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
WARM Ups:
Section 4.1: p. 151 #4
Section 4.2: p. 152 #13
Section 4.3: Day One: p. 152 #18 (practice Hamilton’s Method) Day Two: Homework Quiz since done in class
Section 4.4: Day One: p. 152 #26 Day Two: p. 152 #28
Section 4.5: Day One: p. 153 #36 Day Two: p. 153 #40
Section 4.6: p. 153 #48
REVIEW MATERIAL:
Section 4.1 Apportionment Problem
APPORTIONMENT Problem:
Dividing (sharing) proportionally based on more information than number of players
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)? 10 for each child
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).
Minutes worked / 150 / 78 / 173 / 204 / 295 / 900
Chocolate Pieces /
8.333 /
4.333 /
9.611 /
11.333 /
16.388 / 50
For how many minutes of chores worked would a child earn one piece of candy?
900 minutes of chores over 50 pieces of candy = 18 minutes per 1 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)? 15 per student
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.
TEST SCORE / 85 / 88 / 92 / 95 / 360
DVDs Received / / / / / 60
For how many points earned on a test would a student earn one DVD?
360points over 60 pieces of candy = 6 pointsper 1 DVD
Based on examples 1 and 2 Are there any challenge to trying to share items proportionally to all individuals?
Not all items can be shared in fraction/ decimal increments.
Apportionment methods will determine what whole number of items will be given to each player
- Elements of Apportionment Problem:
- “States”: players involved in apportionment
- NOTATION for N States: A1, A2, …, AN
- “Seats”: set of M identical, indivisible objects to be divided
More Seats than States; INDIVISIBLE = one seat cannot be shared by multiple states
- “Populations”: set of N positive numbers which are the basis for the apportionment of seats to state (proportionality criterion)
- NOTATION for N Populations: p1, p2, …, pN
Total Population = P = p1 + p2 + … + pN
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.
States = Children, Populations = GPA, Seats = Allowance
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.
States = Grade Levels, Populations = Students, Seats = Teachers
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.
States = Grade Levels, Populations = Students, Seats = Teachers
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.
States = Grade Levels, Populations = Students, Seats = Teachers
Ratios (fractions) are the important measurements for apportionment
- STANDARD DIVISOR, SD: ratio of total population to seats
Standard divisors represents the # of people (population unit) per 1 seat
- STANDARD QUOTAS, q: ratio of the state population, p, to the standard divisor
(same as in the example )
Standard Quota represents the EXACT number (including decimal) of seats that each state should get (if ONE seat could be divided into smaller parts)
TOTAL OF STANDARD QUOTAS = SEATS
- 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 / 12,500,000/ 250 = 50,000
Standard Quota / 32.92 / 138.72 / 3.08 / 41.82 / 13.70 / 19.76 / 250 = # of seats
Normal Rounding
(0.5 Rule) / 33 / 139 / 3 / 42 / 14 / 20 / 251
SPECIAL ROUNDING OF QUOTAS(0.5 rule doesn’t apply)
UPPER Quota, ↑: rounding up the standard quota to nearest integer
- Remove the Decimal and add one
a. 45.6↑ = 46b. 9.2↑ = 10c. 17.5↑ = 18d. 108↑ = 108
LOWER Quota, ↓:rounding down the standard quota to nearest integer
- Remove the decimal
a. 35.6↓ = 35b. 99.2↓ = 99c. 16.5↓ = 16d. 108↓ = 108
EXAMPLE #3b: find the upper and lower quotas for the 6 nations from Example 3.
State / A / B / C / D / E / F / TotalUpper Quota / 33 / 139 / 4 / 42 / 14 / 20 / 252
Lower Quota / 32 / 138 / 3 / 41 / 13 / 19 / 246
Will the lower and upper quotas be equal?
If the standard quotas are ALL whole numbers.
APPORTIONMENT METHODS will use some concept of upper / lower quotas.
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
Standard Divisor / 1800/30= 60 students per 1 teaching assistant
Standard Quota / 6.25 teaching assistants / 4.95 teaching assistants / 6.8 teaching assistants / 5.66
Teaching assistants / 6.33 teaching assistants
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 / TotalPopulation / 125 / 150 / 350 / 275 / 900
Standard Divisor / 900/200 = 4.5
Standard Quota / 27.77 / 33.33 / 77.77 / 61.11 / 200
Normal
Rounding / 28 / 33 / 78 / 61 / 200 (exact)
Upper Quotas / 28 / 34 / 78 / 62 / 202 (over)
Lower Quotas / 27 / 33 / 77 / 61 / 198 (under)
“Good” Apportionments:
- Produce a valid apportionment (exactly M seats given)
- Produce a “fair” apportionment
HOMEWORK: p.150 # 2, 3, 5
Section 4.2: Hamilton’s Method and Quota Rule
- Every state will get at least its lower quota.
