7 Apportionment & Voting

Chapter
 7 Apportionment & Voting
Chapter

Apportionment is the process of allocating political power among a set of

constituencies based on population. Apportionment is a method of dividing a whole
into various parts.
Apportionment Methods:
1. Hamilton’s Method
2. Jefferson’s Method
3. Adam’s Method
Definitions:
Total Population
4. Webster’s Method 1. Standard Divisor =
Number of Allocated Items

Population of a particular Group

2. Standard Quota =
Standard Divisor

3. Lower Quota = standard quota rounded down to the nearest

whole number

4. Upper Quota = standard quota rounded up to the nearest whole

number
Apportionment 7 Apportionment & Voting
Chapter

Example: The provincial bord allots 100 seats, divided among five cities depending on their
respective populations. Data are given in the table.
State City A City B City C City D City E Total
Population
144 236 260 362 398 1400
(in thousands)
1. What is the standard divisor?
Standard Divisor =

2. Find each state’s standard quota.

State Population Solution Standard
Quota
A 144
B 236
C 260
D 362
E 398
Apportionment 7 Apportionment & Voting
Chapter

Example: The provincial bord allots 100 seats, divided among five cities depending on their
respective populations. Data are given in the table.
State City A City B City C City D City E Total
Population
144 236 260 362 398 1400
(in thousands)
1. What is the standard divisor?
1400
Standard Divisor = = 14
100
2. Find each state’s standard quota.
State Population Solution Standard
Quota
A 144 144/14 10
B 236 236/14 17
C 260 260/14 19
D 362 362/14 26
E 398 398/14 28
Apportionment Methods
1. Hamilton’s Method (method of largest remainders)
Steps to follow
a) Calculate each group’s standard quota
b) Round each standard quota down to the nearest whole number by
dropping all decimal places and retaining only the whole number. This
yields the lower quota.
c) Add surplus one at a time, to the groups with the largest decimal part in
the standard quota until there are no more surplus.
Apportionment Methods
1. Hamilton’s Method (method of largest remainders)

Example:1
A university is composed of five colleges. The enrolment in each college is given in the
table. Use the Hamilton’s Method to find each college’s apportionment of 180 notebooks.

College Enrolment
Art & Sciences 280
Business & Accountancy 398 Standard Divisor =
Computer Studies 181
Engineering 516
Tourism 785
Total 2160
Apportionment Methods
1. Hamilton’s Method (method of largest remainders)

Example:1
A university is composed of five colleges. The enrolment in each college is given in the
table. Use the Hamilton’s Method to find each college’s apportionment of 180 notebooks.

College Enrolment
Art & Sciences 280
2160
Business & Accountancy 398 Standard Divisor = = 12
180
Computer Studies 181
Engineering 516
Tourism 785
Total 2160
Apportionment Methods
1. Hamilton’s Method (method of largest remainders)
Example:1 Continuation
Standard Lower
College Enrollment Apportionment
Quota Quota
Arts and Sciences 280
Business & Accountancy 398
Computer Studies 181
Engineering 516
Tourism 785
Total 2160
Apportionment Methods
1. Hamilton’s Method (method of largest remainders)
Example:1 Continuation
Standard Lower
College Enrollment Apportionment
Quota Quota
Arts and Sciences 280 23.33 23 23
Business & Accountancy 398 33.17 33 33
Computer Studies 181 15.08 15 15
Engineering 516 43.00 43 43
Tourism 785 65.42 65 66
Total 2160 180.00 179 180

Using the lower quota, the sum is 179 which means there is a surplus of 1. the surplus
is added to the highest decimal value in the standard quota. If there are more surplus, the next
is allocated to the second highest decimal value and so on.
Apportionment Methods
2. Jefferson’s Method (method of greatest divisors)
1. The standard divisor using the formula is calculated.
2. Each standard Quota is Computed.
3. The lower quota is initially assigned.
4. The sum of the lower quotas must be equal to the correct number of notebooks to be
apportioned.
5. If the sum of the quota is not equal to the total number of notebooks, a Modified Divisor is
determined. Each modified standard quota is recomputed and the lower quota is assigned until the
total number is equal to the total number of notebooks.
Apportionment Methods
2. Jefferson’s Method (method of greatest divisors)
Example: Using Ex. # 1
Standard Lower Modified Modified Lower
College Enrollment
Quota Quota Standard Quota Quota
Arts and Sciences 280
Business & Accountancy 398
Computer Studies 181
Engineering 516
Tourism 785
Total 2160
 Apportionment Methods
2. Jefferson’s Method (method of greatest divisors)
Example: Using Ex. # 1
Standard Lower Modified Modified Lower
College Enrollment
Quota Quota Standard Quota Quota
Arts and Sciences 280 23.33 23 23.73 23
Business & Accountancy 398 33.17 33 33.73 33
Computer Studies 181 15.08 15 15.34 15
Engineering 516 43.00 43 43.73 43
Tourism 785 65.42 65 66.53 66
Total 2160 180.00 179 180

