## 3 Results  
To simplify the results, the procedure for obtaining the probability that the system is in each of the states described in Table 1 for one of the five maintenance strategies is detailed. For the rest of the maintenance policies the process is the same, therefore only the results will be shown. Maintenance strategy number 2 has been chosen. The maintenance matrix _[M]_ of this example is shown in Figure 3.  
By applying the proposed maintenance policy, the maintenance would be carried out in such a way that the system would never reach states higher than 4 and always returns to state 2, creating a regular sequence of states 2, 3 and 4, as illustrated in Figure 4.  
The maintenance strategy transition graph shows two parts of the process life cycle. The first goes from the initial state (1) to the first state of the periodic stage (2) and the second is a regular sequence of states 2, 3, 4 and 5. The real interest of the study lies in the analysis of this second part. The analysis of the behavior in the stationary state of the process probability distributions is one of the main problems of Markov chains. In particular, the boundary of each process probability distribution should be calculated for each state \(i\) when n tends to infinity and this boundary belongs to the distribution of probability in the space of states. If this boundary exists, it must be unique and the distribution boundary (\(\pi\)) must satisfy the properties shown in Equation (5) [17]:  
\[ \begin{array}{l}\pi(i)\geq 0\;i\;\in S\\ \sum\pi(i)=1\\ \pi(j)=\pi(i).P(i,j)\end{array}  \tag{5}\]  
Figure 4: Maintenance strategy transition graph. Figure 3: Maintenance matrix. By implementing the maintenance strategy in the form of a matrix, Equation (5) becomes Equation (6):  
\[\pi(j)=\pi(i)\cdot[M(i,j)\cdot P(i,j)] \tag{6}\]  
where each component of the vector \(\pi(j)\) corresponds with the time in which the system remains in state \(j\) after reaching the periodic behavior. Furthermore, for the solution to correspond to a process boundary distribution, the Markov chain must have been proven irreducible and aperiodic and the chain states recurrent positive. As for the maintenance policy analysis, the state vector for 20, 40 and 80 observations are first evaluated by Equation (1), which leads to Equations (7)-(9). \[E_{20} = [0.4420,\ 0.3478,\ 0.1919,\ 0.0164,\ 0.0018,\ 0,\ 0,\ 0,\ 0] \tag{7}\] \[E_{40} = [0.1954,\ 0.4780,\ 0.2980,\ 0.0256,\ 0.0029,\ 0,\ 0,\ 0,\ 0]\] (8) \[E_{80} = [0.0382,\ 0.5589,\ 0.3677,\ 0.0316,\ 0.0037,\ 0,\ 0,\ 0,\ 0] \tag{9}\]  
The results lead the authors to conclude that the higher the number of observations, the higher the probability of the system being in states 2 and 3, whereas the probability of the system being in state 1 is less. First element of the state vector of Equation (7) has a value of 0.442. Therefore, there is a 44.2% probability that the system is in the state of degradation 1. When the observations increase, the probability that the system is in state 1 decrease to 19.54% and 3.82% according to Equations (8) and (9) respectively. In the same way, it is proved as the value of the second and third element of the state vector increases. Finally, the system behavior was evaluated in a stationary state, i.e., when n tends to infinity. This was determined by solving Equation (6), which translates into a system of equations made up of nine equations with nine unknowns, the results of which are:  
\[\pi(j)=[0,\ 0.5785,\ 0.3846,\ 0.0331,\ 0.0038,\ 0,\ 0,\ 0,\ 0] \tag{10}\]  
The authors found that the probability of the system being in a stationary state while in state 1, 6, 7, 8 or 9 was nil since they are all outside the periodic sequence. This procedure has been carried out with 5 different maintenance strategies to compare them and determine which of them is more suitable for the system. The results are shown in Table 2.  
The optimal solution was obtained by comparing the results of the Markov chains and relating them to the number of incidents in the corresponding healthcare centers.