Timeline for Linearization the product of three variables (two binary & one continuous)
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2023 at 19:43 | vote | accept | Ahmed | ||
| Mar 21, 2023 at 23:51 | answer | added | RobPratt | timeline score: 5 | |
| Mar 21, 2023 at 23:43 | history | edited | Ahmed | CC BY-SA 4.0 |
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| Mar 21, 2023 at 23:22 | comment | added | Ahmed | @prubin you are correct, sorry for the lack of clarity, I re-wrote the question | |
| Mar 21, 2023 at 23:21 | history | edited | Ahmed | CC BY-SA 4.0 |
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| Mar 21, 2023 at 23:21 | comment | added | Ahmed | @RobPratt I have re-written the question. Sorry about the lack of clarity and thanks your engagement | |
| Mar 21, 2023 at 23:19 | comment | added | RobPratt | Your rewritten question looks better. I see three logical implications that you want to enforce, but I don’t think you want $f(x)=$. | |
| Mar 21, 2023 at 23:12 | history | edited | Ahmed | CC BY-SA 4.0 |
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| Mar 21, 2023 at 22:23 | comment | added | RobPratt | It looks like your correction is backwards. Please see the comment from @prubin. It might help to just write the logical conditions you want and leave the algebraic expressions for the community to provide. | |
| Mar 21, 2023 at 21:41 | comment | added | Ahmed | @RobPratt Thank you. You are correct regarding the copy-and-paste error it was a typo, and I fixed it. In the latter expressions, I have $xyz$ multiplication occurring a couple of times on each side, and I assumed the linearization of such an expression would be similar. and I started with $xyz \le p \le xyz$ for brevity. | |
| Mar 21, 2023 at 20:23 | history | edited | Ahmed | CC BY-SA 4.0 |
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| Mar 21, 2023 at 15:41 | comment | added | prubin♦ | If $x=1$ and $z=1,$ your inequality on $p$ becomes $2y \le p \le y.$ Since $z=1 \implies y\ge 0,$ that will force $y=0=p.$ If $x=1$ and $z=0,$ the inequality becomes $y \le p \le 2y.$ Since $z=0 \implies y\le 0,$ that will again force $y=0=p.$ I'm guessing that is not what you want. | |
| Mar 21, 2023 at 15:26 | history | edited | prubin♦ | CC BY-SA 4.0 |
Fixed a typo.
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| Mar 21, 2023 at 15:00 | comment | added | RobPratt | Welcome to ORSE. Your question needs some clarification. Your first inequality $xyz \le p \le xyz$ is the same as $p = xyz$, and that matches your title, but then you write different nonlinear constraints later. Also, your "$z$ is $1$ when $y$ is positive, and $z$ is $0$ when $y$ is positive" looks like a copy-and-paste error. Finally, the usual linearization requires lower and upper bounds on $y$. | |
| Mar 21, 2023 at 14:34 | answer | added | Sutanu Majumdar | timeline score: 0 | |
| Mar 21, 2023 at 10:25 | answer | added | A.Omidi | timeline score: 1 | |
| Mar 21, 2023 at 7:46 | history | asked | Ahmed | CC BY-SA 4.0 |