Chapter 4: Mathematics of Appotionment
Chapter 4: Mathematics of Apportionment
ESSENTIAL QUESTIONS:
Section 4.1: What are the concepts and the goal of an apportionment problem?
Section 4.2: How’s does Hamilton’s Method attempt to solve issue of standard divisors?
Section 4.3: What are the 3 paradoxes that can occur in Hamilton’s Method?
Section 4.4: How does Jefferson’s Method treat quotas for apportionment?
Section 4.5: How does Adam’s Method treat quotas for apportionment?
Section 4.6: How does Webster’s Method treat quotas for apportionment?
WORD WALL:
ADAM’S METHOD
ALABAMA PARADOX
APPORTIONMENT
BALINSKI AND YOUNG’S IMPOSSIBILITY THEOREM
HAMILTON’S METHOD
JEFFERSON’S METHOD
NEW-STATES PARADOX
POPULATION PARADOX
QUOTA RULE
STANDARD QUOTA
LOWER QUOTA
UPPER QUOTA
WEBSTER’S METHOD
STANDARD DIVISOR
MODIFIED DIVISOR
Section 4.1 Apportionment Problem
APPORTIONMENT Problem:
EXAMPLE #1: Mrs. Phillips has 50 identical pieces of chocolate which she is planning to divide among her 5 children (division part). She wants to do this fairly.
What would one fair way to divide the chocolate be (Chapter 3)?
However, to teach her children a lesson about hard work she decides to give her children candy based on how much time they work doing chores in the next week (proportionality criterion).
| |Child #1 |Child #2 |Child #3 |Child #4 |Child #5 |Total |
|Minutes worked |150 |78 |173 |204 |295 |900 |
|Chocolate Pieces | | | | | | |
For how many minutes of chores worked would a child earn one piece of candy?
EXAMPLE #2: Dr. Williams has 60 blank DVDs to give to 4 students in a class.
What would one fair way to divide the DVDs be (Chapter 3)?
Dr. Williams thinks that the DVDs should be given to the students when you also consider their score on the last test they all took.
| |Child #1 |Child #2 |Child #3 |Child #4 |Total |
|TEST SCORE |85 |88 |92 |95 |360 |
|DVDs Received | | | | | |
For how many points earned on a test would a student earn one DVD?
Based on examples 1 and 2 Are there any challenge to trying to share items proportionally to all individuals?
• Elements of Apportionment Problem:
o “STATES”: players involved in apportionment
NOTATION for N States: A1, A2, …, AN
o “SEATS”: set of M identical, indivisible objects to be divided
o “POPULATIONS”: set of N positive numbers which are the basis for the apportionment of seats to state (proportionality criterion)
NOTATION for N Populations:
For #1 – 4: IDENTIFY who or what are the seats, populations, and states.
Exp #1: Mr. Gates plan to split $3500 allowances between his 4 children at the end of each quarter. The children will receive their allowances in proportion to their GPA.
Exp #2: There are 40 teachers in an elementary school and the principal plans to apportion teachers to the 5 grade levels (1st through 5th) in the school based on the current enrollment of each grade level.
Exp#3: There are 75 administrative assistants for the entire college which has 10 academic departments. Each department will receive assistants based on the number of students that have declared that department as a major.
Exp #4: The city is planning to reorganize their 5 major bus routes around the city. The average number of passengers on each route will determine how the city’s 36 buses will be apportioned.
Ratios (fractions) are the important measurements for apportionment
o STANDARD DIVISOR, SD: ratio of total population to seats
Standard divisors represents the _________________________ (population unit) per __________________
o STANDARD QUOTAS, q: ratio of the state population, p, to the standard divisor
Standard Quota represents the ___________________ number (including decimal) of seats that each state should get (if ONE seat could be divided into smaller parts)
• EXAMPLE #3: Consider a nation of 6 states with only 250 seats in their congress with different populations.
|State |A |B |C |D |E |F |Total |
|Population |1,646,000 |6,936,000 |154,000 |2,091,000 |685,000 |988,000 |12,500,000 |
|Standard Divisor | |
|Standard Quota | | | | | | | |
|Normal Rounding | | | | | | | |
|(0.5 Rule) | | | | | | | |
SPECIAL ROUNDING OF QUOTAS (0.5 rule doesn’t apply)
UPPER Quota, ↑: rounding up the standard quota to nearest integer
a. 45.6↑ = b. 9.2↑ = c. 17.5↑ = d. 108↑ =
LOWER Quota, ↓: rounding down the standard quota to nearest integer
a. 35.6↓ = b. 99.2↓ = c. 16.5↓ = d. 108↓ =
EXAMPLE #3b: find the upper and lower quotas for the 6 nations from Example 3.
|State |A |B |C |D |E |F |Total |
|Upper Quota | | | | | | | |
|Lower Quota | | | | | | | |
Will the lower and upper quotas be equal?