Example #1 Hamilton’s Method: 6 nations with 250 seats for congress. (Calculations from 4.1 notes)
State / A / B / C / D / E / F / TotalPopulation / 1,646,000 / 6,936,000 / 154,000 / 2,091,000 / 685,000 / 988,000 / 12,500,000
Standard Divisor / 512,500,000/ 250 = 50,000
1)Calculate each state’s STANDARD QUOTAS, q
Step #1:Standard Quota / 32.92 / 138.72 / 3.08 / 41.82 / 13.70 / 19.76 / 250
2)Give each state its LOWER quota.
Step #2:Lower Quota / 32 / 138 / 3 / 41 / 13 / 19 / 246
UNDER by 4
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 / .92
+1 / .72
+1 / .08 / .82
+1 / .70 / .76
+1 / Surplus of 4 seats
FinalApportionment for HAMILTON:Assign each state it’s lower quota of seats and any surplus seat.
State
/ A = 33 / B = 139 / C = 3 / D = 42 / E = 13 / F = 20 / 250***Hamilton’s Method is NOT 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 / TotalMinutes worked / 150 / 78 / 173 / 204 / 295 / 900
S1: Standard Quota / 8.333 / 4.333 / 9.611 / 11.333 / 16.388 / 50
S2: Lower Quota / 8 / 4 / 9 / 11 / 16 / 48
S3: Standard Quota Decimal / .33 / .33 / .61
+1 / .33 / .39
+1 / + 2 surplus
Apportionment / 8 / 4 / 10 / 11 / 17 / 50
EXAMPLE 3:Perform Hamilton’s Method for 4 players and 150 seats
State A / State B / State C / State D / TotalPopulations / 300 / 450 / 800 / 650 / 2200
Standard Divisor / 2200/150 = 14 2/3 = 14 2/3 people per 1 seat
S1: Standard Quota / 20.45 / 30.681 / 54.54 / 44.318 / 150
S2: Lower Quota / 20 / 30 / 54 / 44 / 148
S3: Standard Quota Decimal / .45 / .681
+1 / .54
+1 / .318 / Under by 2
Apportionment / 20 / 31 / 55 / 44 / 150
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
Standard Divisor / 1540/40 = 38.5
Standard Quota / 7.117 / 9.662 / 8.598 / 7.896 / 6.727
LOWER / 7 / 9 / 8 / 7 / 6
Surplus / +1
FINAL / 7 / 9 / 8 / 8 / 6
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.
Minimum = lower quota and Maximum = Upper Quota
The Hamilton Method SATISFIES the quota rule
Lower-quota Violation: a state is apportioned number of seats LESS THAN lower quota
Upper-quota Violation: a state is apportioned number of seats MORE THAN upper quota
Section 4.3 Alabama and Other Paradoxes
Paradoxes or Illogical 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 / TOTALPopulation / 940 / 9030 / 10,030 / 20,000
Suppose the country decides to use 200 representatives. Use Hamilton’s Method to apportion the seats.
SD = 100 people per 1 representative
State
/ A / B / C / TOTALStep #1:
SQ / 9.4 / 90.3 / 100.3 / 200
Step #2:
LQ / 9 / 90 / 100 / 199
Step #3: Surplus / .4
+1 / .3 / .3 / +1
Apportionment / 10 / 90 / 100 / 200
Suppose state C makes the request to have 201 seats instead. Where do you think the extra seat will go? Why?
ANSWERS WILL VARY
Part B: Use Hamilton’s Method to apportion the 201 seats.
SD = 99.50248756
State / A / B / C / TOTALStep #1 / 9.45 / 90.75 / 100.80 / 201
Step #2 / 9 / 90 / 100 / 199
Step #3 / .4 / .75
+1 / .80
+1
Apportionment / 9 / 91 / 101 / 201
Does this seem fair? NO
- 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.
