a mathematical view of our world

A Mathematical View A Mathematical View of Our Worldof Our World

11stst ed. ed.

Parks, Musser, Trimpe, Parks, Musser, Trimpe, Maurer, and MaurerMaurer, and Maurer

Chapter 5Chapter 5

ApportionmentApportionment

Section 5.1Section 5.1Quota MethodsQuota Methods

• GoalsGoals• Study apportionmentStudy apportionment

• Standard divisorStandard divisor• Standard quotaStandard quota

• Study apportionment methodsStudy apportionment methods• Hamilton’s methodHamilton’s method• Lowndes’ methodLowndes’ method

5.1 Initial Problem5.1 Initial Problem• The number of campers in each group at The number of campers in each group at

a summer camp is shown below.a summer camp is shown below.

5.1 Initial Problem, cont’d5.1 Initial Problem, cont’d

• The camp organizers will assign 15 The camp organizers will assign 15 counselors to the groups of campers.counselors to the groups of campers.

• How many of the 15 counselors should How many of the 15 counselors should be assigned to each group?be assigned to each group?• The solution will be given at the end of the The solution will be given at the end of the

section.section.

ApportionmentApportionment• The verb The verb apportionapportion means means

• ““Assign to as a due portion.”Assign to as a due portion.”• ““To divide into shares which may not be equal.”To divide into shares which may not be equal.”

• Apportionment problems arise when what is Apportionment problems arise when what is being divided cannot be divided into being divided cannot be divided into fractional parts.fractional parts.• An example of apportionment is the process of An example of apportionment is the process of

assigning seats in the House of Representatives assigning seats in the House of Representatives to the states.to the states.

Apportionment, cont’dApportionment, cont’d

• The The apportionment problemapportionment problem is to is to determine a method for rounding a determine a method for rounding a collection of numbers so that: collection of numbers so that: • The numbers are rounded to whole The numbers are rounded to whole

numbers.numbers.• The sum of the numbers is unchanged.The sum of the numbers is unchanged.

Apportionment, cont’dApportionment, cont’d• The Constitution does not specify a method The Constitution does not specify a method

for apportioning seats in the House of for apportioning seats in the House of Representatives.Representatives.

• Various methods, named after their authors, Various methods, named after their authors, have been used: have been used: • Alexander HamiltonAlexander Hamilton• Thomas JeffersonThomas Jefferson• Daniel WebsterDaniel Webster• William LowndesWilliam Lowndes

The Standard DivisorThe Standard Divisor• Suppose the total population is Suppose the total population is PP and and

the number of seats to be apportioned the number of seats to be apportioned is is MM. .

• The The standard divisorstandard divisor is the ratio is the ratio DD = = P/MP/M. . • The standard divisor gives the number of The standard divisor gives the number of

people per legislative seat. people per legislative seat.

Example 1Example 1• Suppose a country Suppose a country

has 5 states and 200 has 5 states and 200 seats in the seats in the legislature.legislature.

• The populations of the The populations of the states are given states are given below.below.

• Find the standard Find the standard divisor. divisor.

Example 1, cont’dExample 1, cont’d

• Solution: The total population is found Solution: The total population is found by adding the 5 state populations.by adding the 5 state populations.• PP = 1,350,000 + 1,500,000 + 4,950,000 + = 1,350,000 + 1,500,000 + 4,950,000 +

1,100,000 + 1,100,000 = 10,000,000.1,100,000 + 1,100,000 = 10,000,000.• The number of seats is The number of seats is MM = 200 = 200


• Solution, cont’d: The standard divisor is Solution, cont’d: The standard divisor is DD = 10,000,000/200 = 50,000. = 10,000,000/200 = 50,000.

• Each seat in the legislature represents Each seat in the legislature represents 50,000 citizens.50,000 citizens.

The Standard QuotaThe Standard Quota

• Let Let DD be the standard divisor. be the standard divisor.• If the population of a state is If the population of a state is pp, then , then

Q Q = = p/Dp/D is called the is called the standard quotastandard quota. . • If seats could be divided into fractions, we If seats could be divided into fractions, we

would give the state exactly would give the state exactly QQ seats in the seats in the legislature. legislature.

Example 2Example 2• In the previous In the previous

example, the example, the standard divisor was standard divisor was found to be found to be DD = = 50,000.50,000.

• Find the standard Find the standard quotas for each quotas for each state in the country. state in the country.

Example 2, cont’dExample 2, cont’d• Solution: Divide the population of each Solution: Divide the population of each

state by the standard divisor.state by the standard divisor.• State A: State A: QQ = 1,350,000/50,000 = 27 = 1,350,000/50,000 = 27• State B: State B: QQ = 1,500,000/50,000 = 30 = 1,500,000/50,000 = 30• State C: State C: QQ = 4,950,000/50,000 = 99 = 4,950,000/50,000 = 99• State D: State D: QQ = 1,100,000/50,000 = 22 = 1,100,000/50,000 = 22• State E: State E: QQ = 1,100,000/50,000 = 22 = 1,100,000/50,000 = 22

Example 2, cont’dExample 2, cont’d• Solution, cont’d: This solution, with all Solution, cont’d: This solution, with all

standard quotas being whole numbers, is not standard quotas being whole numbers, is not typical.typical.

• Note that the sum of the quotas is 27 + 30 + Note that the sum of the quotas is 27 + 30 + 99 +22 + 22 = 200, the total number of 99 +22 + 22 = 200, the total number of seats.seats.

• The standard quotas indicate how many The standard quotas indicate how many seats each state should be assigned. seats each state should be assigned.

