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1. Apportionment and Voting

2. Apportionment Has its roots in the U.S. Constitution. The apportionment of seats in the House of Representatives is based on the relative population of each state . A method of dividing a whole into various parts.

3. Apportionment Apportionment Method: The Hamilton Plan (Alexander Hamilton) The Jefferson Plan (Thomas Jefferson) Huntington-Hill Apportionment Method

4. Class Number of Students Kinder 38 Grade 1 39 Grade 2 35 Grade 3 27 Grade 4 21 Grade 5 31 Grade 6 33 Total 224 Illustration: How can we apportion a total of 25 teacher aides among seven classes at a new elementary school?

5. Apportionment The Hamilton Plan Standard Divisor (D) t he number of people/voters represented by each representative  

6. Apportionment Hamilton’s Method: Using the standard divisor (D=8.96), calculate the quotient of each class. For Kinder the quotient is .  

7. Class Number of Students Quotient Kinder 38 Grade 1 39 4.353 Grade 2 35 3.906 Grade 3 27 3.013 Grade 4 21 2.344 Grade 5 31 3.460 Grade 6 33 3.683 Total 224 Class Number of Students Quotient Kinder 38 Grade 1 39 4.353 Grade 2 35 3.906 Grade 3 27 3.013 Grade 4 21 2.344 Grade 5 31 3.460 Grade 6 33 3.683 Total 224

8. Apportionment Hamilton’s Method: Using the standard divisor (D=8.96), calculate the quotient of each class. Next, compute for the standard quota (Q) rounded to the nearest whole number (lower quota). Initially, each sub-group receives a number of seats equal to its lower quota.

9. Class Number of Students Quotient Std. Quota Kinder 38 4 Grade 1 39 4.353 4 Grade 2 35 3.906 3 Grade 3 27 3.013 3 Grade 4 21 2.344 2 Grade 5 31 3.460 3 Grade 6 33 3.683 3 Total 224 22 Class Number of Students Quotient Std. Quota Kinder 38 4 Grade 1 39 4.353 4 Grade 2 35 3.906 3 Grade 3 27 3.013 3 Grade 4 21 2.344 2 Grade 5 31 3.460 3 Grade 6 33 3.683 3 Total 224 22

10. Apportionment Hamilton’s Method: Using the standard divisor (D), calculate the standard quota (Q) rounded to the nearest whole number (lower quota). Initially, each sub-group receives a number of seats equal to its lower quota. If the sum of the lower quotas equals the number of seats to be apportioned, the apportionment process is complete.

11. Class Number of Students Quotient Std. Quota Kinder 38 4 Grade 1 39 4.353 4 Grade 2 35 3.906 3 Grade 3 27 3.013 3 Grade 4 21 2.344 2 Grade 5 31 3.460 3 Grade 6 33 3.683 3 Total 224 22 Class Number of Students Quotient Std. Quota Kinder 38 4 Grade 1 39 4.353 4 Grade 2 35 3.906 3 Grade 3 27 3.013 3 Grade 4 21 2.344 2 Grade 5 31 3.460 3 Grade 6 33 3.683 3 Total 224 22

12. Apportionment Hamilton’s Method: If the sum of the lower quotas equals the number of seats to be apportioned, the apportionment process is complete. But in our case, it is not. So, we revisit the calculation of the quotients and assign an additional representative to the class with the largest decimal remainder. This process is continued until the number of teacher aides equals the number required.

13. Class Number of Students Quotient Std. Quota Number of Teacher Aides Kinder 38 4 4 Grade 1 39 4.353 4 4 Grade 2 35 3.906 3 4 Grade 3 27 3.013 3 3 Grade 4 21 2.344 2 2 Grade 5 31 3.460 3 4 Grade 6 33 3.683 3 4 Total 224 22 25 Class Number of Students Quotient Std. Quota Number of Teacher Aides Kinder 38 4 4 Grade 1 39 4.353 4 4 Grade 2 35 3.906 3 4 Grade 3 27 3.013 3 3 Grade 4 21 2.344 2 2 Grade 5 31 3.460 3 4 Grade 6 33 3.683 3 4 Total 224 22 25

14. Apportionment Summary of the Hamilton’s Method: Using the standard divisor (D), calculate the standard quota (Q) rounded to the nearest whole number (lower quota). Initially, each sub-group receives a number of seats equal to its lower quota. If the sum of the lower quotas equals the number of seats to be apportioned, the apportionment process is complete.