EXAMPLE #4: The school board wants to assign 30 new teaching assistants among 5 elementary schools based on the current number of students in the schools. Complete the calculations for Standard Divisor and Standard Quota for the table given.
| |North |South |East |West |Central |
|# of Students |375 |297 |408 |340 |380 |
EXAMPLE #5: For 4 players and 200 seats. Complete the table by finding the standard divisor, standard quotas and appropriately round the standard quotas.
|State |A |B |C |D |Total |
|Population |125 |150 |350 |275 |900 |
|Standard Divisor | |
|Standard Quota | | | | | |
|Normal | | | | | |
|Rounding | | | | | |
|Upper Quotas | | | | | |
|Lower Quotas | | | | | |
“Good” Apportionments:
HOMEWORK: p.150 # 2, 3, 5
Section 4.2: Hamilton’s Method and Quota Rule
• Every state will get _______________ ______its lower quota.
Example #1 Hamilton’s Method: 6 nations with 250 seats for congress. (Calculations from 4.1 notes)
|State |A |
1) Calculate each state’s STANDARD QUOTAS, q
|Step #1: |32.92 |138.72 |3.08 |41.82 |13.70 |19.76 |250 |
|Standard Quota | | | | | | | |
2) Give each state its LOWER quota.
|Step #2: | | | | | | | |
|Lower Quota | | | | | | | |
3) SURPLUS: Find the decimal for each state’s standard quota. Give one seat to each state from the largest to smallest decimal until out of surplus seats.
|Step #3: | | | | | | | |
|Standard Quota | | | | | | | |
|Decimal | | | | | | | |
Final Apportionment for HAMILTON: Assign each state it’s lower quota of seats and any surplus seat.
|State | | | | | | | |
***Hamilton’s Method is _________ Population Neutral, so it is biased in favor of larger than smaller states***
EXAMPLE 2: Mrs. Phillips has 50 pieces of chocolate to divide among her 5 children. (4.1 Notes)
| |Child #1 |Child #2 |Child #3 |Child #4 |Child #5 |Total |
|S2: Lower Quota | | | | | | |
|S3: Standard Quota Decimal| | | | | | |
|Apportionment | | | | | | |
EXAMPLE 3: Perform Hamilton’s Method for 4 players and 150 seats
| |State A |
|S1: Standard Quota | | | | | |
|S2: Lower Quota | | | | | |
|S3: Standard Quota Decimal| | | | | |
|Apportionment | | | | | |
EXAMPLE #4: The school board wants to assign 40 new teaching assistants among 5 elementary schools based on the current number of students in the schools. Complete performing all steps of Hamilton’s Method to determine the apportionment of all teaching assistants.
| |North |South |East |West |Central |
|# of Students |274 |372 |331 |304 |259 |
FAIRNESS CRITERION – QUOTA RULE: a state shouldn’t be apportioned a number of seats smaller than its lower quota or larger than its upper quota.
The Hamilton Method _________________________ the quota rule
Lower-quota Violation: a state is apportioned number of seats ________________ THAN lower quota
Upper-quota Violation: a state is apportioned number of seats ________________ THAN upper quota
Section 4.3 Alabama and Other Paradoxes
Paradoxes or _______________________ Outcomes can occur (seat assignment for state’s) when applying Hamilton’s Method and then a change is made to the basic elements of the apportionment problem.
A change to the seats, states, or population of known states can directly cause a change in the standard divisor which will then affect the standard quotas (particularly the decimals) and possibly affect the surplus statement
EXAMPLE #1: 3 states, 20,000 total people and the following data for population per state:
|State |A |B |C |TOTAL |
|Population |940 |9030 |10,030 |20,000 |
Suppose the country decides to use 200 representatives. Use Hamilton’s Method to apportion the seats.
Standard Divisor =
|State |A |B |C |TOTAL |
|Step #1: | | | | |
|SQ | | | | |
|Step #2: | | | | |
|LQ | | | | |
|Step #3: Surplus | | | | |
|Apportionment | | | | |
Suppose state C makes the request to have 201 seats instead. Where do you think the extra seat will go? Why?
Part B: Use Hamilton’s Method to apportion the 201 seats.