Question:Question:A country consists of 3 states with populations A country consists of 3 states with populations shown in the table below. Find the standard quotas shown in the table below. Find the standard quotas if 200 legislative seats are to be apportioned. if 200 legislative seats are to be apportioned. Round the quotas to the nearest hundredth. Round the quotas to the nearest hundredth.

a. State A: 0.01, State B: 0.02, State C: 0.02a. State A: 0.01, State B: 0.02, State C: 0.02b. State A: 89.75, State B: 44.31, State C: 65.45b. State A: 89.75, State B: 44.31, State C: 65.45c. State A: 90.91, State B: 42.12, State C: 67.29c. State A: 90.91, State B: 42.12, State C: 67.29d. State A: 90.91, State B: 43.64, State C: 65.45d. State A: 90.91, State B: 43.64, State C: 65.45

State A State B State CPopulation

250,000 120,000 180,000

Apportionment, cont’dApportionment, cont’d

• Typically, the standard quotas will not Typically, the standard quotas will not all be whole numbers and will have to all be whole numbers and will have to be rounded.be rounded.

• The various apportionment methods The various apportionment methods provide procedures for determining how provide procedures for determining how the rounding should be done. the rounding should be done.

Hamilton’s MethodHamilton’s Method1)1) Find the standard divisor.Find the standard divisor.2)2) Determine each state’s standard quota.Determine each state’s standard quota.

• Round each quota down to a whole number.Round each quota down to a whole number.• Each state gets that number of seats, with a Each state gets that number of seats, with a

minimum of 1 seat.minimum of 1 seat.

3)3) Leftover seats are assigned one at a time Leftover seats are assigned one at a time to states according to the size of the to states according to the size of the fractional parts of the standard quotas. fractional parts of the standard quotas.

• Begin with the state with the largest fractional Begin with the state with the largest fractional part. part.

Example 3Example 3• A country has 5 A country has 5

states and 200 states and 200 seats in the seats in the legislature.legislature.

• Apportion the seats Apportion the seats according to according to Hamilton’s method.Hamilton’s method.

Example 3, cont’dExample 3, cont’d• Solution: the standard divisor is found:Solution: the standard divisor is found:

• Then the standard quota for each state is Then the standard quota for each state is found.found.

• For example:For example:

10000000 50,000200

PDM

1320000 26.450000

AA

pQD


• Solution, cont’d: All of the standard Solution, cont’d: All of the standard quotas are shown below.quotas are shown below.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: The integer parts of Solution, cont’d: The integer parts of

the standard quotas add up to 26 + the standard quotas add up to 26 + 30 + 98 + 22 + 22 = 198.30 + 98 + 22 + 22 = 198.

• A total of 198 seats have been A total of 198 seats have been apportioned at this point.apportioned at this point.

• There are 2 seats left to assign There are 2 seats left to assign according to the fractional parts of the according to the fractional parts of the standard quotas.standard quotas.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: Consider the size of Solution, cont’d: Consider the size of

the fractional parts of the standard the fractional parts of the standard quotasquotas..

• State C has the largest fractional part, State C has the largest fractional part, of 0.7of 0.7

• State A has the second largest State A has the second largest fractional part, of 0.4.fractional part, of 0.4.

• The 2 leftover seats are apportioned to The 2 leftover seats are apportioned to states C and A.states C and A.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: The final Solution, cont’d: The final

apportionment is shown below.apportionment is shown below.

Question:Question:

A country consists of 3 states with A country consists of 3 states with populations shown in the table below. populations shown in the table below. Hamilton’s method is being used to Hamilton’s method is being used to apportion the 200 legislative seats. The apportion the 200 legislative seats. The standard quotas are State A: 90.91, State B: standard quotas are State A: 90.91, State B: 43.64, State C: 65.45.43.64, State C: 65.45.What is the final apportionment?What is the final apportionment?


250,000 120,000 180,000

Question cont’dQuestion cont’d

a. State A: 91 seats, State B: 44 a. State A: 91 seats, State B: 44 seats, State C: 65 seatsseats, State C: 65 seatsb. State A: 91 seats, State B: 43 b. State A: 91 seats, State B: 43 seats, State C: 66 seatsseats, State C: 66 seatsc. State A: 92 seats, State B: 43 c. State A: 92 seats, State B: 43 seats, State C: 65 seatsseats, State C: 65 seatsd. State A: 90 seats, State B: 44 d. State A: 90 seats, State B: 44 seats, State C: 66 seatsseats, State C: 66 seats

Quota RuleQuota Rule

• Any apportionment method which Any apportionment method which always assigns the whole number just always assigns the whole number just above or just below the standard quota above or just below the standard quota is said to satisfy the is said to satisfy the quota rulequota rule..

• Any apportionment method that obeys Any apportionment method that obeys the quota rule is called a the quota rule is called a quota methodquota method..• Hamilton’s method is a quota method. Hamilton’s method is a quota method.

Relative Fractional PartRelative Fractional Part

• For a number greater than or equal to For a number greater than or equal to 1, the 1, the relative fractional partrelative fractional part is the is the fractional part of the number divided by fractional part of the number divided by the integer part.the integer part.• Example: The relative fractional part of 5.4 Example: The relative fractional part of 5.4

is 0.4/5 = 0.08. is 0.4/5 = 0.08.

Lowndes’ MethodLowndes’ Method1)1) Find the standard divisor.Find the standard divisor.2)2) Determine each state’s standard Determine each state’s standard

quota.quota.• Round each quota Round each quota downdown to a whole to a whole

number.number.• Each state gets that number of seats, with a Each state gets that number of seats, with a

minimum of 1 seat.minimum of 1 seat.

