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1662537753 June23 2022

Esame completo di GAME THEORY per il corso di Management Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

GAME THEORYEsame completo

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Esame completo di GAME THEORY per il corso di Management Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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GAME THEORY, Management June 23, 2022 Surname: Name: Matricola: Only these papers will be accepted back for evaluation; The exam, or the single question, will not be evaluated if: • The surname (last name) and name are not clearly written in the right place; • The box for the answer either does not contain the answer, or contains answer followed by explanations; • The explanations are not clearly written and contain cancellations. Grades: Exercise 1: 2+2+3+2, Exercise 2: 3+3+3 Exercise 1 Given the following bimatrix game   (3, 4) (2 , 5) (5 , 2) (2, 0) (4 , 4) (6 , 0) (1, a) (5 , 6) ( b, 8)   with a, b two real parameters: 1. find a, b such that the game is a potential game; 2. for every a, b∈ R find the best response of the second player to strategy ( 1 2 , 0, 1 2 ) of the first player; 3. find all mixed Nash equilibria profiles for a < 5 and b = 7; 4. find all mixed Nash equilibria profiles for a < 5 and b = 0. Answer of exercise 1 1. Looking for a potential such that 4 is in the top left position and the first row copying the utilities of the second player in the first row, then the columns are built according to the utilities of the first player: 4 5 2 3 7 3 2 8 b-3 verifying the differences of utilities of the second player in the second and third row: in the second one it is ok, in the third one the conditions become 6− a = 8− 2, b− 3 = 10 providing a = 0, b = 13; 2. twice the utilities of the second player are: 4 + a, first column, 11 second column, 10 third column. Thus the best response multifunction is BR2( 1 2 , 0, 1 2 ) =    (1, 0, 0) if a > 7 (0, 1, 0) if a < 7 (p, 1− p, 0) if a = 7 with 0≤ p≤ 1; 3. For a < 5 the first column is strictly dominated by the second one and the game reduces to:   (2, 5) (5 , 2) (4, 4) (6 , 0) (5, 6) ( b, 8)   . Now the first row…

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