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class+January+28 - Math 103, Section 11, January 28, 2011...

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Math 103, Section 11, January 28, 2011 1. Give out Instructions on Getting your Homework to Sakai sheet. 2. Sum of consecutive integers formula 3. Go over homework. Exercises 41,55,56, and 57 4. Submitted Homework due dates: Exercises 12,18,20,24 due on Friday, January 28. Exercises 34,38,42,60 due on Tuesday, February 1. 5. Quiz on Friday, January 28. Quiz will cover chapter 1 and will last about 15 minutes. 6. Insincere or strategic voting 7. Arrow’s Impossibility Theorem 8 . Review of satisfying and violating fairness criteria. The sum of consecutive integers formula is 1+2+3+4+5+…+L=L*(L+1) 2 where L is the last integer to be added. In an election with N candidates, we substitute L=N-1 in the sum formula above. This gives the formula for pairwise comparisons for N candidates: 1+2+3+. ..+ N - 1 = (N-1)(N ) pairwise comparisons. 2 Examples. 1. In an election with 10 candidates,  a. How many pairwise comparisons are there?             Answer:  45 pairwise comparisons Solution: N=10.       L=N-1=9, Use the formula: 1+2+3+. ..+ N - 1 = (N-1)(N) 2 Then 1+2+3+…+9=(9*10)/2=45 pairwise comparisons 2. Given an election with 10 candidates, how many Borda points are given out by 1 ballot? 3.  In an election with 15 candidates,       a. How many pairwise comparisons are there?  b. How many Borda points are given out by 1 ballot?
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4. a. In an election with 54 candidates,  How many pairwise comparisons are there?  Answer:  1431 pairwise comparisons Solution: N=54.       L=N-1=53, Use the formula: 1+2+3+. ..+ N - 1 = (N-1)(N) 2 Then 1+2+3+…+53=(53*54)/2=1431 pairwise comparisons b. If one of the candidates is a Condorcet candidate, how many points does this candidate  receive?  Answer is 53 points. The Condorcet candidate wins each head-to-head. That is, the Condorcet candidate is preferred  over each of the other 53 candidates in a head-to-head comparison.  5. How many matches are played in a round robin tournament with 12 teams? Answer: 66 matches Solution: N=12. N-1=11. 1+2+3+…+11=(11*12)/2=66 6. How many handshakes occur in a group of 30 people, if each person shakes hands once with every other? Answer: 435 handshakes Solution: N=30 N-1=29 1+2+3+…+29=(29*30)/2=435 7 . An election is held among five candidates (A, B, C, D, and E). There are 37 voters. Using the Method of Pairwise Comparisons it is found that A, B, and C each win one pairwise comparison while D wins three. If E wins the remaining comparisons in this election, then which statement is true? Explain. (a) D is the Condorcet candidate . (b) E is the Condorcet candidate . (c) Both D and E are Condorcet candidates.
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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+January+28 - Math 103, Section 11, January 28, 2011...

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