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Class February 11 - Math 103 Section 11 Friday 1 Submitted Homework Assignment#1 for Chapter 2 is at Sakai It is in the Assignments section and is

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Math 103, Section 11, Friday, February 11, 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 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 Our final exam will be on Wednesday, May 11 from 8-11 am On that day, you will show me all that you have learned in this course! We have a long way until then. So enjoy the course. As a preview to Chapter 4: What was the first Presidential veto in U.S. history? Dear Students, Thank you Seth for your excellent observation regarding the weighted voting system: [8:8,4,3,1]. Seth pointed out that P1 is a dictator and hence everyone else should be a dummy. But {P2,P3,P4} is a winning coalition in which P2 is critical, P3 is critical and P4 is critical. The problem with this weighted voting system is that the quota is too low. The quota must always be more than half of the total number of votes and it must always be less than or equal to the total number of votes. So the quota must be greater than 9 and less than or equal to 16. Because the quota is set to 8, we are having a problem. In summary, the weighted voting system [8:8,4,3,1] is not allowed. The quota is too small, causing problems. Please see the bottom of page 45 of our text. This situation is the mathematical version of anarchy in which we have enough Yes and enough No votes to meet the quota. We have a stalemate. Here is another example with no dictator in which we have the same problem.: the weighted voting system [10:8,7,3,2]
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If P1 and P4 vote Yes and if P2 and P3 vote No, we have a stalemate. The problem occurs since the quota is too small. The quota needs to be larger than 10 and less than or equal to 20. Please see page 45, example 2.2 in your text. Thank you again Seth for raising this excellent point! Thank you Ron for your excellent connection between veto power and Banzhaf power. The weight of a player is not a good predictor of a player’s Banzhaf power. As Ron pointed out in class, the key is not the weight, but the number of times a player is a critical player in a winning coalition. Ron reminded us that since a player P with veto power is critical in every winning coalition, then player P will be critical the maximum number of times possible, giving player P the largest amount of Banzhaf power. In summary, if a player has veto power, then this player will have the largest amount of
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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 11 - Math 103 Section 11 Friday 1 Submitted Homework Assignment#1 for Chapter 2 is at Sakai It is in the Assignments section and is

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