Skip to content
Textbook Guides

Solution Manual for Excursions in Modern Mathematics, 9th Edition

Need help with textbook problems? Solution Manual for Excursions in Modern Mathematics, 9th Edition has all the answers and solutions to get you through.

Preview 16 of 272 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

     S OLUTIONS M ANUAL D ALE B USKE , P H .D. St. Cloud State University E XCURSIONS IN M ODERN M ATHEMATICS N INTH E DITION Peter Tannenbaum California State University–Fresno

  2. Page 2
  3. Page 3
    Read page 3 as text

    iii Table of Contents Solutions Chapter 1 1 Chapter 2 21 Chapter 3 44 Chapter 4 70 Chapter 5 91 Chapter 6 104 Chapter 7 120 Chapter 8 130 Chapter 9 148 Chapter 10 160 Chapter 11 173 Chapter 12 191 Chapter 13 211 Chapter 14 224 Chapter 15 233 Chapter 16 247 Chapter 17 261

  4. Page 4
    Read page 4 as text

    Chapter 1 WALKING 1.1. Ballots and Preference Schedules 1. Number of voters 5 3 5 3 2 3 1st choice A A C D D B 2nd choice B D E C C E 3rd choice C B D B B A 4th choice D C A E A C 5 th choice E E B A E D This schedule was constructed by noting, for example, that there were five ballots listing candidate C as the first preference, candidate E as the second preference, candidate D as the third preference, candidate A as the fourth preference, and candidate B as the last preference. 2. Number of voters 4 5 6 2 1st choice A B C A 2nd choice D C A C 3rd choice B D D D 4 th choice C A B B 3. (a) 5 + 5 + 3 + 3 + 3 + 2 = 21 (b) 11. There are 21 votes all together. A majority is more than half of the votes, or at least 11. (c) Chavez. Argand has 3 last-place votes, Brandt has 5 last-place votes, Chavez has no last-place votes, Dietz has 3 last-place votes, and Epstein has 5 + 3 + 2 = 10 last-place votes. 4. (a) 202 + 160 + 153 + 145 + 125 + 110 + 108 + 102 + 55 = 1160 (b) 581; There are 1160 votes all together. A majority is more than half of the votes, or at least 581. (c) Alicia. She has no last-place votes. Note that Brandy has 125 + 110 + 55 = 290 last-place votes, Cleo has 202 + 145 + 102 = 449 last-place votes, and Dionne has 160 + 153 + 108 = 421 last-place votes. 5. Number of voters 37 36 24 13 5 1st choice B A B E C 2nd choice E B A B E 3rd choice A D D C A 4th choice C C E A D 5 th choice D E C D B Here Brownstein was listed first by 37 voters. Those same 37 voters listed Easton as their second choice, Alvarez as their third choice, Clarkson as their fourth choice, and Dax as their last choice. 6. Number of voters 14 10 8 7 4 1st choice B B A D E 2nd choice A D B C B 3rd choice E A E B A 4th choice C E D E C 5 th choice D C C A D

  5. Page 5
    Read page 5 as text

    2 Chapter 1: The Mathematics of Elections 7. Number of voters 14 10 8 7 4 A 2 3 1 5 3 B 1 1 2 3 2 C 5 5 5 2 4 D 4 2 4 1 5 E 3 4 3 4 1 Here 14 voters had the same preference ballot listing B as their first choice, A as their second choice, E as their third choice, D as their fourth choice, and C as their fifth and last choice. 8. Number of voters 37 36 24 13 5 A 1 2 5 2 4 B 3 1 2 4 1 C 2 4 3 1 5 D 5 3 1 5 2 E 4 5 4 3 3 9. Number of voters 255 480 765 1st choice L C M 2nd choice M M L 3rd choice C L C (0.17)(1500) = 255; (0.32)(500) = 480; The remaining voters (51% of 1500 or 1500-255-480=765) prefer M the most, C the least, so that L is their second choice. 10. Number of voters 450 900 225 675 1st choice A B C C 2nd choice C C B A 3rd choice B A A B 100% - 20% - 40% = 40% of the voters number 225 + 675 = 900. So, if N represents the total number of voters, then (0.40) 900 N . This means there are N = 2250 total voters. 20% of 2250 is 450 (these voters have preference ballots A , C , B ). 40% of 2250 is 900 (these voters have preference ballots B , C , A ). 1.2. Plurality Method 11. (a) C. A has 15 first-place votes. B has 11 + 8 + 1 = 20 first-place votes. C has 27 first-place votes. D has 9 first-place votes. C has the most first-place votes with 27 and wins the election. (b) C , B , A , D . Candidates are ranked according to the number of first-place votes they received (27, 20, 15, and 9 for C, B, A , and D respectively). 12. (a) D. A has 21 first-place votes. B has 18 first-place votes. C has 10 + 1 = 11 first-place votes. D has 29 first-place votes. D has the most first-place votes with 29 and wins the election. (b) D , A , B , C . 13. (a) C. A has 5 first-place votes. B has 4 + 2 = 6 first-place votes. C has 6 + 2 + 2 + 2 = 12 first-place votes. D has no first-place votes. C has the most first-place votes with 12 and wins the election. (b) C , B , A , D . Candidates are ranked according to the number of first-place votes they received (12, 6, 5, and 0 for C, B, A , and D respectively). 14. (a) B. A has 6 + 3 = 9 first-place votes. B has 6 + 5 + 3 = 14 first-place votes. C has no first-place votes. D has 4 first-place votes. B has the most first-place votes with 14 and wins the election.