The standard divisor is set at 12 obtained by dividing 2,160 by 180. The standard quota is determined
by dividing each enrolment date with the standard divisor 12. In Arts and Sciences, the standard quota of
23.33 is obtained by dividing 280 by 12. the other standard quota is calculated in the same manner. Each
standard quota is rounded down to the nearest whole number and is shown in the column of lower quota.

Using the standard quota, the total of the lower quota is 179 which is lower than the standard quota of
180. the standard divisor was set at 11.8 and yields the modified standard quota for Tourism higher by 1.
this gives a total of 180.
Apportionment Methods
3. Adam’s Method
1) Find a modified divisor, d, such that when each group’s modified quota is rounded up to the
nearest whole number, the sum of the whole numbers for all group is the number of items to
be apportioned.

2) Calculate each group’s

𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝐺𝑟𝑜𝑢𝑝
Modified Upper Quota =
𝑀𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝐷𝑖𝑣𝑖𝑠𝑜𝑟

3) Apportion to each group its Modified Upper Quota.

Apportionment Methods
3. Adam’s Method
Example: Using Ex. # 1
Standard Upper Modified Rounded Up
College Enrollment
Quota Quota Standard Quota Quota
Arts and Sciences 280
Business & Accountancy 398
Computer Studies 181
Engineering 516
Tourism 785
Total 2160
 Apportionment Methods
3. Adam’s Method
Example: Using Ex. # 1
Standard Upper Modified Rounded Up
College Enrollment
Quota Quota Standard Quota Quota
Arts and Sciences 280 23.33 24 23.14 24
Business & Accountancy 398 33.17 34 32.89 33
Computer Studies 181 15.08 16 14.96 15
Engineering 516 43.00 44 42.64 43
Tourism 785 65.42 66 64.88 65
Total 2160 180.00 184 180

The standard divisor is set at 12 obtained by dividing 2,160 by 180. The standard quota is determined
by dividing each enrolment date with the standard divisor 12. In Arts and Sciences, the standard quota of
23.33 is obtained by dividing 280 by 12. the other standard quota is calculated in the same manner. Each
standard quota is rounded up to the nearest whole number and is shown in the column of Upper Quota.

Using the standard quota, the total of the Upper Quota is 184 which is greater than the standard quota
of 180. the standard divisor was adjusted to 12.1 instead of 12. The modified standard quota is calculated
for each college and rounded up. Rounding up the modified standard quota yields a sum of 180.
Apportionment Methods
4. Webster’s Method
1) Find a modified divisor, d, such that when each group’s modified quota is rounded off to the
nearest whole number, the sum of the whole numbers for all group is the number of items to
be apportioned.

2) Calculate each group’s

𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝐺𝑟𝑜𝑢𝑝
Modified Rounded Quota =
𝑀𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝐷𝑖𝑣𝑖𝑠𝑜𝑟

3) Apportion to each group its Modified Rounded Quota.

Apportionment Methods
4. Webster’s Method
Example: Using Ex. # 1
Standard Rounded Modified
College Enrollment Rounded Quota
Quota Off Rounded Quota
Arts and Sciences 280
Business & Accountancy 398
Computer Studies 181
Engineering 516
Tourism 785
Total 2160
 Apportionment Methods
4. Webster’s Method
Example: Using Ex. # 1
Standard Rounded Modified
College Enrollment Rounded Quota
Quota Off Rounded Quota
Arts and Sciences 280 23.33 23 23.43 23
Business & Accountancy 398 33.17 33 33.31 33
Computer Studies 181 15.08 15 15.15 15
Engineering 516 43.00 43 43.18 43
Tourism 785 65.42 65 65.69 66
Total 2160 180.00 179 180

The standard divisor is set at 12 obtained by dividing 2,160 by 180. The standard quota is determined
by dividing each enrolment date with the standard divisor 12. In Arts and Sciences, the standard quota of
23.33 is obtained by dividing 280 by 12. The other standard quota is calculated in the same manner. Each
standard quota is rounded off to the nearest whole number and is shown in the column of Rounded Off.

