§ 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.” - AMERICA (THE BOOK)

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 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 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.

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. PLANET ANDORIA EARTH TELLAR VULCAN TOTAL POPULATION in billions 16.2 16.1 28.3 8.9 69.5

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. District A B C D Total Crimes 162 161 283 89 695

Apportionment Problems: Basic Concepts 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 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 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.

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. PLANET ANDORIA EARTH TELLAR VULCAN TOTAL POPULATION in billions 16.2 16.1 28.3 8.9 69.5 HERE THE STANDARD DIVISOR IS: 69.5/139 = .5 BILLION (OR 500 MILLION) BEINGS PER SEAT.

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. PLANET ANDORIA EARTH TELLAR VULCAN TOTAL POPULATION in billions 16.2 16.1 28.3 8.9 69.5 STANDARD QUOTA 32.4 32.2 56.6 17.8 139

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. District A B C D Total Crimes 162 161 283 89 695 The Standard Divisor is: 695 / 139 = 5 crimes per officer

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. District A B C D Total Crimes 162 161 283 89 695 Standard Quota 32.4 32.2 56.6 19.8 139

Example: The Republic of Parador has a population of 12 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. State A B C D E F Total Standard Quota 32.92 138.72 3.08 41.82 13.70 19.76 250

State Population (est.) Standard Quota Rounded Connecticut 237,000 6.88 7 Delaware 56,000 1.62 2 Georgia 71,000 2.06 Kentucky 69,000 2.00 Maryland 279,000 8.09 8 Massachusetts 475,000 13.78 14 New Hampshire 142,000 4.12 4 New Jersey 180,000 5.22 5 New York 332,000 9.63 10 North Carolina 354,000 10.27 Pennsylvania 433,000 12.56 13 Rhode Island 68,000 1.97 South Carolina 206,000 5.98 6 Vermont 86,000 2.50 3 Virginia 631,000 18.31 18 Total 3,619,000 105 106 Example: 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.