Timeline for How to resolve this issue in multi-objective optimization?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2019 at 10:13 | comment | added | KGM | thanks for your very illustrative answer. It helps a lot to grasp. You mentioned that max-min ais a specific case of OWA. Any comment/example or reference for this would be highly appreciated. | |
| Sep 19, 2019 at 7:43 | history | edited | TheSimpliFire♦ | CC BY-SA 4.0 |
Improved formatting
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| Sep 19, 2019 at 6:46 | comment | added | drskd | I have edited the answer to elaborate a bit more on the max min approach! | |
| Sep 19, 2019 at 6:45 | history | edited | drskd | CC BY-SA 4.0 |
added 1611 characters in body
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| Sep 19, 2019 at 6:19 | comment | added | drskd | Since the value of $f(y^1)=12+1$ and $f(y^2)=3+10$ is 13 and $f(y^3)$ is 12, then $y^3$ cannot be optimal by definition (in a maximization problem) because $13>12$. | |
| Sep 18, 2019 at 15:24 | comment | added | KGM | Solution $y^3$ is preferred to solutions $y^1$ and $y^2$. Note that I do have any preference. All the objectives have equal priority. Would you please comment a bit more on "you could only maximize the minimum objective (specific case of the OWA)." Which one is more appropriate in my case, you think? | |
| Sep 18, 2019 at 14:57 | comment | added | KGM | Thank you for the reply. I appreciate it. As I choose all the weights to be 1, does my solver outputs the solution $y^3$ even if the sum is lower than the solutions $y^1$ and $y^2$? In my problem, the expected values of all the objective functions are close to 1 or around 1. | |
| Sep 18, 2019 at 14:51 | vote | accept | KGM | ||
| Sep 18, 2019 at 14:38 | vote | accept | KGM | ||
| Sep 18, 2019 at 14:51 | |||||
| Sep 18, 2019 at 14:37 | history | bounty ended | KGM | ||
| Sep 18, 2019 at 14:25 | review | First posts | |||
| Sep 18, 2019 at 14:47 | |||||
| Sep 18, 2019 at 14:25 | history | answered | drskd | CC BY-SA 4.0 |