  6. Page 6
    Read page 6 as text

    ISM: Excursions in Modern Mathematics, 9E 3 (b) B, A , D , C . Candidates are ranked according to the number of first-place votes they received (14, 9, 4 and 0 for B, A, D , and C respectively). 15. (a) D. A has 11% of the first-place votes. B has 14% of the first-place votes. C has 24% of the first-place votes. D has 23% + 19% + 9% = 51% of the first-place votes. E has no first-place votes. D has the largest percentage of first-place votes with 51% and wins the election. (b) D, C , B , A , E . Candidates are ranked according to the percentage of first-place votes they received (51%, 24%, 14%, 11% and 0% for D, C, B, A , and E respectively). 16. (a) C. A has 12% of the first-place votes. B has 15% of the first-place votes. C has 25% + 10% + 9% + 8% = 52% of the first-place votes. D has no first-place votes. E has 21% of the first-place votes. C has the largest percentage of first-place votes with 52% and wins the election. (b) C, E , B , A , D . Candidates are ranked according to the percentage of first-place votes they received (52%, 21%, 15%, 12%, and 0% for C, E, B, A and D respectively). 17. (a) A. A has 5 + 3 = 8 first-place votes. B has 3 first-place votes. C has 5 first-place votes. D has 3 + 2 = 5 first-place votes. E has no first-place votes. A has the most first-place votes with 8 and wins the election. (b) A , C, D , B, E. Candidates are ranked according to the number of first-place votes they received (8, 5, 5, 3, and 0 for A, C, D, B , and E respectively). Since both candidates C and D have 5 first-place votes, the tie in ranking is broken by looking at last-place votes. Since C has no last-place votes and D has 3 last- place votes, candidate C is ranked above candidate D . 18. (a) A. A has 153 + 102 + 55 = 310 first-place votes. B has 202 + 108 = 310 first-place votes. C has 160 + 110 = 270 first-place votes. D has 145 + 125 = 270 first-place votes. Both A and B have the most first- place votes with 310 so the tie is broken using last-place votes. A has no last-place votes. B has 125 + 110 + 55 = 290 last-place votes. So A wins the election. (b) A , B, D, C. Candidates are ranked according to the number of first-place votes they received (310, 310, 270 and 270 for A, B, D , and C respectively). In part (a), we saw that the tie between A and B was broken in favor of A . Since both candidates C and D have 270 first-place votes, the tie in ranking is broken by looking at last-place votes. Since C has 202 + 145 + 102 = 449 last-place votes and D has 160 + 153 + 108 = 421 last-place votes, candidate D is ranked above candidate C . 19. (a) A. A has 5 + 3 = 8 first-place votes. B has 3 first-place votes. C has 5 first-place votes. D has 3 + 2 = 5 first-place votes. E has no first-place votes. A has the most first-place votes with 8 and wins the election. (Note: This is exactly the same as in Exercise 17(a).) (b) A , C, D , B, E. Candidates are ranked according to the number of first-place votes they received (8, 5, 5, 3, and 0 for A, C, D, B , and E respectively). Since both candidates C and D have 5 first-place votes, the tie in ranking is broken by a head-to-head comparison between the two. But candidate C is ranked higher than D on 5 + 5 + 3 = 13 of the 21 ballots (a majority). Therefore, candidate C is ranked above candidate D . 20. (a) B. A has 153 + 102 + 55 = 310 first-place votes. B has 202 + 108 = 310 first-place votes. C has 160 + 110 = 270 first-place votes. D has 145 + 125 = 270 first-place votes. Both A and B have the most first- place votes with 310 so the tie is broken by head-to-head comparison. But candidate A is ranked higher than B on 153 + 125 + 110 + 102 + 55 = 545 of the 1160 ballots (less than a majority). So B wins the tiebreaker and the election.

