nossi ch 5 updated

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

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 the and the

number of seats to be apportioned is number of seats to be apportioned is MM. . • The The standard divisorstandard divisor is the ratio is the ratio DD = = P/MP/M. .

(handout calls this the (handout calls this the ideal ratioideal ratio.).)• The standard divisor gives the number of people The standard divisor gives the number of people

per legislative seat. 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 (ideal ratio). divisor (ideal ratio).

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 1,100,000 + 1,100,000

= 10,000,000.= 10,000,000.• The number of seats is The number of seats is MM = 200 = 200


• Solution, cont’d: The standard divisor (ideal Solution, cont’d: The standard divisor (ideal ratio) is ratio) 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 (ideal ratio). be the standard divisor (ideal ratio).

• 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 would If seats could be divided into fractions, we would

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

Example 2Example 2• In the previous In the previous

example, the example, the standard divisor standard divisor (ideal ratio) was (ideal ratio) 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 state Solution: Divide the population of each state

by the standard divisor.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

Wow! They all came out whole numbers!Wow! They all came out whole numbers!

Example 2, cont’dExample 2, cont’d• This solution, with all standard quotas being This solution, with all standard quotas being

whole numbers, is whole numbers, is notnot 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.

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 (ideal ratio).Find the standard divisor (ideal ratio).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 (ideal ratio) Solution: the standard divisor (ideal ratio)

is found: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.

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.

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 (ideal ratio) is:The standard divisor (ideal ratio) 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

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.

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. Note: total delegates = 200method. Note: total delegates = 200

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: Recall the standard divisors and Solution: Recall the standard divisors and

the apportionment found using the rounded the apportionment found using the rounded down method. Note: total delegates = 198down method. Note: total delegates = 198

Rounded-Rounded-down down methodmethod

26263030989822222222

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. Sum of delegate = 200calculated. Sum of delegate = 200

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.

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 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 3Example 3• Solution: First determine the standard Solution: First determine the standard

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

(27,500+40,000+65,000)/54 = 2454(27,500+40,000+65,000)/54 = 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 Jefferson’s MethodExample 3 Jefferson’s Method

• 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 Jefferson’s methodExample 3 Jefferson’s method

• 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 = ≈+

Example 3 Jefferson’s methodExample 3 Jefferson’s method

• 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.

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.

(72+43.5+34.5)/85 coins = 1.76(72+43.5+34.5)/85 coins = 1.76

You should each get about 1.76 coins You should each get about 1.76 coins for every hour you have worked. for 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.

(This is a guess, but a method to find a guess is (This is a guess, but a method to find a guess is largest pop./(quota +1))largest pop./(quota +1))


• 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

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..

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.

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.

Ch 5 AssignmentCh 5 Assignment The Chapter 5 assignment is on the 2 The Chapter 5 assignment is on the 2

handouts given in class. Please pay handouts given in class. Please pay attention to the handwritten notes about attention to the handwritten notes about questions that only need to be read and questions that only need to be read and ones that may be omitted.ones that may be omitted.

Reminder: for the Midterm you may use Reminder: for the Midterm you may use quizzes, but not cell phones or computers. quizzes, but not cell phones or computers. You will want to bring a calculator.You will want to bring a calculator.

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