Figure 1.
Example of band splitting: the bandwidth is split into 36 different time-frequency zones corresponding to the different colors. They are designed to accommodate an FMCW chirp with a bandwidth of 300 MHz, and a chirp time of 20 s with a 4 s reset time and with a receiver bandwidth of 30 MHz (up to 40 MHz). Time/frequency zones are spaced apart by 4 s in time and 333 MHz in frequency, to fill the 1 GHz available bandwidth. With a duty cycle of 50%, the bandwidth is split into 6 × 3 × 2 = 36 zones.
Figure 1.
Example of band splitting: the bandwidth is split into 36 different time-frequency zones corresponding to the different colors. They are designed to accommodate an FMCW chirp with a bandwidth of 300 MHz, and a chirp time of 20 s with a 4 s reset time and with a receiver bandwidth of 30 MHz (up to 40 MHz). Time/frequency zones are spaced apart by 4 s in time and 333 MHz in frequency, to fill the 1 GHz available bandwidth. With a duty cycle of 50%, the bandwidth is split into 6 × 3 × 2 = 36 zones.
Figure 2.
Example of band splitting: the bandwidth is split into 9 different time-frequency zones corresponding to the different colors. They are designed to accommodate a FMCW frame with a bandwidth of around 300 MHz, an emitting time of 22 msm and a duty-cycle of 33% to accommodate 2 other radars in the same frequency band without overlapping emissions. This splits the radar band into 3 × 3 = 9 zones.
Figure 2.
Example of band splitting: the bandwidth is split into 9 different time-frequency zones corresponding to the different colors. They are designed to accommodate a FMCW frame with a bandwidth of around 300 MHz, an emitting time of 22 msm and a duty-cycle of 33% to accommodate 2 other radars in the same frequency band without overlapping emissions. This splits the radar band into 3 × 3 = 9 zones.
Figure 3.
Illustration of line-of-sight (left) where two radars R1 and R2 see each other directly. Illustration of indirect line-of-sight (right) where the signal from radar R1 bounces on the back of vehicle V3, generating an echo at 180° that is in line-of-sight of radar R2.
Figure 3.
Illustration of line-of-sight (left) where two radars R1 and R2 see each other directly. Illustration of indirect line-of-sight (right) where the signal from radar R1 bounces on the back of vehicle V3, generating an echo at 180° that is in line-of-sight of radar R2.
Figure 4.
Snapshot from the simulation. Each line-of-sight, direct, (solid green) or indirect (dashed red) between two radars of two cars (blue) are represented with a line.
Figure 4.
Snapshot from the simulation. Each line-of-sight, direct, (solid green) or indirect (dashed red) between two radars of two cars (blue) are represented with a line.
Figure 5.
Example of coloring graph with 3 radars, 2 colors, and 3 timesteps, with the following adjacency matrices (denoted with the dotted lines): . By using the order [,,] at the start, the shortest path for is choosing color 1 and keeping it as there are no conflicts. The shortest path for is then to pick color 2 to avoid conflicting with at timesteps 1 and 2. Finally, the shortest path for is to chose color 2 until timestep 2 to avoid conflicts with , then change to color 1 to avoid conflicting with . These paths are denoted by the thick red arrows. In total, this solution yields 0 conflict and 1 color change.
Figure 5.
Example of coloring graph with 3 radars, 2 colors, and 3 timesteps, with the following adjacency matrices (denoted with the dotted lines): . By using the order [,,] at the start, the shortest path for is choosing color 1 and keeping it as there are no conflicts. The shortest path for is then to pick color 2 to avoid conflicting with at timesteps 1 and 2. Finally, the shortest path for is to chose color 2 until timestep 2 to avoid conflicts with , then change to color 1 to avoid conflicting with . These paths are denoted by the thick red arrows. In total, this solution yields 0 conflict and 1 color change.
Figure 6.
Small example graph to illustrate the necessity of gates. Without adding the second gate, the first sub-graph to compute its shortest path picks the red color and keeps it, and the second sub-graph picks the green color to avoid conflict with the 1st one. The third will have a conflict when it’s adjacent to the two other. By adding a gate in position 2, it is possible to reshuffle the order of shortest path computation, forcing another sub-graph to change its color. In this example, the first gate has the order [1,2,3] and the second one has the order [3,1,2]. The temporal adjacency matrix is illustrated by the small graphs, left of the coloring graph.
Figure 6.
Small example graph to illustrate the necessity of gates. Without adding the second gate, the first sub-graph to compute its shortest path picks the red color and keeps it, and the second sub-graph picks the green color to avoid conflict with the 1st one. The third will have a conflict when it’s adjacent to the two other. By adding a gate in position 2, it is possible to reshuffle the order of shortest path computation, forcing another sub-graph to change its color. In this example, the first gate has the order [1,2,3] and the second one has the order [3,1,2]. The temporal adjacency matrix is illustrated by the small graphs, left of the coloring graph.
Figure 7.
Solutions are in the form of a list of gates. In this example, the list is empty except for timesteps 0 and k, which contain gates with different permutations.
Figure 7.
Solutions are in the form of a list of gates. In this example, the list is empty except for timesteps 0 and k, which contain gates with different permutations.
Figure 8.
A neighbor of a solution is found by randomly applying one of four different actions. From left to right: changing the order of a gate by selecting a radar and moving it earlier in the order, deleting a gate, adding a random gate, and swapping a gate to another timestep.
Figure 8.
