Informazioni sul documento
- Università
- Politecnico di Milano
- Corso di laurea
- Computer Engineering
- Materia
- Game Theory
- Anno accademico
- 2022-2023
- Classificazione
- Esame · Esame completo
- Contenuto
- Testo d’esame
- Formato originale
- Testo
- Testo ricercabile
Esame completo di Game Theory per il corso di Computer Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.
Esame completo di Game Theory per il corso di Computer Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.
Qualità dell’importazione: il testo è stato estratto direttamente dal documento originale.
Passaggi rappresentativi riconosciuti nelle diverse parti del materiale. Il testo completo resta presente nella pagina per la ricerca, mentre l’anteprima compatta rende più semplice la lettura.
GAME THEORY - January 23, 2023 Last name: First name: ID #: SOLVE THE EXERCISES AND ANSWER THE QUESTIONS ON THESE SHEETS Exercise 1 5 points Four identical items are auctioned among 6 players using a VCG scheme. The marginal valuation for each additional item received by each player are shown in the following table Item P1 P2 P3 P4 P5 P6 1 75 90 95 90 70 60 2 75 40 80 x 50 60 3 50 20 75 40 50 50 4 30 0 30 40 50 20 Thus, the valuation of player 1 if she receives 1 item is v1 = 75, if she gets 2 items is v1 = 150, and so on. Find the smallest value x ∈ N such that P4 gets two items for sure (no tie), and compute how much she has to pay. Answer of exercise 1 In the example the 3 highest marginal valuations are 95,90,90. For x to be the 4-th highest, we need x ≥ 81. So P4 chooses 81 as her second marginal valuation, and she gets two items for a total value 171. If P4 were not there, her items would both go to P1 and P3, for a total value 155. Therefore P4 pays 155. 1 Exercise 2 5 points Given the bargaining problem C = {(x, y) : 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , x + ky ≤ 2} , d = (0, 0) , find the Nash solution for all k ≥ 0. Explain your answer. Answer of exercise 2 If 0 ≤ k ≤ 1, then C = {(x, y) : 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1}, which is symmetric, so that the Nash solution is (1, 1). If k > 1, then the solution must be on the segment S defined by x + ky = 2, x ≤ 1, y ≤ 1. We need to find the largest value of c such that the hyperbola xy = c intersects S. Since the line x + ky = 2 has slope strictly larger than −1, then the intersection is on the line x = 1, and therefore the intersection point, that is the Nash solution, is (1, 1/k). 2 Theory Questions Answer one and only one question (7 points). Only the question 2 may lead to the top grade 30 e lode. 1. Given a symmetric game defined…
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