## 5 KKT Conditions and the Fast Solution of Problem (2)  
Replacing the expression of the vector \(x\) in (2) within the objective function, we easily obtain the equivalent problem  
\[\min_{\alpha,\beta}\ \ \varphi(\alpha,\beta) \tag{14}\] \[\mathcal{P}\ :\ \left\{ \begin{array}{l}a_{1}\leq\alpha\leq b_{1} \\ a_{2}\leq\beta\leq b_{2}\\ e_{1}\alpha+e_{2}\beta\leq e_{3}\text{,}\end{array} \right.\]  
where  
\[\left\{ \begin{array}{l}\varphi(\alpha,\beta)=\frac{1}{2}\left(\begin{array}[ ]{cc}\alpha\\ \beta\end{array}\right)^{T}\left(\begin{array}{cc}t&u\\ u&w\end{array}\right)\left(\begin{array}{c}\alpha\\ \beta\end{array}\right)+\left(\begin{array}{c}y\\ h\end{array}\right)^{T}\left(\begin{array}{c}\alpha\\ \beta\end{array}\right)+q\\ t=d^{T}Qd\\ u=d^{T}Qz=z^{T}Qd\\ w=z^{T}Qz\\ y=(Q\bar{x}+b)^{T}d\\ h=(Q\bar{x}+b)^{T}z\\ q=\left(\frac{1}{2}Q\bar{x}+b\right)^{T}\bar{x}+c\text{.}\end{array} \right.\]  
Observe that transforming (2) into (14) only requires the computation of two additional matrix-vector products (i.e., \(Qd\) and \(Qz\)), along with six inner products. The problem (14) is a constrained quadratic problem, such that first-order Fritz-John optimality conditions do not require additional constraint qualifications (since all the constraints are linear). Thus, after considering its Lagrangian function  
\[\mathcal{L}(\alpha,\beta,\mu_{1},\mu_{2},\mu_{3},\mu_{4},\mu_{5}) = \tag{15}\] \[\varphi(\alpha,\beta)-\mu_{1}(\alpha-a_{1})+\mu_{2}(\alpha-b_{1} )-\mu_{3}(\beta-a_{2})+\mu_{4}(\beta-b_{2})+\mu_{5}(e_{1}\alpha+e_{2}\beta-e_ {3})\]  
we have the next set of equalities/inequalities representing the associated KKT conditions:  
\[\left\{ \begin{array}{l}\left(\begin{array}{cc}t&u\\ u&w\end{array}\right)\left(\begin{array}{c}\alpha^{*}\\ \beta^{*}\end{array}\right)+\left(\begin{array}{c}y\\ h\end{array}\right)+\left(\begin{array}{c}-\mu_{1}^{*}+\mu_{2}^{*}+\epsilon_{ 1}\mu_{5}^{*}\\ -\mu_{3}^{*}+\mu_{4}^{*}+\epsilon_{2}\mu_{5}^{*}\end{array}\right)=\left( \begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{c}\alpha^{*}\\ \beta^{*}\end{array}\right)\in\mathcal{P}\\ \mu_{1}^{*}[\alpha^{*}-a_{1}]=0\\ \mu_{2}^{*}[\alpha^{*}-b_{1}]=0\\ \mu_{3}^{*}[\beta^{*}-a_{2}]=0\\ \mu_{4}^{*}[\beta^{*}-b_{2}]=0\\ \mu_{5}^{*}[\epsilon_{1}\alpha^{*}+e_{2}\beta^{*}-\epsilon_{3}]=0\\ \mu_{i}^{*}\geq 0\text{, \ \ \ \ \ }i=1,\ldots,5\text{.}\end{array} \right.\]  
The remaining part of the present section will be devoted to analyze all the possible solutions of (15), with the aim of possibly computing a global minimum for (2). In this regard, exploiting the solutions of (15) evidently undergoes a reduction, allowing us to analyze the cases (I)-(XII) in Figure 3.  
Observing that in (15) the multipliers \(\mu_{i}^{*}\) and \(i=1,\ldots,5\) must fulfill nonnegativity conditions, it is not difficult to realize that computing all the KKT points satisfying (15) can turn out to be a burdensome task, including a number of sub-cases depending on the possible combinations of signs for the parameters \(a_{1}\), \(b_{1}\), \(a_{2}\), \(b_{2}\), \(e_{1}\), \(e_{2}\) and \(e_{3}\). Conversely, a global minimizer for (2) can be equivalently exploited by analyzing all the possible solutions of (15) uniquely in terms of \(\alpha^{*}\) and \(\beta^{*}\), without requiring the computation of the multipliers as well. Hence, we limit our analysis to consider the computation of \(\alpha^{*}\) and \(\beta^{*}\) in the cases (I)-(XII) of Figure 3, where  
* Cases (I), (II), (III), (IV) are associated with possible solutions in the vertices of the box constraints;
* Cases (V), (VI), (VII), (VIII) are associated with possible solutions on the edges of the box constraints;
* Case (IX) represents a possible feasible unconstrained minimizer for the objective function in (2);
* Cases (X), (XI), (XII) are associated with possible solutions, making the last inequality constraint in (14) active. Then, in Lemma 2, we will provide a simple theoretical result which justifies our simplification, with respect to computing all the KKT points.