15. Apportionment c ontinuation … If the sum of the lower quotas is less than the total number of seats to be apportioned, then assign a seat to the sub-group that has the highest decimal part in its standard quota. Repeat step 3 (using the next highest decimal part) until the total number of seats has been apportioned.

16. Apportionment The Jefferson Plan uses a modified standard divisor that yields to the correct number of representatives by trial and error so that the sum of the standard quotas is equal to the number of representatives the modified standard divisor is always less than the standard divisor  

17. Apportionment Jefferson’s Method: Using the modified standard divisor, calculate the standard quota and the lower quotas of each sub-group. If the sum of the lower quotas equals the total number of seats to be apportioned, the apportionment process is complete, that is, each subgroup receives a number of seats equal to its lower quota.

18. Apportionment Jefferson’s Method: Using the modified standard divisor, calculate the standard quota and the lower quotas of each sub-group. Let: What is the sum of the lower quotas?  

19. Apportionment Jefferson’s Method: If the sum of the lower quotas equals the total number of seats to be apportioned, the apportionment process is complete, that is, each subgroup receives a number of seats equal to its lower quota.

20. Class Number of Students Quotient Std. Quota Number of Teacher Aides Kinder 38 4 Grade 1 39 4.875 4 Grade 2 35 4.375 4 Grade 3 27 3.375 3 Grade 4 21 2.625 2 Grade 5 31 3.875 3 Grade 6 33 4.125 4 Total 224 24 Class Number of Students Std. Quota Number of Teacher Aides Kinder 38 4 Grade 1 39 4.875 4 Grade 2 35 4.375 4 Grade 3 27 3.375 3 Grade 4 21 2.625 2 Grade 5 31 3.875 3 Grade 6 33 4.125 4 Total 224 24

21. Apportionment c ontinuation … If the sum of the lower quotas does not equal to the number of seats to be apportioned, choose a modified divisor less than the standard divisor and calculate the modified quotas and lower modified quotas. Repeat step 3 until you find a modified divisor such that the sum of the lower modified quotas equals the total number of seats to be apportioned. Each subgroup receives a number of seats equal to its lower modified quota, and the apportionment process is complete.

22. Class Number of Students Quotient Std. Quota Number of Teacher Aides Kinder 38 5 Grade 1 39 5.200 5 Grade 2 35 4.667 4 Grade 3 27 3.600 3 Grade 4 21 2.800 2 Grade 5 31 4.133 4 Grade 6 33 4.400 4 Total 224 27 Class Number of Students Std. Quota Number of Teacher Aides Kinder 38 5 Grade 1 39 5.200 5 Grade 2 35 4.667 4 Grade 3 27 3.600 3 Grade 4 21 2.800 2 Grade 5 31 4.133 4 Grade 6 33 4.400 4 Total 224 27

23. Class Number of Students Quotient Std. Quota Final Number of Teacher Aides Kinder 38 4 4 Grade 1 39 5.00 5 5 Grade 2 35 4.487 4 4 Grade 3 27 3.461 3 3 Grade 4 21 2.692 2 2 Grade 5 31 3.974 3 3 Grade 6 33 4.230 4 4 Total 224 25 25 Class Number of Students Std. Quota Final Number of Teacher Aides Kinder 38 4 4 Grade 1 39 5.00 5 5 Grade 2 35 4.487 4 4 Grade 3 27 3.461 3 3 Grade 4 21 2.692 2 2 Grade 5 31 3.974 3 3 Grade 6 33 4.230 4 4 Total 224 25 25 Final

24. Apportionment Summary of Jefferson’s Method: Using the modified standard divisor, calculate the standard quota and the lower quotas of each sub-group. If the sum of the lower quotas equals the total number of seats to be apportioned, the apportionment process is complete, that is, each subgroup receives a number of seats equal to its lower quota.