Using the standard quota, the total of the Rounded Off is 179 which is less than the standard quota of
180. the standard divisor was adjusted to 11.95 instead of 12. The Modified Rounded Quota is calculated
for each college and rounded off. Rounding off the Modified Rounded Quota yields a sum of 180.
Summary of the Apportionment Methods

Method Value of the Divisor Rounding Rules Apportionment

The excess is
Standard Divisor = distributed one at a
Rounded Down to the
time to the groups that
Hamilton’s 𝑇𝑜𝑡𝑎𝑙 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 nearest whole number
has the largest decimal
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐴𝑙𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝐼𝑡𝑒𝑚𝑠
points until all excess
are distributed

Rounded Down to the Apportionment is made

Modified Divisor is less
Jefferson’s nearest whole number to the group in its
than standard divisor
modified lower quota

Modified Divisor is Rounded Up to the Apportionment is made

Adam’s Greater than standard nearest whole number to the group in its
divisor modified upper quota

Modified is less than, Apportionment is made

Rounded Off to the
Webster’s greater than or equal to to the group in its
nearest whole number
standard divisor modified rounded quota
MATHEMATICS OF VOTING
There are two different types of voting methods: preferential and non-preferential.
In the preferential system, the voter simply states a preference by ranking
alternatives using a preference table or preference schedule.

Example:
A group of student council members would like to order pizza for the meeting. Three available
toppings should be added one-topping on one family, one large, and one single sized pizza. The
members should decide the most popular topping that should go on each size. The most popular
topping will be on the family size, the second choice will be on the large size and the third choice
will be on the single size. The toppings are All meat, Hawaiian, Sausage and Pepperoni. Each
member should fill out a preference ballot and the results are shown in the table.
MATHEMATICS OF VOTING
Example: Graph
Choice First Second Third Fourth

Member 1 All Meat (AM) Hawaiian (H) Sausage (S) Pepperoni (P)

Member 2 Hawaiian (H) All Meat (AM) Pepperoni (P) Sausage (S)

Member 3 All Meat (AM) Hawaiian (H) Pepperoni (P) Sausage (S)

Member 4 Pepperoni (P) Hawaiian (H) All Meat (AM) Sausage (S)

Member 5 All Meat (AM) Sausage (S) Hawaiian (H) Pepperoni (P)

Member 6 All Meat (AM) Pepperoni (P) Sausage (S) Hawaiian (H)

Considering each person’s first choice, majority prefers all meat. However, Hawaiian can also
be considered since it appears as first and second choice. in this case, analysis is done in the
voting systems.
Voting systems can be analyzed using plurality, plurality with runoff, plurality with elimination,
Borda count and pairwise comparison. For a simpler look at the table, the terms are changed into
representatives letters.
Plurality
By plurality of votes, the candidate that has the most first-place
votes is declared the winner.
Example: A group of 30 student council members would like to order pizza for the meeting. Three
available toppings should be added one-topping on one family, one large, and one single sized
pizza. The members should decide the most popular topping that should go on each side. The
most popular topping will be on the family size, the second choice will be on the large size and
the third choice will be on the single size. The toppings are All meat, Hawaiian, Sausage and
Pepperoni. After the survey on the preferred toppings, the following data were reported. (the result is
not the same with the 1st given example since there are 30 members on this example)

Rankings
Number of voters
6 7 7 10
All meat (1) 1 2 1 3
Hawaiian (2) 2 1 3 2
Sausage (3) 3 4 4 1
Pepperoni (4) 4 3 2 4
Plurality
After summarizing the data in the given table we came up with the following result.
Toppings First Place Votes
All Meat 6 + 7 = 13
Hawaiian 7
Sausage 10
Pepperoni 0