Lowndes’ Method, cont’dLowndes’ Method, cont’d3)3) Determine the relative fractional part of Determine the relative fractional part of

each state’s standard quota. each state’s standard quota. 4)4) Any leftover seats are assigned one at a Any leftover seats are assigned one at a

time according to the size of the relative time according to the size of the relative fractional parts.fractional parts.

• Begin with the state with the largest relative Begin with the state with the largest relative fractional part.fractional part.

• Note that Lowndes’ method is a quota Note that Lowndes’ method is a quota method.method.

Example 4Example 4• Apportion the seats from the previous example Apportion the seats from the previous example

using Lowndes’ method.using Lowndes’ method.• Solution: The standard quotas were already Solution: The standard quotas were already

found and are shown below. found and are shown below.

Example 4, cont’dExample 4, cont’d• Solution, cont’d: As with Hamilton’s Solution, cont’d: As with Hamilton’s

method, 198 seats have been apportioned method, 198 seats have been apportioned so far, based on the whole number part of so far, based on the whole number part of the standard quotas.the standard quotas.

• To apportion the remaining 2 seats, To apportion the remaining 2 seats, calculate the relative fractional part of calculate the relative fractional part of each state’s standard quota. each state’s standard quota.

Example 4, cont’dExample 4, cont’d• Solution, cont’d: Solution, cont’d:

States D and A States D and A have the largest have the largest relative fractional relative fractional parts for their parts for their standard quotas.standard quotas.

• States D and A States D and A each get one more each get one more seat.seat.

Example 4, cont’dExample 4, cont’d• Solution, cont’d: The final apportionment Solution, cont’d: The final apportionment

is shown below. is shown below.

Question:Question:A country consists of 3 states with populations A country consists of 3 states with populations shown in the table below. Lowndes’ method is shown in the table below. Lowndes’ method is being used to apportion the 200 legislative being used to apportion the 200 legislative seats. The standard quotas are State A: seats. The standard quotas are State A: 90.91, State B: 43.64, State C: 65.45.90.91, State B: 43.64, State C: 65.45.

A total of 198 seats are apportioned by the A total of 198 seats are apportioned by the integer parts of the standard quotas. To which integer parts of the standard quotas. To which state is the first leftover seat assigned? state is the first leftover seat assigned? a. State Aa. State A b. State Bb. State B c. State Cc. State C


250,000 120,000 180,000

Method ComparisonMethod Comparison• Hamilton’s method and Lowndes’ Hamilton’s method and Lowndes’

method gave different method gave different apportionments in the previous apportionments in the previous example.example.

• Hamilton’s method is biased toward Hamilton’s method is biased toward larger states.larger states.

• Lowndes’ method is biased toward Lowndes’ method is biased toward smaller states.smaller states.

5.1 Initial Problem Solution5.1 Initial Problem Solution• A camp needs to assign 15 counselors A camp needs to assign 15 counselors

among 3 groups of campers.among 3 groups of campers.• The groups are shown in the table below.The groups are shown in the table below.

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• First the counselors will be apportioned using First the counselors will be apportioned using

Hamilton’s method.Hamilton’s method.• The standard divisor is:The standard divisor is:

• This indicates that 1 counselor should be This indicates that 1 counselor should be assigned to approximately every 12.67 campers.assigned to approximately every 12.67 campers.

42 67 81 12.6715

PDM

Initial Problem Solution, cont’dInitial Problem Solution, cont’d

• Next, calculate the standard quota for each Next, calculate the standard quota for each group of campers.group of campers.


• A total of 3 + 5 + 6 = 14 counselors have A total of 3 + 5 + 6 = 14 counselors have been assigned so far.been assigned so far.

• The 1 leftover counselor is assigned to the 6The 1 leftover counselor is assigned to the 6thth grade group.grade group.


• The final apportionment according to The final apportionment according to Hamilton’s method is:Hamilton’s method is:• 3 counselors for 43 counselors for 4thth grade grade• 5 counselors for 55 counselors for 5thth grade grade• 7 counselors for 67 counselors for 6thth grade grade


• Next the counselors will be apportioned using Next the counselors will be apportioned using Lowndes’ method.Lowndes’ method.

• The standard divisor is The standard divisor is DD = 12.67. = 12.67.• The standard quotas are:The standard quotas are:

• 44thth Grade: 3.31 Grade: 3.31• 55thth Grade: 5.29 Grade: 5.29• 66thth Grade: 6.39 Grade: 6.39

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• As before, 14 counselors have been As before, 14 counselors have been

apportioned so far.apportioned so far.

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• The standard quota for the group of 4The standard quota for the group of 4thth grade grade

campers has the largest relative fractional campers has the largest relative fractional part, so they get the leftover counselor.part, so they get the leftover counselor.

• The final apportionment using Lowndes’ The final apportionment using Lowndes’ method is:method is:• 4 counselors for 44 counselors for 4thth grade grade• 5 counselors for 55 counselors for 5thth grade grade• 6 counselors for 66 counselors for 6thth grade grade

Section 5.2Section 5.2Divisor MethodsDivisor Methods

• GoalsGoals• Study apportionment methodsStudy apportionment methods

• Jefferson’s methodJefferson’s method• Webster’s methodWebster’s method

5.2 Initial Problem5.2 Initial Problem• Suppose you, your sister, and your brother Suppose you, your sister, and your brother

have inherited 85 gold coins.have inherited 85 gold coins.• The coins will be divided based on the The coins will be divided based on the

number of hours each of you have number of hours each of you have volunteered at the local soup kitchen.volunteered at the local soup kitchen.

