{\beta}}(tm+h)}{\sum_{k=1}^{t}(e^{-\theta(k-1)^{\beta}}-e^{-\theta t^{ \beta}})k+e^{-\theta t^{\beta}}t}.\]  
Proof of Proposition 2.: By the definition of \(t_{(s,a)}\) we have  
\[\mathrm{E}(Q_{(s,a)}(t_{(s,a)},s+L_{a}))\leq\mathrm{E}(Q_{(s,a)}(t,s+L_{a})), \quad t=s+1,\ldots,T+1.\]  
To prove the assertion, it is sufficient to verify that  
\[W(u)=R_{1}+\ldots+R_{N(u)},\]0  
for \(t=s+\delta+1,\ldots,T+1\). To derive (12), we will introduce a special notation  
\[W(u)=R_{1}+\ldots+R_{N(u)},\]1 \[W(u)=R_{1}+\ldots+R_{N(u)},\]2  
and notice that for \(t\leq T\)  
\[W(u)=R_{1}+\ldots+R_{N(u)},\]3 \[W(u)=R_{1}+\ldots+R_{N(u)},\]4  
or in other words,  
\[W(u)=R_{1}+\ldots+R_{N(u)},\]5 \[W(u)=R_{1}+\ldots+R_{N(u)},\]6  
On the other hand, we have, similarly,  
\[W(u)=R_{1}+\ldots+R_{N(u)},\]7 \[W(u)=R_{1}+\ldots+R_{N(u)},\]8  
The key observation leading to (12) is that the difference  
\[W(u)=R_{1}+\ldots+R_{N(u)},\]9 \[X_{1}+\ldots+X_{k}\leq u<X_{1}+\ldots+X_{k+1}.\]0  
is independent of \(t\). In view of this fact, it is clear that the earlier observed inequality  
\[X_{1}+\ldots+X_{k}\leq u<X_{1}+\ldots+X_{k+1}.\]1  
entails  
\[X_{1}+\ldots+X_{k}\leq u<X_{1}+\ldots+X_{k+1}.\]2 \[X_{1}+\ldots+X_{k}\leq u<X_{1}+\ldots+X_{k+1}.\]3which immediately implies (12). **Proof of Proposition 3. ** If \(t\leq T\), then, according to (3)  
\[X_{1}+\ldots+X_{k}\leq u<X_{1}+\ldots+X_{k+1}.\]4  
and  
\[ \begin{split}\mathrm{E}(Q_{(s,0)}(t,L))-(T-s)c=g\mathrm{P}(L\leq t -s)+(h+(t-s)m)\mathrm{P}(L>t-s)\\ -c(\mathrm{E}(L1_{\{L\leq t-s\}})+(t-s)\mathrm{P}(L>t-s))\\ =\mathrm{E}(R(t-s))-c\mathrm{E}(X(t-s))=(q_{t-s}-c)\mathrm{E}(X( t-s)),\end{split} \]  
where in the last line we use notation (10) and (11). Since by definition, \(c\leq q_{t-s}\), the obtained equality  
\[\mathrm{E}(Q_{(s,0)}(t,L))=(T-s)c+(q_{t-s}-c)\mathrm{E}(X(t-s))\]  
implies that, provided \(t_{(0,0)}+s\leq T\),  
\[f^{*}_{(s,0)}=\min_{t}\mathrm{E}(Q_{(s,0)}(t,L))=(T-s)c,\]  
and we conclude  
\[f^{*}_{(s,0)}-f^{*}_{(s,0)}=(s_{2}-s_{1})c. \tag{13}\]  
Furthermore, we see that  
\[t_{(s,0)}=t_{(0,0)}+s,\]  
which, together with Proposition 2, yields (7). It remains to prove that \(b_{(s,a)}=b_{a}\), or equivalently,  
\[f^{*}_{(s,a)}=f^{*}_{(0,a)}-sc.\]  
Since  
\[