  7. Page 7
    Read page 7 as text

    4 Chapter 1: The Mathematics of Elections (b) B, A, D, C. Candidates are ranked according to the number of first-place votes they received (310, 310, 270 and 270 for B , A , D , and C respectively). In part (a), we saw that the tie between A and B was broken in favor of B . Since both candidates C and D have 270 first-place votes, the tie in ranking is broken by head-to-head comparison. Candidate C is ranked higher than D on 160 + 153 + 110 + 108 = 531 of the 1160 ballots (less than a majority). So in the final ranking, candidate D is ranked above candidate C . 1.3. Borda Count 21. (a) A has 4 15 3 (9 8 1) 2 11 1 27 163 u  u    u  u points. B has 4 (11 8 1) 3 15 2 (27 9) 1 0 197 u    u  u   u points. C has 4 27 3 0 2 8 1 (15 11 9 1) 160 u  u  u  u    points. D has 4 9 3 (27 11) 2 (15 1) 1 8 190 u  u   u   u points. The winner is B . (b) B , D, A , C. Candidates are ranked according to the number of Borda points they received. 22. (a) A has 4 21 3 18 2 (29 10) 1 1 217 u  u  u   u points. B has 4 18 3 (10 1) 2 21 1 29 176 u  u   u  u points. C has 4 (10 1) 3 (29 21) 2 18 1 0 230 u   u   u  u points. D has 4 29 3 0 2 1 1 (21 18 10) 167 u  u  u  u   points. The winner is C . (b) C, A, B, D. Candidates are ranked according to the number of Borda points they received. 23. (a) A has 4 5 3 2 2 (6 2) 1 (4 2 2) 50 u  u  u   u   points. B has 4 (4 2) 3 (2 2) 2 2 1 (6 5) 51 u   u   u  u  points. C has 4 (6 2 2 2) 3 0 2 (5 4 2) 1 0 70 u     u  u    u points. D has 4 0 3 (6 5 4 2) 2 2 1 (2 2) 59 u  u     u  u  points. The winner is C . (b) C, D, B, A. Candidates are ranked according to the number of Borda points they received. 24. (a) A has 4 (6 3) 3 (4 3) 2 6 1 5 74 u   u   u  u points. B has 4 (6 5 3) 3 3 2 0 1 (6 4) 75 u    u  u  u  points. C has 4 0 3 (6 6 5) 2 (4 3 3) 1 0 71 u  u    u    u points. D has 4 4 3 0 2 (6 5) 1 (6 3 3) 50 u  u  u   u   points. The winner is B. (b) B, A, C, D. Candidates are ranked according to the number of Borda points they received. 25. Here we can use a total of 100 voters for simplicity. A has 5 11 4 (24 23 19) 3 (14 9) 2 0 1 0 388 u  u    u   u  u points. B has 5 14 4 0 3 (24 11) 2 23 1 (19 9) 249 u  u  u   u  u  points. C has 5 24 4 (14 11 9) 3 23 2 19 1 0 363 u  u    u  u  u points. D has 5 (23 19 9) 4 0 3 0 2 14 1 (24 11) 318 u    u  u  u  u  points. E has 5 0 4 0 3 19 2 (24 11 9) 1 (23 14) 182 u  u  u  u    u  points. The ranking (according to Borda points) is A, C, D, B, E .

  8. Page 8
    Read page 8 as text

    ISM: Excursions in Modern Mathematics, 9E 5 26. We use a total of 100 voters for simplicity. A has 5 12 4 0 3 9 2 (25 21 10) 1 (15 8) 222 u  u  u  u    u  points. B has 5 15 4 9 3 (21 12) 2 8 1 (25 10) 261 u  u  u   u  u  points. C has 5 (25 10 9 8) 4 0 3 0 2 15 1 (21 12) 323 u     u  u  u  u  points. D has 5 0 4 (21 15 12 10) 3 (25 8) 2 0 1 9 340 u  u     u   u  u points. E has 5 21 4 (25 8) 3 (15 10) 2 (12 9) 1 0 354 u  u   u   u   u points. The ranking (according to Borda points) is E, D, C, B, A. 27. Cooper had 3 49 2 280 1 316 1023 u  u  u points. Gordon had 3 37 2 432 1 275 1250 u  u  u points. Mariota had 3 788 2 74 1 22 2534 u  u  u points. The ranking (according to Borda points) is Mariota (2534), Gordon (1250), and Cooper (1023). 28. Hernandez had 7 13 4 17 3 0 2 0 1 0 159 u  u  u  u  u points. Kluber had 7 17 4 11 3 2 2 0 1 0 169 u  u  u  u  u points. Lester had 7 0 4 0 3 3 2 15 1 7 46 u  u  u  u  u points. Sale had 7 0 4 2 3 19 2 5 1 3 78 u  u  u  u  u points. Scherzer had 7 0 4 0 3 4 2 6 1 8 32 u  u  u  u  u points. As in Example 1.12, this uses a modified Borda count. In this case, first-place votes count 7 points rather than the usual 5. The ranking (according to modified Borda points) is Kluber (169), Hernandez (159), Sale (78), Lester (46), and Sherzer (32). 29. Each ballot has 4 1 3 1 2 1 1 1 10 u  u  u  u points that are awarded to candidates according to the Borda count. With 110 voters, there are a total of 110 10 1100 u Borda points. So D has 1100 – 320 – 290 – 180 = 310 Borda points. The ranking is thus A (320), D (310), B (290), and C (180). 30. Each ballot has 7 1 4 1 3 1 2 1 1 1 17 u  u  u  u  u points that are awarded to candidates according to the Borda count. With 50 voters, there are a total of 50 17 850 u Borda points. So E has 850 – 152 – 133 – 191 – 175 = 199 Borda points. The ranking is thus E (199), C (191), D (175), A (152), and B (133). 1.4. Plurality-with-Elimination 31. (a) A is the winner. Round 1: Candidate A B C D Number of first-place votes 15 20 27 9 D is eliminated. Round 2: The 9 first-place votes originally going to D now go to A . Candidate A B C D Number of first-place votes 24 20 27 B is eliminated. Round 3: There are 8 + 1 = 9 first-place votes originally going to B that now go to A . There are also 11 first-place votes going to B that would now go to D . But, since D is already eliminated, these 11 first- place votes go to A . Candidate A B C D Number of first-place votes 44 27 Candidate A now has a majority of the first-place votes and is declared the winner. (b) A complete ranking of the candidates can be found by noting in part (a) when each candidate was eliminated. Since D was eliminated first, it is ranked last. Since B was eliminated next, it is ranked next to last. The final ranking is hence A, C, B, D .

