Hamilton’s Method of Apportionment -- A bundle of contradictions?? Examples Problems

Hamilton’s Method: Example #1 We will begin these applications by creating a chart with 6 columns Label the columns as seen here... State population SQ LQ Rank Apmt The ORANGE type indicates that these columns will be labeled appropriately for the given application. apportionment The number of rows will depend on the number of “states”.

Hamilton’s Method: Vocabulary & Example APPLICATION: Mathland is a small country that consists of 3 states; Algebra, Geometry, and Trigonometry. The populations of the 3 states will be given in the chart. Suppose that one year, their government’s representative body allows 176 seats to be filled. The number of seats awarded to each state will be determined using Hamilton’s Method of apportionment.

Hamilton’s Method: Vocabulary & Example *handout Hamilton’s Method: Vocabulary & Example SD: TOTAL: State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 176

Hamilton’s Method: Vocabulary & Example State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 176 Use the given information to calculate the TOTAL population and the “standard divisor” TOTAL: 17600 SD: 100 The Standard Divisor is the “number of people per representative” [calculated by dividing TOTAL by #seats] Here, the SD means that for every 100 people a state will receive 1 representative

Hamilton’s Method: Vocabulary & Example The STANDARD QUOTA is the exact quotient upon dividing a state’s population by the SD SD: 100 TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 176 The SQ is the exact # of seats that a state would be allowed if fractional seats could be awarded. Notice that the “# of seats awarded” can also be thought of as the “# of representatives” for a state.

Hamilton’s Method: Vocabulary & Example SD: 100 TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 176 The STANDARD QUOTA is the exact quotient upon dividing a state’s population by the SD 92.3 82.31 1.39 For example… The SQ for Algebra is: 9230/100 = 92.3 The SQ for Geometry is: 8231/100 = 82.31 etc… You will note that, very often, the values upon division must be rounded. There is a danger in rounding to too few decimal places, as the decimal values will be used to determine which states are awarded the remaining seats.

Hamilton’s Method: Vocabulary & Example 92.3 82.31 1.39 SD: 100 TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 176 The LOWER QUOTA is the integer part of the SQ 92 82 1 For example… The LQ for Algebra is: 92 etc… It may be thought of as “rounding down” to the lower of the two integers the SQ lies between.

Hamilton’s Method: Vocabulary & Example If each state was awarded its LQ, What would the total # of seats apportioned be? 92 82 1 92.3 82.31 1.39 SD: 100 TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 176 175 That leaves 1 seat empty! And we certainly can’t have that!

Hamilton’s Method: Vocabulary & Example The 1 empty seat will be awarded by ranking the decimal portions of the SQ. 92 82 1 92.3 82.31 1.39 SD: 100 TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 176 1st 175 Since we only have 1 seat to fill, we need only rank the highest decimal portion.

Hamilton’s Method: Vocabulary & Example So, in this application, the state with the highest decimal portion is awarded its UPPER QUOTA 92 82 1 92.3 82.31 1.39 SD: 100 TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 176 92 82 1st 2 175 176 The Upper Quota is the first integer that is higher that the SQ. It may be thought of as “rounding up” to the higher of the two integers the SQ lies between. And now the TOTAL # reps is equal to the allowable #seats!

Try the next one on your own! HAMILTON’S METHOD Try the next one on your own!

State Population SQ LQ Rank Rep A 45,300 45.3 45 B 31,070 31.07 31 C 20,490 20.49 20 2nd 21 D 14,160 14.16 14 E 10,260 10.26 10 F 8720 8.72 8 1st 9 TOTAL 130,000 # Seats 130 128 2 extra seats SD 1000

HAMILTON’S METHOD The ALABAMA PARADOX

Hamilton’s Method seems simple enough... So, what’s the problem? The only confusion might occur if two decimal values are the same and both can not be ranked. In that case, usually the state with the higher integer value would be awarded the extra seat.. Problems may occur when Hamilton’s Method is used repeatedly in an application which has certain changes (in # of seats, populations, and/or # of states) over a period of time.

Hamilton’s Method seems simple enough... APPARENT CONTRADICTIONS... So, with populations unchanged, a re-apportionment is done. What results would make sense??? Think about this… As our country was continuing to grow in the 18th & 19th centuries, the number of SEATS in the USHR continued to increase. Suppose that one election year, though a new census had not been taken, Congress decided to ADD a seat in the HR Some state would gain a seat, while all others remain the same.

Hamilton’s Method seems simple enough... APPARENT CONTRADICTIONS... A contradiction to the logical solution of a problem is called a PARADOX. Of course, that would make perfect sense. And much of the time, that’s what occurs… However, in 1881, this very thing happened But... upon re-apportionment, the state of Alabama actually LOST a seat, even though its population had not been recalculated!1 1. For All Practical Purposes, 4th ed., COMAP, W.H. Freeman, 1997

Hamilton’s Method: ALABAMA PARADOX example *handout SD: TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 177 Reapportionment of Mathland representative body with change in # seats.

Hamilton’s Method: ALABAMA PARADOX example SD: TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 177 92.82 92 1st 93 82.77 82 2nd 83 1.40 1 1 Used to have 2 reps! 175 177 99.44 Remember: SD = Total population/ #seats SQ = state population/ SD Values were rounded to fit in the chart after determining that it would not change the rankings.