• How should the coins be apportioned? How should the coins be apportioned? • The solution will be given at the end of the The solution will be given at the end of the

section.section.

Apportionment MethodsApportionment Methods• Section 5.1 covered two quota Section 5.1 covered two quota

methods.methods.• Hamilton’s methodHamilton’s method• Lowndes’ method.Lowndes’ method.

• This section will consider two This section will consider two apportionment methods that do not apportionment methods that do not follow the quota rule. follow the quota rule.

Jefferson’s MethodJefferson’s Method

• Suppose Suppose MM seats will be apportioned. seats will be apportioned.1)1)

a)a) Choose a number, Choose a number, dd, called the , called the modified modified divisordivisor..

b)b) For each state, compute the For each state, compute the modified quotamodified quota, , which is the ratio of the state’s population to which is the ratio of the state’s population to the modified divisor:the modified divisor: pmQ

d

Jefferson’s Method, cont’dJefferson’s Method, cont’d

1)1) Cont’d:Cont’d:c)c) If the integer parts of the modified quotas If the integer parts of the modified quotas

for all the states add to for all the states add to MM, then go on to , then go on to Step 2. Otherwise go back to Step 1, Step 2. Otherwise go back to Step 1, part (a) and choose a different value for part (a) and choose a different value for dd. .

2)2) Assign to each state the integer part Assign to each state the integer part of its modified quota.of its modified quota.

Example 1Example 1• Use Jefferson’s Use Jefferson’s

method to method to apportion 200 apportion 200 seats to the 5 seats to the 5 states in the states in the example from example from Section 5.1.Section 5.1.

Example 1, cont’dExample 1, cont’d• Solution: Recall the standard divisors and Solution: Recall the standard divisors and

the apportionment found using Hamilton’s the apportionment found using Hamilton’s method.method.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: In Jefferson’s Solution, cont’d: In Jefferson’s

method all the (modified) quotas will method all the (modified) quotas will be rounded down.be rounded down.

• Note that if all the standard quotas were Note that if all the standard quotas were rounded down, the total would be only rounded down, the total would be only 198 seats.198 seats.

• The modified quotas need to be slightly The modified quotas need to be slightly larger than the standard ones.larger than the standard ones.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: For the modified quotas to Solution, cont’d: For the modified quotas to

be larger, the modified divisor needs to be be larger, the modified divisor needs to be smaller than the standard divisor of smaller than the standard divisor of 50,000.50,000.

• A good guess for a modified divisor can be A good guess for a modified divisor can be found by dividing the largest state’s found by dividing the largest state’s population by 1more, 2 more, 3 more,…, population by 1more, 2 more, 3 more,…, than the integer part of its standard quota.than the integer part of its standard quota.

• Start with 1 more, and keep going until you Start with 1 more, and keep going until you find one that works. find one that works.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: The largest state has Solution, cont’d: The largest state has

a population of 4,935,000 and its a population of 4,935,000 and its standard quota has an integer part of standard quota has an integer part of 98.98.

• A possible modified divisor is:A possible modified divisor is:

4935000 49,84898 1

d

Example 1, cont’dExample 1, cont’d• Solution, cont’d: We will try a modified Solution, cont’d: We will try a modified

divisor of divisor of dd = 49,848. = 49,848.• The modified quota for each state is The modified quota for each state is

calculated.calculated.• For example, the modified quota of state For example, the modified quota of state

A is: A is: 1320000 26.4849848

pmQd


• Solution, cont’d: When the modified quotas are Solution, cont’d: When the modified quotas are rounded down they add to 199.rounded down they add to 199.

• The modified divisor needs to be even smaller.The modified divisor needs to be even smaller.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: The largest state has Solution, cont’d: The largest state has

a population of 4,935,000 and its a population of 4,935,000 and its standard quota has an integer part of standard quota has an integer part of 98.98.

• A second possible modified divisor is:A second possible modified divisor is:

4935000 49,35098 2

d

Example 1, cont’dExample 1, cont’d• Solution, cont’d: New modified quotas are Solution, cont’d: New modified quotas are

calculated.calculated.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: Now when the Solution, cont’d: Now when the

modified quotas are rounded down modified quotas are rounded down they add to 26 + 30 + 100 + 22 + 22 = they add to 26 + 30 + 100 + 22 + 22 = 200.200.

• Since the sum of the rounded Since the sum of the rounded modified quotas equals the number of modified quotas equals the number of seats, the apportionment is complete.seats, the apportionment is complete.

Question:Question:The standard quotas and the apportionment The standard quotas and the apportionment under Jefferson’s method from the previous under Jefferson’s method from the previous example are recalled in the table below. Which example are recalled in the table below. Which state’s apportionment illustrates that Jefferson’s state’s apportionment illustrates that Jefferson’s method is not a quota method? method is not a quota method?

a. A a. A b. B b. B c. Cc. C d. Dd. De. None of e. None of the abovethe above

Divisor MethodsDivisor Methods

• Any apportionment method that uses a Any apportionment method that uses a divisor other than the standard divisor divisor other than the standard divisor is called a is called a divisor methoddivisor method..• Jefferson’s method is a divisor method.Jefferson’s method is a divisor method.• Webster’s method, which will be studied Webster’s method, which will be studied

next, is also a divisor method. next, is also a divisor method.

Webster’s MethodWebster’s Method

• Suppose Suppose MM seats are to be seats are to be apportioned.apportioned.