  9. Page 9
    Read page 9 as text

    6 Chapter 1: The Mathematics of Elections 32. (a) B is the winner. Round 1: Candidate A B C D Number of first-place votes 21 18 11 29 C is eliminated. Round 2: The 11 first-place votes originally going to C would next go to B . Candidate A B C D Number of first-place votes 21 29 29 Round 3: The 21 first-place votes going to A would next go to C . But C has been eliminated. So these 21 first-place votes go to B . Candidate A B C D Number of first-place votes 50 29 Candidate B now has a majority of the first-place votes and is declared the winner. (b) A complete ranking of the candidates can be found by noting in part (a) when each candidate was eliminated. Since C was eliminated first, it is ranked last. Since A was eliminated next, it is ranked next to last. The final ranking is hence B, D, A, C . 33. (a) C is the winner. Round 1: Candidate A B C D Number of first-place votes 5 6 12 0 Candidate C has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that C wins and at the end of round 1, D is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D Number of first-place votes 5 6 12 A is eliminated. Round 3: There are 5 first-place votes originally going to A that now go to C . Candidate A B C D Number of first-place votes 6 17 The final ranking is C, B, A, D . 34. (a) B is the winner. Round 1: Candidate A B C D Number of first-place votes 9 14 0 4 Candidate B has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that B wins and at the end of round 1, C is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D Number of first-place votes 9 14 4 D is eliminated. Round 3: There are 4 first-place votes originally going to D that now go to A . Candidate A B C D Number of first-place votes 13 14 The final ranking is B, A, D, C .

  10. Page 10
    Read page 10 as text

    ISM: Excursions in Modern Mathematics, 9E 7 35. Round 1: Candidate A B C D E Number of first-place votes 8 8 7 6 0 E is eliminated. Round 2: No first-place votes change. Candidate A B C D E Number of first-place votes 8 8 7 6 D is eliminated. Round 3: There are 4 first-place votes for D that move to candidate C and there are 2 first-place votes for D that move to candidate B . Candidate A B C D E Number of first-place votes 8 10 11 A is eliminated. Round 4: There are 5 first-place votes for A that move to candidate B and there are 3 first-place votes for A that move to candidate C (since D and E have both been eliminated). Candidate A B C D E Number of first-place votes 15 14 B now has a majority of the first-place votes and is declared the winner. The final ranking is B, C, A, D, E . 36. Round 1: Candidate A B C D E Number of first-place votes 8 4 11 12 5 B is eliminated. Round 2: There are 4 first-place votes for B that move to candidate E . Candidate A B C D E Number of first-place votes 8 11 12 9 A is eliminated. Round 3: There are 5 first-place votes for A that move to candidate E (since B is eliminated), 2 first-place votes for A that move to candidate D (since B is eliminated), and 1 first-place vote for A that moves to C . Candidate A B C D E Number of first-place votes 12 14 14 C is eliminated. Round 4: There are 6 first-place votes for C that move to candidate E (since A is eliminated), 5 first-place votes for C that move to candidate D (since both A and B are eliminated), and there is 1 first-place vote for C that moves to candidate E ( C earned this vote earlier when candidate A was eliminated). Candidate A B C D E Number of first-place votes 19 21 E now has a majority of the first-place votes and is declared the winner. The final ranking is E, D, C, A, B. 37. (a) D is the winner. Round 1: Candidate A B C D E Percentage of first-place votes 11 14 24 51 0 Candidate D has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that D wins and at the end of round 1, E is the first candidate eliminated. Round 2: No first-place votes are changed.

