Fair Division & Apportionment
Chapters 13 & 14
Austin Cole
Fair Division Outline
• Adjusted Winner Procedure
• Knaster Inheritance Procedure
• Taking Turns
• Bottom-Up Strategy
• Divide and Choose
▫ Steinhaus Proportional Procedure
▫ Banach-Knaster Proportional Procedure
▫ Selfridge-Conway Envy-Free Procedure
Adjusted Winner Procedure
• Allows two parties to settle a dispute involving
issues or objects with a certain mathematical
degree of fairness
Glaxo Wellcome & SmithKline
Beecham Merger
• Assume there were five social issues for
compromise:
▫
▫
▫
▫
▫
Name of the company
Location of headquarters
Person to serve as Chairman
Person to serve as CEO
Where layoffs would come from
Steps in Adjusted Winner Procedure
1. Each party distributes 100 points over items to
reflect their relative worth
Issue
Glaxo Wellcome
SmithKline Beecham
Name
5
10
Headquarters
25
10
Chairman
35
20
CEO
15
35
Layoffs
20
25
2. Each item is given to party that assigned it
more points. Each party tallies number of points
received. The party with fewest points is given
items on which both parties placed the same
number of points.
Issue
Glaxo
SmithKline
Glaxo
SmithKline
Name
5
10
HQ (25)
Name (10)
HQ
25
10
Chairman (35) CEO (35)
Chairman
35
20
CEO
15
35
Layoffs
20
25
Layoffs (25)
Points=60
Points=70
3. If point totals aren’t equal, let A denote party
with more points and B be the other party.
Transfer items from A to B until point totals are
equal (can involve fractional transfer).
4. Order is determined by increasing point ratio:
A’s point value of the item
B’s point value of the item
Layoffs Point Ratio: 25/20=1.25
Name Point Ratio: 10/5=2
CEO Point Ratio: 2.33
Glaxo
SmithKline
HQ (25)
Name (10)
Chairman (35) CEO (35)
Layoffs (25)
Points=60
Points=70
Glaxo
SmithKline
HQ (25)
Name (10)
Chairman (35) CEO (35)
Layoffs (25)
Points=60
•
•
•
•
Points=70
So we transfers layoffs, but not the whole value
25 + 35 + 20(1-X) = 10 + 35 + 25X
X = 7/9
So, 60 + 20(2/9) = 45 + 25(7/9) = 64
Glaxo
SmithKline
HQ (25)
Name (10)
Chairman (35) CEO (35)
2/9 Layoffs (4) 7/9 Layoffs (19)
Points=64
Points=64
Theorem: Properties of the Adjusted
Winner Allocation
• Allocation is equitable: both players receive
same number of points
• Allocation is envy-free: neither player would be
happier with what the other received
• Allocation is Pareto-optimal: no other allocation
can make one party better off without making
the other worse off
Knaster Inheritance Procedure
• A house has four heirs-Bob, Carol, Ted, & Alice
BOB
CAROL
TED
ALICE
$120,000
$200,000
$140,000
$180,000
• Since Carol is the highest bidder, she gets the
house
▫ But since her fair-share is ¼, she puts $150K up
for grabs
▫ Each person withdraws ¼ of their bid
▫ Bob: $30K Ted: $35K Alice: $45K
▫ Then the surplus is split four ways
• So the final settlement is
BOB
CAROL
TED
ALICE
$40K
House - $140K
$45K
$55K
• What if there were more than one object?
BOB
CAROL
TED
ALICE
House
$120K
$200K
$140K
$180K
Cabin
$60K
$40K
$90K
$50K
Boat
$30K
$24K
$20K
$20K
• We just do the same for each object
• Example 1
BOB
CAROL
TED
ALICE
Boat-$20,875
$7,625
$6,625
$6,625
Taking Turns
• How do we decide who chooses first?
• Because choosing first is often an advantage,
shouldn’t we compensate the other party in
some way?
• Should a player always choose the object he
most favors from those that remain, or are their
strategic considerations to take into account?
