Towards railway traffic management using switching Max-plus-linear systems

Abstract

In this paper we present a railway traffic model and a model predictive controller for online railway traffic management of railway networks with a periodic timetable. The main aim of the controller is to recover from delays in an optimal way by changing the departure of trains, by breaking connections, by splitting joined trains, and - in the case of multiple tracks between two stations - by redistributing the trains over the tracks. The railway system is described by a switching max-plus-linear model. We assume that measurements of current running and dwell times and estimates of future running times and dwell times are continuously available so that they can be taken into account in the optimization of the system’s control variables. The switching max-plus-linear model railway model is used to determine optimal dispatching actions, based on the prediction of the future arrival and departure times of the trains, by recasting the dispatching problem as a Mixed Integer Linear Programming (MILP) problem and solving it. Moreover, we use properties from max-plus algebra to rewrite and reduce the model such that the MILP problem can be solved in less time. We also apply the algorithm to a model of the Dutch railway network.

Appendix

1.1 A.1 Headway constraints with control variables

If u _{i l} (k − μ _i,l ) = 0 in Eqs. 37–40, then $\overline {u_{il}(k-\mu _{i,l})}=\varepsilon$ and the equations become:

$$\begin{array}{@{}rcl@{}} d_{i}(k)\geq d_{l}(k-\mu_{i,l})\otimes\tau_{\mathrm{h,d},i,l} (k)\otimes 0&=&d_{l}(k)\otimes\tau_{\mathrm{h,d},i,l} (k)\\ a_{i}(k)\geq a_{l}(k-\mu_{i,l})\otimes \tau_{\mathrm{h,a},{i,l}}(k)\otimes 0&=&a_{l}(k)\otimes \tau_{\mathrm{h,a},{i,l}}(k)\\ d_{l}(k-\mu_{i,l})\geq d_{i}(k)\otimes \tau_{\mathrm{h,d},{l,i}}(k-\mu_{i,l})\otimes \varepsilon&=&\varepsilon\\ a_{l}(k-\mu_{i,l})\geq a_{i}(k)\otimes \tau_{\mathrm{h,d},{l,i}}(k-\mu_{i,l})\otimes \varepsilon&=&\varepsilon. \end{array} $$

The first two equations are identical to Eqs. 5 and 6, and the last two equations simply state that d _l (k − μ _i,l ) and a _l (k − μ _i,l ) should be larger than ε, which they always are. This results in the default order of the train runs: first l, then i.

If u _i,l (k − μ _i,l ) = ε, then $\overline {u_{i,l}(k-\mu _{i,l})}=0$ and the equations become:

$$\begin{array}{@{}rcl@{}} d_{i}(k)&\geq& d_{l}(k-\mu_{i,l})\otimes\tau_{\mathrm{h,d},i,l} (k)\otimes \varepsilon=\varepsilon\\ a_{i}(k)&\geq& a_{l}(k-\mu_{i,l})\otimes \tau_{\mathrm{h,a},{i,l}}(k)\otimes \varepsilon=\varepsilon\\ d_{l}(k-\mu_{i,l})&\geq& d_{i}(k)\otimes \tau_{\mathrm{h,d},l,i} (k-\mu_{i,l})\otimes 0=d_{i}(k)\otimes \tau_{\mathrm{h,d},l,i} (k-\mu_{i,l})\\ a_{l}(k-\mu_{i,l})&\geq& a_{i}(k)\otimes \tau_{\mathrm{h,d},{l,i}}(k-\mu_{i,l})\otimes 0=a_{i}(k)\otimes \tau_{\mathrm{h,a},{l,i}}(k-\mu_{i,l}) \end{array} $$

Now the first two equations simply state that d _i (k) and a _i (k) should be larger than ε, which they always are, and the last two equations are equal to Eqs. 35 and 36. This results in the changed order of train runs: first i, then l.

