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

Hamilton’s Method Examples Problems. Hamilton’s Method of Apportionment -- A bundle of contradictions??. We will begin these applications by creating a chart with 6 columns Label the columns as seen here. The number of rows will depend on the number of “ states ”. State. p opulation. SQ.

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

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

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

  3. 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

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

  5. Use the given information to calculate the TOTAL population and the “standard divisor” 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 State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 TOTAL: 17600 #seats 176 SD: 100 Hamilton’s Method: Vocabulary & Example

  6. The STANDARD QUOTA is the exact quotient upon dividing a state’s population by the SD 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. State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 TOTAL: 17600 #seats 176 SD: 100 Hamilton’s Method: Vocabulary & Example

  7. The STANDARD QUOTA is the exact quotient upon dividing a state’s population by the SD 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. State population SQ LQ Rank #reps Algebra 9230 Geometry 8231 Trig 139 TOTAL: 17600 #seats 176 SD: 100 Hamilton’s Method: Vocabulary & Example 92.3 82.31 1.39

  8. The LOWER QUOTA is the integer part of the SQ 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. State population SQ LQ Rank #reps Algebra 9230 92.3 Geometry 8231 82.31 Trig 139 1.39 TOTAL: 17600 #seats 176 SD: 100 Hamilton’s Method: Vocabulary & Example 92 82 1

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

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

  11. So, in this application, the state with the highest decimal portion is awarded its UPPER QUOTA 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! State population SQ LQ Rank #reps Algebra 9230 92.3 92 Geometry 8231 82.31 82 Trig 139 1.39 1 TOTAL: 17600 #seats 176 SD: 100 Hamilton’s Method: Vocabulary & Example 92 82 1st 2 175 176

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

  13. The ALABAMA PARADOX HAMILTON’S METHOD

  14. 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... So, what’s the problem?

  15. 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 So, with populations unchanged, a re-apportionment is done. What results would make sense??? Hamilton’s Method seems simple enough... APPARENT CONTRADICTIONS... Some state would gain a seat, while all others remain the same.

  16. 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 A contradiction to the logical solution of a problem is called a PARADOX. Hamilton’s Method seems simple enough... APPARENT CONTRADICTIONS... 1. For All Practical Purposes, 4th ed., COMAP, W.H. Freeman, 1997

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

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

  19. 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 State population SQ LQ Rank #reps Algebra 9230 92.82 92 1st 93 Geometry 8231 82.77 82 2nd 83 Trig 139 1.40 1 1 TOTAL: 17600 #seats 177 175 177 SD: 99.44 Used to have 2 reps! Hamilton’s Method: ALABAMA PARADOX example

  20. The POPULATION PARADOX HAMILTON’S METHOD

  21. 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 MethodPOPULATION PARADOX EXAMPLE...

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

  23. *solution#1 Mathland U. Graduate courseapportionment for Fall Semester 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

  24. 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 MethodPOPULATION PARADOX EXAMPLE...

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

  26. *solution#2 Mathland U. Graduate courseapportionment for SPRING Semester Gained the most students, but lost a section! 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

  27. *solution#2 The POPULATION PARADOX occurs when: The “population” of a “state” increases, but it loses a seat upon re-apportionment. 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

  28. The NEW-STATES PARADOX HAMILTON’S METHOD

  29. 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 schedule28 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 MethodNEW-STATES PARADOX EXAMPLE...

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

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

  32. 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: New-States Paradox

  33. 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. Hamilton’s Method… then what?

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