New Standard Divisor =
|State |A |B |C |TOTAL |
|Step #1 |9.45 |90.75 |100.80 |201 |
|Step #2 |9 |90 |100 |199 |
|Step #3 |.4 |.75 |.80 | |
| | |+1 |+1 | |
|Apportionment |9 |91 |101 |201 |
Does this seem fair?
• Alabama Paradox: an increase in the total number of seats being apportioned, in and of itself, forces a state to lose one of its seat.
EXAMPLE #2: We have the following data for a continent with 5 countries with a population of 900 (in millions). And there are 50 seats to be apportioned.
SD =
|State |Alamos |Brandura |Canton |Dexter |Elexion |Total |
|Step #2 | | | | | | |
|Step #3 | | | | | | |
|Apportionment | | | | | | |
Part B: Suppose 10 years go by and the country recounts its population to get the following chart:
|State |Alamos |Brandura |Canton |Dexter |Elexion |Total |
|Step #2 | | | | | | |
|Step #3 | | | | | | |
|Apportionment | | | | | | |
Does this seem fair?
• Population Paradox: A state could potentially lose seats because its population grew.
o State A loses a seat to state B even through the population of A grew at a higher rate than the population of state B.
EXAMPLE #3: Suppose we have the following data for the population of two high schools. The school board needs to determine how many counselors to apportion to the two high schools. There are 100 counselors available. Use Hamilton’s Method to apportion the seats.
|State |North HS |South HS |Total |
|Population |1045 |8955 |10,000 |
|Step #1 | | | |
|SD = 100 | | | |
|Step #2 | | | |
|Step #3 | | | |
|Apportionment | | | |
Suppose a new high school comes along and is added to the new district. The new high school, called New High School, has an enrollment of 525 students. The school board decides to hire 5 new counselors and assign them to the New High School. Does 5 make sense? Why or why not?
Part B: The school board decides to add the New High School and 5 new counselors. Find the apportionment using Hamilton’s Method for the 3 high schools assuming North and South maintain the same number of students.
|State |North HS |South HS |New HS |Total |
|Population |1045 |8955 |525 |10,525 |
|Step #1: | | | | |
|SD = | | | | |
|Step #2 | | | | |
| | | | | |
|Step #3 | | | | |
| | | | | |
|Apportionment | | | | |
Does this seem fair?
• New – States Paradox: the addition of a new state with its FAIR SHARE of seats an, in and of itself, affect the apportionment of other states.
o Fair share is based on original SD and new state’s population.
o Expect states to maintain same apportionment if fair share added
HOMEWORK: p.152 #12, 17, 19- 22
Section 4.4 Jefferson’s Method (Divisor Method)
• DIVISOR METHOD: manipulate the divisor (modified) in the apportionment.
▪ GENERAL DIVISION RULE for modified divisor:
[pic];
• To create BIGGER quotas, you need to use a ____________________________ divisor.
• To create SMALLER quotas, you need to use _____________________________ divisor
Examples: Circle all that apply
1) Consider the expression, [pic] where x = 6, which new value of x will result in a larger answer. (A) x = 5 (B) x = 6 (C) x = 7 (D) x = 8
2) Consider the expression, [pic] where x = 9, which new value of x will result in a smaller answer. (A) x = 6 (B) x = 10 (C) x = 8 (D) x =12
3) Consider the expression, [pic] where y = 125, which new value of x will result in a smaller answer. (A) y = 120 (B) y = 123 (C) y = 125 (D) y = 127
4) Consider the expression, [pic] where z = 0.35, which new value of x will result in a larger answer. (A) z = 0.15 (B) x = .275 (C) x = 3.5 (D) x = 0.09
Goal of Jefferson’s Method: make quotas bigger so that LOWER QUOTAS will be an exact apportionment.
JEFFERSON’S METHOD:
1) Calculate Standard Divisor, SD
Standard quotas will work in Jefferson’s Method ONLY when they are INTEGERS
2) Find a “suitable” modified divisor, MD such that the sum of LOWER quotas is the number of seats. (exact apportionment with no surplus)
Finding MD: Originally guess should be reasonably < Standard Divisor
• Current Total LESS than exact number of seats, then a SMALLER modified divisor is needed
• Current Total GREATER than exact number of seats, then a LARGER modified divisor is needed
GUESSING DIVISORS: Use previous divisor statements to assist
1) If divisor of 100 created OVER apportionment and divisor of 105 created a BELOW apportionment. Next guess should be:
2) If divisor of 64 created OVER apportionment and divisor of 66.5 created an OVER apportionment. Next guess should be:
3) If divisor of 135 created BELOW apportionment and divisor of 117 created a BELOW apportionment. Next guess should be:
EXAMPLE #1: There are 15 scholarships to be apportioned among 231 English majors, 502 History majors, and 355 Psychology majors.
