Homework 4 solutions

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1 Homework 4 solutions ASSIGNMENT: exercises 2, 3, 4, 8, and 17 in Chapter 2, (pp ). Solution to Exercise 2. A coalition that has exactly 12 votes is winning because it meets the quota. This coalition is minimal because if any country is removed from the coalition, the coalition has fewer than 12 votes, and is losing. So coalitions with exactly 12 votes are minimal winning coalitions. A coalition with fewer than 12 votes is losing, and so cannot be a minimal winning coalition. All that remains is to show that a coalition with more than 12 votes is not a minimal winning coalition. Suppose we have a coalition with v votes, and v > 12. Then there are only a few possibilities. v = 13: There is only one country with an odd number of votes: Luxembourg. So if v is odd, the coalition must contain Luxembourg. Removing Luxembourg from the coalition results in a coalition with 12 votes, which is still a winning coalition. Hence, the original coalition was not minimal. v = 14: Since v is even, and Luxembourg is the only country with an odd number of votes, Luxembourg is not in the coalition. The total number of votes held by France, Germany, and Italy is 12, so the coalition must contain these three countries, plus either Belgium or the Netherlands. Whichever one of these two is in the coalition, if we remove that country then the resulting coalition has 12 votes. Hence, the original coalition was not minimal. v = 15: Since v is odd, the coalition must contain Luxembourg, and removing Luxembourg yields a winning coalition. Hence, the original coalition was not minimal. v 16: Since the maximum number of votes that any country has is 4, if we remove any country from the coalition than we still have at least 16 4 = 12 votes, which is enough to be a winning coalition. Hence, the original coalition was not minimal. Solution to Exercise 3. Luxembourg s vote, and hence its support, is completely irrelevant to any election in the European Economic Community, as will now show. Consider any winning coalition containing Luxembourg, and let v be the number of votes in this coalition. Since the coalition is winning, we know v 12. Since Luxembourg is the only

2 country with an odd number of votes, we know that v must be an odd number. The total number of votes in the European Economic Community is 17, so v must be an odd number between (or equal to) 12 and 17, which means v is either 13, 15, or 17. If Luxembourg withdraws its support from this coalition, then the coalition only loses 1 vote, so it still has v 1 votes. Since v 1 is either 12, 14, or 16, all of which are greater than or equal to the quota 12, we conclude that, even without Luxembourg, this coalition is winning. Therefore Luxembourg s vote is irrelevant. Solution to Exercise 4. Each of the three permanent members are functionally identical in this voting system, and each of the eight non-permanent members are also functionally identical. So, it s reasonable to assume that all of the permanent members have the same weight, and that all of the non-permanent members have the same weight. Since the permanent members have veto power and the non-permanent members do not, the weight of each permanent member must be larger than the weight of each nonpermanent member. Since we could always multiply the weights and quota of a weighted system by a positive number to make any particular weight 1, we ll set the weight of each non-permanent member to be 1. The weight of each permanent member and the quota will be left as the variables x and q, respectively. The minimal winning coalitions in this system contain the three permanent members and four of the seven non-permanent members, yielding the inequality 3x + 4 q. Each losing coalition either contains 2 or fewer permanent members or 3 or fewer nonpermanent members, yielding the inequalities: 2x + 8 < q and 3x + 3 < q. Putting the first and second of these inequalities together yields 2x + 8 < q 3x + 4, so 3x + 4 > 2x + 8, x > 4. Based on this, we set x = 5. (Since there are infinitely many possible correct combinations of weights for x and q, we will need to make some choices. Making x the smallest integer possible is probably the simplest choice.) Given that x = 5, the first and the third

3 inequalities tell us that 3x + 3 <q 3x + 4, 18 <q 19. There s only one integer that is greater than 18 but less than or equal to 19, so we set q = 19. Since we obtained this value for q without taking the second inequality into account, we really ought to check that this value satisfies it that inequality also. Plugging x = 5 and q = 19 into 3x + 3 < q yields 18 < 19, which is true. These calculations lead us to believe that choosing 1 as the weight of each non-permanent member, 5 as the weight of each permanent member, and 19 as the quota gives us a weighted voting system that is identical to the given voting system. But we still have to check. Claim. The given voting system is identical to the weighted voting system where each non-permanent member has weight 1, each permanent member has weight 5, and the quota is 19. Proof. Each minimal winning coalition in the given voting system contains three permanent members and four non-permanent members. The total weight of such a coalition is (3 5) + (4 1) = 19, meeting the quota of 19, so it is a winning coalition in the weighted voting system. Since both voting systems are monotonic, this proves that every coalition that is winning in the given voting system is also winning in the weighted voting system. Each coalition that is losing in the given voting system satisfies at least one of the following conditions. (a) The coalition contains 2 or fewer permanent members. (b) The coalition contains 3 or fewer non-permanent members. In case (a), the weight of the coalition is less than (2 5) + (8 1) = 18, and so the total weight of the coalition does not meet the quota. In case (b), the weight of the coalition is less than (3 5) + (3 1) = 18, and so the total weight of the coalition does not meet the quota. Since every coalition that is winning in the given voting system has weight that meets or exceeds the quota, and every coalition that is losing in the given voting system has weight that fails to meet the quota, the two voting systems are identical.

