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Econ经济作业代写Economics代考|Five non-simple games

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Econ经济作业代写Economics代考|Five non-simple games

Buying a car|Econ经济作业代写Economics代考

Following Morris (1994, S. 162), we consider three agents involved in a car deal. Andreas (A) has a used car he wants to sell, Frank (F) and Tobias (T) are potential buyers with willingness to pay of 700 and 500 , respectively. This leads to the coalition function $v$ given by
$$
\begin{aligned}
v(A) &=v(F)=v(T)=0, \
v(A, F) &=700, \
v(A, T) &=500, \
v(F, T) &=0 \text { and } \
v(A, F, T) &=700 .
\end{aligned}
$$
One-man coalitions have the worth zero. For Andreas, the car is useless (he believes in cycling rather than driving). Frank and Tobias cannot obtain the car unless Andreas cooperates. In case of a deal, the worth is equal to the (maximal) willingness to pay.

The Maschler game|ECON经济作业代写ECONOMICS代考

According to the Shapley value, the players 1 and 2 obtain less than their common worth. Therefore, they can block the payoff vector suggested by the Shapley value. Indeed, for any efficient payoff vector, we can find a two-player coalition that can be made better off. Differently put: the core is empty.

This can be seen easily. We are looking for vectors $\left(x_{1}, x_{2}, x_{3}\right)$ that fulfill both
$$
x_{1}+x_{2}+x_{3}=72
$$
and
$$
\begin{aligned}
x_{1} & \geq 0, x_{2} \geq 0, x_{3} \geq 0, \
x_{1}+x_{2} & \geq 60 \
x_{1}+x_{3} & \geq 60 \text { and } \
x_{2}+x_{3} & \geq 60 .
\end{aligned}
$$
Summing the last three inequalities yields
$$
2 x_{1}+2 x_{2}+2 x_{3} \geq 3 \cdot 60=180
$$
and hence a contradiction to efficiency.

The gloves game, once again|Econ经济作业代写Economics代考

$$
\begin{aligned}
&L={1,2, \ldots, 100} \
&R={101, \ldots, 199}
\end{aligned}
$$
If a payoff vector
$$
\left(x_{1}, \ldots, x_{100}, x_{101}, \ldots, x_{199}\right)
$$
is to belong to the core, we have
$$
\sum_{i=1}^{199} x_{i}=99
$$
by the efficiency axiom. We now pick any left-glove holder $j \in{1,2, \ldots, 100}$. We find
$$
v(L \backslash{j} \cup R)=99
$$

  1. FIVE NON-SIMPLE GAMESand hence
    $$
    \begin{aligned}
    x_{j} &=99-\sum_{\substack{i=1 \
    i \neq j}}^{199} x_{i}(\text { efficiency) }\
    &\leq 99-99 \text { (blockade by coalition } L \backslash{j} \cup R) \
    &=0
    \end{aligned}
    $$
    Therefore, we have $x_{j}=0$ for every $j \in L$.
    Every right-glove owner can claim at least 1 because he can point to coalitions where he is joined by at least one left-glove owner. Therefore, every right-glove owner obtains the payoff 1 and every left-glove owner the payoff zero. Inspite of the fact that the scarcity is minimal, the right-glove owners get evervthing.

The chess game|Econ经济作业代写Economics代考

DEFINITION XV.23 (CHESS GAME). The chess game $v$ is defined by
$$
v(K)= \begin{cases}\frac{|K|}{2}, & |K| \text { is even } \ \frac{|K|-1}{2}, & |K| \text { is odd }\end{cases}
$$
Find the core!
I copied this game from lectures notes by Chris Wallace (Trinity College, Oxford) who calls this game a treasure hunt.

Cost-division gamess|ECON经济作业代写ECONOMICS代考

Following Young (1994, pp. 1195), we consider a specific example. Two towns $A$ and $B$ plan a water-distribution system. Town $A$ could build such a system for itself at a cost of 11 million Euro and town $B$ would need 7 million Euro for a system tailor-made to its needs. The cost for a common water-distribution system is 15 million Euro. The cost function is given by
$$
\begin{aligned}
c({A}) &=11, c({B})=7 \text { and } \
c({A, B}) &=15 .
\end{aligned}
$$
The associated cost-savings game is $v: 2^{{A, B}} \rightarrow \mathbb{R}$ defined by
$$
\begin{aligned}
v({A}) &=0, c({B})=0 \text { and } \
v({A, B}) &=7+11-15=3 .
\end{aligned}
$$
$v$ ‘s core is obviously given by
$$
\left{\left(x_{A}, x_{B}\right) \in \mathbb{R}{+}^{2}: x{1}+x_{2}=3\right} .
$$
The cost savings of $3=11+7-15$ can be allotted to the towns such that no town is worse off compared to going it alone. Turning to cost division, the set of “undominated” cost allocations is
$$
\left{\left(c_{A}, c_{B}\right) \in \mathbb{R}^{2}: c_{A}+c_{B}=15, c_{A} \leq 11, c_{B} \leq 7\right} .
$$

