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class+January+21 - Math 103 Section 11 1 Give Gateway...

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Math 103, Section 11, January 21, 2010 1. Give Gateway Quiz-last 15 minutes of class. 2. Go over homework. Fill in Voting Methods and Fairness Criteria table. 3. Homework to be done and submitted to Sakai. Due next Tuesday, January 25 Chapter 1: Exercises 12,18,20,24. 4. Section 1.4 The Plurality-with-Elimination Method 5. Section 1.5 The Method of Pairwise Comparisons 6. Homework not to be submitted. Will be discussed on Tuesday, January 25: Exercises 33,37 and 59 Exercise #11: a. A has 153+102+55= 310 votes B has 202+108+20= 330 votes C has 110+160= 270 votes D has 175+155= 330 votes B and D tie with 330 votes b. B has 55+110+175= 340 votes D has 153+20+160= 333 votes
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D has fewer last place votes than B. D wins. D would be the Homecoming Queen. c. Who wins in the head-to-head match between B and D? 153+102+202+108+20+160=745 prefer B 55+110+175+155=495 prefer D. B wins. B would be the Homecoming Queen. Discuss how to make table: Insert, Table. Exercise #17. # of voters 5 3 5 3 2 3 1 ST choice A:25 A:15 C:25 D:15 D:10 B:15 2 ND choice B:20 D:12 E:20 C:12 C:8 E:12 3 rd choice C:15 B:9 D:15 B:9 B:6 A:9 4 th choice D:10 C:6 A:10 E:6 A:4 C:6 5 th choice E:5 E:3 B:5 A:3 E:2 D:3 A has 25+15+10+3+4+9=66 Borda points B has 20+9+5+9+6+15=64 Borda points C has 15+6+25+12+8+6=72 Borda points D has 10+12+15+15+10+3=65 Borda points E has 5+3+20+6+2+12=48 Borda points C wins with 72 Borda points. Preference table without candidate E: # of voters 5 3 5 3 2 3 1 ST choice A:20 A:12 C:20 D:12 D:8 B:12 2 ND choice B:15 D:9 D:15 C:9 C:6 A:9 3 rd choice C:10 B:6 A:10 B:6 B:4 C:6 4 th choice D:5 C:3 B:5 A:3 A:2 D:3
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A has 20+12+10+3+2+9=56 Borda points. B has 15+6+5+6+4+12=48 Borda points. C has 10+3+20+9+6+6=54 Borda points. D has 5+9+15+12+8+3=52 Borda points. A wins with 56 Borda points. d. Fairness criterion: Independence-of-irrelevant-alternatives (IIA). 19. Preference schedule for the election: # of voters 8 7 6 2 1 1 ST choice A D D C E 2 nd choice B B B A A 3 rd choice C A E B D 4 th choice D C C D B 5 th choice E E A E C A. Find the Borda count winner. # of voters 8 7 6 2 1 1 ST choice A:40 D:35 D:30 C:10 E:5 2 nd choice B:32 B:28 B:24 A:8 A:4
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3 rd choice C:24 A:21 E:18 B:6 D:3 4 th choice D:16 C:14 C:12 D:4 B:2 5 th choice E:8 E:7 A:6 E:2 C:1 A has 40+21+6+8+4=79 Borda points. B has 32+28+24+6+2=92 Borda points C has 24+14+12+10+1=61 Borda points. D has 16+35+30+4+3=88 Borda points. E has 8+7+18+2+5=40 Borda points. B is the Borda winner with 92 points. b. There are 8+7+6+2+1=24 voters. A majority candidate needs 13 or more votes. Candidate D has 7+6=13 votes. Candidate D is the “should be” winner according to the majority criterion. Candidate B is the actual winner using the Borda count method. Because the actual winner is not candidate D, there is a violation of the majority criterion. c. Is there a Condorcet candidate? Note that since candidate D has a majority of first-place votes, then D is automatically a Condorcet candidate . D is not the winner of the election. Hence a violation of the Condorcet criterion.
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