Timeline for How can I formulate this specific if-then constraint?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Nov 22, 2020 at 0:11 | vote | accept | MAHER | ||
| Nov 22, 2020 at 0:11 | vote | accept | MAHER | ||
| Nov 22, 2020 at 0:11 | |||||
| Nov 22, 2020 at 0:10 | vote | accept | MAHER | ||
| Nov 22, 2020 at 0:10 | |||||
| Nov 22, 2020 at 0:10 | vote | accept | MAHER | ||
| Nov 22, 2020 at 0:10 | |||||
| Nov 20, 2020 at 4:49 | vote | accept | MAHER | ||
| Nov 22, 2020 at 0:10 | |||||
| Nov 19, 2020 at 13:30 | vote | accept | MAHER | ||
| Nov 20, 2020 at 4:47 | |||||
| Nov 17, 2020 at 15:31 | vote | accept | MAHER | ||
| Nov 18, 2020 at 1:20 | |||||
| Nov 17, 2020 at 4:35 | comment | added | MAHER | I just saw this. It's correct now for future readers. | |
| Nov 17, 2020 at 2:02 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Nov 17, 2020 at 1:59 | comment | added | RobPratt | By the way, the sum should be over $d$ in the desired rule. You want, for each worker $i$, to enforce $\sum_d x_{i,d} \ge 6 \implies y_i=1$. | |
| Nov 17, 2020 at 1:55 | comment | added | RobPratt | The point raised by @prubin is that you could have $y_i=1$ even if all $x_i=0$. The objective discourages that, but it is still feasible unless you explicitly prevent it, as I did in my updated answer. | |
| Nov 17, 2020 at 1:51 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Nov 17, 2020 at 1:47 | comment | added | MAHER | @RobPratt So when formulating the problem all constraints should be put together !? The way i see it is that if the following is added xi+(1−yi)≥1for i∈{1,…,6}xi≤yifor i∈{1,…,6} It goes against the first formulation ∑i=16xi−yi≤5 . Please correct if I'm wrong. I'm still learning. | |
| Nov 17, 2020 at 0:32 | comment | added | RobPratt | @prubin I added the converse just now. | |
| Nov 17, 2020 at 0:31 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Nov 16, 2020 at 22:48 | comment | added | prubin♦ | Just for future readers of this question, note that this enforces the "if" part but not the "only if" (sum less than 6 implies $y = 0$). From the context of the original question, $y=1$ increases a cost that is presumably being minimized, so the solver will set $y=0$ whenever it is allowed to. | |
| Nov 16, 2020 at 12:57 | comment | added | MAHER | Thanks a lot! you are amazing! you made it sound too easy. | |
| Nov 16, 2020 at 4:32 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Nov 16, 2020 at 4:23 | history | answered | RobPratt | CC BY-SA 4.0 |