  11. Page 11
    Read page 11 as text

    8 Chapter 1: The Mathematics of Elections Candidate A B C D E Percentage of first-place votes 11 14 24 51 A is eliminated. Round 3: The 11% of the first-place votes that went to A now go to C . Candidate A B C D E Percentage of first-place votes 14 35 51 B is eliminated. Round 4: The 14% of the first-place votes that went to B now go to C . Candidate A B C D E Percentage of first-place votes 49 51 A complete ranking of the candidates can be found by noting when each candidate was eliminated. The final ranking is hence D, C, B, A, E . 38. (a) C is the winner. Round 1: Candidate A B C D E Percentage of first-place votes 12 15 52 0 21 Candidate C has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that C wins and at the end of round 1, D is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D E Percentage of first-place votes 12 15 52 21 A is eliminated. Round 3: The 12% of the first-place votes that went to A now go to B (since D has already been eliminated). Candidate A B C D E Percentage of first-place votes 27 52 21 E is eliminated. Round 4: The 14% of the first-place votes that went to E now go to B (since D has already been eliminated). Candidate A B C D E Percentage of first-place votes 48 52 A complete ranking of the candidates can be found by noting when each candidate was eliminated. The final ranking is hence C, B, E, A, D . 39. Round 1: Candidate A B C D E Number of first-place votes 8 8 7 6 0 Candidates E, D, and C are all eliminated. Round 2: There are 4 first-place votes for D that go to B (since C has been eliminated). There are 2 first- place votes for D that go to B . There are 7 first-place votes for C that go to A (since both D and E are eliminated). Candidate A B C D E Number of first-place votes 15 14 A now has a majority of the first-place votes and is declared the winner.

  12. Page 12
    Read page 12 as text

    ISM: Excursions in Modern Mathematics, 9E 9 40. Round 1: Candidate A B C D E Number of first-place votes 8 4 11 12 5 Candidates B, E, and A are all eliminated. Round 2: There are 4 first-place votes for B that go to C (since E has been eliminated). There are 5 first-place votes for E that go to D (since A has been eliminated). There are 5 first-place votes for A that go to C (since both B and E are eliminated), there are 2 first-place votes for A that go to D (since B is eliminated), and 1 first-place vote for A that goes to C . Candidate A B C D E Number of first-place votes 21 19 C now has a majority of the first-place votes and is declared the winner. 1.5. Pairwise Comparisons 41. (a) Candidate D is the winner. A versus B : 15 + 9 = 24 votes to 27 + 11 + 8 + 1 = 47 votes ( B wins). B gets 1 point. A versus C : 15 + 11 + 9 + 8 + 1 = 44 votes to 27 votes ( A wins). A gets 1 point. A versus D : 15 + 8 + 1 = 24 votes to 27 + 11 + 9 = 47 votes ( D wins). D gets 1 point. B versus C : 15 + 11 + 9 + 8 + 1 = 44 votes to 27 votes ( B wins). B gets 1 point. B versus D : 15 + 11 + 8 + 1 = 35 votes to 27 + 9 = 36 votes ( D wins). D gets 1 point. C versus D : 27 + 8 = 35 votes to 15 + 11 + 9 + 1 = 36 votes. ( D wins). D gets 1 point. The final tally is 1 point for A , 2 points for B , 0 points for C , and 3 points for D . (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is D (3 points), B (2 points), A (1 point), and C (0 points). 42. (a) Candidate C is the winner. A versus B : 29 + 21 = 50 votes to 18 + 10 + 1 = 29 votes ( A wins). A gets 1 point. A versus C : 21 + 18 = 39 votes to 29 + 10 + 1 = 40 votes ( C wins). C gets 1 point. A versus D : 21 + 18 + 10 = 49 votes to 29 + 1 = 30 votes ( A wins). A gets 1 point. B versus C : 18 votes to 29 + 21 + 10 + 1 = 61 votes ( C wins). C gets 1 point. B versus D : 21 + 18 + 10 + 1 = 50 votes to 29 votes ( B wins). B gets 1 point. C versus D : 21 + 18 + 10 + 1 = 50 votes to 29 votes. ( C wins). C gets 1 point. The final tally is 2 points for A , 1 point for B , 3 points for C , and 0 points for D . (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is C (3 points), A (2 points), B (1 point), and D (0 points). 43. (a) Candidate C is the winner. A versus B : 6 + 5 = 11 votes to 4 + 2 + 2 + 2 + 2 = 12 votes ( B wins). B gets 1 point. A versus C : 5 + 2 = 7 votes to 6 + 4 + 2 + 2 + 2 = 16 votes ( C wins). C gets 1 point. A versus D : 5 + 2 + 2 = 9 votes to 6 + 4 + 2 + 2 = 14 votes ( D wins). D gets 1 point. B versus C : 4 + 2 = 6 votes to 6 + 5 + 2 + 2 + 2 = 17 votes ( C wins). C gets 1 point. B versus D : 4 + 2 + 2 + 2 = 10 votes to 6 + 5 + 2 = 13 votes ( D wins). D gets 1 point. C versus D : 6 + 2 + 2 + 2 + 2 = 10 votes to 5 + 4 = 9 votes. ( C wins). C gets 1 point. The final tally is 0 points for A , 1 point for B , 3 points for C , and 2 points for D . (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is C (3 points), D (2 points), B (1 point), and A (0 points).

