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Final Examination Mathematics 167, Game Theory
Ferguson Tues. June 14, 2005
- (a) Consider a game of nim with 3 piles of sizes 9, 17 and 21. Is this a P-position or an N-position? If an N-position, what is a winning first move?
(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.
- (a) Solve the following two-person zero-sum matrix game. ( 5 3 0 2 0 2 6 4
(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.)
- Player II receives one card at random from a deck of three cards consisting of Ace, King andJoker, probability 1/3 each. If she receives an Ace or King, she must announce that card. If she receives a Joker, she may announce either the Ace or the King, whichever she wants. Then, after hearing the announcement, Player I must either accept or challenge. If he accepts, he wins 6 if II has the Ace and3 if II has the King. If he challenges, he wins 18 if II has the Joker andhas announcedthe Ace, andhe wins 9 if II has the Joker and has announcedthe King. There is no payoff otherwise.
(a) Draw the Kuhn Tree.
(b) Findthe equivalent strategic form of the game.
- Consider the non-cooperative bimatrix game:
(a) Findthe safety levels, andthe maxmin strategies for both players. (b) Findas many strategic equilibria as you can.
- Suppose in the Cournot duopoly model that the two firms have different production costs and different set-up costs. Suppose Player I’s cost of producing x is x + 2, and II’s cost of producing y is 3y + 1. Suppose also that the price function is P (x, y) = 17 − x − y, where x and y are the amounts producedby I andII respectively. What is the equilibrium production, and what are the players’ equilibrium profits?
- Consider the cooperative TU bimatrix game:
(a) Findthe TU-values. (b) Findthe associatedside payment. (c) Findthe optimal threat strategies.
- (a) Define what it means for a vector (¯u, ¯v) ∈ S, where S is the NTU-feasible set, to be Pareto optimal in a two-player NTU game.
(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.
- Consider the 3-person game in coalitional form with characteristic function, v, satisfying
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.)
- (a) Define a Simple Game. (b) Findthe Shapley value (the Shapley-Shubik Index) for the weightedvoting game with four players in which player A holds 10 shares, player B holds 9 shares, player C holds 7 shares andplayer D holds 6 shares, when 18 or more shares are requiredpass a measure.
- Consider the three-person game in coalitional form with characteristic function,
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).
- The players’ profits are u 1 (x, y) = x(17 − x − y) − (x + 2) and u 2 (x, y) = y(17 − x − y) − (3y + 1). To find the equilibrium production, we set the partial derivatives to zero:
∂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.
- The set of imputations is the set on points {x : x 1 + x 2 + x 3 = 3, x 1 ≥ − 2 , x 2 ≥ − 1 , x 3 ≥^0 }.^ This is the equilateral triangle with vertices (−^2 ,^ −^1 ,^ 6), (−^2 ,^5 ,^ 0) and (4, − 1 , 0). The core is the subset of this set that satisfies x 1 + x 2 ≥ 2, x 1 + x 3 ≥ 1, and x 2 + x 3 ≥ 1. This is the four-sided figure with vertices (1, 2 , 0), (2, 1 , 0), (2, 0 , 1) and (0, 2 , 1).
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