- ONLY THE NUMBER OF SEATS IS CHANGING
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 = 18
State / Alamos / Brandura / Canton / Dexter / Elexion / TotalPopulation / 150 / 78 / 173 / 204 / 295 / 900
Step #1 / 8.33 / 4.33 / 9.61 / 11.33 / 16.38 / 50
Step #2 / 8 / 4 / 9 / 11 / 16 / 48
Step #3 / .33 / .33 / .61
+1 / .33 / .38
+1
Apportionment / 8 / 4 / 10 / 11 / 17 / 50
Part B: Suppose 10 years go by and the country recounts its population to get the following chart:
State / Alamos / Brandura / Canton / Dexter / Elexion / TotalPopulation / 150 / 78 / 181 / 204 / 296 / 909
Which state’s had changes to their population? How will the new apportionment will turn out (still using only 50 seats)?
CANTON and ELEXION
Perform Hamilton’s Method on new populations: SD = 18.18
Step #1 / 8.25 / 4.29 / 9.96 / 11.22 / 16.28 / 50Step #2 / 8 / 4 / 9 / 11 / 16 / 48
Step #3 / .25 / .29
+1 / .96
+1 / .22 / .28
Apportionment / 8 / 5 / 10 / 11 / 16 / 50
Does this seem fair? NO, Elexion lost a seat
- Population Paradox:A state could potentially lose seats because its population grew.
- 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.
ONLY THE STATE POPULATIONS CAN OCCUR
EXAMPLE #3:Suppose we have the following data for the population of two highschools. 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 / TotalPopulation / 1045 / 8955 / 10,000
Step #1
SD = 100
/ 10.45 / 89.55 / 100Step #2 / 10 / 89 / 99
Step #3 / .45 / .55
+1
Apportionment / 10 / 90 / 100
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?
SD for original is 100 so 525/ 100 = 5.25 so lower quota is 5
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 / TotalPopulation / 1045 / 8955 / 525 / 10,525
Step #1:
SD = 100.2380952 / 10.42 / 89.34 / 5.24 / 105
Step #2
/ 10 / 89 / 5 / 104Step #3 / .42
+1 / .34 / .24
Apportionment / 11 / 89 / 5 / 105
Does this seem fair? No, South Lost a counselor
- 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.
- Fair share is based on original SD and new state’s population.
- Expect states to maintain same apportionment if fair share added
NUMBER OF STATES, POPULATIONS, AND SEATS ARE ALL CHANGING
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:
; for a divisor method the population is fixed for each state
- To create BIGGER quotas, you need to use a SMALLER divisor.
- To create SMALLER quotas, you need to use BIGGER divisor
Examples:Circle all that apply
1) Consider the expression, 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, where x = 9, which new value of x will result in a smalleranswer.(A) x = 6 (B) x = 10 (C) x = 8 (D) x =12
3) Consider the expression, where y = 125, which new value of x will result in a smalleranswer. (A) y = 120 (B) y = 123 (C) y = 125 (D) y = 127
4) Consider the expression, 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, MDsuch 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 thanexact 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: BETWEEN 100 and 105.
2) If divisor of 64 created OVER apportionment and divisor of 66.5 created an OVER apportionment. Next guess should be: OVER 66.5
3) If divisor of 135 created BELOW apportionment and divisor of 117 created a BELOW apportionment. Next guess should be: BELOW 117
EXAMPLE #1: There are 15 scholarships to beapportioned among 231 English majors,502 History majors, and 355 Psychology majors.
Majors / English / History / Psychology / TotalPopulation / 231 / 502 / 355 / 1088
Standard Divisor / 1088/15 = 72.53
Standard Quotas / 3.185 / 6.921 / 4.895 / 15
Lower Quotas / 3 / 6 / 4 / 13 = SMALLER
Modified Divisor
/ New modified divisor should be smaller than 72.53. TRY 72.Modified Quotas / 3.208 / 6.97 / 4.93
Lower Quotas / 3 / 6 / 4 / 13 = SMALLER
Modified Divisor
/ New modified divisor should be smaller than 72. TRY 71.Modified Quotas / 3.254 / 7.07 / 5
Lower Quotas / 3 / 7 / 5 / 15 = EXACT
EXAMPLE #2: Try 40 scholarships on different subjects
Majors / Math / Biology / Political Science / TotalPopulation / 245 / 481 / 654 / 1380
Standard Divisor / 1380/40 = 34.5
Standard Quotas / 7.101 / 13.942 / 18.957 / 40
Lower Quotas / 7 / 13 / 18 / 38 = smaller
BE SELECTIVE TO GUESS: Take all populations and divide by the next whole number you want
Math / Biology / Political Science245/8 = 30.625 / 481/14 =34.357 / 654/19 = 34.421
Modified Divisor
/ Try 34.35 the second smallest divisorModified Quotas / 7.13 / 14.003 / 19.039 / Modified Total not important
Lower Quotas / 7 / 14 / 19 / 30
EXAMPLE #3: Previous example of the 6 nations with 250 seats for congress.