1.2 A.2 Infeasible train order

Consider the headway constraints of three trains running over the same track in the same cycle. Let us assume that the headway times between all three trains are 3 minutes, then the headway constraints between the departures are given by

$$\begin{array}{@{}rcl@{}} A_{0,4,\mathrm{d}}(u(k),k)=\left[\begin{array}{llll} \quad\varepsilon\quad\quad\quad 3\otimes \overline{u_{1}(k)}\quad3\otimes \overline{u_{3}(k)}\\ 3\otimes u_{1}(k)\quad\,\,\,\quad \varepsilon\quad\quad\,\, 3\otimes \overline{u_{2}(k)}\\ 3\otimes u_{3}(k)\,\, 3\,\,\otimes u_{2}(k)\quad\quad\,\,\,\, \varepsilon \end{array}\right] \end{array} $$

where u ₁(k) determines the order between train 1 and 2, u ₂ determines the order between train 2 and 3, and u ₃(k) determines the order between train 1 and 3. Now if we set u ₁(k) = 0, then train 1 traverses the track before train 2; if we set u ₂(k) = 0, then train 2 traverses the track before train 3; and if we set u ₃(k) = ε, then train 3 traverses the track before train 1. But this is impossible because if train 1 traverses the track before train 2 and 2 traverses the track before 3 then u ₁ = 0 together with u ₂ = 0 implies that train 1 traverses the track before train 3, which is exactly the opposite of what u ₃ = ε implies. Clearly the combination u ₁ = 0, u ₂ = 0, and u ₃ = ε is an infeasible combination and can be removed from the model without having an effect on the feasible train orders.

These infeasible train orders can be derived from the matrix powers of A _0,4,d:

$$\begin{array}{@{}rcl@{}} A_{0,4,\mathrm{d}}^{\otimes^{2}}(u(k),k)=\left[\begin{array}{ccc} 6\otimes \overline{u_{1}(k)}\otimes u_{1}(k) &6\otimes u_{2}(k)\otimes \overline{u_{3}(k)} & 6\otimes \overline{u_{1}(k)}\otimes \overline{u_{2}(k)} \\ 6\otimes \overline{u_{2}(k)}\otimes u_{3}(k) &6\otimes u_{2}(k)\otimes \overline{u_{2}(k)} &6\otimes u_{1}(k)\otimes \overline{u_{3}(k)} \\ 6\otimes u_{1}(k)\otimes u_{2}(k) &6\otimes \overline{u_{1}(k)}\otimes u_{3}(k) &6\otimes u_{3}(k)\otimes \overline{u_{3}(k)} \end{array}\right]\\ =\left[\begin{array}{ccc} \varepsilon & 6\otimes u_{2}(k)\otimes \overline{u_{3}(k)} & 6\otimes \overline{u_{1}(k)}\otimes \overline{u_{2}(k)} \\ 6\otimes \overline{u_{2}(k)}\otimes u_{3}(k) & \varepsilon &6\otimes u_{1}(k)\otimes \overline{u_{3}(k)} \\ 6\otimes u_{1}(k)\otimes u_{2}(k) & 6\otimes \overline{u_{1}(k)}\otimes u_{3}(k) & \varepsilon \end{array}\right] \end{array} $$

The diagonal elements in $A_{0,4,\mathrm {d}}^{\otimes ^{2}}(u(k),k)$ are ε by definition since $\overline {u_{i}(k)}\otimes u_{i}(k)=\varepsilon$ and ε⊗a = ε. The next matrix power of A _0,4,d is:

