## 5 KKT Conditions and the Fast Solution of Problem (2)
In this regard, we preliminarily set \(i=1\) and consider the next cases from Figure 3, with \(\{y_{i}\}\) being the sequence of _tentative_ solution points of (14):  
* Case (I): We set \(\bar{\alpha}=b_{1}\), \(\bar{\beta}=b_{2}\). If \(\boxed{e_{1}\bar{\alpha}+e_{2}\bar{\beta}\leq e_{3}}\), then set \[P_{1}=\left( \begin{array}{c}b_{1}\\ b_{2}\end{array} \right),\qquad\varphi_{i}=\varphi(b_{1},b_{2}),\qquad y_{i}=P _{1},\qquad i=i+1;\] (16)
* Case (II): We set \(\bar{\alpha}=a_{1}\), \(\bar{\beta}=b_{2}\). If \(\boxed{e_{1}\bar{\alpha}+e_{2}\bar{\beta}\leq e_{3}}\), then set \[P_{2}=\left( \begin{array}{c}a_{1}\\ b_{2}\end{array} \right),\qquad\varphi_{i}=\varphi(a_{1},b_{2}),\qquad y_{i}=P _{2},\qquad i=i+1;\] (17)  
Figure 3: Overview of possible solutions (I)–(XII) for KKT conditions in (15). Case (VIII): We set \(\bar{\beta}=b_{2}\) and possibly compute the solution \(\bar{\alpha}=-(ub_{2}+y)/t\) of the equation \[\frac{d\varphi(\alpha,b_{2})}{d\alpha}=t\alpha+ub_{2}+y=0,\] so that: - if \(\framebox{$t\
eq 0$ AND $\epsilon_{1}\bar{\alpha}+\epsilon_{2}\bar{\beta}\leq\epsilon_{3}$}\) then set \[P_{8}=\left( \begin{array}{c}\bar{\alpha}\\ \bar{\beta}\end{array} \right),\qquad\varphi_{i}=\varphi(\bar{\alpha},\bar{ \beta}),\qquad y_{i}=P_{8},\qquad i=i+1;\] (23) - if \(\framebox{$t=0$ AND $ub_{2}+y\
eq 0$}\) then there is no solution for Case (VIII); - if \(\framebox{$t=0$ AND $ub_{2}+y=0$}\) then set \(\bar{\alpha}\in[a_{1},b_{1}]\) as any value satisfying \(\epsilon_{1}\bar{\alpha}+\epsilon_{2}\bar{\beta}\leq\epsilon_{3}\), and compute \(P_{8}\) as in (23);
* \(\framebox{$\varepsilon_{2}t-\epsilon_{1}u>0$}\) admitting other three sub-cases, namely
* \(\framebox{$u>0$}\), so that we set \[\beta_{1}=\max\biggl{\{}a_{2},-\frac{b_{1}t+y}{u}\biggr{\}}\leq\beta\leq\min \biggl{\{}b_{2},\frac{\epsilon_{3}t+\epsilon_{1}y}{\epsilon_{2}t-\epsilon_{1} u},-\frac{a_{1}t+y}{u}\biggr{\}}=\beta_{2}\]
* \(\framebox{$u=0$}\) so that we set \[\beta_{1}=a_{2}\leq\beta\leq\min\biggl{\{}b_{2},\frac{\epsilon_{3}t+\epsilon_{ 1}y}{\epsilon_{2}t}\biggr{\}}=\beta_{2}\]
* \(u<0\), so that we set \[\beta_{1}=\max\biggl{\{}a_{2},-\frac{a_{1}t+y}{u}\biggr{\}}\leq\beta\leq\min \biggl{\{}b_{2},\frac{\varepsilon_{3}t+\varepsilon_{1}y}{\varepsilon_{2}t- \varepsilon_{1}u},-\frac{b_{1}t+y}{u}\biggr{\}}=\beta_{2}\]
* \(\varepsilon_{2}t-\varepsilon_{1}u=0\), admitting no solution for Case (IX) as long as the condition \(\varepsilon_{3}t+\varepsilon_{1}y<0\) holds. Conversely, in case \(\varepsilon_{3}t+\varepsilon_{1}y\geq 0\), we have the three cases:
* \(u>0\), so that we set $ub_{2}+y\
eq 0$0
* \(u=0\), so that we set $ub_{2}+y\
eq 0$1
* \(u<0\), so that we set $ub_{2}+y\
eq 0$2
* \(\varepsilon_{2}t-\varepsilon_{1}u<0\), corresponding to the three cases:
* \(u>0\), so that we set $ub_{2}+y\
eq 0$3
* \(u=0\), so that we set $ub_{2}+y\
eq 0$4
* \(u<0\), so that we set $ub_{2}+y\
eq 0$5
* \(u=0\), then, recalling that we are in the sub-case where equations \(\varphi_{\alpha}(\alpha,\beta)=0\) and \(\varphi_{\beta}(\alpha,\beta)=0\) yield the same information, with \(tw-u^{2}=0\), we exploit equation \(\varphi_{\alpha}(\alpha,\beta)=0\) with \(t=u=0\). Therefore, we have \[\left\{ \begin{array}{l}a_{2}\leq\beta\leq b_{2}\\ y=0\\ a_{1}\leq\alpha\leq b_{1}\end{array} \right.\] which yield the next two cases:
* \(y=0\) This case implies that the objective function is constant (i.e., \(\varphi(\alpha,\beta)=q\)), so that we set \[\beta_{1}=a_{2}\leq\beta\leq b_{2}=\beta_{2}\]
* \(y\
eq 0\): admitting no solution for Case (IX)
3. \(\boxed{t<0}\) then, recalling that we are again in the sub-case where equations \(\varphi_{\alpha}(\alpha,\beta)=0\) and \(\varphi_{\beta}(\alpha,\beta)=0\) yield the same information, we exploit equation \(\varphi_{\alpha}(\alpha,\beta)=0\) and we set \(\alpha=-(u\beta+y)/t\). Thus, from the bounds and the last inequality in (14), we obtain \[\left\{ \begin{array}{l}a_{2}\leq\beta\leq b_{2}\\ \\ (\varepsilon_{2}t-\varepsilon_{1}u)\beta\geq\varepsilon_{3}t+\varepsilon_{1} y\\ \\ b_{1}t+y\leq-u\beta\leq a_{1}t+y\end{array}