Econ经济作业代写Economics代考|Five non-simple games

BUYING A CAR|ECON经济作业代写ECONOMICS代考

在 Morris (1994, S. 162) 之后,我们考虑了涉及汽车交易的三个代理人。Andreas (A) 有一辆他想要出售的二手车,Frank (F) 和 Tobias (T) 是潜在买家,愿意支付的价格分别为 700 和 500 。这导致了联合函数v由
v(一种)=v(F)=v(吨)=0, v(一种,F)=700, v(一种,吨)=500, v(F,吨)=0 和  v(一种,F,吨)=700.
单人联盟的价值为零。对于安德烈亚斯来说,汽车毫无用处(他相信骑自行车而不是开车)。除非 Andreas 合作,否则 Frank 和 Tobias 无法获得汽车。在交易的情况下,价值等于(最大)支付意愿。

THE MASCHLER GAME|ECON经济作业代写ECONOMICS代考

根据 Shapley 值,玩家 1 和 2 获得的价值低于他们的共同价值。因此,他们可以阻止 Shapley 值建议的支付向量。事实上,对于任何有效的回报向量,我们都可以找到一个可以变得更好的两人联盟。换一种说法:核心是空的。

这很容易看出。我们正在寻找载体(X1,X2,X3)既满足
X1+X2+X3=72

X1≥0,X2≥0,X3≥0, X1+X2≥60 X1+X3≥60 和  X2+X3≥60.
对最后三个不等式求和
2X1+2X2+2X3≥3⋅60=180
因此与效率相矛盾。

THE GLOVES GAME, ONCE AGAIN|ECON经济作业代写ECONOMICS代考

一世=1,2,…,100 R=101,…,199
如果一个支付向量
(X1,…,X100,X101,…,X199)
是属于核心的,我们有
∑一世=1199X一世=99
由效率公理。我们现在选择任何左手套架j∈1,2,…,100. 我们发现
v(一世∖j∪R)=99

  1. 五个非简单游戏因此
    Xj=99−∑一世=1 一世≠j199X一世( 效率)  ≤99−99 (联军封锁 一世∖j∪R) =0
    因此,我们有Xj=0对于每个j∈一世.
    每个右手套拥有者至少可以声明 1,因为他可以指出至少有一名左手套拥有者加入的联盟。因此,每个右手套拥有者获得收益 1,每个左手套拥有者获得收益 0。尽管稀缺性很小,但正确的手套所有者却得到了一切。

THE CHESS GAME|ECON经济作业代写ECONOMICS代考

定义 XV.23(国际象棋游戏)。国际象棋游戏v定义为
v(到)={|到|2,|到| 甚至  |到|−12,|到| 很奇怪 
找到核心!
我从克里斯·华莱士(牛津三一学院)的讲义中复制了这个游戏,他称这个游戏为寻宝游戏。

COST-DIVISION GAMESS|ECON经济作业代写ECONOMICS代考

在 Young (1994, pp. 1195) 之后,我们考虑一个具体的例子。两个城镇一种和乙规划供水系统。镇一种可以为自己建立这样一个系统,成本为 1100 万欧元和城镇乙将需要 700 万欧元用于为其需求量身定制的系统。一个普通的配水系统的成本是 1500 万欧元。成本函数由下式给出
C(一种)=11,C(乙)=7 和  C(一种,乙)=15.
相关的成本节约游戏是v:2一种,乙→R被定义为
v(一种)=0,C(乙)=0 和  v(一种,乙)=7+11−15=3.
v的核心显然是由
$$
\left{\left(x_{A}, x_{B}\right) \in \mathbb{R} {+}^{2}: x {1}+x_{2 }=3\right} 。
吨H和C○s吨s一种v一世nGs○F$3=11+7−15$C一种nb和一种一世一世○吨吨和d吨○吨H和吨○在nss你CH吨H一种吨n○吨○在n一世s在○rs和○FFC○米p一种r和d吨○G○一世nG一世吨一种一世○n和.吨你rn一世nG吨○C○s吨d一世v一世s一世○n,吨H和s和吨○F“你nd○米一世n一种吨和d”C○s吨一种一世一世○C一种吨一世○ns一世s
\left{\left(c_{A}, c_{B}\right) \in \mathbb{R}^{2}: c_{A}+c_{B}=15, c_{A} \leq 11, c_{B} \leq 7\right} 。
$$

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