# Rendezvous on the Line with Different Speeds and Markers That Can Be Dropped at Chosen Time
## 4 Solution of the Problem without Marker
For this strategy the expressions for the rendezvous times and the rendezvous value of the game are given by:  
\[o_{1}+d_{1}b_{1}t_{1}=a_{1}vt_{1}\]3  
\[o_{1}+d_{1}b_{1}t_{1}=a_{1}vt_{1}\]4  
Since \(v=0.619\) the optimal strategy is a switch to \(a_{1}=1,a_{2}=-1,a_{3}=1\) and \(a_{4}=-1\) for player I and \(d_{1}=1,d_{2}=1,d_{3}=-1\) and \(d_{4}=-1\). In terms of the optimal strategy of Corollary 1 the speeds of player I are \(v_{1}=-v\), \(v_{2}=v\), \(v_{3}=-v\) and \(v_{4}=v\). The speed of player I is large enough to allow the improvement of the rendezvous value by moving from  
\[o_{1}+d_{1}b_{1}t_{1}=a_{1}vt_{1}\]5 Optimal strategy for \(v\in[0.001,0.618]\). \[o_{1}+d_{1}b_{1}t_{1}=a_{1}vt_{1}\]6 Plot of the solution of the rendezvous problem vs. the maximum speed of player I. The computed solution is a plot a 1001 points evaluated at \(n/1000\), \(n\in[0,1000]\). The exact solution has two algebraic forms for \(v<(\sqrt{5}-1)/2\) (exactl\(<\), (13)) or \(v>(\sqrt{5}-1)/2\) (exact \(>\), (14)). the start. The expressions for the rendezvous times and rendezvous value of the game are given by:  
\[o_{1}+d_{1}b_{1}t_{1}=a_{1}vt_{1}\]7 \[o_{1}+d_{1}b_{1}t_{1}=a_{1}vt_{1}\]8  
With direct computations, we see that (13) is better that (14) for \(v\leq(\sqrt{5}-1)/2\), see \[o_{1}+d_{1}b_{1}t_{1}=a_{1}vt_{1}\]9  
It is stated in [26], Chapter 17.1, that the optimal solution is given by (13) for \(v\leq(\sqrt{5}-1)/2\) and by (14) for \(v\geq(\sqrt{5}-1)/2\). With our linear programming approach we first conclude that (13) is optimal for \(v=n/1000\leq(\sqrt{5}-1)/2\) and by (14) for \(v=n/1000\geq(\sqrt{5}-1)/2\), \(n=0,1,2,\ldots\) (discrete values). However, we can say more. Let us denote \(opt(v)\) the function that returns the optimal value of the game when the speed of player I is bounded by \(v\) and \(nextToopt(v)\) the function that returns the next-to-optimal value of the game when the speed of player I is bounded by \(v\). These two functions are decreasing since a strategy for \(v\) is always a strategy for \(v^{\prime}\geq v\). Hence, if we have that \(opt(v)<nextToopt(v+dv)\) and the strategy at \(opt(v)\) is the same as that at \(opt(v+dv)\), it must be that this strategy is optimal in the interval \([v,v+dv]\). By computing \(opt(n/1000)\) and \(nextToopt(n/1000)\), \(n=0,\ldots,1000\), we detect that the condition stated above is satisfied for \(v\in[1/1000,0.618]\) and for \(v\in[0.619,0.990]\). To summarize, we have proved the following theorem. **Theorem 1** ([26], Chapter 17.1). : _For \(v\in[1/1000,0.618]\), the rendezvous value of the game is given by (13) and the optimal strategy is plotted in Figure 3 and for \(v\in[0.619,0.990]\) the rendezvous value of the game is given by (14) and the optimal strategy is plotted in \[o_{2}+d_{1}b_{2}t_{1}+d_{2}b_{2}(t_{2}-t_{1})=a_{1}vt_{1}+a_{2}vt _{2}\]0_  
The interval in which Theorem 1 is stated to be true may be enlarged by computing the numerical solutions on a finer mesh, i.e., increasing the number values of \(n\) for which we solve the LP.