25. Apportionment c ontinuation … If the sum of the lower quotas does not equal to the number of seats to be apportioned, choose a modified divisor less than the standard divisor and calculate the modified quotas and lower modified quotas. Repeat step 3 until you find a modified divisor such that the sum of the lower modified quotas equals the total number of seats to be apportioned. Each subgroup receives a number of seats equal to its lower modified quota, and the apportionment process is complete.

26. Apportionment Example: A total of 25 teacher aides are to be apportioned among seven classes at a new elementary school. The enrolments in the seven classes are shown in the following table. Class Number of Students Kinder 38 Grade 1 39 Grade 2 35 Grade 3 27 Grade 4 21 Grade 5 31 Grade 6 33

27. Apportionment a. Determine the standard divisor. What is the meaning of the standard divisor in the context of this exercise? b. Use the Hamiltonian method to determine the number of teacher aides to be apportioned to each class. c. Use the Jefferson method to determine the number of teacher aides to be apportioned to each class. Is this apportionment in violation of the quota rule? d. How do the apportionment results produced using the Jefferson method compare with the results produced using the Hamiltonian method?

28. It is the number of pupils t o be served by each t eacher aide. a. Determine the standard divisor. What is the meaning of the standard divisor in the context of this exercise?  

29. Apportionment a. Determine the standard divisor. What is the meaning of the standard divisor in the context of this exercise? b. Use the Hamiltonian method to determine the number of teacher aides to be apportioned to each class. c. Use the Jefferson method to determine the number of teacher aides to be apportioned to each class. Is this apportionment in violation of the quota rule? d. How do the apportionment results produced using the Jefferson method compare with the results produced using the Hamiltonian method?

30. Class Number of Students Apportionment By Hamilton Method Apportionment By Jefferson Method Kinder 38 4 4 Grade 1 39 4 5 Grade 2 35 4 4 Grade 3 27 3 3 Grade 4 21 2 2 Grade 5 31 4 3 Grade 6 33 4 4 Total 224 25 25

31. Apportionment a. Determine the standard divisor. What is the meaning of the standard divisor in the context of this exercise? b. Use the Hamiltonian method to determine the number of teacher aides to be apportioned to each class. c. Use the Jefferson method to determine the number of teacher aides to be apportioned to each class. Is this apportionment in violation of the quota rule? d. How do the apportionment results produced using the Jefferson method compare with the results produced using the Hamiltonian method?

32. There is unfairness of the Apportionment

33. Apportionment Fairness in Apportionment One criterion of fairness for an apportionment plan is that it should satisfy the quota rule Quota Rule – The number of representatives apportioned to a state is the standard quota or one more than the standard quota

34. Class Number of Students Quotient Std. Quota Number of Teacher Aides By Hamiltonian Method Kinder 38 4 4 Grade 1 39 4.353 4 4 Grade 2 35 3.906 3 4 Grade 3 27 3.013 3 3 Grade 4 21 2.344 2 2 Grade 5 31 3.460 3 4 Grade 6 33 3.683 3 4 Total 224 22 25 Class Number of Students Quotient Std. Quota Number of Teacher Aides By Hamiltonian Method Kinder 38 4 4 Grade 1 39 4.353 4 4 Grade 2 35 3.906 3 4 Grade 3 27 3.013 3 3 Grade 4 21 2.344 2 2 Grade 5 31 3.460 3 4 Grade 6 33 3.683 3 4 Total 224 22 25

35. Class Number of Students Quotient Std. Quota Final Number of Teacher Aides by Jefferson Method Kinder 38 4 4 Grade 1 39 5.00 5 5 Grade 2 35 4.487 4 4 Grade 3 27 3.461 3 3 Grade 4 21 2.692 2 2 Grade 5 31 3.974 3 3 Grade 6 33 4.230 4 4 Total 224 25 25 Class Number of Students Std. Quota Final Number of Teacher Aides by Jefferson Method Kinder 38 4 4 Grade 1 39 5.00 5 5 Grade 2 35 4.487 4 4 Grade 3 27 3.461 3 3 Grade 4 21 2.692 2 2 Grade 5 31 3.974 3 3 Grade 6 33 4.230 4 4 Total 224 25 25 Final