  13. Page 13
    Read page 13 as text

    10 Chapter 1: The Mathematics of Elections 44. (a) Candidate B is the winner. A versus B : 6 + 4 + 3 = 13 votes to 6 + 5 + 3 = 14 votes ( B wins). B gets 1 point. A versus C : 6 + 4 + 3 + 3 = 16 votes to 6 + 5 = 11 votes ( A wins). A gets 1 point. A versus D : 6 + 6 + 3 + 3 = 18 votes to 5 + 4 = 9 votes ( A wins). A gets 1 point. B versus C : 6 + 5 + 3 = 14 votes to 6 + 4 + 3 = 13 votes ( B wins). B gets 1 point. B versus D : 6 + 5 + 3 = 14 votes to 6 + 4 + 3 = 13 votes ( B wins). B gets 1 point. C versus D : 6 + 6 + 5 + 3 + 3 = 23 votes to 4 votes. ( C wins). C gets 1 point. The final tally is 2 points for A , 3 points for B , 1 point for C , and 0 points for D . (b) The final ranking is B (3 points), A (2 points), C (1 point), and D (0 points). 45. Candidate D is the winner. A versus B : 24% + 23% + 19% + 11% + 9% = 86% of the votes to 14% of the votes ( A wins). A versus C : 23% + 19% + 11% = 53% of the votes to 24% + 14% + 9% = 47% of the votes ( A wins). A versus D : 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes ( D wins). A versus E : 24% + 23% + 19% + 14% + 11% + 9% = 100% of the votes to 0% of the votes ( A wins). B versus C : 14% of the votes to 24% + 23% + 19% + 11% + 9% = 86% of the votes ( C wins). B versus D : 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes ( D wins). B versus E : 24% + 23% + 14% + 11% = 72% of the votes to 19% + 9% = 28% of the votes ( B wins). C versus D : 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes ( D wins). C versus E : 24% + 23% + 14% + 11% + 9% = 81% of the votes to 19% of the votes ( C wins). D versus E : 23% + 19% + 14% + 9% = 65% of the votes to 24% + 11% = 35% of the votes ( D wins). The final tally is 3 points for A , 1 point for B , 2 points for C , 4 points for D , and 0 points for E . 46. Candidate C is the winner. A versus B : 25% + 12% + 10% = 47% of the votes to 21% + 15% + 9% + 8% = 53% of the votes ( B wins). A versus C : 21% + 12% = 33% of the votes to 25% + 15% + 10% + 9% + 8% = 67% of the votes ( C wins). A versus D : 12% + 9% = 21% of the votes to 25% + 21% + 15% + 10% + 8% = 79% of the votes ( D wins). A versus E : 12% + 9% = 21% of the votes to 25% + 21% + 15% + 10% + 8% = 79% of the votes ( E wins). B versus C : 21% + 15% + 12% = 48% of the votes to 25% + 10% + 9% + 8% = 52% of the votes ( C wins). B versus D : 15% + 9% = 24% of the votes to 25% + 21% + 12% + 10% + 8% = 76% of the votes ( D wins). B versus E : 15% + 12% + 9% = 36% of the votes to 25% + 21% + 10% + 8% = 64% of the votes ( E wins). C versus D : 25% + 10% + 9% + 8% = 52% of the votes to 21% + 15% + 12% = 48% of the votes ( C wins). C versus E : 25% + 10% + 9% + 8% = 52% of the votes to 21% + 15% + 12% = 48% of the votes ( C wins). D versus E : 15% + 12% + 10% = 37% of the votes to 25% + 21% + 9% + 8% = 63% of the votes ( E wins). The final tally is 0 points for A , 1 point for B , 4 points for C , 2 points for D , and 3 points for E . 47. A versus B : 7 + 5 + 3 = 15 votes to 8 + 4 + 2 = 14 votes ( A wins). A gets 1 point. A versus C : 8 + 5 + 3 = 16 votes to 7 + 4 + 2 = 13 votes ( A wins). A gets 1 point. A versus D : 8 + 5 + 3 = 16 votes to 7 + 4 + 2 = 13 votes ( A wins). A gets 1 point. A versus E : 5 + 3 + 2 = 10 votes to 8 + 7 + 4 = 19 votes ( E wins). E gets 1 point. B versus C : 8 + 5 + 2 = 15 votes to 7 + 4 + 3 = 14 votes ( B wins). B gets 1 point. B versus D : 8 + 5 = 13 votes to 7 + 4 + 3 + 2 = 16 votes ( D wins). D gets 1 point. B versus E : 8 + 5 + 4 + 2 = 19 votes to 7 + 3 = 10 votes ( B wins). B gets 1 point. C versus D : 8 + 7 + 5 = 20 votes to 4 + 3 + 2 = 9 votes ( C wins). C gets 1 point. C versus E : 7 + 5 + 4 + 2 = 18 votes to 8 + 3 = 11 votes ( C wins). C gets 1 point. D versus E : 5 + 4 + 3 + 2 = 14 votes to 8 + 7 = 15 votes ( E wins). E gets 1 point. The final tally is 3 points for A , 2 points for B , 2 points for C , 1 point for D , and 2 points for E . Now B, C , and E each have 2 points. In head-to-head comparisons, B beats C and B beats E so that B is ranked higher than C and E . Also, C beats E so C is ranked higher than E as well. The final ranking is thus A, B, C, E, D .