|Majors |English |History |Psychology |Total |
|Population |231 |502 |355 |1088 |
|Standard Divisor | |
|Standard Quotas | | | | |
|Lower Quotas | | | | |
|Modified Divisor | |
|Modified Quotas | | | | |
|Lower Quotas | | | | |
|Modified Divisor | |
|Modified Quotas | | | | |
|Lower Quotas | | | | |
EXAMPLE #2: Try 40 scholarships on different subjects
|Majors |Math |Biology |Political Science |Total |
|Population |245 |481 |654 |1380 |
|Standard Divisor | |
|Standard Quotas | | | | |
|Lower Quotas | | | | |
BE SELECTIVE TO GUESS: Take all populations and divide by the next whole number you want
|Math |Biology |Political Science |
|245 |481 |654 |
|Modified Divisor | |
|Modified Quotas | | | | |
|Lower Quotas | | | | |
EXAMPLE #3: Previous example of the 6 nations with 250 seats for congress.
|State |A |
|Q |32.92 |138.72 |3.08 |41.82 |13.70 |19.76 |250 |
|LQs | | | | | | | |
Should your next divisor be bigger or smaller than 50,000?
SHORT HAND NOTATION: Saves space and you round immediately after calculating quota
| |1,646,000 |6,936,000 |154,000 |2,091,000 |685,000 |988,000 |12,500,000 |
What do you notice about the apportionment and the standard quotas?
Does there have to be exactly one MD for every apportionment problem?
HOMEWORK: p. 152 #24, 25, 30, 31
Section 4.4 Jefferson’s Method Part 2
Use the calculator to assist in the repetitive divisor of the state populations to find quotas to round
Calculator [Y =] method: You will need to use the correct quotas depending on the method and check the total of them to determine your next divisor guess .
1) [pic]and X = divisor
2) [2nd] [WINDOW: TBLSET] TblStart = Divisor you want (Next Guess)
3) [2nd] [GRAPH: TABLE] … shows all of the modified quotas for that divisor.
EXAMPLE #4: Banana Republic has states Apure, Barinas, Carabobo, and Dolores with populations given in millions and 160 seats in the legislature.
|State |A |
|Standard Quotas | | | | | |
|Lower Quota | | | | | |
SHORT HAND NOTATION: You may need more than the ones provided
|Modified Divisor |Lower Quota |Lower Quota |Lower Quota |Lower Quota |TOTAL |
| |of State A |of State B |of State C |of State D | |
EXAMPLE #5: Ms. Gambill has 150 colored pencils to give to 6 students in her class. She plans to give students the pencils based on their exam scores. Find the apportionment of Ms. Gambill’s 6 students under Jefferson’s method.
| |Zac |Kate |Eileen |AJ |Tommy |Megan |Total |
|Exams |89 |75 |82 |97 |78 |93 |514 |
|SD | |
|SQ | | | | | | | |
|LQs | | | | | | | |
SHORT HAND NOTATION: Saves space and you round immediately after calculating quota
|Modified Divisor |Zac |Kate |Eileen |AJ |Tommy |Megan |Total |
Pros and Cons of Jefferson’s Method:
Section 4.5 Adam’s Method (Divisor Method)
Goal of Adam’s Method: make quotas smaller so that when rounding up (UPPER QUOTAS) there is NO SURPLUS.
ADAM’S METHOD: SAME RULES FOR GUESSING GAME FOR MODIFIED DIVISOR
1) Calculate Standard Divisor, SD
▪ Standard quotas will work in Adam’s Method ONLY when they are INTEGERS
2) Find a “suitable” modified divisor, MD such that the sum of UPPER quotas is the number of seats. (exact apportionment with no surplus)
Finding MD: Originally guess should be reasonably < Standard Divisor
• Current Total LESS than exact number of seats, then a SMALLER modified divisor is needed
• Current Total GREATER than exact number of seats, then a LARGER modified divisor is needed
EXAMPLE #1: There are 15 scholarships to be apportioned among 231 English majors, 502 History majors, and 355 Psychology majors.