36. Apportionment Average Constituency (C) – the average constituencies of the sub-group are approximately the same  

37. Class Number of Students Apportionment By Hamilton Method Average Constituency Apportionment By Jefferson Method Average Constituency Kinder 38 4 9.50 4 9.50 Grade 1 39 4 9.75 5 7.80 Grade 2 35 4 8.75 4 8.75 Grade 3 27 3 9.00 3 9.00 Grade 4 21 2 10.50 2 10.50 Grade 5 31 4 7.75 3 10.33 Grade 6 33 4 8.25 4 8.25 Total 224 25 25

38. Apportionment Apportionment Principle When adding a new representative to a sub-group, the representative is assigned to the group in such a way as to give the smallest relative unfairness of apportionment   where R is the relative unfairness of apportionment A is the absolute unfairness of apportionment = C is the average constituency of the sub-group receiving the new representative  

39. Apportionment Absolute Unfairness of an Apportionment The absolute unfairness of an apportionment is the absolute value of the difference between the average constituency of state A and the average of state B.  

40. Apportionment Example 2 A company is planning to add a new sales associate to one of its stores after noting a daily increase in the number of customers. Use the apportionment principle to determine which store should receive the new employee. Shopping mall location Number of sales associates Average number of customers per day Summer Hill Galleria 587 5289 Seaside Malle Galleria 614 6215

41. Apportionment Solution: Shopping mall location Summer Hill receives the new employee Seaside receives the new employee Summer Hill Galleria Seaside Malle Galleria

42. Apportionment Solution: Shopping mall location Summer Hill receives the new employee Seaside receives the new employee Summer Hill Galleria Seaside Malle Galleria Shopping mall location Summer Hill receives the new employee Seaside receives the new employee Summer Hill Galleria Seaside Malle Galleria

43. Apportionment Solution: Shopping mall location Summer Hill receives the new employee Seaside receives the new employee Absolute unfairness Remarks Summer Hill Galleria Seaside Malle Galleria Seaside Malle receives the new employee Shopping mall location Summer Hill receives the new employee Seaside receives the new employee Absolute unfairness Remarks Summer Hill Galleria Seaside Malle Galleria Seaside Malle receives the new employee

44. Apportionment Huntington-Hill Apportionment Method Method of equal proportions Used by the House of Representatives since 1940 Implemented by calculating what is called Huntington-Hill number which is derived from the apportionment principle.

45. Apportionment When there is a choice of adding one representative to a number of sub-groups, the representative should be added to the sub-group with the greatest Huntington-Hill number denoted by   w here is the population of the sub-group A a is the current number of representatives of sub-group A  

46. Apportionment Example: The table below shows the number of computers that are assigned to four different schools and the number of students in those schools. Use the Hungtington -Hill apportionment principle to determine which school a new computer should be assigned. School Number of Computers Number of students Rose 26 625 Lincoln 22 532 Midway 26 620 Valley 31 754

47. Apportionment Solution School Number of Computers (a) Number of students ( ) Huntington-Hill Number Rose 26 625 Lincoln 22 532 Midway 26 620 Valley 31 754 School Number of Computers (a) Rose 26 625 Lincoln 22 532 Midway 26 620 Valley 31 754

48. Apportionment Solution School Number of Computers (a) Number of students ( ) Huntington-Hill Number Rose 26 625 556.45 Lincoln 22 532 559.34 Midway 26 620 547.58 Valley 31 754 573.10 School Number of Computers (a) Rose 26 625 556.45 Lincoln 22 532 559.34 Midway 26 620 547.58 Valley 31 754 573.10

49. Apportionment Which of the following apportionment methods can violate the quota rule? Hamilton plan Jefferson plan Huntington-Hill method

50. Apportionment Paradoxes Alabama Paradox an increase in the total number of items to be apportioned results in the loss of an item for a group Population Paradox Group A loses items to Group B, even though the population of group A grew at a faster rate than that of group B

51. New States Paradox The addition of a new group changes the apportionments of other methods.

52. Exercises The following table shows how the average constituency changes for two regional governing boards, Joshua and Salinas, when a new representative is added to each board. Joshua’s average constituency Salinas’s average constituency Joshua receives new board member 1215 1547 Salinas receives new board member 1498 1195

53. Determine the relative unfairness of an apportionment that gives a new board member to Joshua rather than to Salinas. Round to the nearest thousandth. Determine the relative unfairness of an apportionment that gives a new board member to Salinas rather than to Joshua. Round to the nearest thousandth. Using the apportionment principle, determine which regional governing board should receive the new board member.

