Management Engineering - Game Theory

Full exam

GAME THEORY 5 cfu February 1, 2022Surname: Name: Matricola: The exam, or the single exercise, will not be evaluated if: ˆThe surname (last name) and name are not clearly written in the right place; ˆThe explanations are not clearly written and contain cancellations; Grades: Exercise 1:2+3+3, Exercise 2: 2+2+3+3Exercise 1 Given the two player game, described by the following bimatrix, wherea; bare real parameters and digits represent utilities: 0 @(1 ;1) (3;2) (2;3) (6; b) (5;2) (1;6) (3;5) (a;5) (2;2)1 A 1.Finda; b(if any) such that the game is a potential game; 2. ndqsuch that the third row is the rst player's best reaction to the strategy(q;0;1q)of the second player; 3. nda; b(if any) such that there is a NEp with support the second and third row for the rst player and rst and second column for the second player. Write all NEp in such cases. Answer 1The game is never a potential game. Answer 2 0 q14 Answer 3 a > 5; b= 2the NEp are[(0; p;1p);(a 5a 2)] with0< p < > > > > :x 1+ x 2+ x 3= 5 xi 1 x1+ x 2 4 x1+ x 3 2 x3+ x 2 2 3.Observe that for anyj >2, it isv(A[ fjg)v(A) = 1, so that for every semivaluesit iss j= 1 . For Shapley, again, symmetry and eciency allow to compute 1and  2and conclude. Instead to compute 1= 2it holds 1=12 n 1 2n 2
1 + 2n 2
3 = 2 where the term2n 2
1depends from the fact that there are2n 2 coalitionsAnot containing player 2 for which v(A[ f1g)v(A) = 1and the second term is relative to the other coalitions that contain players 2 for whichv(A[ f1g)v(A) = 3. 4.Using eciency and the fact thatx 1+ x 2+ x 3+


+x j1+ x j+1+


+x n n1 + 2we have thatx j= 1 for all j >2, to conclude we have to consider eciency again and the conditionsx 1+ x 2 4; x 1 1; x 2 1. 2