In [1], I discussed a model for the Euclidean Streiner Tree problem. A non-convex integer programming model from the literature was reformulated into a MISOCP (not a version of a Japanese dish, but a Mixed-Integer Second-Order Cone Program). Together with a symmetry-breaking constraint and some high-performance solvers, this can lead to being able to solve larger models.
One of the disadvantages of the model in [1] is that it only works with the full number of Steiner points \(n-2\) where \(n\) is the number of given (original) points. Here I want to discuss how we can use Minimum Spanning Tree (MST) models [2], as a building block for a model where we can add just a few Steiner points. One reason this is interesting: the first few Steiner points have the most impact on the objective.
We can just pick any model from [2]. Let's use the single-commodity flow model:
| MST: Single-commodity flow formulation |
|---|
| \[\begin{align}\min\>&\color{darkred}z = \sum_{(i,j)\in \color{darkblue}A} \color{darkblue}c_{i,j}\color{darkred}x_{i,j}\\& \sum_{j|(i,j)\in \color{darkblue}A}\color{darkred}f_{i,j}=\sum_{j|(j,i)\in\color{darkblue}A}\color{darkred}f_{j,i} + \color{darkblue}b_{i} && \forall i \\ & \color{darkblue}M\cdot \color{darkred}x_{i,j} \ge \color{darkred}f_{i,j} && \forall (i,j) \in \color{darkblue}A \\ & \color{darkred}x_{i,j}\in \{0,1\}\\ & \color{darkred}f_{i,j} \in \{0,1,2,\dots,\color{darkblue}n-1\}\end{align}\] |
To add some Steiner points to this model, we split the set of arcs \(\color{darkblue}A\) into \(\color{darkblue}A_1\) (only given points) and \(\color{darkblue}A_2\) (at least one Steiner point for each arc). This means the lengths of arcs in \(\color{darkblue}A_1\) are constant but in \(\color{darkblue}A_2\) are varying depending on where the Steiner points will be placed. Using a similar SOCP formulation as in [2], we can write:
| MISOCP Model |
|---|
| \[\begin{align}\min\>&\color{darkred}z = \sum_{(i,j)\in \color{darkblue}A_1} \color{darkblue}c_{i,j}\color{darkred}x_{i,j} + \sum_{(i,j)\in \color{darkblue}A_2} \color{darkred}d_{i,j}\\ & \color{darkred}d_{i,j}\ge \color{darkred}\Delta_{i,j}- \color{darkblue}M_1(1-\color{darkred}x_{i,j}) && \forall (i,j)\in \color{darkblue}A_2 \\ & \color{darkred}\Delta^2_{i,j} \ge \sum_c \color{darkred}\delta^2_{i,j,c} && \forall (i,j) \in \color{darkblue}A_2 \\ & \color{darkred}\delta_{i,j,c} = \color{darkred}p_{i,c}-\color{darkred}p_{j,c} && \forall (i,j)\in \color{darkblue}A_2, c \in \{x,y\} \\& \sum_{j|(i,j)\in \color{darkblue}A}\color{darkred}f_{i,j}=\sum_{j|(j,i)\in\color{darkblue}A}\color{darkred}f_{j,i} + \color{darkblue}b_{i} && \forall i \\ & \color{darkblue}M_2\cdot \color{darkred}x_{i,j} \ge \color{darkred}f_{i,j} && \forall (i,j) \in \color{darkblue}A \\ & \color{darkred}p_{i,c}=\color{darkblue}p^0_{i,c} && \text{for given points} \\ & \color{darkred}p_{i,c} \in [L,U] && \text{for Steiner points} \\ & \color{darkred}p_{i,x} \le \color{darkred}p_{i+1,x} && \text{for Steiner points}\\ & \color{darkred}d_{i,j},\color{darkred}\Delta_{i,j}\ge 0 \\ & \color{darkred}x_{i,j}\in \{0,1\}\\ & \color{darkred}f_{i,j} \in \{0,1,2,\dots,\color{darkblue}n-1\}\end{align}\] |
The first three constraints can be combined into \[\color{darkred}d_{i,j} \ge \sqrt{\sum_c\left(\color{darkred}p_{i,c}-\color{darkred}p_{j,c}\right)^2}-\color{darkblue}M(1-\color{darkred}x_{i,j})\] Splitting it in three components can help the solver recognize this as a convex quadratic constraint. Note that the variables \(\color{darkred}p_{i,c}, c\in \{x,y\}\), indicating the position of point \(i\), are constant for the original points and are unknown for the Steiner points. The Steiner points must lie in the convex hull of the given points, so we can deduce lower- and upper-bounds for them. The lowerbound is the smallest \(x/y\) coordinate of th given points, and the upperbound is the largest. We order the Steiner points by \(x\) coordinate.
When we run this model for a random problem with 6 given points, allowing for 0,1,2,3, and 4 Steiner points we see:
The Steiner point is really helpful. After that, they do contribute little if anything. We can see this also in the reporting:
---- 198 PARAMETER Objs
iter0 157.441, iter1 151.967, iter2 151.426, iter3 151.426, iter4 151.426
---- 198 PARAMETER Points
INDEX 1 = iter0
x y
p1.given 85.28326.044
p2.given 97.18232.402
p3.given 21.66333.972
p4.given 86.10168.526
p5.given 44.48571.240
p6.given 59.00826.009
INDEX 1 = iter1
x y
p1.given 85.28326.044
p2.given 97.18232.402
p3.given 21.66333.972
p4.given 86.10168.526
p5.given 44.48571.240
p6.given 59.00826.009
s1.steiner 39.97141.171
INDEX 1 = iter2
x y
p1.given 85.28326.044
p2.given 97.18232.402
p3.given 21.66333.972
p4.given 86.10168.526
p5.given 44.48571.240
p6.given 59.00826.009
s1.steiner 39.97241.171
s2.steiner 93.59933.559
INDEX 1 = iter3
x y
p1.given 85.28326.044
p2.given 97.18232.402
p3.given 21.66333.972
p4.given 86.10168.526
p5.given 44.48571.240
p6.given 59.00826.009
s1.steiner 39.97141.171
s2.steiner 39.97141.171
s3.steiner 93.59933.559
INDEX 1 = iter4
x y
p1.given 85.28326.044
p2.given 97.18232.402
p3.given 21.66333.972
p4.given 86.10168.526
p5.given 44.48571.240
p6.given 59.00826.009
s1.steiner 21.66333.972
s2.steiner 39.97141.171
s3.steiner 93.59933.559
s4.steiner 93.59933.559
Note: The Steiner points are never fixed. So in principle, they can be moved around from one iteration to the next. We see that does not happen here. A good Steiner point stays put (the ordering constraint may give it a different label).
References
- Euclidean Steiner tree problems as MISOCP, https://yetanothermathprogrammingconsultant.blogspot.com/2021/03/euclidean-steiner-tree-problems-as.html
- Minimum Spanning Trees in Math Programming models, https://yetanothermathprogrammingconsultant.blogspot.com/2021/03/minimum-spanning-trees-in-math.html