54. Voting

55. Voting Plurality Method of Voting E ach voter votes for one candidate, and the candidate with the most votes wins. The winning candidate does not have to have a majority of the votes. m ajority vote: over 50% of the people voting must vote for the candidate

56. Voting Example Fifty people were asked to rank their preferences of five varieties of chocolate candy, using 1 for their favorite and 5 for their least favorite. The results are shown below. Rankings Caramel 5 4 4 4 2 4 Vanilla 1 5 5 5 5 5 Almond 2 3 2 1 3 3 Toffee 4 1 1 3 4 2 Solid 3 2 3 2 1 1 Number of Voters 17 11 9 8 3 2

57. Voting Solution: Rankings First –place votes Caramel 5 4 4 4 2 4 Vanilla 1 5 5 5 5 5 17 Almond 2 3 2 1 3 3 8 Toffee 4 1 1 3 4 2 11+ 9 = 20 Solid 3 2 3 2 1 1 3+2=5 Number of Voters 17 11 9 8 3 2

58. Voting Borda Count Method of Voting Each voter ranks all of the candidates; that is, each voter selects his or her first choice, second choice, third choice, and so on. If there are k candidates, each candidate receives k points for each first-choice vote, ( k- 1 ) points for each second-choice vote, ( k- 2 ) points for each third-choice vote, and so on. The candidate with the most total points is declared the winner.

59. Voting Example What is the most favorite fruit using the Borda count method? Fruit Rankings apple 1 2 1 orange 2 1 3 mango 3 3 2 No. of voters 20 35 2

60. Voting Solution: Fruit Rankings Total Votes apple 1 2 1 136 orange 2 1 3 147 mango 3 3 2 59 No. of voters 20 35 2

61. Voting Plurality with Elimination Method e ach person votes for his or her favorite candidate. If a candidate receives a majority of votes, that candidate is declared the winner. If no candidate receives a majority, then the candidate with the fewest votes is eliminated and a new election is held. This process continues until a candidate receives a majority of the votes.

62. Voting Problem A company is planning its annual summer retreat and has asked its employees to rank five different choices of recreation in order of preference. The results are given in the table.

63. Voting Rankings Picnic in a Park 1 2 1 3 4 Water skiing at a lake 3 1 2 4 3 Amusement Park 2 5 5 1 2 Riding horses at a ranch 5 4 3 5 1 Dinner cruise 4 3 4 2 5 Number of Votes 10 18 6 28 16

64. Voting Using the plurality voting system, what activity should be planned for the retreat? Use the plurality with elimination method to determine which activity should be chosen? Using the Borda Count method of voting, which activity should be planned?

65. Voting Example A service club is going to sponsor a dinner to raise money for a charity. The club has decided to serve Italian, Mexican, Japanese, Chinese, or Indian food. The members of the club were surveyed to determine their preferences. The results are shown in the table.

66. Voting Use the plurality with elimination method to determine the food preference of the club Rankings Italian 2 5 1 4 3 Mexican 1 4 5 2 1 Japanese 3 1 4 5 2 Chinese 4 2 3 1 4 Indian 5 3 2 3 5 Number of Ballots 33 30 25 20 18

67. Voting Pairwise Comparison Voting Method s ometimes referred to as the head-to-head method Each voter ranks all of the candidates; that is, each voter selects his or her first choice, second choice, third choice, and so on. For each possible pairing of candidates, the candidate with the most votes receives 1 point; if there is a tie, each candidate receives 1 2 point. The candidate who receives the most points is declared the winner.

68. Voting Condorcet Criterion a candidate who wins all possible head-to-head matchups should win an election when all candidates appear on the ballot.