  14. Page 14
    Read page 14 as text

    ISM: Excursions in Modern Mathematics, 9E 11 48. A versus B : 6 + 5 + 5 + 5 + 5 + 2 + 1 = 29 votes to 11 votes ( A wins). A gets 1 point. A versus C : 7 + 5 + 5 + 2 + 1 = 20 votes to 20 votes (tie). A and C each get ½ point. A versus D : 6 + 5 + 5 + 5 + 2 + 1 = 24 votes to 16 votes ( A wins). A gets 1 point. A versus E : 7 + 6 + 5 + 5 + 5 + 2 + 1 = 31 votes to 9 votes ( A wins). A gets 1 point. B versus C : 7 + 5 + 5 + 4 + 2 = 23 votes to 17 votes ( B wins). B gets 1 point. B versus D : 6 + 5 + 5 + 4 + 2 + 1 = 23 votes to 17 votes ( B wins). B gets 1 point. B versus E : 7 + 5 + 5 + 4 + 2 = 23 votes to 17 votes ( B wins). B gets 1 point. C versus D : 6 + 5 + 5 + 4 + 1 = 21 votes to 19 votes ( C wins). C gets 1 point. C versus E : 7 + 6 + 5 + 5 + 1 = 24 votes to 16 votes ( C wins). C gets 1 point. D versus E : 7 + 5 + 5 + 2 = 19 votes to 21 votes ( E wins). E gets 1 point. The final tally is 3.5 points for A , 3 points for B , 2.5 points for C , 0 points for D , and 1 point for E . The final ranking is thus A, B, C, E, D . 49. (a) With five candidates, there are a total of 4 + 3 + 2 + 1 = 10 pairwise comparisons. Each candidate is part of 4 of these (one against each other candidate). So, to find the number of points each candidate earns, we simply subtract the losses from 4. The 10 points are distributed as follows: E wins 1 1 2 points, D wins 1 2 2 points, C gets 3 points, B gets 2 points, and A gets the remaining 1 1 10 1 2 3 2 1 2 2     point. So A loses 3 pairwise comparisons. (b) Candidate C , with 3 points, is the winner. [The complete ranking is C, D, B, E, A .] 50. (a) Since there are a total of (6 5) / 2 15 u pairwise comparisons, F must have won 15 – 1 – 2 – 2 - 1 3 2 - 1 2 2 = 4 of them ( A earned 1 point, B and C each earned 2 points, D earned 1 3 2 , and E earned 1 2 2 points). This means F lost 1 pairwise comparison. (b) Candidate F , with 4 points, is the winner. 1.6. Fairness Criteria 51. First, we determine the winner using the Borda count. A has 4 6 3 0 2 0 1 (2 3) 29 u  u  u  u  points. B has 4 2 3 6 2 3 1 0 32 u  u  u  u points. C has 4 3 3 2 2 6 1 0 30 u  u  u  u points. D has 4 0 3 3 2 2 1 6 19 u  u  u  u points. So candidate B is the winner using Borda count. However, candidate A has 6 of the 11 votes (a majority) and beats all three other candidates ( B , C , and D ) in head-to-head comparisons. That is, candidate A is a Condorcet candidate. Since this candidate did not win using the Borda count, this is a violation of the Condorcet criterion. 52. First, we determine the winner is candidate B using the plurality-with-elimination method (see Exercise 32(a)). In Exercise 42, however, we saw that candidate C was a Condorcet candidate (beating each other candidate in head-to-head comparisons and earning 3 points in the process). Since candidate C did not win using plurality-with-elimination, this is a violation of the Condorcet criterion.