The highest first-place vote is 13 for all meat. However, it was the first choice of 43% (13/30) of
the voters and which does not represent the majority.
Plurality with Runoff
Each voter votes for one candidate. If no candidate receives a majority vote, the
top two candidates will have a runoff election. The other candidates with lower votes
are eliminated. The candidate that receives a majority in the runoff election is the
winner
In the example, since the top two choices are All Meat and Sausage, the other two (Hawaiian
and Pepperoni) are eliminated.
In the 1st and 3rd column, AM is
Number of voters
6 7 7 10 preferred over S by 13 (6 + 7) voters.
First Choice AM H AM S
In the 4th column, S is preferred over
Second Choice H AM H AM
AM by 10 voters.
Third Choice S P P H
Fourth Choice P S S P AM is the winner.
The Borda Count
Voters rank candidates from most to least favorable. The candidate in the last place
gets (0) point. The next-to-last-place candidate gets one (1) point. The 3rd-from-last-
place candidate gets (2) points, and so on. The candidate who receives the most
points is the winner.

Add the points

Points AM = (3*6) + (2*7) + (3*7) + (2*10) = 73
6 7 7 10
H = (2*6) + (3*7) + (2*7) + (1*10) = 50
3 AM H AM S
S = (1*6) + (0*7) + (0*7) + (3*10) = 36
2 H AM H AM P = (0*6) + (1*7) + (2*7) + (0*10) = 21
1 S P P H
0 P S S P The Highest point garnered is 73, Thus All Meat (AM) is
the winner
Plurality with elimination (Hare Method)
Each voter votes for one candidate. If no candidate receives a majority, candidate
with the fewest votes is eliminated and the voting is done again. In case of tie for the
fewest votes, all of them are eliminated. The process is repeated until a candidate
receives a majority.
Example: the table shows the number of votes for each candidate.
6 7 7 10 On row 1, AM has 13 votes, H has 7 votes, S has 10
R1 AM H AM S votes and P has no votes.
R2 H AM H AM Since there are 30 voters (6+7+7+10), a majority
need 15 votes. Since no one got the majority vote, one of
R3 S P P H
them will be eliminated. Since P has no votes, P will be
R4 P S S P eliminated. All votes in each column below P will move
up one place.

6 7 7 10
R1 AM H AM S On row 1, AM has 13 votes, H has 7 votes, S has 10
votes
R2 H AM H AM Since H has the fewest votes, H will be eliminated
R3 S S S H and all votes in each column below H will move up.
Plurality with elimination (Hare Method)
6 7 7 10
On row 1, AM has 20 votes, and S has 10 votes
R1 AM AM AM S
R2 S S S AM AM has the majority of vote, so AM is the winner
Pairwise Comparison
The pairwise comparison method, also known as “head-to-head” method compares
each candidate one-on-one with each of the candidates. A candidate receives one (1) point
if it wins, half (0.5) point for a tie and zero (0) point for a loss. The candidate who receives
the most over all points is the winner.
AM-H pair
Rankings AM favored over H on 6 + 7 + 10 = 23 votes
Number of voters
6 7 7 10 H was favored over AM on 7 votes
1st Choice AM H AM S The winner is AM
2nd Choice H AM S AM
AM-S pair
3rd Choice S P P H
AM is favored over S on 6 + 7 + 7 = 20 votes
4th Choice P S H P S is favored over AM on 10 votes
AM is the winner

AM-P pair
H-S pair
AM is favored over P on 6 + 7 + 7 + 10 = 30 votes
H is favored over S on 6 + 7 = 13 votes
P is favored over AM in 0 votes
S is favored over H on 7 + 10 = 17 votes
AM is the winner
S is the winner
Pairwise Comparison
Rankings H-P pair
Number of voters H is favored over P on 6 + 7 + 10 = 23 votes
6 7 7 10
P is favored over H on 7 votes
1st Choice AM H AM S H is the winner
2nd Choice H AM S AM
3rd Choice S P P H S-P pair
4th Choice P S H P

AM has 3 wins, S has 2 and H has one. Therefore, AM is the winner

Activity – Chapter 7
Data were collected to determine the most preferred fast-food chain in the country.
Choices were given as: Jollibee (J)
McDonalds (M) Number of voters 43 32 28 17
KFC (K) First Choice J M K S
Shakey’s (S) Second Choice K S M K
Determine the fast-food that wins using: Third Choice M J J M
a) Plurality Method
Fourth Choice S K S J
b) Plurality with runoff method
c) The Borda Count
d) Plurality with elimination method
e) Pairwise Comparison
Include a short explanation and conclusion on each method used.
End of Chapter 7