Dividing up possessions for Divorce
Bob’s Ranking
Pension
House
Investments
Vehicles
1: Pension
3: Investments
1: House
3: Pension
Carol’s Ranking
House
Investments
Pension
Vehicles
2: House
4: Vehicles
2: Investments
4: Vehicles
Bottom-Up Strategy
Bob
Carol
A
C
B
E
C
D
D
A
E
B
• Rational player will never choose
least preferred alternative
• Rational player will avoid wasting a
choice on an object that he knows
will remain available and can be
chosen later
Bob
Carol
C
A
D
B
E
Divide-and-Choose
• http://www.youtube.com/watch?v=AdYFVN35h5
w
• One party divides the object into two parts and
the other party chooses whichever part he wants
Cake-Division
• Cake-Division Procedure: n players allocate a
cake among themselves so that each player
has a strategy that will guarantee that player
a piece with which he is satisfied, even in the
face of collusion by others
• Procedure is proportional if each player’s
strategy guarantees that player a piece of size
at least 1/n of the whole in his estimation
• Procedure envy-free if each player’s strategy
guarantees that player a piece considered to be
at least tied for largest
Steinhaus Proportional Procedure for 3
Players
• Bob divides a cake into three pieces
• Carol & Ted must individually approve a piece to
be of size at least 1/3
• Case 1: Carol & Ted approve different pieces.
They each get their piece and Bob gets the
piece left over.
• Case 2: Carol & Ted approve the same piece A
and disapprove of piece C. Give Bob piece C.
Put A & B back together and let Carol & Ted
divide and choose on AB.
Banach-Knaster Proportional Procedure
for 4+ Players
• Bob cuts a ¼ piece of cake and gives to Carol
• If Carol thinks piece is too big, she trims it &
places trimmings back on cake & passes piece
to Ted
• Ted proceeds as Carol did & passes piece to
Alice
• Alice does the same but then holds on to piece
• The piece goes to the last person that trimmed it
• The Bob, Carol, & Alice resume this process
with the rest of the cake
• Bob cuts a ¼ piece of cake
• Carol & Alice each get a chance to trim it and
the piece goes to the last person that trimmed it
• Final two players use
divide-and choose
method
Problem with Envy
• Both of these proportional procedures are not
envy-free
• 3 person: What if Carol & Ted both find one
piece unacceptable that is given to Bob?
• 4+ person: What if Bob receives his first cut
piece without any trimmings?
Selfridge-Conway Envy-Free Procedure
for 3 Players
1. Player 1 cuts cake into 3 pieces of same size. He
hands 3 pieces to player 2.
2. Player 2 trims at most 1 of 3 pieces to create at
least 2-way tie for largest. Set the trimmings aside
& hand 3 pieces to player 3.
3. Player 3 chooses a piece
4. Player 2 chooses from 2 remaining pieces. (If he
trimmed a piece in step 2 & player 3 didn’t choose
it, he must choose it)
5. Player 1 receives remaining piece
6. From trimmings, player 2 cuts into 3 pieces and
players choose in order of 3, 1, 2.
Apportionment Outline
•
•
•
•
•
•
Hamilton Method
Jefferson Method
Webster Method
Hill-Huntington Method
Districts
Discussion
Apportionment
• Apportionment problem
▫ to round a set of fractions so that their sum is
maintained at its original value
• Apportionment method
▫ the rounding procedure which must be able to be
applied constantly
State
Population
Quota
Apportionment
Virginia
630,560
18.310
18
Massachusetts
475,327
13.803
14
Pennsylvania
432,879
12.570
13
North Carolina
353,523
10.266
10
New York
331,589
9.629
10
Maryland
278,514
8.088
8
Connecticut
236,841
6.878
7
South Carolina
206,236
5.989
6
New Jersey
179,570
5.214
5
New Hampshire
141,822
4.118
4
Vermont
85,533
2.484
2
Georgia
70,835
2.057
2
Kentucky
68,705
1.995
2
Rhode Island
68,446
1.988
2
Delaware
55,540
1.613
2
Totals
3,615,920
105
105
Original Plan
for
Congressional
Apportionment
(1790)
High School Math Teacher
Course
Population
Quota
Rounded
Geometry
52
52/20=2.6
3
Pre-Calc
33
33/20=1.65
2
Calculus
15
15/20=.75
1
Totals
100
5
6
• Standard Divisor: total population divided by
house size (100/5=20)
• Quota: a population divided by the standard
divisor
Hamilton Method
• 1. Calculate each state’s quota
• 2. Tentatively assign each state
its lower quota of representatives.
This leaves additional seats.