  15. Page 15
    Read page 15 as text

    12 Chapter 1: The Mathematics of Elections 53. The winner of this election is candidate R using the plurality method. Now F is clearly a nonwinning candidate. Removing F as a candidate leaves the following preference table. Number of voters 49 48 3 1st choice R H H 2nd choice H S S 3rd choice O O O 4th choice S R R In a recount, candidate H would be the winner using the plurality method. This is a violation of the IIA criterion. 54. First, we determine the winner using the Borda count. A has 4 14 3 0 2 0 1 (10 8 4 1) 79 u  u  u  u    points. B has 4 4 3 (14 10) 2 (8 1) 1 0 106 u  u   u   u points. C has 4 (10 1) 3 8 2 (14 4) 1 0 104 u   u  u   u points. D has 4 8 3 (4 1) 2 10 1 14 81 u  u   u  u points. So candidate B (Boris) is the winner using Borda count. Now D (Dave) is clearly a nonwinning (irrelevant) candidate. Removing D as a candidate leaves the following preference table. Number of voters 14 10 8 4 1 1st choice A C C B C 2nd choice B B B C B 3rd choice C A A A A We now recount using the Borda count. A has 3 14 2 0 1 (10 8 4 1) 65 u  u  u    points. B has 3 4 2 (14 10 8 1) 1 0 78 u  u     u points. C has 3 (10 8 1) 2 4 1 14 79 u    u  u points. In a recount, candidate C (Carmen!) would be the winner using Borda count, a violation of the IIA criterion. 55. First, we determine the winner using plurality-with-elimination. Round 1: Candidate A B C D E Number of first-place votes 8 3 5 5 0 Candidate E is eliminated. Round 2: No votes are shifted. Candidate A B C D E Number of first-place votes 8 3 5 5 Candidate B is eliminated. Round 3: The 3 first-place votes for B now go to A (since E has been eliminated). Candidate A B C D E Number of first-place votes 11 5 5 Candidate A now has a majority (11 of the 21 votes) and is declared the winner. Now C is clearly a nonwinning (irrelevant) candidate. Removing C as a candidate leaves the following preference table. Number of voters 5 5 3 3 3 2 1st choice A E A D B D 2nd choice B D D B E B 3rd choice D B B E A A 4th choice E A E A D E We now recount using the plurality-with-elimination.

  16. Page 16
    Read page 16 as text

    ISM: Excursions in Modern Mathematics, 9E 13 Round 1: Candidate A B D E Number of first-place votes 8 3 5 5 Candidate B is eliminated. Round 2: The 3 first-place votes that went to B now shift to E . Candidate A B D E Number of first-place votes 8 5 8 Candidate D is eliminated. Round 3: D had 5 first-place votes. Of these, 3 go to E and 2 go to A (since B was eliminated). Candidate A B D E Number of first-place votes 10 11 Candidate E now has a majority (11 of the 21 votes) and is declared the winner. Remember that candidate A was the winner before candidate C was removed. This is a violation of the IIA criterion. 56. If X has a majority of the first-place votes, then X will win every pairwise comparison (it is ranked above all other candidates on more than half the ballots) and is, therefore, the winner under the method of pairwise comparisons. 57. If X is the Condorcet candidate, then by definition X wins every pairwise comparison and is, therefore, the winner under the method of pairwise comparisons. 58. When a voter moves a candidate up in his or her ballot the number of first place votes for that candidate either increases or stays the same. It follows that if X had a plurality of the first place votes and a voter changes his or her ballot to rank X higher, then X still has a plurality. 59. When a voter moves a candidate up in his or her ballot, that candidate’s Borda points increase. It follows that if X had the most Borda points and a voter changes his or her ballot to rank X higher, then X still has the most Borda points. 60. When a voter moves a candidate up in his or her ballot it can’t hurt the candidate in a pairwise comparison— the candidate wins the same pairwise comparisons as before and possibly a few more. It follows that if X won the most pairwise comparisons and a voter changes his or her ballot to rank X higher, then X still wins the most pairwise comparisons. JOGGING 61. Suppose the two candidates are A and B and that A gets a first-place votes and B gets b first-place votes and suppose that a > b . Then A has a majority of the votes and the preference schedule is Number of voters a b 1st choice A B 2nd choice B A It is clear that candidate A wins the election under the plurality method, the plurality-with-elimination method, and the method of pairwise comparisons. Under the Borda count method, A gets 2 a + b points while B gets 2 b + a points. Since a > b , 2 a + b > 2 b + a and so again A wins the election. 62. In this variation of the Borda count each candidate gets 1 less point per ballot. It follows that if N is the number of voters, each candidate gets N fewer points than he or she would under the standard Borda count method. Since each candidate’s total points gets decreased by the same number N , the ranking of the candidates remains the same. 63. The number of points under this variation is complementary to the number of points under the standard Borda count method: a first place is worth 1 point instead of N , a second place is worth 2 points instead of N – 1,…, a last place is worth N points instead of 1. It follows that having the fewest points here is equivalent to having the most points under the standard Borda count method, having the second fewest is equivalent to having the second most, and so on.

256 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Mathematics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.