Hamilton’s Method: ALABAMA PARADOX example SD: TOTAL: 17600 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 #seats 177 92.82 92 1st 93 82.77 82 2nd 83 1.40 1 99.44 175 Used to have 2 reps! The ALABAMA PARADOX occurs when: The number of seats is increased, and, although there has been no re-calculation of populations, a state loses a seat upon re-apportionment

The POPULATION PARADOX HAMILTON’S METHOD The POPULATION PARADOX

Hamilton’s Method POPULATION PARADOX EXAMPLE... EXAMPLE part 1: The Mathematics Department at Mathland U. plans to offer 4 graduate courses during the Fall semester. They are allowed to schedule 25 class sections, and wish to apportion these sections using Hamilton’s Method. The projected enrollments are given in the chart on the following slide...

Hamilton’s Method: Population Paradox *handout#1 Hamilton’s Method: Population Paradox total#sect.: 25 Course Proj. Enrol. SQ LQ Rank #sect. Math A 400 Math B 90 Math C 225 Math D 200 TOTAL: SD: Mathland U. Graduate course apportionment for Fall Semester

Hamilton’s Method: Population Paradox *solution#1 Hamilton’s Method: Population Paradox Course Proj. Enrol. SQ LQ Rank #sect. Math A 400 10.929 10 1st 11 Math B 90 2.459 2 2 Math C 225 6.148 6 6 Math D 200 5.464 5 2nd 6 TOTAL: 915 total#sect.: 25 23 25 SD: 36.6 Mathland U. Graduate course apportionment for Fall Semester

Hamilton’s Method POPULATION PARADOX EXAMPLE... EXAMPLE part 2: The Mathematics Department at Mathland U. also plans to offer 4 graduate courses during the Spring semester. Once again, they are allowed to schedule 25 class sections, and wish to apportion these sections using Hamilton’s Method. The Spring semester projected enrollments are given in the chart on the following slide...

Hamilton’s Method: Population Paradox *handout#2 Hamilton’s Method: Population Paradox total#sect.: 25 Course Proj. Enrol. SQ LQ Rank #sect. Math A 400 Math B 99 Math C 225 Math D 211 TOTAL: SD: Mathland U. Graduate course apportionment for Spring Semester

Hamilton’s Method: Population Paradox *solution#2 Hamilton’s Method: Population Paradox Course Proj. Enrol. SQ LQ Rank #sect. Math A 400 10.695 10 1st 11 Math B 99 2.647 2 2nd 3 Math C 225 6.106 6 6 Math D 211 5.642 5 5 Gained the most students, but lost a section! TOTAL: 935 total#sect.: 25 23 25 SD: 37.4 Mathland U. Graduate course apportionment for SPRING Semester

Hamilton’s Method: Population Paradox *solution#2 Hamilton’s Method: Population Paradox Course Proj. Enrol. SQ LQ Rank #sect. Math A 400 10.695 10 1st 11 Math B 99 2.647 2 2nd 3 Math C 225 6.106 6 6 Math D 211 5.642 5 5 TOTAL: 935 total#sect.: 25 23 25 SD: 37.4 The POPULATION PARADOX occurs when: The “population” of a “state” increases, but it loses a seat upon re-apportionment.

The NEW-STATES PARADOX HAMILTON’S METHOD The NEW-STATES PARADOX

Hamilton’s Method NEW-STATES PARADOX EXAMPLE... EXAMPLE part 3: During the Fall semester following the one mentioned in EXAMPLE part 1, the Mathematics Department at Mathland U. plans to offer 5 graduate courses. They decide to schedule 28 class sections, due to the addition of the new course, and, once again, wish to apportion these sections using Hamilton’s Method. The projected enrollments for the 5 classes are given in the chart on the following slide...

Hamilton’s Method: New-States Paradox *handout#3 Hamilton’s Method: New-States Paradox Math E 120 Course Proj. Enrol. SQ LQ Rank #sect. Math A 400 Math B 90 Math C 225 Math D 200 total#sect.: 28 TOTAL: SD: Mathland U. Graduate course apportionment for New Fall Semester

Hamilton’s Method: New-States Paradox *solution#3 Hamilton’s Method: New-States Paradox Math E 120 Course Proj. Enrol. SQ LQ Rank #sect. Math A 400 Math B 90 Math C 225 Math D 200 total#sect.: 28 TOTAL: SD: 10.821 10 1st 11 Used to have 2 sections 2.435 2 2nd 3 6.087 6 6 Used to have 6 sections 5.411 5 5 3.246 3 3 1035 26 28 36.964 Mathland U. Graduate course apportionment for New Fall Semester

Hamilton’s Method: New-States Paradox The NEW-STATES PARADOX occurs when: The total “population” changes due to the addition of a new “state” The “# of seats” changes appropriately using the old apportionment And, though the populations of the original “states” remain unchanged, An original state loses a “seat” upon re-apportionment and/or another original state gains a seat upon re-apportionment.

Hamilton’s Method… then what? Remember that Hamilton’s Method of apportionment is very useful and easy. Even though these paradoxes can occur, they happen infrequently and do not invalidate the method. Next class we will explore two other methods of apportionment that were proposed.