69. Flaws of Voting Systems Fairness Criteria Majority Criterion The candidate who receives a majority of the first-place votes is the winner. Monotonicity Criterion If candidate A wins an election, then candidate A will also win the election if the only change in the voters’ preferences is that supporters of a different candidate change their votes to support candidate A.

70. Flaws of Voting Systems c ontinuation … Condorcet criterion A candidate who wins all possible head-to-head matchups should win an election when all candidates appear on the ballot. Independence of irrelevant alternatives If a candidate wins an election, the winner should remain the winner in any recount in which losing candidates withdraw from the race.

71. Weighted Voting Systems

72. Weighted Voting System A weighted voting system of n voters is written , where q is the quota and through represent the weights of each of the n voters. A quota is t he number of votes that are required to pass a measure The weight of vote is the number of votes  

73. Weighted Voting One person, one vote Each person has one vote and five votes, a majority, are required to pass a measure. Dictatorship in this system, the person with 21 votes can pass any measure. Even if the remaining four people get together, their votes do not total the quota of 20.  

74. Weighted Voting Null System i f all the members of this system vote for a measure, the total number of votes is 16, which is less than the quota. Therefore , no measure can be passed. Veto Power System the sum of all the votes is 21, the quota. Therefore, if any one voter does not vote for the measure , it will fail . Each voter is said to have a veto power.  

75. Weighted Voting More Definitions: Coalition – set of voters each of whom votes the same way, either for or against a resolution’ Winning Coalition – a set of voters the sum of whose votes is greater or equal to the quota Losing Coalition – a set of voters the sum of whose votes is less than the quota

76. Weighted Voting c ontinuation … Critical Voter – a voter who leaves a winning coalition and thereby turns into a losing coalition Dictator – a voter who has a weight that is greater or equal to the quota Dummy – a voter who is never a critical voter and has no power

77. Weighted Voting Number of Possible Coalition of n Voters The number of possible coalitions of n voters is  

78. Weighted Voting A weighted voting system is given by What is the quota? How many voters are in the system? What is the weight of voter B? What is the weight of the coalition {A, C}? Is {A, D} a winning coalition? Which voters are critical voters in the coalition {A, C, D}? How many coalitions can be formed? How many coalitions consists of exactly three voters?  

79. Weighted Voting Banzhaf Power Index (BPI) derived b y John F. Banzhaf III in 1965 the purpose of this index is to determine the power of a voter in a weighted voting system the banzhaf power index of a voter v, is given by  

80. Weighted Voting Example Suppose a government is composed of four political parties, A, B, C, and D. The voting system for this government is Determine the B anzhaf power index for each voter.  

81. Thank you for listening.

82. Apportionment Standard Quota (Q) The whole number part of the quotient of a population divided by the standard divisor  

83. Class Number of Students Quotient Lower Quota Kinder 38 4 Grade 1 39 4.353 4 Grade 2 35 3.906 3 Grade 3 27 3.013 3 Grade 4 21 2.344 2 Grade 5 31 3.460 3 Grade 6 33 3.683 3 Total 224 22 Class Number of Students Quotient Lower Quota Kinder 38 4 Grade 1 39 4.353 4 Grade 2 35 3.906 3 Grade 3 27 3.013 3 Grade 4 21 2.344 2 Grade 5 31 3.460 3 Grade 6 33 3.683 3 Total 224 22

84. Apportionment Example 1 A university is composed of four schools. There are 350 new computers to be apportioned among the four schools according to their respective enrolments. The enrolment in each school is given in the following table. Find the standard divisor and standard quota of each school. School Humanities Business Education Science and Math Enrolment 1250 985 1420 1595

85. Apportionment Example 3 A University has a president’s council that is composed of students from each of the undergraduate classes. If a new student representative is added to the council, use the Huntington-Hill apportionment principle to determine which class the new student council member should represent.

86. Apportionment Class Number of Representatives Number of students 1 st Year 12 2015 2 nd Year 10 1755 3 rd Year 9 1430 4 th Year 8 1309

87. Voting Example Who will win the presidency using the Borda count method? Number of Voters 13 11 9 5 1 st choice A C B C 2 nd choice B A A B 3 rd choice C B C A