Timeline for 11 cyclists in a race
Current License: CC BY-SA 4.0
18 events
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| Jan 26, 2023 at 15:11 | comment | added | justhalf | Ah ok I just got what Rob means by 176 possible teams. It's 11 teams of 1 cyclist, and (11 3) teams of 3 cyclists. Then we take (disjoint) combinations of these teams to find the best one. | |
| Jan 26, 2023 at 14:55 | comment | added | RobPratt | @JacobRaihle The number of solutions to constraint (1) is $$1+\binom{11}{3}+\binom{11}{3}\binom{8}{3}/2!+\binom{11}{3}\binom{8}{3}\binom{5}{3}/3! = 20186,$$ but the MILP solver does not explicitly generate them all. | |
| Jan 26, 2023 at 12:23 | comment | added | Jacob is on Codidact | @justhalf I think I had the same hangup, but just caught on. Your concern for double-counting would apply if we were calculating the total number of ways to arrange all three teams (and two individuals) at once. But we're just counting the number of possible "teams" (groups of three or one). Via constraint (1) we can then pick from those possible teams making sure each cyclist is "used" exactly once. There are 2^176 possible arrangements, but only a small fraction (I think 55440, but I'm rusty) satisfy that constraint. | |
| Jan 26, 2023 at 7:37 | comment | added | justhalf | I mean, for the purpose of counting different combinations, it's double counting some, right? (it doesn't affect the optimization result of course) | |
| Jan 25, 2023 at 20:08 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Jan 25, 2023 at 17:20 | comment | added | Jaap Scherphuis | Also, the time that such a leftover person takes must also be taken into account, so it is easier to treat them as a chosen team of size 1 with an associated time. | |
| Jan 25, 2023 at 15:55 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Jan 25, 2023 at 13:20 | comment | added | RobPratt | I have a constraint for each person to appear in exactly one selected team (of one or three). | |
| Jan 25, 2023 at 8:35 | comment | added | justhalf | Yes, but we don't select them, right? It's just a leftover from selecting 9. | |
| Jan 25, 2023 at 8:23 | comment | added | Jaap Scherphuis | @justhalf Because 11 does not split evenly into triplets, there will be riders that are not in a triplet and form a "team" of size 1. | |
| Jan 25, 2023 at 5:11 | comment | added | justhalf | What is the (11 1) for in the total count of combinations? | |
| Jan 25, 2023 at 3:29 | vote | accept | Dmitry Kamenetsky | ||
| Jan 25, 2023 at 3:17 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Jan 25, 2023 at 3:11 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Jan 25, 2023 at 3:07 | comment | added | RobPratt | Yes, 13 optimal solutions. | |
| Jan 25, 2023 at 3:04 | comment | added | Dmitry Kamenetsky | This is correct! Actually there are many equivalent solutions. | |
| Jan 25, 2023 at 2:58 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Jan 25, 2023 at 2:42 | history | answered | RobPratt | CC BY-SA 4.0 |