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§ 4.1 - 4.2 The Apportionment Problem; Basic Concepts “The baby legislature would be cut in half, creating a unique bicameral structure, consisting of.

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Presentation on theme: "§ 4.1 - 4.2 The Apportionment Problem; Basic Concepts “The baby legislature would be cut in half, creating a unique bicameral structure, consisting of."— Presentation transcript:

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2 § 4.1 - 4.2 The Apportionment Problem; Basic Concepts “The baby legislature would be cut in half, creating a unique bicameral structure, consisting of two distinct chambers: the Senate or ‘Upper House,’ and the House of Representatives, or ‘Lower House.’ One late night session led the Founders to consider the addition of a third chamber or ‘Waffle House,’ though by morning most agreed the idea, seemingly so brilliant at 3 A.M., now seemed stupid.” A MERICA (THE BOOK) - A MERICA (THE BOOK)

3 The Apportionment Problem: History  The U.S. Constitution dictates that the seats in the House of Representatives are to be allocated to each state on the basis of their respective populations.  The document does not, however, say how the allocation is to be done. Aside from its proportional nature, the only constitutional demand on the apportionment is that it be done using a systematic method. The Apportionment Problem: History

4  The first instance of a presidential veto blocked Alexander Hamilton’s “act of apportionment” in 1792. Washington did this at Thomas Jefferson’s behest-- Jefferson’s own method for apportionment was eventually selected.  In 1872 no systematic method was used--in violation of laws passed by Congress and the Constitution. When the presidential election of 1876 ended up in the House, Rutherford B. Hayes was given victory. If the apportionment had been done legally his opponent, Samuel Tilden would have prevailed. The Apportionment Problem: History

5 The Apportionment Problem  Apportionment problems are a type of fair-division problem.  Earlier we dealt with games in which we divide different objects among players who are entitled to an equal share.  Now we will examine the reverse: the objects to be divided will be the same, but each player is entitled to a different share. The Apportionment Problem

6 PLANETANDO RIA EART H TELLA R VULCA N TOTAL POPULA TION in billions 16.216.128.38.969.5 Example: THE PLANETS OF ANDORIA, EARTH, TELLAR AND VULCAN HAVE DECIDED TO FORM A UNITED FEDERATION OF PLANETS. THE RULING BODY OF THIS GOVERNMENT WILL BE THE 139 MEMBER FEDERATION COUNCIL.

7 Example: Example: Chief Wiggum needs to assign the 139 officers under his command to four different districts of Springfield (we will just call them A, B, C and D). He will make the assignments on the basis of how many crimes were committed in the district last year.DistrictABCDTotal Crimes16216128389695

8  The natural way to determine the number of people per seat of the legislature is divide the total population, P, by the number of seats to be apportioned, M. This number is called the Standard Divisor. Or: SD = P M Apportionment Problems: Basic Concepts

9  Now that we have the standard divisor we can use it to calculate the total number of seats each ‘state’ would be entitled to (if partial seats were possible). This is called the state’s Standard Quota: State X’s Standard Quota = State’s Population SD Apportionment Problems: Basic Concepts

10  There are two other quotas we will make use of: 1. The Lower Quota: the standard quota rounded down. 2. The Upper Quota: the standard quota rounded up. Apportionment Problems: Basic Concepts

11 PLANETANDO RIA EART H TELLA R VULCA N TOTAL POPULA TION in billions 16.216.128.38.969.5 Example: THE PLANETS OF ANDORIA, EARTH, TELLAR AND VULCAN HAVE DECIDED TO FORM A UNITED FEDERATION OF PLANETS. THE RULING BODY OF THIS GOVERNMENT WILL BE THE 139 MEMBER FEDERATION COUNCIL. FIND EACH PLANET’S STANDARD QUOTA. HERE THE STANDARD DIVISOR IS: 69.5/139 =.5 BILLION (OR 500 MILLION) BEINGS PER SEAT.

12 PLANETANDO RIA EART H TELLA R VULCA N TOTAL POPULA TION in billions 16.216.128.38.969.5 STANDA RD QUOTA 32.432.256.617.8139 Example: THE PLANETS OF ANDORIA, EARTH, TELLAR AND VULCAN HAVE DECIDED TO FORM A UNITED FEDERATION OF PLANETS. THE RULING BODY OF THIS GOVERNMENT WILL BE THE 139 MEMBER FEDERATION COUNCIL. FIND EACH PLANET’S STANDARD QUOTA.

13 Example: Example: Chief Wiggum needs to assign the 139 officers under his command to four different districts of Springfield (we will just call them A, B, C and D). He will make the assignments on the basis of how many crimes were committed in the district last year. Find each district`s standard quota.DistrictABCDTotal Crimes16216128389695 The Standard Divisor is: 695 / 139 = 5 crimes per officer

14 Example: Example: Chief Wiggum needs to assign the 139 officers under his command to four different districts of Springfield (we will just call them A, B, C and D). He will make the assignments on the basis of how many crimes were committed in the district last year. Find each district`s standard quota.DistrictABCDTotal Crimes16216128389695 Standard Quota 32.432.256.619.8139

15 Example: Example: The Republic of Parador has a population of 12.5 million citizens. Seats in the country’s legislature are distributed amongst it’s six states according to their populations. The standard quota for each state is given below: (a) Find the number of seats in the legislature. (b) Find the Standard Divisor. (c) Find the population of each state. StateABCDEFTotal Standa rd Quota 32.9 2 138. 72 3.0841.8213.7019.76250

16 StatePopulation (est.) Standard Quota Round ed Connecticu t 237,0006.887 Delaware56,0001.622 Georgia71,0002.062 Kentucky69,0002.002 Maryland279,0008.098 Massachus etts 475,00013.7814 New Hampshire 142,0004.124 New Jersey 180,0005.225 New York332,0009.6310 North Carolina 354,00010.2710 Pennsylva nia 433,00012.5613 Rhode Island 68,0001.972 South Carolina 206,0005.986 Vermont86,0002.503 Virginia631,00018.3118 Total3,619,000105106Example: The population’s of the first 15 states to join the United States are given to the right. The first apportioned House of Representatives had 105 members. Notice that using conventional rounding results in more delegates than there are seats. We will need another method if we are to apportion a set number of representatives.


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