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Class February 8 - Math 103 Section 11 Tuesday February 8 2011 1 Submitted Homework Assignment#1 for Chapter 2 is at Sakai It is in the Assignments

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Unformatted text preview: Math 103, Section 11, Tuesday, February 8, 2011 1. Submitted Homework Assignment #1 for Chapter 2 is at Sakai. It is in the Assignments section and is called: Assignment Chapter 2A Please do Exercises 18b,18c,20 and 24 on page 68 and upload to Sakai. Homework is due by Friday, February 11 at midnight The Exercises are at Sakai, Resources, Chapter 2 Exercises from text.doc Quiz on Friday, February 11 Submitted Homework Assignment #2 for Chapter 2 is at Sakai. It is in the Assignments section and is called: Assignment Chapter 2B Please do Exercises 28d, 34, and 36 on page 69 and upload to Sakai. Homework is due by Friday, February 18 at midnight The Exercises are at Sakai, Resources, Chapter 2 Exercises from text.doc Homework to be discussed in class on Friday, February 11: The following 6 problems: Exercises 20, 27b, 33, and 35 and 1. UN Problem Consider the following simplified version of the UN Security Council: There are 4 players, each with one vote, and a majority of votes is needed to pass a resolution. Moreover, P 1 has veto power. (a) Find the Banzhaf power distribution of this weighted voting system. (b) In order to find the Shapley-Shubik power distribution, how many sequential coalitions must you examine? Since there are 4 players, there are 4!= (4)(3)(2)(1) = 24 sequential coalitions. (c) Find the Shapley-Shubik power distribution of this weighted voting system. Start by listing all the sequential coalitions. Then fill in the following table: Banzhaf power index Shapley-Shubik power index P1 P2 P3 P4 2. Given the weighted voting system {4: 3, 2, 2} Fill in the following table: Banzhaf power index Shapley-Shubik power index P1 P2 P3 Correction to the handout from our last class: 3. Consider the weighted voting system [q:7, 5, 3, 2]. 3d. At least how much would the quota have to be so that all the players have veto power? In class I said 17, but the correct quota is 16. In fact, For P1 to have veto power, the quota must be at least: 11 Why? For P2 to have veto power, the quota must be at least: 13 Why? For P3 to have veto power, the quota must be at least: 15 Why? For P4 to have veto power, the quota must be at least: 16 Why? Definition. A player P has veto power if and only if the sum of the weights of the other players is less than the quota. Consider [the weighted voting system: [ q:P1,P2,P3,P4] V/2<q<=V where the weight of P1 is w1. The weight of P2 is w2. The weight of P3 is w3. The weight of P4 is w4. and V=w1+w2+…+w4 P1 has veto power means that w2+w3+w4<q When listing winning coalitions, remember that For 3 players, there are 2^3-1-3=4 coalitions to consider: {P1,P2} {P1,P3} {P2,P3} and {P1,P2,P3} For 4 players, there are 2^4-1-4=11 coalitions to consider....
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This note was uploaded on 01/20/2012 for the course MATH 103 taught by Professor Berkowitz during the Spring '07 term at Rutgers.

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Class February 8 - Math 103 Section 11 Tuesday February 8 2011 1 Submitted Homework Assignment#1 for Chapter 2 is at Sakai It is in the Assignments

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