1)1)

a)a) Choose a number, Choose a number, dd, called the modified , called the modified divisor.divisor.

b)b) For each state, calculate the modified For each state, calculate the modified quota, quota, mQmQ..

Webster’s Method, cont’dWebster’s Method, cont’d1)1)

c)c) If when the modified quotas are rounded If when the modified quotas are rounded normally, their sum is normally, their sum is MM, then go on to , then go on to Step 2. Otherwise go back to Step 1, Step 2. Otherwise go back to Step 1, part (a) and choose a different value for part (a) and choose a different value for dd..

2)2) Assign to each state the integer nearest its Assign to each state the integer nearest its modified quota. modified quota.

Example 2Example 2• Use Webster’s Use Webster’s

method to method to apportion 200 apportion 200 seats to the 5 seats to the 5 states in the states in the previous previous example. example.

Example 2, cont’dExample 2, cont’d• Solution: Rounding the standard quotas Solution: Rounding the standard quotas

normally gives a sum of 199, one short of normally gives a sum of 199, one short of the desired total.the desired total.

• Unlike when using Jefferson’s method, it is Unlike when using Jefferson’s method, it is not clear whether the modified divisor not clear whether the modified divisor should be larger or smaller than the should be larger or smaller than the standard divisor.standard divisor.

• In general, try to create an apportionment like In general, try to create an apportionment like the one obtained from Hamilton’s method. the one obtained from Hamilton’s method.


• Solution, cont’d:Solution, cont’d:• State A needs a modified quota of 26.5 or State A needs a modified quota of 26.5 or

greater so it will round up.greater so it will round up.• We do not want any other states besides A We do not want any other states besides A

and C to have quotas that round up.and C to have quotas that round up.


• Solution, cont’d: To make the modified Solution, cont’d: To make the modified quota for A larger, we need a modified quota for A larger, we need a modified divisor that is slightly smaller.divisor that is slightly smaller.

• The calculations done for Jefferson’s method The calculations done for Jefferson’s method show us that show us that dd = 49,848 is too large and = 49,848 is too large and d d = = 49,350 is too small.49,350 is too small.

Example 2, cont’dExample 2, cont’d• Solution, cont’d: Try a modified divisor Solution, cont’d: Try a modified divisor

in between those two values. For in between those two values. For example, try example, try dd = 49,700. = 49,700.

• Check the modified quota for State A to Check the modified quota for State A to see if it is 26.5 or greater:see if it is 26.5 or greater:

• Next check the rest of the modified quotas.Next check the rest of the modified quotas.

1320000 26.5649700

mQ


• Solution, cont’d: All of the modified Solution, cont’d: All of the modified quotas are shown below.quotas are shown below.


• Solution, cont’d: The rounded Solution, cont’d: The rounded modified quotas add to 27 + 30 + 99 + modified quotas add to 27 + 30 + 99 + 22 + 22 = 200.22 + 22 = 200.

• This is the correct total, so the modified This is the correct total, so the modified divisor was appropriate.divisor was appropriate.

• The apportionment is complete.The apportionment is complete.

Question:Question:The standard quotas and the apportionment The standard quotas and the apportionment under Webster’s method from the previous under Webster’s method from the previous example are recalled in the table below. example are recalled in the table below. Which state’s apportionment illustrates that Which state’s apportionment illustrates that Webster’s method is not a quota method? Webster’s method is not a quota method?

a. Aa. A b. Bb. Bc. Cc. C d. Dd. De. None of e. None of the abovethe above

Example 3Example 3• Suppose a retail store needs to apportion Suppose a retail store needs to apportion

54 sales associates to three stores.54 sales associates to three stores.

Example 3, cont’dExample 3, cont’d• Solution: First determine the standard Solution: First determine the standard

divisor by dividing the total customer divisor by dividing the total customer base by the number of sales base by the number of sales associates.associates.

• The standard divisor is The standard divisor is DD = 2454. = 2454.• There needs to be approximately 1 sales There needs to be approximately 1 sales

associate for every 2454 customers. associate for every 2454 customers.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: The standard quotas are Solution, cont’d: The standard quotas are

calculated, as shown in the table below. calculated, as shown in the table below. • Note that the rounded quotas add to 53.Note that the rounded quotas add to 53.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: The sum of the rounded Solution, cont’d: The sum of the rounded

standard quotas was too small, so the standard quotas was too small, so the modified divisor needs to be a little larger modified divisor needs to be a little larger than the standard divisor. than the standard divisor.

• Find a new modified divisor using the guess-Find a new modified divisor using the guess-and-check method.and-check method.

• First, try First, try 65000 240726 1

d


• Solution, cont’d: The modified divisor is appropriate Solution, cont’d: The modified divisor is appropriate because the rounded modified quotas add to 54. because the rounded modified quotas add to 54.


• Solution, cont’d: The final Solution, cont’d: The final apportionment is:apportionment is:

• Northside store: 11 sales associatesNorthside store: 11 sales associates• Westside store: 16 sales associatesWestside store: 16 sales associates• Eastside store: 27 sales associatesEastside store: 27 sales associates

5.2 Initial Problem Solution5.2 Initial Problem Solution• You, your sister, and your brother will divide You, your sister, and your brother will divide

85 gold coins based on the number of hours 85 gold coins based on the number of hours you have each volunteered at the soup you have each volunteered at the soup kitchen.kitchen.


• The standard divisor is found by The standard divisor is found by dividing the total number of hours dividing the total number of hours worked by the number of gold coins.worked by the number of gold coins.• DD is approximately 1.76. is approximately 1.76.• You should each get about 1.76 coins for You should each get about 1.76 coins for

every hour you have worked. every hour you have worked.


