1. © 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 14 Voting and Apportionment

2. © 2010 Pearson Prentice Hall. All rights reserved. 2 14.3 Apportionment Methods

3. © 2010 Pearson Prentice Hall. All rights reserved. 3 Objectives 1.Find standard divisors and standard quotas. 2.Understand the apportionment problem. 3.Use Hamilton’s method. 4.Understand the quota rule. 5.Use Jefferson’s method. 6.Use Adam’s method. 7.Use Webster’s method.

4. © 2010 Pearson Prentice Hall. All rights reserved. 4 Standard Divisors and Standard Quotas Standard Divisors The standard divisor is found by dividing the total population under consideration by the number of items to be allocated. The standard quota for a particular group is found by dividing that group’s population by the standard divisor.

5. © 2010 Pearson Prentice Hall. All rights reserved. 5 Example 1: Finding Standard Quotas The Republic of Margaritaville is composed of four states, A, B, C, and D. The table of each state’s population (in thousands) is given below. StateABCDTotal Population2753834657671890 Standard Quota

6. © 2010 Pearson Prentice Hall. All rights reserved. 6 The standard quotas are obtained by dividing each state’s population by the standard divisor. We previously computed the standard divisor and found it to be 63. Example 1: Finding Standard Quotas

7. © 2010 Pearson Prentice Hall. All rights reserved. 7 Standard quota for state A Standard quota for state B Standard quota for state C Standard quota for state D Example 1: Finding Standard Quotas

8. © 2010 Pearson Prentice Hall. All rights reserved. 8 StateABCDTotal Population2753834657671890 Standard Quota4.376.087.3812.1730 Notice that the sum of all the standard quotas is 30, the total number of seats in the congress. Example 1: Finding Standard Quotas

9. © 2010 Pearson Prentice Hall. All rights reserved. 9 Apportionment Problem The standard quotas represent each state’s exact fair share of the 30 seats for the congress of Margaritaville. –Can state A have 4.37 seats in congress? The apportionment problem is to determine a method for rounding standard quota into whole numbers so that the sum of the numbers is the total number of allocated items. –Can we round standard quotas up or down to the nearest whole number and solve this problem? The lower quota is the standard quota rounded down to the nearest whole number. The upper quota is the standard quota rounded up to the nearest whole number.

10. © 2010 Pearson Prentice Hall. All rights reserved. 10 Here’s the apportionment problem for Margaritaville. There are four different apportionment methods called Hamilton’s method, Jefferson’s method, Adam’s method, and Webster’s method. Apportionment Problem

11. © 2010 Pearson Prentice Hall. All rights reserved. 11 Hamilton’s Method 1.Calculate each group’s standard quota. 2.Round each standard quota down to the nearest whole number, thereby finding the lower quota. Initially, give to each group its lower quota. 3.Give the surplus items, one at a time, to the groups with the largest decimal parts until there are no more surplus items.

12. © 2010 Pearson Prentice Hall. All rights reserved. 12 Example: Consider the lower quotas for Margaritaville. We have a surplus of 1 seat. The surplus seat goes to the state with the greatest decimal part. The greatest decimal part is 0.38 for state C. Thus, state C receives the additional seat, and is assigned 8 seats in congress. Hamilton’s Method

13. © 2010 Pearson Prentice Hall. All rights reserved. 13 A group’s apportionment should be either its upper quota or its lower quota. An apportionment method that guarantees that this will always occur is said to satisfy the quota rule. A group’s final apportionment should either be the nearest upper or lower whole number of its standard quota. The Quota Rule

14. © 2010 Pearson Prentice Hall. All rights reserved. 14 Jefferson’s Method 1.Find a modifier divisor, d, such that each group’s modified quota is rounded down to the nearest whole number, the sum of the quotas is the number of items to be apportioned. The modified quotients that are rounded down are called modified lower quotas. 2.Apportion to each group its modified lower quota.

15. © 2010 Pearson Prentice Hall. All rights reserved. 15 A rapid transit service operates 130 buses along six routes, A, B, C, D, E, and F. The number of buses assigned to each route is based on the average number of daily passengers per route. Use Jefferson’s method with d = 486 to apportion the buses. Example 3: Using Jefferson’s Method Route ABCDEFTotal Average Number of Passengers 43605130708010,24515,5352265065,000

16. © 2010 Pearson Prentice Hall. All rights reserved. 16 Solution: Using d = 486, the table below illustrates Jefferson’s method. Example 3: Using Jefferson’s Method

17. © 2010 Pearson Prentice Hall. All rights reserved. 17 Adam’s Method 1.Find a modifier divisor, d, such that when each group’s modified quota is rounded up to the nearest whole number, the sum of the quotas is the number of items to be apportioned. The modified quotients that are rounded up are called modified upper quotas. 2.Apportion to each group its modified upper quota.

18. © 2010 Pearson Prentice Hall. All rights reserved. 18 A rapid transit service operates 130 buses along six routes, A, B, C, D, E, and F. The number of buses assigned to each route is based on the average number of daily passengers per route. Use Adam’s method to apportion the buses. Example 4: Using Adam’s Method Route ABCDEFTotal Average Number of Passengers 43605130708010,24515,53522,65065,000

19. © 2010 Pearson Prentice Hall. All rights reserved. 19 Solution: We begin by guessing at a possible modified divisor, d, that we hope will work. We guess d = 512. Example 4: Using Adam’s Method

20. © 2010 Pearson Prentice Hall. All rights reserved. 20 Because the sum of modified upper quotas is too high, we need to lower the modified quotas. In order to do this, we must try a higher divisor. If we let d = 513, we obtain 129 for the sum of modified upper quotas. Because this sum is too low, we decide to increase the divisor just a bit to d = 512.7 instead. Example 4: Using Adam’s Method

21. © 2010 Pearson Prentice Hall. All rights reserved. 21 Webster’s Method 1.Find a modifier divisor, d, such that when each group’s modified quota is rounded to the nearest whole number, the sum of the quotas is the number of items to be apportioned. The modified quotients that are rounded down are called modified rounded quotas. 2.Apportion to each group its modified rounded quota.

22. © 2010 Pearson Prentice Hall. All rights reserved. 22 A rapid transit service operates 130 buses along six routes, A, B, C, D, E, and F. The number of buses assigned to each route is based on the average number of daily passengers per route. Use Webster’s method to apportion the buses. Example 5: Using Webster’s Method Route ABCDEFTotal Average Number of Passengers 43605130708010,24515,5352265065,000

23. © 2010 Pearson Prentice Hall. All rights reserved. 23 Solution: We begin by guessing at a possible modified divisor, d, that we hope will work. We guess d = 502, a divisor greater than the standard divisor, 500. We would obtain 129 for the sum of modified rounded quotas. Because 129 is too low, this suggests that d is too large. So, we guess a lower d than 500. Say, d = 498. Example 5: Using Webster’s Method