Solution to Tutorial 11

Transcription

1 Solution to Tutorial 202/203 Semeter I MA4264 Game Theory Tutor: Xiang Sun November 5, 202 Exercie. Conider the following three-peron game: v( = 0, v(} = 0.2, v(2} = v(3} = 0, v(, 2} =.5, v(, 3} =.6, v(2, 3} =.8, v(, 2, 3} = 2. (a Find the core of thi game. (b Find the Shapley value of thi game. (c Find an imputation dominating the imputation (, /2, /2. Solution. (a Suppoe (x, x 2, x 3 i in the core, then it hould atify: x 0.2 ( x 2, x 3 0 (2 x + x 2.5 (3 x + x 3.6 (4 x 2 + x 3.8 (5 x + x 2 + x 3 = 2 (6 By (3, (4 and (5, we have x + x 2 + x , which contradict to (6. Therefore, the core i empty. (b Since n = 3, we have γ(0 = γ(2 = /3 and γ( = /6. For Player, S 2} 3} 2, 3} v(s } v(s Hence φ (v = = For Player 2, S } 3}, 3} v(s 2} v(s xiangun@nu.edu.g. Suggetion and comment are alway welcome.

2 MA4264 Game Theory 2/6 Solution to Tutorial Hence φ 2 (v = = For Player 3, by efficiency, φ 3 (v = v(n φ (v φ 2 (v = 0.7. Therefore, the Shapley value i (0.65, 0.65, 0.7. (c It uffice to find an allocation y = (y, y 2, y 3, uch that y 0.2, y 2, y 3 0, y + y 2 + y 3 = 2, and y dominate x. Let y = (0.2, 0.9, 0.9, and S = 2, 3}. y trictly dominate x = (, 0.5, 0.5 via S. Exercie 2. The managing board of a corporation conit of three tock-holder who have repectively 20, 30 and 50 hare of tock and the chairman who ha no hare. Any deciion can be ettled by approval of board member holding a imple majority of the hare and the chairman can decide tie vote. Thu, we define a characteritic function a follow. It ha value if a coalition hold > 50 hare or hold 50 hare and ha the chairman in the coalition. It ha value 0 otherwie. Find the Shapley value. Solution. We number the player a follow: tock-holder with 20 hare Player, tockholder with 30 hare Player 2, tock-holder with 50 hare Player 3, chairman Player 4. The characteritic function i a follow: v( = v(} = v(2} = v(3} = v(4} = 0, v(, 2} = v(, 4} = v(2, 4} = 0, v(, 3} = v(2, 3} = v(3, 4} =, v(, 2, 3} = v(, 2, 4} = v(, 3, 4} = v(2, 3, 4} = v(, 2, 3, 4} =. Since n = 4, we have γ(0 = γ(3 = 4, and γ( = γ(2 = 2. For Player, S 2} 3} 4} 2, 3} 2, 4} 3, 4} 2, 3, 4} v(s } v(s Hence φ (v = = 6. For Player 2, S } 3} 4}, 3}, 4} 3, 4}, 3, 4} v(s 2} v(s Hence φ 2 (v = = 6. For Player 3, S } 2} 4}, 2}, 4} 2, 4}, 2, 4} v(s 3} v(s 0 0 Hence φ 3 (v = 2. For Player 4, S } 2} 3}, 2}, 3} 2, 3}, 2, 3} v(s 4} v(s It ha better to computer the Shapley value for Player 3 by tandard way, and to apply efficiency to check whether your calculating i correct.

3 MA4264 Game Theory 3/6 Solution to Tutorial Hence φ 4 (v = 6. Therefore, the Shapley value i ( 6, 6, 2, 6. Exercie 3. Three doctor have banded together to form a joint practice: the Port Charle Trio. The overhead for the practice i $40,000 per year. Each doctor bring in annual revenue and incur annual variable cot a follow: doctor $55,000 in revenue, $40,000 in variable cot; doctor 2 $60,000 in revenue, $35,000 in variable cot; doctor 3 $40,000 in revenue, $38,000 in variable cot. The Port Charle Trio want to ue game theory to determine how much each doctor hould be paid. Determine the relevant characteritic function and how that the core of the game conit of an infinite number of point. Alo determine the Shapley value of the game. Doe the Shapley value give a reaonable diviion of the practice profit? Proof and Solution. (i Characteritic function v: v( = 0 v(} = ( = 75, v(2} = 85, v(3} = 62; v(, 2} = ( ( = 200, v(, 3} = 77, v(2, 3} = 87; v(, 2, 3} = 302. (ii Core. To olve (x, x 2, x 3, uch that, Solving them, we have x 75, x 2 85, x 3 62, x + x 2 200, x + x 3 77, x 2 + x 3 87, x + x 2 + x 3 = 302. C(N, v = (x, x 2, x 3 : x [75, 5], x 2 [85, 25], x 3 [62, 02], x +x 2 +x 3 = 302}. Since (75, 25, 02 and (76, 24, 02 are in the core, and for any λ [0, ], λ(75, 25, 02+ ( λ(76, 24, 02 i alo in the core, that i, there are infinite many element in the core. (iii Shapley value. Since n = 3, we have γ(0 = γ(2 = /3, and γ( = /6. For Player. S 2} 3} 2, 3} v(s } v(s Hence φ (v = = For Player 2. S } 3}, 3} v(s 2} v(s Hence φ 2 (v = For Player 3, by efficiency, φ 3 (v = v(n φ (v φ 2 (v = 266 Therefore, the Shapley value i ( 305 3, 335 3,