• The standard quotas for each person are The standard quotas for each person are shown in the table.shown in the table.


• Consider the apportionment for Jefferson’s Consider the apportionment for Jefferson’s method:method:• Rounding down the standard quotas yields a Rounding down the standard quotas yields a

sum of 40 + 24 + 19 = 83, which is too small.sum of 40 + 24 + 19 = 83, which is too small.• The modified divisor must be smaller. The modified divisor must be smaller.

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• Suppose we try a modified divisor of Suppose we try a modified divisor of

dd = 1.74. = 1.74.


• The rounded-down modified quotas The rounded-down modified quotas add to 85.add to 85.

• The final apportionment, using The final apportionment, using Jefferson’s method, is: Jefferson’s method, is: • You will receive 41 coins.You will receive 41 coins.• Your sister will receive 25 coins.Your sister will receive 25 coins.• Your brother will receive 19 coins.Your brother will receive 19 coins.


• Now apportion the coins using Now apportion the coins using Webster’s method:Webster’s method:• When the standard quotas are rounded When the standard quotas are rounded

normally, they add to 86.normally, they add to 86.• The modified divisor needs to be slightly The modified divisor needs to be slightly

larger than the standard divisor so that the larger than the standard divisor so that the sum of the rounded quotas will be smaller. sum of the rounded quotas will be smaller.


• Try using a modified divisor of Try using a modified divisor of dd = 1.77. = 1.77.


• The normally-rounded modified quotas add The normally-rounded modified quotas add to 85.to 85.

• The final apportionment, using Webster’s The final apportionment, using Webster’s method, is: method, is: • You will receive 41 coins.You will receive 41 coins.• Your sister will receive 25 coins.Your sister will receive 25 coins.• Your brother will receive 19 coins.Your brother will receive 19 coins.

• In this case, both apportionments are the In this case, both apportionments are the same.same.

Section 5.3Section 5.3Flaws of the Flaws of the

Apportionment MethodsApportionment Methods• GoalsGoals

• Study the Alabama paradoxStudy the Alabama paradox• Study the population paradoxStudy the population paradox• Study the new-states paradox Study the new-states paradox

5.3 Initial Problem5.3 Initial Problem• A total of 25 computers will be divided A total of 25 computers will be divided

among the schools in a district.among the schools in a district.• Initially, your school is to receive 6 Initially, your school is to receive 6

computers.computers.• After the total number of computers After the total number of computers

increases to 26, it is discovered that your increases to 26, it is discovered that your school will now only get 5 computers.school will now only get 5 computers.

• How is that possible? How is that possible? • The solution will be given at the end of the section.The solution will be given at the end of the section.

Apportionment ProblemsApportionment Problems• No apportionment method is free of No apportionment method is free of

flaws.flaws.• Circumstances that can cause Circumstances that can cause

apportionment problems include:apportionment problems include:• A reapportionment based on population A reapportionment based on population

changes.changes.• A change in the total number of seats.A change in the total number of seats.• The addition of one or more new states.The addition of one or more new states.

The Quota RuleThe Quota Rule• Recall that the quota rule says that each Recall that the quota rule says that each

state’s apportionment should be equal to the state’s apportionment should be equal to the whole number just below or just above the whole number just below or just above the state’s standard quota.state’s standard quota.• Every quota method satisfies the quota rule.Every quota method satisfies the quota rule.• No divisor method can always satisfy the quota No divisor method can always satisfy the quota

rule.rule.• Both quota and divisor methods may have Both quota and divisor methods may have

flaws. flaws.

The Alabama ParadoxThe Alabama Paradox• In 1880 it was discovered that if the number In 1880 it was discovered that if the number

of seats in the House of Representatives was of seats in the House of Representatives was increased from 299 to 300 then Alabama increased from 299 to 300 then Alabama would be apportioned one fewer seat than would be apportioned one fewer seat than before, using Hamilton’s method.before, using Hamilton’s method.

• The possibility that the addition of one The possibility that the addition of one legislative seat will cause a state to lose a legislative seat will cause a state to lose a seat is called the seat is called the Alabama paradoxAlabama paradox..

The Alabama Paradox, cont’dThe Alabama Paradox, cont’d

• When the total number of seats is increased, When the total number of seats is increased, each standard quota must also increase.each standard quota must also increase.• For a state to lose a seat (under a quota For a state to lose a seat (under a quota

method), the decrease must be due to a change method), the decrease must be due to a change from rounding up the state’s quota to rounding it from rounding up the state’s quota to rounding it down.down.

• The lost seat must be apportioned to another The lost seat must be apportioned to another state, so that state’s quota had to change from state, so that state’s quota had to change from being rounded down to rounded up. being rounded down to rounded up.

Question:Question:A country had 299 legislators for 4 states, A country had 299 legislators for 4 states, with an apportionment of State A: 50 with an apportionment of State A: 50 seats, State B: 85 seats, State C: 91 seats, seats, State B: 85 seats, State C: 91 seats, and State D: 73 seats. After the number of and State D: 73 seats. After the number of seats in the legislature was increased to seats in the legislature was increased to 300, the apportionment changed to State 300, the apportionment changed to State A: 50 seats, State B: 86 seats, State C: 91 A: 50 seats, State B: 86 seats, State C: 91 seats, and State D: 73 seats.seats, and State D: 73 seats.