4 Solution to Exercise 8. (a) A coalition of members of the Bennet family is winning if it contains at least one parent, and at least two children. One way to generate the list of winning coalitions is to start with the grand coalition (the coalition containing everyone) and figure out who we can remove from the coalition without losing its winningness. The answer is that we can only remove up to one parent and up to one child. Therefore, the winning coalitions are the following. Winning Coalitions grand coalition: remove one parent: remove one child: remove one parent and one child: Tom, Jane, Rachel,, Rob Jane, Rachel,, Rob Tom, Rachel,, Rob Tom, Jane,, Rob Tom, Jane, Rachel, Rob Tom, Jane, Rachel, Jane,, Rob Jane, Rachel, Rob Jane, Rachel, Tom,, Rob Tom, Rachel, Rob Tom, Rachel, (b) This voting system is NOT swap robust. Informally, the reason is that the parents and children do not have equal power, (just as Senators and Representatives don t have equal power in the U.S. Federal System). To prove it is not swap robust, we need only provide a counterexample. Consider the following two winning coalitions. X Tom Rachel Y Jane Rob

5 Notice that Tom is in X but not Y, and Rob is in Y but not X, so we can swap them, yielding the following coalitions. X Y Rob Rachel Jane Tom The coalition X is losing because it doesn t contain any parents, and Y is losing because it only contains one child. Therefore this is an example of two winning coalitions which, after performing a swap, both become losing coalitions. Hence this voting system is not swap robust. Solution to Exercise 17. (a) Claim. The given voting system is swap robust. Proof. Let X and Y be arbitrary winning coalitions, x be a voter in X that is not in Y, and y be a voter in Y that is not in X. Let X be the coalition X with y replacing x, and Y be the coalition Y with x replacing y. Note that X contains the same number of voters as X, and Y contains the same number of voters Y. Hence, X and Y each have at least five of the eight voters. Since each of x and y are either minority voters or majority voters, we have four cases. We will prove that, in each case, at least one of X and Y has two or more minority voters, and is hence a winning coalition. Case 1: Suppose x and y are both majority voters. Then X has the same number of minority voters as X and Y has the same number of minority voters as Y, so both X and Y contain at least two of the three minority voters. Case 2: Suppose x is a majority voter and y is a minority voter. Then X has one more minority voter than X. Since X is winning, it contains at least two minority voters. Therefore X contains all three minority members and is therefore a winning coalition. Case 3: Suppose x is a minority voter and y is a majority voter. Then Y has one more minority voter than Y. Since Y is winning, it contains at least two minority voters. Therefore Y contains all three minority members and is therefore a winning coalition.

6 Case 4: Suppose x and y are both minority voters. Then X has the same number of minority voters as X and Y has the same number of minority voters as Y, so both X and Y contain at least two of the three minority voters. In each case, at least one of X and Y is a winning coalition, and hence this voting system is swap robust. (b) Claim. The given voting system is NOT trade robust. The actual proof of this claim should not include how we found a counterexample; it just has to show the counterexample, and prove that it really is counter. So before proceeding with the proof, we will discuss the logic behind constructing a counterexample to trade robustness in this case. There is more freedom in making a trade than in making a swap, and since we know this voting system is swap robust, it follows that a counterexample to trade robustness will have to use some of this additional freedom. One property that trades have but swaps don t is that a trade can involve an exchange of an unequal number of voters, i.e. an unbalanced exchange. In seeking a counterexample, we should probably start with coalitions that are almost losing, in the sense that they have a minimal number of voters (5), and perhaps also a minimal number of minority voters (2). With all of this in mind, we can construct a counterexample by performing a trade wherein one coalition gets many majority voters, but doesn t keep enough minority voters, and the other coalition has plenty of minority voters, but not enough voters in total. Proof of Claim. Let X = {A, B, C, F, G} and Y = {C, D, E, G, H}. Note that both X and Y are winning coalitions, since they each contain 5 voters and 2 minority voters. If we trade A and B for H, the result is the coalitions X = {C, F, G, H} and Y = {A, B, C, D, E, G}. Both are losing coalitions, because X only has 4 voters and Y only has 1 minority voter.