• 3. Allot the remaining seats (one each) to states
whose quotas have the largest fractional parts
until house is filled
State
Population
Quota
Apportionment
Virginia
630,560
18.310
18
Massachusetts
475,327
13.803
14
Pennsylvania
432,879
12.570
13
North Carolina
353,523
10.266
10
New York
331,589
9.629
10
Maryland
278,514
8.088
8
Connecticut
236,841
6.878
7
South Carolina
206,236
5.989
6
New Jersey
179,570
5.214
5
New Hampshire
141,822
4.118
4
Vermont
85,533
2.484
2
Georgia
70,835
2.057
2
Kentucky
68,705
1.995
2
Rhode Island
68,446
1.988
2
Delaware
55,540
1.613
2
Totals
3,615,920
105
105
High School Math Teacher with
Hamilton’s Method
Course
Population
Quota
Lower Quota
Geometry
52
52/20=2.6
2
Pre-Calc
33
33/20=1.65
1
Calculus
15
15/20=.75
0
Totals
100
5
3
• Calculus has largest fraction +1
• Pre-Calc has second-largest fraction +1
• So totals would be:
▫ Geometry 2, Pre-Calc 2, Calculus 1
• Alabama Paradox: a state loses a seat as the
result of an increase in house size
• Apportioning 30/31 Teaching Assistants:
Course
Enrollment
Quota
Lower Quota
Apportionment
A
188
7.52/7.771
7 /7^
7/8
B
142
5.68/5.869
5^ /5^
6/6
C
138
5.52/5.704
5 /5^
5/6
D
64
2.56/2.645
2^ /2
3/2
E
218
8.72/9.011
8^ /9
9/9
Totals
750
30/31
27/28
30/31
Jefferson Method
• Divisor method: determines each state’s
apportionment by dividing its population by a
common divisor d and rounding the quotient
• Apportionment for a state i is
▫ Ai = pi rounded down
d
Jefferson Method
• 1.Determine the standard divisor s and quota qi for
each state
• 2.Assign each state i its tentative apportionment:
ni= pi rounded down
s
• 3. Find the critical divisor for state i, di= pi
ni + 1
• 4.The state with the largest critical divisor receives
another seat
• 5. If there are extra seats, recompute the critical
divisor
• 6.When house is filled, the last critical divisor is
divisor d, representing the minimum district
population
High School Math Teacher
Course
Population
Lower Quota
Critical Divisor
Geometry
52
2
52/3=17.333
Pre-Calc
33
1
33/2=16.5
Calculus
15
0
15/1=15
Totals
100
3
• Geometry with greatest critical divisor adds a
section (then new critical value is 13)
• So then Pre-Calc adds a section
• Final Apportionment: Geometry 3, Pre-Calc 2,
Calculus 0
• Minimum section size is 16.5
Hamilton vs. Jefferson Method
• 1820 Census: NY Population-1,368,775
US Population-8,969,878
• House size 213
• Standard divisor = 42,112 NY’s quota = 32.503
• Hamilton method→33 seats
• Jefferson method→ with d=39,900
▫ 1,368,775/39,900 rounded down awards 34 seats
Webster Method
• The divisor method that rounds the quota to
nearest whole number
1. Calculate standard divisor and find each state’s
quota
2. The tentative apportionment ni is the rounded
quota
3. Calculate sum of tentative apportionments
4. If tentative apportionments don’t fill the house,
the critical divisor for state i is di+= pi
ni + ½
The state with largest critical divisor receives a
seat.
5.If tentative apportionments overfill the house,
the critical divisor for state i is di-= pi
ni - ½
The state with the smallest critical divisor loses a
seat
High School Math Teacher
Course
Population
Quota
Tentative Apportionment
Geometry
52
2.6
3
Pre-Calc
33
1.65
2
Calculus
15
0.75
1
Totals
100
6
• Calculate di- for each class.
• Geo-20.8; Pre-Calc-22; Calc-30
• Geometry loses a section
Hill-Huntington Method
• Used to apportion House of Reps since 1940
• Find standard divisor and quotas
• If quota is greater than geometric mean, round
tentative apportionment up
• Critical divisor is di± = pi/√ ni(ni±1)
• Exercise 2
Districts
• Representative share: apportionment/population
• District population: state pop./apportionment
• Relative difference: given positive A,B and A>B,
▫ It is (A-B)/B X 100%
77th Congress
• Michigan was given 17 seats with a population
of 5.256106 million (rep share=3.234)
• Arkansas was given 7 seats with a population of
1.949387 million (rep share=3.591)
• So the relative difference was
3.591-3.234
3.234
x100% = 11.04%
Discussion
• In what instances can you think of that have
used proportional procedures? Were they envyfree?
• Can you think of other uses for apportionment?
Which method would be best?
• Homework (7th Edition)
▫ Chapter 13 #3
▫ Chapter 14 #19