# How do I solve this exercise (My attempt is included)

Suppose that each wind turbine needs an area of one acre and produces electricity that can be sold for one money unit. Our partial market has six agents $$1, \ldots, 6$$. Agent $$i, i=1,2,3$$, has $$i$$ acres of land (i.e., Agent 1 has one, Agent 2 has two, and Agent 3 has three acres) and no turbine. Each of the remaining agents 4,5,6 has two wind turbines, but no land. The payoff of a coalition is the number of turbines that the members can run. For instance, if Agent 3 agrees with one turbine owner to produce energy, then two wind turbines can run and the profit of two units is generated, but if the same agent convinces two or three of the turbine owners to produce energy exclusively on her land, then the resulting profit would be three units of money.

1. Model the foregoing decision problem as a cooperative transferable utility game $$(N, v)$$ that is the minimum of two inessential games and compute and describe the core of $$(N, v)$$ and its vertices.

My attempt:

The game can be modeled as $$\begin{gathered} N=\{1,2,3,4,5,6\} \\ v(1)=0 \\ v(2)=1 \\ v(3)=2 \\ v(4)=v(5)=v(6)=0 \end{gathered}$$ Game 1 is inessential because it is a zero-sum game, and Game 2 is inessential because it is a game where every coalition is winning. The game $$(N, v)$$ is the minimum of these two games because the payoff of any coalition in $$(N, v)$$ is less than or equal to the payoff of the same coalition in either Game 1 or Game 2.

The core of the game $$(N, v)$$ is empty. To see this, suppose that there is a core allocation. Then this allocation must satisfy the following conditions:

Each player must receive a nonnegative payoff. The total payoff to all players must be equal to the total value of the game, which is 6 . However, it is impossible to satisfy both conditions.

1. Describe and compute the nucleolus and the prenucleolus of $$(N, v)$$.

The nucleolus and the prenucleolus of the game $$(N,v)$$. are also empty. This is because the core of the game is empty, and the nucleolus and the prenucleolus are always contained in the core.

• Hi, please show us your attempt first per our self-study question policy
– 1muflon1
Aug 25, 2023 at 14:23
• Voted to reopen since the question now contains OP’s attempt to solve the problem. Aug 28, 2023 at 13:08

## 1 Answer

Let $$S$$ be a coalition (i.e. a subset of $$\{1,\ldots, 6\}$$).

Then $$v(S)$$ should equal the value the coalition $$S$$ can generate. For example if $$S = \{1,2\}$$ then 1 and 2 provide together 3 units of land, but there are no turbines, so $$V(S) = 0$$. For another example, if $$S = \{1,3\}$$ then 1 provides 1 unit of land and 3 provides 2 turbines, so they can run at most one turbine. This gives $$v(S) = 1$$. Notice that the grand coalition can reach value $$6$$, so $$v(N) = 6$$.

In general, $$v(S)$$ will be equal to the minimum of the number of units of land and the number of turbines that coalition $$S$$ provides. So: $$v(S) = \min\left\{ \begin{array}{l}1 \,I(1 \in S) + 2 \, I(2 \in S) + 3 \,I(3 \in S),\\ 2 \,I(4 \in S) + 2\, I(5 \in S) + 2 \,I(6 \in S)\end{array}\right\}$$ Where $$I(.)$$ is the indicator function that equals 1 if the formula between brackets is true and 0 if not.

Let $$\hat v(S) = 1 \,I(1 \in S) + 2 \,I(2 \in S) + 3 \,I(3 \in S).$$ and $$\tilde v(S) = 2\, I(4 \in S) + 2\, I(5 \in S) + 2\, I(6 \in S)$$ Then, $$v(S) = \min\{\hat v(S), \tilde v(S)\},$$ and both $$\hat v(.)$$ and $$\tilde v(.)$$ are inessential (i.e. additive).

Now, in order to determine the core, we need to find an imputation $$x_1, \ldots, x_6$$, such that: $$\sum_{i = 1}^N x_i = 6,\\ \sum_{i \in S} x_i \ge v(S), \forall S \subset N$$ There are 71 non-empty coalitions, so a large number of inequalities to check, although some of them are easily redundant. For example, any coalition of $$\{1,2,3\}$$ will have empty value, as is any subcoalition of $$\{4,5,6\}$$.

I think the core is non-empty. For this, we can show that the game is balanced. Let $$\lambda_S$$ be the weight for coalition $$S$$ and assume that: $$\sum_{S \subset N} \lambda_S 1_S = 1_N$$ where $$1_S$$ is the $$N$$-dimensional vector for which $$(1_S)_i = 1$$ if $$i \in S$$ and $$0$$ otherwise. Then \begin{align*} \sum_{S \subset N} \lambda_S v(S) &= \sum_{S \subset N} \lambda_S \min\{\hat v(S), \bar v(S)\},\\ &\le \sum_{S \subset N} \lambda_S \bar v(S),\\ &= \sum_{S \subset N} 2 \lambda_S I(4 \in S) + \sum_{S \subset N} 2 \lambda_S I(5 \in S) + \sum_{S \subset N} 2 \lambda_S I(6 \in S),\\ &= 2 \sum_{S \subset N} \lambda_S (1_S)_4 + 2 \sum_{S \subset N} \lambda_S (1_S)_5 + 2 \sum_{S \subset N} \lambda_S (1_S)_6 = 2 + 2 + 2 = 6 = v(N). \end{align*} This shows that the core is non-empty. For example, I think $$(0,0,0,2,2,2)$$ is in the core.

Given that the core is non-empty, the nucleolus also exists.