# Rendezvous on the Line with Different Speeds and Markers That Can Be Dropped at Chosen Time
## 5 Solution of the Problem with a Marker Held by the Slow Player
However, in this case, the optimal solution is to continue in the same direction, i.e., the equation would be  
\[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]0  
where \(k=1\) if player II does not find the marker and \(k=0\) else, and \(t_{02}\) is the time at which player II finds the marker. This equation reduces to (*). In (*), if a player finds the marker in the interval \([t_{1},t_{2}]\) the optimal strategy is to continue in the same direction and the marker is useless, and so on for (***) and (****) if the marker is found in the interval \([t_{2},t_{3}]\) and \([t_{3},t_{4}]\), respectively. The three remaining sets of equations (**), (***) and (****) are composed of two equations. The first one accounts for the rendezvous of agent \((o_{i},b_{i})\) with player I and the second one is valid only if the marker is used \((k_{1},\ k_{21},\ k_{22},\ k_{31},\ k_{32},\ k_{33}\) equal 1) and we define the times when the marker is found as \(t_{12},t_{22},t_{32}\). The next set of equations is composed of the speed constraints. The variable \(vz\) is the product of the speed of player I and the time \(z\) at which the marker is dropped; this product is bounded by \(v\cdot z\) since the speed of player I is bounded by \(v\). In the results, we observed that the speed of player I is \(v\) (maximal) or 0 but we obtain no solution with \(v\) in between. \[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]1 \[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]2 \[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]3 \[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]4 \[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]5  
The family of linear programs was generated by assigning values to the parameters of Equations (15)-(17). These values must satisfy the constraints (the constraints \(t_{i}\geq 0\) were included in all linear programs)  
\[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]6 \[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]7  
The families of linear programs were solved for a maximal speed of player I ranging from 0 to 1 with a step size of \(1/1000\), i.e., the optimal solutions \(opt(v)\) were computed for \(v=n/1000\), \(n=0,1,\ldots,1000\). The result was that the same strategy was used, see \[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]8 The speed of player I was maximal along the trajectory and the best solution was obtained for \(z\in[0,t_{1}]\). With respect to the notation of Corollary 2 the speeds of player I were \((v_{0}=-v,\ v_{1}=v,\ v_{2}=-v,\ v_{3}=v)\). The marker reduced the rendezvous value even when the speed of player I was very slow. The rendezvous times are given by:  
\[o_{1}+d_{1}b_{1}t_{1}=a_{0}vz+a_{1}vt_{1}\]9 \[k_{1}(o_{2}+d_{1}b_{2}t_{1}+d_{2}b_{2}\Delta t_{2})+\]0  
\[k_{1}(o_{2}+d_{1}b_{2}t_{1}+d_{2}b_{2}\Delta t_{2})+\]1The optimal solution function \(opt(v)\) is decreasing because a strategy for \(v\) is a strategy for \(v^{\prime}\geq v\) as well. Hence, if the value of the next-to-optimal strategy \(nextToopt(v+dv)\) for a speed \(v+dv\) is larger than \(opt(v)\), the strategy that leads to \(opt(v)\) is optimal for speeds in \([v,v+dv)\). With our mesh size of \(1/1000\), we numerically observed that this occurred since \(v\geq 17/1000\). Hence, we have a computer-assisted proof summarized in the following Theorem. **Theorem 2**. : _For \(v\in[17/1000,1]\), the rendezvous value of the game is given by (18). The optimal strategy is plotted in Figure 5 and the optimal rendezvous values in \[k_{1}(o_{2}+d_{1}b_{2}t_{1}+d_{2}b_{2}\Delta t_{2})+\]2_  
The interval in which the Theorem is stated to be true may be enlarged by computing the numerical solutions on a finer mesh. It is relevant to point out that for the optimal strategy, the time at which the marker is found is \(t_{1}\), i.e., the same time as the first rendezvous occurs.