|Majors |English |History |Psychology |Total |
|Population |231 |502 |355 |1088 |
|Standard Divisor |1088/15 = 72.53 |
|Standard Quotas | | | | |
|Upper Quotas | | | | |
|Modified Divisor |UPPER Quota English |UPPER Quota History |UPPER Quota Psychology |TOTAL |
| | | | | |
EXAMPLE #2: Try 20 scholarships on four different sciences
|Majors |Biology |
|Standard Quotas |5.238 |6.603 |5.778 |2.381 |20 |
|Upper Quotas |6 |7 |6 |3 |22 Above |
|Modified Divisor |UPPER Quota Biology |UPPER Quota Chemistry |UPPER Quota Physics |UPPER Quota Astronomy |TOTAL |
EXAMPLE #3: Previous example problem of the 6 nations with 250 seats for congress.
|State |A |
|Standard Quotas |32.92 |138.72 |3.08 |41.82 |13.70 |19.76 |250 |
|Upper Quota | | | | | | | |
|Modified Divisor |UPPER Quota of A |UPPER Quota of B |UPPER Quota of C |UPPER Quota of D |UPPER Quota of E |UPPER Quota of F |TOTAL |
What do you notice about the apportionment and the standard quotas?
• EXAMPLE #4 Banana Republic has states Apure, Barinas, Carabobo, and Dolores with populations given in millions and 160 seats in the legislature.
|State |A |B |C |D |Total |
|Population |3.31 |2.67 |1.33 |.69 |8 |
|Standard Divisor |.05 |
|Standard Quotas |66.2 |53.4 |26.6 |13.8 |160 |
|Upper Quota | | | | | |
|Modified Divisor |Upper Quota |Upper Quota |Upper Quota |Upper Quota |TOTAL |
| |of State A |of State B |of State C |of State D | |
Pros and Cons of Adam’s Method
HOMEWORK: p. 153 #33, 36, 37, 40, 41
Section 4.6 Webster’s Method (Divisor Method)
Goal of Websters Method: is to make quotas so that when rounding up and down (CONVENTIONAL/NORMAL Rounding) there is NO SURPLUS.
WEBSTER’S Method: (SAME MD GUESSING GAME and USE CALCULATOR TO HELP)
1) Calculate Standard Divisor, SD this is the starting point
2) Find a “suitable” modified divisor, MD such that the sum of conventionally rounded quotas is the number of seats. (exact apportionment with no surplus)
Finding MD: Originally guess should be reasonably < Standard Divisor
• Current Total LESS than exact number of seats, then a SMALLER modified divisor is needed
• Current Total GREATER than exact number of seats, then a LARGER modified divisor is needed
EXAMPLE #1: There are 15 scholarships to be apportioned among 231 English majors, 502 History majors, and 355 Psychology majors.
|Majors |English |History |Psychology |Total |
|Population |231 |502 |355 |1088 |
|Standard Divisor |1088/15 = 72.53 |
|Standard Quotas |3.185 |6.921 |4.895 |15 |
|Normal Rounding | | | | |
EXAMPLE #2: Previous example of the 6 nations with 250 seats for congress.
|State |A |
|Standard Quotas |32.92 |138.72 |3.08 |41.82 |13.70 |19.76 |250 |
|Normal Rounding | | | | | | | |
|Modified Divisor |Normal Rounding A |Normal Rounding B |Normal Rounding C |Normal Rounding D |Normal Rounding E |Normal Rounding F |TOTAL |
EXAMPLE #3: Banana Republic has states Apure, Barinas, Carabobo, and Dolores with populations given in millions and 180 seats in the legislature.
|State |A |
|Standard Quotas | | | | | |
|Normal Rounding | | | | | |
EXAMPLE #3: The police department was able to hire 20 new officers and plans to deploy them in 4 precincts based on the previous year’s crimes.
| |Precinct #1 |
|Standard Quotas | | | | | |
|Upper Quotas | | | | | |
|Modified Divisor |Precinct #1 |Precinct #2 |Precinct #3 |Precinct #4 |TOTAL |
Pros and Cons of WEBSTER’S METHOD:
BALINSKI AND YOUNG’S IMPOSSIBILITY THEOREM:
An apportionment method that does not violate the quota rule and does not produce any paradoxes is a mathematical impossibility.
HOMEWORK: p.153 44 - 47
-----------------------
SMALLER MD
T > M
Modified Divisor
1) Calculate the lower quotas with Modified Divisor.
(*[pic]h
pð5?CJ(\?h
pð
h
pð5?\?h`„h`j:5?CJ( h`„\? hß E6? h`j:6?h
ö
h`j:5?2) Compare total (T) v. # of seats (M)
T < M
You’ve found a SUITABLE MD!!!
T = M
LARGER MD
................
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