State / A / B / C / D / E / F / TotalPopulation / 1,646,000 / 6,936,000 / 154,000 / 2,091,000 / 685,000 / 988,000 / 12,500,000
SD / 12,500,000/ 250 = 50,000
Q / 32.92 / 138.72 / 3.08 / 41.82 / 13.70 / 19.76 / 250
LQs / 32 / 138 / 3 / 41 / 13 / 19 / 246
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 / 12,500,000Modified Divisor / A’s Lower Quota / B’s Lower Quota / C’s Lower Quota / D’s Lower Quota / E’s Lower Quota / F’s Lower Quota / TOTAL
49,000 / 33 / 141 / 3 / 42 / 13 / 20 / 252 = BIGGER
49,500 / 33 / 140 / 3 / 42 / 13 / 19 / 250
What do you notice about the apportionment and the standard quotas?
B 140 > standard quotas upper value at 139 it is ***upper quota violation**
Does there have to be exactly one MD for every apportionment problem?
***NO, for example 49,501 would work… There is a range of Modified Divisors that will work for a 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)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 / B / C / D / TotalPopulation / 3.31 / 2.67 / 1.33 / .69 / 8
Standard Divisor / .05
Standard Quotas / 66.2 / 53.4 / 26.6 / 13.8 / 160
Lower Quota / 66 / 53 / 26 / 13 / 158 = Smaller
SHORT HAND NOTATION: You may need more than the ones provided
Modified Divisor / Lower Quotaof State A / Lower Quota
of State B / Lower Quota
of State C / Lower Quota
of State D / TOTAL
.049 / 67 / 54 / 27 / 14 / 162 = Bigger
.0495 / 66 / 53 / 26 / 13 / 158 = Smaller
.0494 / 67 / 54 / 26 / 13 / 160
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 / TotalExams / 89 / 75 / 82 / 97 / 78 / 93 / 514
SD / 514/150 = 3.4266666
SQ / 25.973 / 21.887 / 23.93 / 28.307 / 22.763 / 27.14 / 150
LQs / 25 / 21 / 23 / 28 / 22 / 27 / 146
SHORT HAND NOTATION:Saves space and you round immediately after calculating quota
Modified Divisor / Zac / Kate / Eileen / AJ / Tommy / Megan / Total3.4 / 26 / 22 / 24 / 28 / 22 / 27 / 149
smaller
3.39 / 26 / 22 / 24 / 28 / 23 / 27 / 150
Pros and Cons of Jefferson’s Method:
- CON: Can cause upper-quota violations
- CON: Biased toward larger states
- PRO: Paradoxes not possible
- Right now it can be a LONG or SHORT guessing game!!!!!!!!!
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, MDsuch 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 thanexact number of seats, then a LARGER modified divisor is needed
EXAMPLE #1: There are 15 scholarships to beapportioned among 231 English majors,502 History majors, and 355 Psychology majors.
Majors
/ English / History / Psychology / TotalPopulation / 231 / 502 / 355 / 1088
Standard Divisor / 1088/15 = 72.53
Standard Quotas / 3.185 / 6.921 / 4.895 / 15
Upper Quotas / 4 / 7 / 5 / 16 = BIGGER
Modified Divisor should be bigger than 72.53
SHORTHAND
Modified Divisor / UPPER Quota English / UPPER Quota History / UPPER Quota Psychology / TOTAL77 / 3 / 7 / 5 / 15 = EXACT
EXAMPLE #2: Try 20 scholarships on four different sciences
Majors / Biology / Chemistry / Physics / Astronomy / TotalPopulation / 165 / 208 / 182 / 75 / 630
Standard Divisor
/ 630/20 = 31.5Standard Quotas / 5.238 / 6.603 / 5.778 / 2.381 / 20
Upper Quotas / 6 / 7 / 6 / 3 / 22 Above
SHORTHAND
Modified Divisor / UPPER Quota Biology / UPPER Quota Chemistry / UPPER Quota Physics / UPPER Quota Astronomy / TOTAL32 / 6 / 7 / 6 / 3 / 22 above
32.5 / 6 / 7 / 6 / 3 / 22 above
34 / 5 / 7 / 6 / 3 / 21
35 / 5 / 6 / 6 / 3 / 20
EXAMPLE #3: Previous example of the 6 nations with 250 seats for congress.