# Modeling and Optimizing the System Reliability Using Bounded Geometric Programming Approach
## 3 Geometric Programming Problem: Basic Concepts
### Geometric Programming Problems
According to the Karush-Kuhn-Tucker (K-K-T) conditions, there is a vector \(\lambda^{*}=(\lambda^{*}_{i1},\ldots,\lambda^{*}_{il})\) such that  
\[\lambda^{*}_{ir}\geq 0 \tag{10}\]  
\[\lambda^{*}_{ir}(h_{ir}(p_{*})-1)=0 \tag{11}\]  
\[\frac{\partial h_{i0}(p^{*})}{\partial p_{ij}}+\sum_{r=1}^{l}\lambda^{*}_{ir} \frac{\partial h_{ir}(p^{*})}{\partial p_{ij}}=0 \tag{12}\]  
Because \(x_{ij}=e^{p_{ij}}\) for \(i=1,2,\ldots,n\), \(j=1,2,\ldots,m\), it follows that \(r=1,2,\ldots,l\)  
\[\frac{\partial h_{ir}(p)}{\partial p_{ij}}=\frac{\partial h_{ir}(p)}{\partial x _{ij}}\frac{\partial x_{ij}}{\partial p_{ij}}=\frac{\partial g_{ir}(p)}{ \partial x_{ij}}e^{p_{ij}}\]  
Therefore, the condition (12) is equivalent to  
\[\frac{\partial h_{i0}(p^{*})}{\partial x_{ij}}+\sum_{r=1}^{l}\lambda^{*}_{ir} \frac{\partial h_{ir}(p^{*})}{\partial x_{ij}}=0 \tag{13}\]  
since \(e^{p_{ij}}>0\) and \(x_{ij}>0\). Hence (13) is equivalent to  
\[x^{*}_{ij}\frac{\partial h_{i0}(p^{*})}{\partial x_{ij}}+\sum_{r=1}^{l} \lambda^{*}_{ir}x^{*}_{ij}\frac{\partial h_{ir}(p^{*})}{\partial x_{ij}}=0 \tag{14}\]  
Now, the terms of \(F_{ir}(p)\) are of the form  
\[u_{ir}(p)=c_{rik}\prod_{i=1}^{n}x^{\alpha_{rikj}}_{ij}\]  
It is clear that  
\[x^{*}_{ij}\frac{\partial h_{i0}(p^{*})}{\partial x_{ij}}=\sum_{k=T_{r-1}+1}^{ T_{r}}\alpha_{rikj},\,(i=1,2,\ldots,n;j=1,2,\ldots,n;r=1,2,\ldots,l)\]  
Therefore, (14) implies  
\[\sum_{k=1}^{T_{r}}\alpha_{0 kj}u_{ik}(p^{*})+\sum_{r=1}^{l}\sum_{k=T_{r-1}+1}^{ T_{r}}\lambda^{*}_{ir}u_{ik}(p^{*})=0,(i=1,2,\ldots,n;j=1,2,\ldots,n) \tag{15}\]  
If we divide the last equation by  
\[F_{i0}(p^{*})=\sum_{k=1}^{T_{0}}u_{ik}p^{*}\]  
we obtain  
\[\lambda^{*}_{ir}(h_{ir}(p_{*})-1)=0 \tag{11}\]0Define the vector \(y^{*}_{ik}\) by  
\[y^{*}_{ik}=\left\{ \begin{array}{ll}\frac{u_{ik}(p^{*})}{F_{i0}(p^{*})},&(i=1,2, \ldots,n;k=1,2,\ldots,T_{0})\\ \frac{\lambda_{ir}u_{ik}(p^{*})}{F_{i0}(p^{*})}&(i=1,2,\ldots,n;k=T_{r-1}+1,2, \ldots,T_{r};r=1,2,\ldots,l)\end{array} \right.\]  
Note that \(y^{*}_{ik}>0(i=1,2,\ldots,n;k=1,2,\ldots,T_{0})\) and \(r\geq 1\), either \(y^{*}_{ik}>0\) for all \(k\) with \(T_{r-1}+1\leq k\leq T_{r}\) or \(y_{ik}=0\) for all \(k\) with \(T_{r-1}+1\leq k\leq T_{r}\); according to the corresponding Karush-Kuhn-Tucker multipliers \(y^{*}_{ir}\), \((i=1,2,\ldots,n;r=1,2,\ldots,l)\) is positive or zero. Furthermore, observe that vector \(y^{*}\) satisfies all of the \(m\) exponent constraint equations in DP, as well as the constraint  
\[\sum_{k=1}^{T_{r}}y^{*}_{ik}=\sum_{k=1}^{T_{0}}\frac{u_{ik}(p^{*})}{F_{i0}(p^{ *})}=\frac{F_{i0}(p^{*})}{F_{i0}(p^{*})}=1\]  
Therefore, \(y^{*}=(y_{11},\ldots,y^{*}_{iT_{0}})\) is a feasible vector for DP. Hence DP is constrained. The Karush-Kuhn-Tucker multipliers \(\lambda^{*}_{ir}\) are related to the corresponding \(\lambda_{ir}(y^{*})\) DP as follows:  
\[\lambda_{ir}(y_{*})=\sum_{k=1}^{T_{r}}y^{*}_{ik}=\sum_{k=1}^{T_{0}}\lambda^{*} _{ir}\frac{u_{ik}(p^{*})}{\delta_{i0}(p^{*})}=\lambda^{*}_{ir}\frac{F_{i0}(p^{ *})}{F_{i0}(p^{*})},(i=1,2,\ldots,n;r=1,2,\ldots,l)\]  
The Karush-Kuhn-Tucker condition (11) becomes  
\[\lambda^{*}_{ir}(F_{ir}(p^{*})-1)=0 \tag{16}\]  
Therefore, we obtain  
\[\lambda^{*}_{ir}F_{ir}(p^{*})=\lambda^{*}_{ir}\]  
Therefore, for \(r=1,2,\ldots,l\) and \(k=T_{r-1}+1,\ldots,T_{r}\), we see that  
\[y^{*}_{ik}=\frac{\lambda^{*}_{ir}u_{ik}(p^{*})}{F_{i0}(p^{*})}=\frac{\lambda^ {*}_{ir}F_{ir}(p^{*})u_{ik}(p^{*})}{F_{i0}(p^{*})}=\lambda_{ir}(y^{*})u_{ik}(p ^{*}) \tag{17}\]  
The fact that \(\delta^{*}\) is a feasible for DP and \(x^{*}\) is a feasible for GP implies that  
\[F_{0}(p^{*})=n\sqrt[n]{F(y^{*})}\]  
because of the primal-dual inequality. Moreover, the values of \(y^{*}_{ik}(i=1.2\ldots,m;r=1,2,\ldots,l;k=1,2,\ldots,T_{r-1}+1,\ldots,T_{r})\) are precisely those that force equality in the arithmetic-geometric mean inequalities that where used to obtain the duality inequality. Finally, Equation (17) shows that either \(F_{ir}(p^{*})=1\) or \(y^{*}_{ir}=0(i=1,2,\ldots,n;r=1,2,\ldots,l)\).