Did the Alabama paradox occur?Did the Alabama paradox occur?a. Yesa. Yes b. Nob. No

Example 1Example 1• A country has a A country has a

population of 100,000 in population of 100,000 in 4 states.4 states.

• Show that the Alabama Show that the Alabama paradox arises under paradox arises under Hamilton’s method if the Hamilton’s method if the number of seats in the number of seats in the legislature is increased legislature is increased from 99 to 100.from 99 to 100.


• Solution: Find the apportionment for 99 Solution: Find the apportionment for 99 seats.seats.• The standard divisor is The standard divisor is DD = 100,000/99 = = 100,000/99 =

1010.10.1010.10.• Calculate theCalculate the standard quota for each standard quota for each

state.state.


• Solution, cont’d: The integer parts of the Solution, cont’d: The integer parts of the standard quotas add up to 98, so there is 1 standard quotas add up to 98, so there is 1 seat leftover.seat leftover.


• Solution, cont’d: Solution, cont’d:

The state with the largest fractional part of its The state with the largest fractional part of its standard quota is state C.standard quota is state C.• State C gets the leftover seat.State C gets the leftover seat.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: The final apportionment of Solution, cont’d: The final apportionment of

99 seats is shown below.99 seats is shown below.


Next, find the Next, find the apportionment for apportionment for 100 seats.100 seats.• The standard The standard

divisor is divisor is DD = 100,000/100 = 100,000/100 = 1000.= 1000.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: The standard quota for each Solution, cont’d: The standard quota for each

state is calculated.state is calculated.• The integer parts of the standard quotas add up The integer parts of the standard quotas add up

to 98, so there are 2 seats leftover.to 98, so there are 2 seats leftover.


• Solution, cont’d: The states with the largest Solution, cont’d: The states with the largest fractional parts of their standard quotas are fractional parts of their standard quotas are states A and B.states A and B.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: The final apportionment of Solution, cont’d: The final apportionment of

100 seats is shown below.100 seats is shown below.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: Note that state C had Solution, cont’d: Note that state C had

11 seats in the first apportionment, but 11 seats in the first apportionment, but only 10 seats in the second only 10 seats in the second apportionment.apportionment.• The bigger states, A and B, have benefited The bigger states, A and B, have benefited

at the expense of the smaller state C.at the expense of the smaller state C.• This is an example of the Alabama This is an example of the Alabama

paradox.paradox.

Population ParadoxPopulation Paradox• The The population paradoxpopulation paradox can occur when can occur when

the population in two states increases.the population in two states increases.• The legislature is reapportioned based on The legislature is reapportioned based on

a new census and a seat is switched from a new census and a seat is switched from one state to the other.one state to the other.

• The paradox occurs when the faster-The paradox occurs when the faster-growing state is the one that loses the growing state is the one that loses the seat.seat.

Question:Question:A census shows that State A is growing A census shows that State A is growing at a rate of 2%, State B at a rate of 0%, at a rate of 2%, State B at a rate of 0%, and State C at a rate of 3.5%. The and State C at a rate of 3.5%. The previous apportionment was State A: 38 previous apportionment was State A: 38 seats, State B: 52 seats, and State C: 60 seats, State B: 52 seats, and State C: 60 seats. The new apportionment after the seats. The new apportionment after the census was State A: 39 seats, State B: census was State A: 39 seats, State B: 52 seats, and State C: 59 seats.52 seats, and State C: 59 seats.

Did the population paradox occur?Did the population paradox occur?a. Yesa. Yes b. Nob. No

Example 2Example 2• A country has 3 states and 100 seats in the A country has 3 states and 100 seats in the

legislature.legislature.• Show that the population paradox occurs Show that the population paradox occurs

when Hamilton’s method is used.when Hamilton’s method is used.


• Solution: Note that states A and B grew, Solution: Note that states A and B grew, while the population of state C remained the while the population of state C remained the same.same.

• Find the rates of increase for states A and B.Find the rates of increase for states A and B.


• The rate of increase for state A is:The rate of increase for state A is:

• The rate of increase for state B is:The rate of increase for state B is:

• State A is the faster-growing state.State A is the faster-growing state.

9651 9555 96 1.0047%9555 9555

19740 19545 195 0.9977%19545 19545


• Solution, cont’d: Calculate the standard divisor for Solution, cont’d: Calculate the standard divisor for the old population: the old population: • DD = 100,000/100 = 1000 = 100,000/100 = 1000

• The standard quotas are shown in the table below.The standard quotas are shown in the table below.

Example 2, cont’dExample 2, cont’d• Solution, cont’d: The integer parts of the Solution, cont’d: The integer parts of the

standard quotas apportion 98 seats.standard quotas apportion 98 seats.• The two remaining seats go to states C The two remaining seats go to states C

and A, which have the largest fractional and A, which have the largest fractional parts. parts.

• The final apportionment for the old The final apportionment for the old population is State A: 10 seats; State B: population is State A: 10 seats; State B: 19 seats; and State C: 71 seats.19 seats; and State C: 71 seats.

Example 2, cont’dExample 2, cont’d• Solution, cont’d: Next, find the apportionment for Solution, cont’d: Next, find the apportionment for

the new population totals.the new population totals.• The standard divisor is The standard divisor is DD = 100,291/100 = 1002.91. = 100,291/100 = 1002.91.• The standard quotas are shown below. The standard quotas are shown below.

Example 2, cont’dExample 2, cont’d• Solution, cont’d: As before, the integer parts Solution, cont’d: As before, the integer parts

of the standard quotas apportion 98 seats.of the standard quotas apportion 98 seats.• The remaining 2 seats are assigned to states The remaining 2 seats are assigned to states

C and B, which have the largest fractional C and B, which have the largest fractional parts.parts.

