Section \[sec:ExaApp\]. Since we don’t know a priori the leader’s optimal solution $x^*$, we proceed by guessing the critical item of $KP^{LP}(x^*)$, namely we formulate an Integer Linear Programming (ILP) model where we impose that a given item $c$ must be critical and evaluate the profit of the corresponding split solution. We consider binary variables $k_j$ $(j= 1, \dots, w_c)$ associated with the weight contribution of the critical item and introduce the following model (denoted as $CRIT_1(c)$).

$$\begin{aligned}
\text{min}\quad & \sum\limits_{i=1}^{c-1} p_{i}(1-x_i) \label{eq:ObjCRIT}\\
    \text{subject to}\quad
    & \sum\limits_{i=1}^{n} v_{i}x_{i} \leq C_u \label{eq:capLeadCRIT}\\
    & \sum\limits_{i=1}^{c-1} w_{i}(1-x_{i}) + \sum\limits_{j=1}^{w_c} jk_j = C_l \label{eq:capFollCRIT}\\
    & \sum\limits_{j=1}^{w_c} k_j = 1 \label{eq:capSOSCRIT}\\
    & x_c = 0 \label{eq:NotInterCRIT}\\
    & x_{i} \in\{0,1\} \qquad i= 1,\dots,n \label{eq:varDefXCRIT} \\
    & k_{j} \in\{0,1\} \qquad j= 1,\dots,w_c \label{eq:varDefKCRIT} \end{aligned}$$

The objective function minimizes the value of the split solution. Constraint represents the leader’s capacity constraint. Constraints and ensure that item $c$ is critical as it is the last item packed, with a weight in the interval $[1, w_c]$. Constraint indicates that item $c$ can be critical only if it is not interdicted by the leader. Constraints and indicate that all variables are binary. We can state the following proposition.

\[firstProp\] If there 