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Data that offers the wanted outcome, it turns out my guess was correct
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Maarten
Maarten
  • 295
  • 1
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I notice you are using FirstSolutionStrategy.Value.PATH_CHEAPEST_ARC. If this name does what it suggests, then it's probably a greedy strategy that chooses between distance 1, 1 and so on.

The clue is that you're using a strategy that does not guarantee optimal solutions. The structure also suggests that a certain random seed is used; there is an iteration above five until the largest node number is reached, an iteration under five until the lowest number is reached and in the end, five is left.

Edit: For any strategy to work, you cannot force it into making random decisions (such as choosing between 1 and 1). This matrix offers enough variation for a simple strategy to work.

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 0, 100, 100, 100, 100, 100, 100, 100, 100],
[2, 100, 0, 100, 100, 100, 100, 100, 100, 100],
[3, 100, 100, 0, 100, 100, 100, 100, 100, 100],
[4, 100, 100, 100, 0, 100, 100, 100, 100, 100],
[5, 100, 100, 100, 100, 0, 100, 100, 100, 100],
[6, 100, 100, 100, 100, 100, 0, 100, 100, 100],
[7, 100, 100, 100, 100, 100, 100, 0, 100, 100],
[8, 100, 100, 100, 100, 100, 100, 100, 0, 100],
[9, 100, 100, 100, 100, 100, 100, 100, 100, 0]

I notice you are using FirstSolutionStrategy.Value.PATH_CHEAPEST_ARC. If this name does what it suggests, then it's probably a greedy strategy that chooses between distance 1, 1 and so on.

The clue is that you're using a strategy that does not guarantee optimal solutions. The structure also suggests that a certain random seed is used; there is an iteration above five until the largest node number is reached, an iteration under five until the lowest number is reached and in the end, five is left.

I notice you are using FirstSolutionStrategy.Value.PATH_CHEAPEST_ARC. If this name does what it suggests, then it's probably a greedy strategy that chooses between distance 1, 1 and so on.

The clue is that you're using a strategy that does not guarantee optimal solutions. The structure also suggests that a certain random seed is used; there is an iteration above five until the largest node number is reached, an iteration under five until the lowest number is reached and in the end, five is left.

Edit: For any strategy to work, you cannot force it into making random decisions (such as choosing between 1 and 1). This matrix offers enough variation for a simple strategy to work.

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 0, 100, 100, 100, 100, 100, 100, 100, 100],
[2, 100, 0, 100, 100, 100, 100, 100, 100, 100],
[3, 100, 100, 0, 100, 100, 100, 100, 100, 100],
[4, 100, 100, 100, 0, 100, 100, 100, 100, 100],
[5, 100, 100, 100, 100, 0, 100, 100, 100, 100],
[6, 100, 100, 100, 100, 100, 0, 100, 100, 100],
[7, 100, 100, 100, 100, 100, 100, 0, 100, 100],
[8, 100, 100, 100, 100, 100, 100, 100, 0, 100],
[9, 100, 100, 100, 100, 100, 100, 100, 100, 0]
I deleted a guess that is probably incorrect, since I am only speculating about how the solver works
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Maarten
Maarten
  • 295
  • 1
  • 9

I notice you are using FirstSolutionStrategy.Value.PATH_CHEAPEST_ARC. If this name does what it suggests, then it's probably a greedy strategy that chooses between distance 1, 1 and so on. If you would change the matrix so that, in symmetric fashion, your distance between the depot and location j equals j, I think the result would be 0 -> 1 -> 0 0 -> 2 -> 0 ... 0 -> 9 -> 0.

The clue is that you're using a strategy that does not guarantee optimal solutions. The structure also suggests that a certain random seed is used; there is an iteration above five until the largest node number is reached, andan iteration under five until the lowest number is reached and in the end, five is left. Let me know if my suggestion worked, good luck.

I notice you are using FirstSolutionStrategy.Value.PATH_CHEAPEST_ARC. If this name does what it suggests, then it's probably a greedy strategy that chooses between distance 1, 1 and so on. If you would change the matrix so that, in symmetric fashion, your distance between the depot and location j equals j, I think the result would be 0 -> 1 -> 0 0 -> 2 -> 0 ... 0 -> 9 -> 0.

The clue is that you're using a strategy that does not guarantee optimal solutions. The structure also suggests that a certain random seed is used; there is an iteration above five until the largest node number is reached, and iteration under five until the lowest number is reached and in the end, five is left. Let me know if my suggestion worked, good luck.

I notice you are using FirstSolutionStrategy.Value.PATH_CHEAPEST_ARC. If this name does what it suggests, then it's probably a greedy strategy that chooses between distance 1, 1 and so on.

The clue is that you're using a strategy that does not guarantee optimal solutions. The structure also suggests that a certain random seed is used; there is an iteration above five until the largest node number is reached, an iteration under five until the lowest number is reached and in the end, five is left.

Source Link
Maarten
Maarten
  • 295
  • 1
  • 9

I notice you are using FirstSolutionStrategy.Value.PATH_CHEAPEST_ARC. If this name does what it suggests, then it's probably a greedy strategy that chooses between distance 1, 1 and so on. If you would change the matrix so that, in symmetric fashion, your distance between the depot and location j equals j, I think the result would be 0 -> 1 -> 0 0 -> 2 -> 0 ... 0 -> 9 -> 0.

The clue is that you're using a strategy that does not guarantee optimal solutions. The structure also suggests that a certain random seed is used; there is an iteration above five until the largest node number is reached, and iteration under five until the lowest number is reached and in the end, five is left. Let me know if my suggestion worked, good luck.