$$\begin{array}{@{}rcl@{}} A_{0,4,\mathrm{d}}^{\otimes^{3}}(u(k),k)=\left[ \begin{array}{c} 9\otimes (u_{1}(k)\otimes u_{2}(k)\otimes \overline{u_{3}(k)}\oplus \overline{u_{1}(k)} \otimes \overline{u_{2}(k)} \otimes u_{3}(k))\\ 9 \otimes u_{1}(k)\otimes u_{2}(k)\otimes \overline{u_{2}(k)}\\ 9 \otimes u_{2}(k)\otimes \overline{u_{2}(k)}\otimes u_{3}(k) \end{array}\ldots\right.\\ \left.\begin{array}{ccc} 9 \otimes \overline{u_{1}(k)}\otimes u_{3}(k)\otimes \overline{u_{3}(k)}\\ 9 \otimes(u_{1}(k)\otimes u_{2}(k)\otimes \overline{u_{3}(k)}\oplus \overline{u_{1}(k)} \otimes \overline{u_{2}(k)} \otimes u_{3}(k))\\ 9 \otimes u_{2}(k)\otimes u_{3}(k)\otimes \overline{u_{3}(k)} \end{array}\right.\ldots\\ \left.\begin{array}{ccc} 9 \otimes u_{1}(k)\otimes \overline{u_{1}(k)}\otimes \overline{u_{3}(k)}\\ 9 \otimes u_{1}(k)\otimes \overline{u_{1}(k)}\otimes \overline{u_{2}(k)}\\ 9 \otimes(u_{1}(k)\otimes u_{2}(k)\otimes \overline{u_{3}(k)}\oplus \overline{u_{1}(k)} \otimes \overline{u_{2}(k)} \otimes u_{3}(k)) \end{array}\right] \end{array} $$

All non-diagonal elements are ε by definition. The diagonal elements are positive if $u_{1}(k)\otimes u_{2}(k)\otimes \overline {u_{3}(k)}=0$ or $\overline {u_{1}(k)} \otimes \overline {u_{2}(k)} \otimes u_{3}(k)=0$. That means these combinations of control variables correspond to infeasible train orders. So by simply determining the matrix powers, which has to be done to determine the explicit model anyway, and looking at the diagonal elements the infeasible combinations of control variables can be found. These combinations of control variables can be removed by replacing $u_{1}(k)\otimes u_{2}(k)\otimes \overline {u_{3}(k)}$ and $\overline {u_{1}(k)} \otimes \overline {u_{2}(k)} \otimes u_{3}(k)$ with ε.

To ensure the MILP does not use these combinations of control variables constraints corresponding to the following max-plus-linear inequalities are added to the explicit MILP problem formulation:

$$\begin{array}{@{}rcl@{}} \overline{u_{1}(k)} \otimes \overline{u_{2}(k)} \otimes u_{3}(k)&\leq\varepsilon\\ u_{1}(k)\otimes u_{2}(k)\otimes \overline{u_{3}(k)}&\leq \varepsilon \end{array} $$

1.3 A.3 Switching between tracks

As an example consider two parallel tracks, track 1 and 2, with two trains on each of the tracks. Trains 1 and 2 traverse track 1 and trains 3 and 4 traverse track 2. All trains are running in the same cycle. The timetable period is 1 hour, and the timetable vector is r(0)=[0 30 5 35 15 50 20 55]^⊤. We can define the following sets and vectors based on this example:

$$\begin{array}{@{}rcl@{}} \begin{array}{llll}T_{1}=[1\, 2]^{\top} &\quad T_{2}=[3 \,4]^{\top} &\quad \tilde{T}_{1}=[1 \,2 \,3 \,4]^{\top} &\quad \\ \mathcal{H}_{1}=\emptyset &\quad \mathcal{H}_{2}=\{1\} &\quad \mathcal{H}_{3}=\emptyset &\quad \mathcal{H}_{4}=\{3\} \end{array} \end{array} $$

Using these sets, vectors and Eq. 47 we can determine the entries of $\tilde {A}_{\mu ,4,\mathrm {d},1}$. Since all trains are in the same cycle only μ = 0 needs to be considered. The first if-condition of Eq. 47 results on the following non-ε elements of $\tilde {A}_{0,4,\mathrm {d},1}$:

$$\begin{array}{@{}rcl@{}} [\tilde{A}_{0,4,\mathrm{d},1}]_{1,2}&=&\tau_{\mathrm{h,d},1,2} (k)\otimes u_{1,2}(k)\otimes (w_{1}(k)\otimes w_{2}(k)\oplus \overline{w_{1}(k)}\otimes \overline{w_{2}(k)})\\ \text{[}\tilde{A}_{0,4,\mathrm{d},1}]_{3,4}&=&\tau_{\mathrm{h,d},3,4} (k)\otimes u_{3,4}(k)\otimes (w_{3}(k)\otimes w_{4}(k)\oplus \overline{w_{3}(k)}\otimes \overline{w_{4}(k)}) \end{array} $$