4 MA4264 Game Theory 4/6 Solution to Tutorial Exercie 4. The management committee of an aociation conit of a preident, a ecretary, a treaurer and two committee member. It require three vote including one from the preident, one from the ecretary or treaurer to approve a propoal. Determine the voting power of each member in the committee. Solution. We number the player a follow: preident Player, ecretary Player 2, Treaurer Player 3, member Player 4, member 2 Player 5. Let the total voting power be. We ue the Shapley value to compute each player voting power. The characteritic function i, if S 3,, 2} S or, 3} S, v(s = 0, otherwie. It i eay to ee that player 2 and player 3 have the ame voting power, and player 4 and player 5 have the ame voting power. Hence, their Shapley value mut be the ame. Since n = 5, we have γ(0 = γ(4 = 5, γ( = γ(3 = 20 and γ(2 = 30. For player, v(s } v(s = if and only if S i one of the following: 2, 3}, 2, 4}, 2, 5}, 3, 4}, 3, 5}, 2, 3, 4}, 2, 3, 5}, 2, 4, 5}, 3, 4, 5}, 2, 3, 4, 5}. Hence the Shapley value of player i φ (v = = For player 2, v(s 2} v(s = if and only if S i one of the following:, 3},, 4},, 5},, 4, 5}. Hence the Shapley value of player 2 i φ 2 (v = = 3 20, which i φ 3(v. For player 4, v(s 4} v(s = if and only if S i one of the following:, 2},, 3}. Hence the Shapley value of player 4 i φ 4 (v = 2 Therefore, the voting power i ( 7 30, 3 20, 3 20, 5, 30 = 5. 5, which i φ 5(v. Exercie 5. Conider an n-peron game in which the only winning coalition are thoe coalition containing player and at leat one other player. If a winning coalition receive a reward of $, find the core and the Shapley value of the game. Solution. When n = 2, the olution i quite eay. (Exercie In the following, we aume n 3. The characteritic function i, if S 2, Player belong to S, v(s = 0, otherwie. (i Core. Suppoe (x, x 2,..., x n i in the core. Then we have n x i =, x i 0, i= x + i S x i, for all S 2, 3,..., n} and S i nonempty. It i eay to ee that the only olution i x =, x 2 = = x n = 0: take S = 3, 4,..., n}, we have x + x x n, and hence x 2 = 0. Similarly x 3 = x 4 = = x n = 0. Therefore the core i (, 0, 0,..., 0}.

5 MA4264 Game Theory 5/6 Solution to Tutorial (ii Shapley value. For Player i, v(s i} v(s = if and only if S = }. Otherwie it i zero. Since γ( = n, Player i Shapley value i For Player, we have φ i (v = φ (v = n(n. n i=2 φ i (v = n n. Therefore, the Shapley value i ( n n, n(n,...,. n(n Exercie 6. Find the Shapley value of the game with N =, 2} and the characteritic function v. Now conider the bargaining game where H = I(N, v and d = (v(}, v(2}. Find the bargaining olution of the game (H, d. Solution. Since n = 2, we have γ(0 = γ( = 2. Denote v = v(n, v = v(} and v 2 = v(2}. (i Shapley value. For Player i, S j} v(s i} v(s v i v v j Hence the Shapley value for Player i i v i+v v j 2. (ii To get the Nah bargaining olution, we olve the following problem max (x v (x 2 v 2. x +x 2 =v,x v,x 2 v 2 The olution i x i = v i+v v j 2. Note that we need to check whether x i v i. Hence, both Nah bargaining olution and the Shapley value given the ame reult. Exercie 7. Conider the following cot allocation problem. Building an airfield will benefit n player. Player j require an airfield that cot c j to build, o to accommodate all the player, the field will be built at a cot of max j n c j. How hould thi cot be plit among the player? Suppoe all the cot are ditinct and let 0 < c < c 2 < < c n. Take the characteritic function of the game to be v(s = max j S c j for S, 2,..., n}. (i Let R k = k, k +,..., n} for k =, 2,..., n, and define the characteritic function v k through the equation (c k c k, if S R k v k (S = 0, if S R k = For convenience, let c 0 = 0. Show that v = n v k.

6 MA4264 Game Theory 6/6 Solution to Tutorial (ii Find the Shapley value of the game v in the form of φ i (v = i α ik(c k c k, i =, 2,..., n, where the coefficient α ik are independent of c, c 2,..., c n. Solution. (i For every coalition S, we have n v k (S = = max(s max(s v k (S + n k=max(s+ max(s v k (S = = c max(s = v(s v k (S (c k c k (ii Since we have v k (S = v k (S i} v k (S = (c k c k, if max(s k 0, if max(s < k (c k c k, if max(s < k i 0, otherwie and hence φ (v k = = φ k (v k = 0, and = n = n = n φ k (v k = = φ n (v k n =0 n =0 n =0 ( n ( n ( n [v k (S i} v k (S] S = S =,max(s<k ( k Therefore, we have [ (c k c k ] φ i (v = [v k (S i} v k (S] + n φ i (v k = i k ( k ( n =0 = n i φ i (v k S =,max(s k (c k c k [v k (S i} v k (S] End of Solution to Tutorial