Document information
- University
- Politecnico di Milano
- Degree programme
- Computer Engineering
- Subject
- Game Theory
- Academic year
- 2016-2017
- Classification
- Exam · First midterm
- Content
- Exam paper only
- Original format
- Text
- Searchable text
First midterm exam for Game Theory in the Computer Engineering degree programme at Politecnico di Milano. The document covers: Game Theory (MID-TERM) 24-11-2016 Surname: Name: Matricola: SOL VE THE EXERCISES AND ANSWER THE QUESTIONS USING ONLY THESE PAPERS Exercise 1 ( 2+1+2+3 ) Given the zero sum game: 0 @ a 3 11 65 1 10 b 1 A , 1. find the conservative values (in pure strategies) of the game; 2. find if
First midterm exam for Game Theory in the Computer Engineering degree programme at Politecnico di Milano. The document covers: Game Theory (MID-TERM) 24-11-2016 Surname: Name: Matricola: SOL VE THE EXERCISES AND ANSWER THE QUESTIONS USING ONLY THESE PAPERS Exercise 1 ( 2+1+2+3 ) Given the zero sum game: 0 @ a 3 11 65 1 10 b 1 A , 1. find the conservative values (in pure strategies) of the game; 2. find if
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Game Theory (MID-TERM) 24-11-2016 Surname: Name: Matricola: SOL VE THE EXERCISES AND ANSWER THE QUESTIONS USING ONLY THESE PAPERS Exercise 1 ( 2+1+2+3 ) Given the zero sum game: 0 @ a 3 11 65 1 10 b 1 A , 1. find the conservative values (in pure strategies) of the game; 2. find if there is any equilibrium in pure strategies; 3. solve the game if b = 1 2 and a 7. 4. solve the game if b = 13 and a 7. Solution 1. The conservative values in pure strategies are vI = 8 >< >: 3i f a> 3 a if 1 a 3 1i f a< 1 vII =5 8a, b. 2. Since vI 6= vII for any value of a and b, there is no equilibrium in pure strategies. 3. If b = 1 2 and a 7 the third row is dominated by the second one and the first column is dominated by the second one. The game is reduced to ✓ 3 11 51 ◆ , and we can find the equilibrium in mixed strategies using the indi↵erence principle: 3p + 5(1 p) = 11 p +( 1 p)= ) p = 1 3 3q + 11(1 q)=5 q +( 1 q)= ) q = 5 6 . The unique equilibrium is therefore {(0, 1 3 , 2 3 ), ( 5 6 , 1 6 , 0)} and the value of the game is v = 13 3 . 4. If b = 13 and a 7 the first column is still dominated by the second one. Thus, the game is reduced to 0 @ 3 11 51 0 13 1 A . By using the graphical representations (see Figure (a)), we note that the third row is never played at the equilibrium, so the solution is the same as in the previous point: {(0, 1 3 , 2 3 ), ( 5 6 , 1 6 , 0)} and the value of the game is v = 13 3 . 1 v 1 q (a) Exercise 1.4 x 2 x x 2 5 2x2x 5 (b) Exercise 2.3 Exercise 2 ( 2+2+2+2 ) Let ( N, v) be the a TU-game where N = {1, 2, 3}, v({1})= v({2}) = 2, v({3}) = 0, v({1, 2})= a, v({1, 3})= v({2, 3}) = 3, v(N ) = 5, where a is a positive real number. 1. Find a such that the core is non empty. 2. For those values of a compute the core of the game. 3. Find the nucleolus for a = 5.…
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