and the second if-condition results in the following non- ε elements of $\tilde {A}_{0,4,\mathrm {d},1}$:

$$\begin{array}{@{}rcl@{}} [\tilde{A}_{0,4,\mathrm{d},1}]_{2,1}&=&\tau_{\mathrm{h,d},2,1} (k)\otimes \overline{u_{1,2}(k)}\otimes (w_{1}(k)\otimes w_{2}(k)\oplus \overline{w_{1}(k)}\otimes \overline{w_{2}(k)})\\ \text{[}\tilde{A}_{0,4,\mathrm{d},1}]_{4,3}&=&\tau_{\mathrm{h,d},4,3}(k)\otimes \overline{u_{3,4}(k)}\otimes (w_{3}(k)\otimes w_{4}(k)\oplus \overline{w_{3}(k)}\otimes \overline{w_{4}(k)}). \end{array} $$

The first two if-conditions result in the headway times between departure events of the trains, that traverse the same track during nominal operation, with the addition of the control variables for track selection. The third if-condition results in the following non- ε elements $\tilde {A}_{0,4,\mathrm {d},1}$:

$$\begin{array}{@{}rcl@{}} [\tilde{A}_{0,4,\mathrm{d},1}]_{1,3}&=&\tau_{\mathrm{h,d},1,3} (k)\otimes u_{1,3}(k)\otimes \left(w_{1}(k)\otimes \overline{w_{3}(k)}\oplus\overline{w_{1}(k)}\otimes w_{3}(k)\right)\\ \text{[}\tilde{A}_{0,4,\mathrm{d},1}]_{1,4}&=&\tau_{\mathrm{h,d},1,4} (k)\otimes u_{1,4}(k)\otimes (w_{1}(k)\otimes \overline{w_{4}(k)}\oplus\overline{w_{1}(k)}\otimes w_{4}(k))\\ \text{[}\tilde{A}_{0,4,\mathrm{d},1}]_{2,4}&=&\tau_{\mathrm{h,d},2,4} (k)\otimes u_{2,4}(k)\otimes (w_{2}(k)\otimes \overline{w_{4}(k)}\oplus\overline{w_{2}(k)}\otimes w_{4}(k))\\ \text{[}\tilde{A}_{0,4,\mathrm{d},1}]_{3,2}&=&\tau_{\mathrm{h,d},{3,2}}(k)\otimes u_{3,2}(k)\otimes (w_{3}(k)\otimes \overline{w_{2}(k)}\oplus\overline{w_{3}(k)}\otimes w_{2}(k)), \end{array} $$

and the fourth if-condition results in the final non- ε elements $\tilde {A}_{0,4,\mathrm {d},1}$:

$$\begin{array}{@{}rcl@{}} [\tilde{A}_{0,4,\mathrm{d},1}]_{3,1}&=\tau_{\mathrm{h,d},3,1} (k)\otimes \overline{u_{1,3}(k)}\otimes (w_{1}(k)\otimes \overline{w_{3}(k)}\oplus\overline{w_{1}(k)}\otimes w_{3}(k))\\ \text{[}\tilde{A}_{0,4,\mathrm{d},1}]_{4,1}&=\tau_{\mathrm{h,d},4,1} (k)\otimes \overline{u_{1,4}(k)}\otimes (w_{1}(k)\otimes \overline{w_{4}(k)}\oplus\overline{w_{1}(k)}\otimes w_{4}(k))\\ \text{[}\tilde{A}_{0,4,\mathrm{d},1}]_{4,2}&=\tau_{\mathrm{h,d},4,2} (k)\otimes \overline{u_{2,4}(k)}\otimes (w_{2}(k)\otimes \overline{w_{4}(k)}\oplus\overline{w_{2}(k)}\otimes w_{4}(k))\\ \text{[}\tilde{A}_{0,4,\mathrm{d},1}]_{2,3}&=\tau_{\mathrm{h,d},2,3} (k)\otimes \overline{u_{3,2}(k)}\otimes (w_{3}(k)\otimes \overline{w_{2}(k)}\oplus\overline{w_{3}(k)}\otimes w_{2}(k)). \end{array} $$

These headway times define the order of the trains if a train from track 1 switches to track 2 or the other way around. For the headway times between arrival events only τ _h,d,i,j (k) needs to be replaced with τ _h,a,i,j (k).