• The final apportionment for the new The final apportionment for the new population totals is State A: 9 seats; State B: population totals is State A: 9 seats; State B: 20 seats; and State C: 71 seats.20 seats; and State C: 71 seats.

Example 2, cont’dExample 2, cont’d• Solution, cont’d: Before the census Solution, cont’d: Before the census

state A had 10 seats in the legislature, state A had 10 seats in the legislature, but after the census it had only 9 seats.but after the census it had only 9 seats.

• After the census, the fastest-growing After the census, the fastest-growing state A lost a seat to the slower-growing state A lost a seat to the slower-growing state B.state B.

• This is an example of the population This is an example of the population paradox. paradox.

New States ParadoxNew States Paradox• If a new state is added to a country, then If a new state is added to a country, then

how many seats must be added to the how many seats must be added to the legislature?legislature?• A reasonable number of seats would seem to be A reasonable number of seats would seem to be

the integer part of the new state’s standard the integer part of the new state’s standard quota.quota.

• The The new-states paradoxnew-states paradox occurs when a occurs when a recalculation of the apportionment results in recalculation of the apportionment results in a change of the apportionment of some of a change of the apportionment of some of the other states, not the new state. the other states, not the new state.

Question:Question:

A fourth state was just added to a A fourth state was just added to a country. The previous apportionment country. The previous apportionment was State A: 48 seats, State B: 52 seats, was State A: 48 seats, State B: 52 seats, and State C: 50 seats. The new and State C: 50 seats. The new apportionment after State D was added apportionment after State D was added is State A: 49 seats, State B: 51 seats, is State A: 49 seats, State B: 51 seats, State C: 50 seats, and State D: 29 seats.State C: 50 seats, and State D: 29 seats.

Did the new-states paradox occur? Did the new-states paradox occur? a. Yesa. Yes b. Nob. No

Example 3Example 3• A country has 2 states and 100 seats in the A country has 2 states and 100 seats in the

legislature.legislature.• The current apportionments under Hamilton’s The current apportionments under Hamilton’s

method are shown in the table below. method are shown in the table below.

Example 3, cont’dExample 3, cont’d• Show that if a third state with a population of Show that if a third state with a population of

10,400 is added, the new-states paradox 10,400 is added, the new-states paradox occurs.occurs.

• Solution: Since the old standard divisor was Solution: Since the old standard divisor was 1000, the new state’s standard quota would 1000, the new state’s standard quota would have been have been QQ = 10,400/1000 = 10.4. = 10,400/1000 = 10.4.• We assume that 10 new seats should be added We assume that 10 new seats should be added

to the legislature.to the legislature.• A total of 110 seats will now be apportioned to A total of 110 seats will now be apportioned to

the 3 states.the 3 states.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: The total population is now Solution, cont’d: The total population is now

110,400.110,400.• The new standard divisor is The new standard divisor is DD = 110,400/110 = = 110,400/110 =

1003.6364.1003.6364.• The new standard quotas are shown below.The new standard quotas are shown below.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: A total of 109 seats are Solution, cont’d: A total of 109 seats are

apportioned according to the integer parts of apportioned according to the integer parts of the standard quotas.the standard quotas.

• The 1 leftover seat is assigned to state A, The 1 leftover seat is assigned to state A, which has the largest fractional part.which has the largest fractional part.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: After the new state, C, Solution, cont’d: After the new state, C,

was added:was added:• State C received 10 seats, as expected.State C received 10 seats, as expected.• State A lost a seat to State B.State A lost a seat to State B.

• The change in apportionment among The change in apportionment among the old states is an example of the new-the old states is an example of the new-states paradox. states paradox.

Paradox SummaryParadox Summary• The three types of paradoxes studied here The three types of paradoxes studied here

are summarized below.are summarized below.

Paradox Summary, cont’dParadox Summary, cont’d• The different types of problems that can The different types of problems that can

occur with the various apportionment occur with the various apportionment methods are summarized below.methods are summarized below.

Paradox Summary, cont’dParadox Summary, cont’d• As seen in the table, all 4 apportionment As seen in the table, all 4 apportionment

methods either violate the quota rule or methods either violate the quota rule or allow paradoxes to occur.allow paradoxes to occur.

• Mathematicians Michel L. Balinski and Mathematicians Michel L. Balinski and H. Peyton Young proved there is no H. Peyton Young proved there is no apportionment method that obeys the apportionment method that obeys the quota rule and always avoids the three quota rule and always avoids the three paradoxes.paradoxes.

5.3 Initial Problem Solution5.3 Initial Problem Solution• When the district was getting 25 When the district was getting 25

computers, your school would get 6 computers, your school would get 6 of them.of them.

• Now that the district will get 26 Now that the district will get 26 computers, but your school will only computers, but your school will only get 5.get 5.

• How is this possible? How is this possible?

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• Without more information, we cannot Without more information, we cannot

check the mathematics behind the check the mathematics behind the apportionment.apportionment.

• However, this situation is possible.However, this situation is possible.• Your school losing 1 computer after the total Your school losing 1 computer after the total

number of computers increased is an example number of computers increased is an example of the Alabama paradox. of the Alabama paradox.

• This can occur if the district uses Hamilton’s or This can occur if the district uses Hamilton’s or Lowndes’ method to apportion the computers.Lowndes’ method to apportion the computers.

a mathematical view of our world

Documents

state d

state b

state c

state populations

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example of apportionment

standard quotalet d