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Material Type: Exam; Class: MATH GAME THEORY; Subject: Mathematics; University: University of California - Los Angeles; Term: Unknown 2005;
Typology: Exams
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Final Examination Mathematics 167, Game Theory
Ferguson Tues. June 14, 2005
(b) Consider the take or break game where a player may remove one or two chips from any pile, or he may break any pile into two non-empty piles (but not both). Findthe Sprague-Grundy function for piles of size less than or equal to 11.
(b) Player II chooses a positive integer andPlayer I tries to guess it. If I guesses too high, he loses 2 to II. If he guesses too low by exactly 1 he loses 1 to II. Otherwise (if he is correct or if he is too low by more than one), he wins 1 from II. Set up the matrix and solve. (Use domination.)
(a) Draw the Kuhn Tree.
(b) Findthe equivalent strategic form of the game.
(a) Findthe safety levels, andthe maxmin strategies for both players. (b) Findas many strategic equilibria as you can.
(a) Findthe TU-values. (b) Findthe associatedside payment. (c) Findthe optimal threat strategies.
(b) Consider the cooperative NTU bimatrix game:
. Let (u∗, v∗) =
(1, 0) be the disagreement point, (threat point, status-quo point). Find the NTU-value.
v({ 1 }) = − 2 v({ 1 , 2 }) = 2 v(∅) = 0 v({ 2 }) = − 1 v({ 1 , 3 }) = 1 v({ 1 , 2 , 3 }) = 3 v({ 3 }) = 0 v({ 2 , 3 }) = 1
Findthe imputations andthe core for this game. (Either graph the core or be fairly explicit in your description.)
v({ 1 }) = 0 v({ 1 , 2 }) = 2 v(∅) = 0 v({ 2 }) = 1 v({ 1 , 3 }) = 3 v({ 1 , 2 , 3 }) = 10 v({ 3 }) = 2 v({ 2 , 3 }) = 6
(a) Findthe Shapley value. (b) Findthe nucleolus.
(b) The top row is strictly dominated, and then the middle column is strictly domi- nated. Removing them does not lose any equilibria (and leads to the Battle of the Sexes). There are two PSE’s, one at (secondrow,first column), andthe other at (thirdrow,third column). There is therefore a thirdSE given by the equalizing strategies in the Battle of the Sexes, namely, (p, q), where p = (0, 2 / 3 , 1 /3) and q = (1/ 3 , 0 , 2 /3).
∂u 1 ∂x
= 16 − 2 x − y = 0
∂u 2 ∂y
= 14 − x − 2 y = 0
which gives (x, y) = (6, 4) as the equilibrium production. The equilibrium profits are (u 1 (6, 4), u 2 (6, 4)) = (34, 15).
6.(a) The maximum total payoff is^ σ^ = 7, with payoff (4,^ 3). The difference matrix is
. The last two and column are dominated, so δ = Val
Therefore, ϕ = ((σ + δ)/ 2 , (σ − δ)/2) = (21/ 5 , 14 /5).
(b) To get to this from (4, 3) requires Player II to pay 1/5 to Player I. (c) The threat strategies are p = (3/ 5 , 2 / 5 , 0) and q = (4/ 5 , 1 / 5 , 0).
7.(a) A vector (¯u, v¯) ∈ S is Pareto optimal if the only point (u, v) ∈ S such that u ≥ u¯ and v ≥ v¯ is (u, v) = (¯u, v¯) itself.
(b) The Pareto optimal boundary is the line segment from (3, 4) to (9, 1). The equation of this line is v − 4 = (− 1 /2)(u − 3) or v = (11 − u)/2. We seek the point on this line that minimizes (u − 1)v = (u − 1)(11 − u)/2. Setting the derivative to zero gives u = 6, which gives v = 5/2. Thus (6, 5 /2) is the NTU solution since it is on the line segment.
9.(a) A coalitional game (N, v) is saidto be simple if for all S ⊂ N, v(S) is either zero or one.
(b) Player A can change a coalition from losing to winning if andonly if that coalition is one of {B}, {B, C}, {B, D} or {C, B}. Therefore,
φA =
Similarly, φB = 1/3, φC = 1/6 and φD = 1/6.
10.(a)
φ{ 1 } = (1/3)0 + (1/6)1 + (1/6)1 + (1/3)4 = 5/ 3 φ{ 2 } = (1/3)1 + (1/6)2 + (1/6)4 + (1/3)7 = 11/ 3 φ{ 3 } = (1/3)2 + (1/6)3 + (1/6)5 + (1/3)8 = 14/ 3
(b) The Shapley value was foundto be (5 / 3 , 11 / 3 , 14 /3) so we might try (2, 4 , 4) as an initial guess at the nucleolus. The largest excess occurs at either of the coalitions { 1 }, { 3 } and { 2 , 3 }. The first andlast cannot be made smaller without making the other larger. So x 1 = 2 in the nucleolus. The excess at { 3 } can be made smaller by making x 3 larger. This increases the excess of { 2 }. These are equal at x 3 = 4.5 and x 2 = 3.5. The nucleolus is (2, 3. 5 , 4 .5).
Coalition Excess (2, 4 , 4) (2, 3. 5 , 4 .5) { 1 } −x 1 − 2 − 2 { 2 } 1 − x 2 − 3 − 2. 5 { 3 } 2 − x 3 − 2 − 2. 5 { 1 , 2 } 2 − x 1 − x 2 = x 3 − 8 − 4 − 3. 5 { 1 , 3 } 3 − x 1 − x 3 = x 2 − 7 − 3 − 3. 5 { 2 , 3 } 6 − x 2 − x 3 = x 1 − 4 − 2 − 2