Timeline for Prenucleolus vs. nucleolus
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13, 2019 at 20:07 | comment | added | Eilon | See the edit to my first response | |
| Nov 13, 2019 at 20:07 | history | edited | Eilon | CC BY-SA 4.0 |
added 755 characters in body
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| Nov 13, 2019 at 19:18 | comment | added | user2925716 | I do not follow what if we take $S=\{1\}$ and $x_1=-1000 000 \ , x_2=x_3=500 000$. Then $x(\{1,2,3\})=0$, BUT $E(S,x)$ is virtually unbounded. What is wrong with this objection here? | |
| Nov 12, 2019 at 19:54 | comment | added | user2925716 | My guess was that $S=\{1,2\}$. But then the maximum is at $x=0$. Is there a better $S$ ? | |
| Nov 11, 2019 at 20:48 | comment | added | Eilon | In the prenucleolus, symmetric players obtain the same payoff, hence the prenucleolus has the form (x,x,-2x) for some real number x. One can draw (on the same graph) the excess function E(S,x) = v(S) - x(S), for every nonempty coalition S. Do it. Find the point x where the maximum of these 7 functions is maximal. You will find that the maximum is attained at x=1/4, and the maximum is 1/2. Therefore the nucleolus is (1/4,1/4,-1/2). | |
| Nov 10, 2019 at 16:29 | comment | added | user2925716 | How did you arrive at the maximal excess $-\frac{1}{2}$ ? | |
| Nov 9, 2019 at 17:15 | vote | accept | user2925716 | ||
| Nov 9, 2019 at 6:11 | history | answered | Eilon | CC BY-SA 4.0 |