1.4 A.4 Matrix multiplication using graph theory

Consider the matrix A ₀

$$A_{0}=\left[\begin{array}{ll}A_{\mathrm{a}}& A_{\mathrm{b}}\\ A_{\mathrm{c}} & A_{\mathrm{d}} \end{array}\right], $$

and its graph $\mathcal {G}(A_{0})$ given in Fig. 10.

Fig. 10

Graph $\mathcal {G}(A_{0})$

According to graph theory [A ^{⊗ m}] _i,j corresponds to the maximum weight of all paths of length m from node j to node i in the precedence graph of A. If we apply this to matrix A ₀ for element $[A_{0}^{\otimes ^{2}}]_{1,1}$, then we should look at all paths of length 2 in the graph that start and end at node 1. There are two paths of length 2 from node 1 to node 1. One consists of taking the edge from node 1 to node 1 twice resulting in a weight of $A_{\mathrm {a}}^{\otimes 2}$. The other path of length two consists of taking the edge from node 1 to node 2 and the edge from node 2 to node 1, resulting in a path weight of A _b⊗A _c. The value of $[A_{0}^{\otimes ^{2}}]_{1,1}$ is the maximum of the weight of both paths: $[A_{0}^{\otimes ^{2}}]_{1,1}=A_{\mathrm {a}}^{\otimes 2}\oplus A_{\mathrm {b}}\otimes A_{\mathrm {c}}$. For the other elements of $A_{\mathrm {a}}^{\otimes 2}$ this results in:

$$\begin{array}{@{}rcl@{}} \left[A_{0}^{\otimes^{2}}\right]_{1,2}&=&A_{\mathrm{a}}\otimes A_{\mathrm{b}}\oplus A_{\mathrm{b}}\otimes A_{\mathrm{d}}\\ \left[A_{0}^{\otimes^{2}}\right]_{2,1}&=&A_{\mathrm{c}}\otimes A_{\mathrm{a}}\oplus A_{\mathrm{d}}\otimes A_{\mathrm{c}}\\ \left[A_{0}^{\otimes^{2}}\right]_{2,2}&=&A_{\mathrm{d}}^{\otimes 2}\oplus A_{\mathrm{c}}\otimes A_{\mathrm{b}} \end{array} $$

For the elements of $A_{0}^{\otimes ^{3}}$ we should look at all paths of length 3. There are four path of length 3 starting at node 1 and ending at node 1

• Take the edge from node 1 to node 1 three times. This path has a weight of $A_{\mathrm {a}}^{\otimes 3}$.

• Take the edge from node 1 to node 1 a single time and then go to node 2 and back. This path has a weight of A _a⊗A _b⊗A _c.

• First go to node 2 and back, then take the edge from 1 node to node 1. This path has a weight of A _b⊗A _c⊗A _a.

• Take the edge from node 1 to node 2, then the edge from node 2 to node 2, and finally the edge from node 2 to node 1. This path has a weight of A _b⊗A _d⊗A _c.

This can also be done for the other elements and for all elements of all other matrix powers. By looking at the graph of A ₀ and the relation between paths in the graph and elements of the matrix powers of the matrix, it is clear only a limited number of paths is possible and that there is a clear structure in these paths. For paths starting and ending in node 1, if the edge from node 1 to node 2 is taken, the edge from node 2 to node 1 must also be in the path. If the edge from node 2 to node 2 is in the path then the edges from node 1 to node 2 and from node 2 to node 1 must also be in the path. The edge from node 1 to node 1 can only be in the path before the edge from node 1 to node 2, after the edge from node 2 to node 1, and before and after itself. And the number of edges is equal to the matrix power. Such rules can be set up for all four of the elements and result in Eqs. 59–62 in Section 4.