# Entropy Analysis of a Flexible Markovian Queue with Server Breakdowns
## 2 Model Formulation and Previous Results
### Entropy Solution Using the First Moment  
In a first step, we will calculate the initial probability vector using the maximum entropy principle and compare it with the initial probability vector obtained in the previous section to make sure they are in agreement. To this end, consider the following nonlinear maximization problem:  
\[\text{max} Z=-\sum_{i=0}^{\infty}p_{i}\ln p_{i}\] \[\text{s.t.} \sum_{n=0}^{c-1}a_{n}p_{n}=1+\frac{\alpha_{1}}{\beta_{1}} \tag{8}\] \[(EP) \sum_{i=0}^{\infty}ip_{i}=E(X)\] (9) \[p_{i}\geq 0,\text{ for all }\ i\]  
Constraint (8) is the summability-to-one condition while constraint (9) is the mean system size equation. This maximization problem can be solved by the method of Lagrange multipliers, see for example [PERSON] and [PERSON] [11]. The Lagrangian function associated with problem \((EP)\) is given by:  
\[\mathcal{L}(P_{i},\lambda) = -\sum_{t=0}^{\infty}p_{i}\ln p_{i}+\lambda_{1}\left(\sum_{n=0}^{c- 1}a_{n}p_{n}-1-\frac{\alpha_{1}}{\beta_{1}}\right)\] \[+\lambda_{2}\left(\sum_{i=0}^{\infty}ip_{i}-E(X)\right),\]  
where the vector \(\lambda=(\lambda_{1},\lambda_{2})\) stands for the Lagrange multipliers. Setting the derivative of \(\mathcal{L}(P_{i},\lambda)\) with respect to \(p_{k}\) to zero yields  
\[p_{k}=e^{-1}e^{a_{k}\lambda_{1}}e^{k\lambda_{2}},\quad k=0,1,2,\cdots. \tag{10}\]  
Substituting (10) in the constraints (8) and (9), we get:  
\[e^{a_{0}\lambda_{1}}+e^{a_{1}\lambda_{1}}e^{\lambda_{2}}+e^{a \lambda_{1}}e^{2\lambda_{2}}\frac{1-e^{(c-2)\lambda_{2}}}{1-e^{\lambda_{2}}}+ \frac{e^{c\lambda_{2}}}{1-e^{\lambda_{2}}}=e \tag{11}\]  
\[e^{a_{1}\lambda_{1}}e^{\lambda_{2}}+e^{a\lambda_{1}}e^{3\lambda_ {2}}\frac{1+(c-3)e^{(c-2)\lambda_{2}}-(c-2)e^{(c-3)\lambda_{2}}}{(1-e^{\lambda _{2}})^{2}}\] \[+2e^{2\lambda_{2}}\frac{1-e^{(c-2)\lambda_{2}}}{1-e^{\lambda_{2} }}+\frac{e^{(c+1)\lambda_{2}}}{(1-e^{\lambda_{2}})^{2}}+\frac{ce^{c\lambda_{2 }}}{1-e^{\lambda_{2}}}=eE(X). \tag{12}\]  
All we need to do now is solve numerically the nonlinear system (11) and (12) to find \(e^{\lambda_{1}}\) and then \(e^{\lambda_{2}}\) and then substitute in (10) to obtain the probabilities \(p_{k}\).  
**Example 1**.: _Some numerical tests are conducted here to see how good is the solution procedure proposed in this section. In this sequel, we will refer to the solution obtained analytically as the exact solution, and to the solution obtained using the entropy approach with the first moment as the approximate solution 1. To compare these solutions, we will use the percentage error (\(PE\ 1\)):_  
\[PE\ 1=\left(\frac{\mid\text{exact value}-\text{ approximate value }\ 1\mid}{\text{ exact value}}\right). \tag{13}\]_Let us take a numerical example where \(c=5\) and calculate the initial probability vector \(P_{i}~{}=~{}(p_{0},\cdots,p_{4})\). Assume \(\mu_{1}=2\), \(\mu_{2}=5.5\), \(\alpha_{1}=0.05\), \(\alpha_{2}=0.08\), \(\beta_{1}=0.07\), and \(\beta_{2}=0.06\). Table 1 shows the exact solution, the approximate solution 1, and the percentage error for two different values of the arrival rate \(\lambda\)._  
_When \(\lambda=0.5\), the average \(PE~{}1\) is \(0.6032\) and when \(\lambda=5.5\), the average \(PE~{}1\) is \(0.5432\). The overall average percentage error is \(0.5732\), which can be greatly improved._