A neighbor of a solution is found by randomly applying one of four different actions. From left to right: changing the order of a gate by selecting a radar and moving it earlier in the order, deleting a gate, adding a random gate, and swapping a gate to another timestep.
Figure 9.
Example of position-based crossover operator (POS). C1 is built by copying a random set from P1 (red) into C1 and filling the blanks following the order of P2 (green).
Figure 9.
Example of position-based crossover operator (POS). C1 is built by copying a random set from P1 (red) into C1 and filling the blanks following the order of P2 (green).
Figure 10.
Example of reverse sequence mutation (RSM). A random section of the order is reversed.
Figure 10.
Example of reverse sequence mutation (RSM). A random section of the order is reversed.
Figure 11.
Number of color changes (log scale) in the best solution for different values of K (does not include solutions with conflicts).
Figure 11.
Number of color changes (log scale) in the best solution for different values of K (does not include solutions with conflicts).
Figure 12.
Distribution of radars using the same resource in the best solution found, depending on their relative position and orientation. This distribution is found by dividing, at each position, the average number of radars using the same color by the average number of radars at this position and multiplying this value by the number of available colors. The resulting value at each position is the ratio of radars with the same color over the expect number of radars with the same color (when colors are random). For example, with the color scale in this figure, an area in yellow is an area where radars are three times more likely to have the same color compared to a random color assignment. Radars use the same resources more often when they are oriented in the same direction (right picture) than in opposite one (left picture). Sharing resources with radars in the opposite direction still happens when they are either far away (>500 m) or multiple lanes to the side. An area of ≈150 m (forward and backward) around a radar will contain very few radars using the same resource.
Figure 12.
Distribution of radars using the same resource in the best solution found, depending on their relative position and orientation. This distribution is found by dividing, at each position, the average number of radars using the same color by the average number of radars at this position and multiplying this value by the number of available colors. The resulting value at each position is the ratio of radars with the same color over the expect number of radars with the same color (when colors are random). For example, with the color scale in this figure, an area in yellow is an area where radars are three times more likely to have the same color compared to a random color assignment. Radars use the same resources more often when they are oriented in the same direction (right picture) than in opposite one (left picture). Sharing resources with radars in the opposite direction still happens when they are either far away (>500 m) or multiple lanes to the side. An area of ≈150 m (forward and backward) around a radar will contain very few radars using the same resource.
Table 1.
Parameters used for the simulated annealing version of the metaheuristic.
Table 1.
Parameters used for the simulated annealing version of the metaheuristic.
Parameter | Value |
---|
| |
| 1 |
| |
| |
| 1000 |
| |
Table 2.
Results from the simulated annealing with the parameters described in
Section 6 for different numbers of available colors. The amount of recoloration in the solutions found by the metaheuristic increases exponentially when lowering the number of available colors
K. Solutions found have no conflict except with
and
.
Table 2.
Results from the simulated annealing with the parameters described in
Section 6 for different numbers of available colors. The amount of recoloration in the solutions found by the metaheuristic increases exponentially when lowering the number of available colors
K. Solutions found have no conflict except with
and
.
| Metaheuristic Results | Benchmark Method |
---|
K | Conflicts | Changes | Conflicts | Changes |
---|
36 | 0 | 6 | 179 | 159 |
35 | 0 | 12 | 190 | 166 |
34 | 0 | 14 | 203 | 177 |
33 | 0 | 18 | 214 | 184 |
32 | 0 | 23 | 226 | 196 |
31 | 0 | 28 | 245 | 208 |
30 | 0 | 38 | 263 | 219 |
29 | 0 | 42 | 288 | 241 |
28 | 0 | 53 | 307 | 252 |
27 | 0 | 69 | 333 | 266 |
26 | 0 | 84 | 368 | 292 |
25 | 0 | 106 | 405 | 317 |
24 | 0 | 126 | 449 | 352 |
23 | 0 | 141 | 491 | 373 |
22 | 0 | 165 | 554 | 416 |
21 | 0 | 213 | 655 | 477 |
20 | 0 | 293 | 767 | 536 |
19 | 0 | 524 | 916 | 610 |
18 | 0 | 1190 | 1148 | 707 |
17 | 10 | 1086 | 1709 | 898 |
16 | 22 | 2357 | 2664 | 1165 |
Table 3.
Results from the genetic algorithm on the sliding window version with the parameters described in
Section 6 for different numbers of available colors. The solutions found have more re-colorations and more conflicts than the non-windowed. The amount of recoloration in the solutions found by the metaheuristic increases exponentially when lowering the number of available colors
K. Most solutions found have no conflict until
.
Table 3.
Results from the genetic algorithm on the sliding window version with the parameters described in
Section 6 for different numbers of available colors. The solutions found have more re-colorations and more conflicts than the non-windowed. The amount of recoloration in the solutions found by the metaheuristic increases exponentially when lowering the number of available colors
K. Most solutions found have no conflict until
.
K | Nb of Conflicts | Nb of Changes |
---|
36 | 0 | 699 |
34 | 0 | 684 |
32 | 0 | 770 |
30 | 0 | 845 |
28 | 0 | 811 |
26 | 0 | 944 |
24 | 0 | 1123 |
22 | 0 | 1060 |
20 | 0 | 1287 |
18 | 123 | 1403 |
16 | 1773 | 1679 |
Table 4.
Parameters used for the genetic algorithm version of the metaheuristic.
Table 4.
Parameters used for the genetic algorithm version of the metaheuristic.
Parameter | Value |
---|
W | 10 |
| 100 |
| |
| 1 |
| |
| |
| 1000 |