Fair Division

1
Fair Division Apportionment

• Chapters 13 14
• Austin Cole

2
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

3
Adjusted Winner Procedure

• Allows two parties to settle a dispute involving
  issues or objects with a certain mathematical
  degree of fairness

4
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

5
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
6

• 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
Name 5 10
HQ 25 10
Chairman 35 20
CEO 15 35
Layoffs 20 25
Glaxo SmithKline
HQ (25) Name (10)
Chairman (35) CEO (35)
Layoffs (25)
Points60 Points70
7

• 3. If point totals arent 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
• As point value of the item
• Bs point value of the item

Glaxo SmithKline
HQ (25) Name (10)
Chairman (35) CEO (35)
Layoffs (25)
Points60 Points70
Layoffs Point Ratio 25/201.25 Name Point Ratio
10/52 CEO Point Ratio 2.33
8
Glaxo SmithKline
HQ (25) Name (10)
Chairman (35) CEO (35)
Layoffs (25)
Points60 Points70

• 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)
Points64 Points64
9
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

10
Knaster Inheritance Procedure

• A house has four heirs-Bob, Carol, Ted, Alice
• 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

BOB CAROL TED ALICE
120,000 200,000 140,000 180,000
11

• So the final settlement is
• What if there were more than one object?
• We just do the same for each object
• Example 1

BOB CAROL TED ALICE
40K House - 140K 45K 55K
BOB CAROL TED ALICE
House 120K 200K 140K 180K
Cabin 60K 40K 90K 50K
Boat 30K 24K 20K 20K
BOB CAROL TED ALICE
Boat-20,875 7,625 6,625 6,625
12
Taking Turns

• How do we decide who chooses first?
• Because choosing first is often an advantage,
  shouldnt 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?

13
Dividing up possessions for Divorce
14
Bottom-Up Strategy

• 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
A C
B E
C D
D A
E B
Bob C A B
Carol D E
15
Divide-and-Choose

• http//www.youtube.com/watch?vAdYFVN35h5w
• One party divides the object into two parts and
  the other party chooses whichever part he wants

16
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 players
  strategy guarantees that player a piece of size
  at least 1/n of the whole in his estimation
• Procedure envy-free if each players strategy
  guarantees that player a piece considered to be
  at least tied for largest

17
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.

18
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

19

• 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

20
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?

21
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 didnt
   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.

22
Apportionment Outline

• Hamilton Method
• Jefferson Method
• Webster Method
• Hill-Huntington Method
• Districts
• Discussion

23
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

24
Original Plan for Congressional Apportionment
(1790)
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
25
High School Math Teacher
Course Population Quota Rounded
Geometry 52 52/202.6 3
Pre-Calc 33 33/201.65 2
Calculus 15 15/20.75 1
Totals 100 5 6

• Standard Divisor total population divided by
  house size (100/520)
• Quota a population divided by the standard
  divisor

26
Hamilton Method

• 1. Calculate each states 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

27
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
28
High School Math Teacher with Hamiltons Method
Course Population Quota Lower Quota
Geometry 52 52/202.6 2
Pre-Calc 33 33/201.65 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

29

• 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
30
Jefferson Method

• Divisor method determines each states
  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
31
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
• 3. Find the critical divisor for state i, di
  pi
• 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

s
ni 1
32
High School Math Teacher
Course Population Lower Quota Critical Divisor
Geometry 52 2 52/317.333
Pre-Calc 33 1 33/216.5
Calculus 15 0 15/115
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

33
Hamilton vs. Jefferson Method

• 1820 Census NY Population-1,368,775 US
  Population-8,969,878
• House size 213
• Standard divisor 42,112 NYs quota 32.503
• Hamilton method?33 seats
• Jefferson method? with d39,900
• 1,368,775/39,900 rounded down awards 34 seats

34
Webster Method

• The divisor method that rounds the quota to
  nearest whole number
• 1. Calculate standard divisor and find each
  states quota
• 2. The tentative apportionment ni is the rounded
  quota
• 3. Calculate sum of tentative apportionments

35

• 4. If tentative apportionments dont fill the
  house, the critical divisor for state i is di
  pi
• 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
• The state with the smallest critical divisor
  loses a seat

ni ½
ni - ½
36
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

37
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/v ni(ni1)
• Exercise 2

38
Districts

• Representative share apportionment/population
• District population state pop./apportionment
• Relative difference given positive A,B and AgtB,
• It is (A-B)/B X 100

39
77th Congress

• Michigan was given 17 seats with a population of
  5.256106 million (rep share3.234)
• Arkansas was given 7 seats with a population of
  1.949387 million (rep share3.591)
• So the relative difference was
• x100 11.04

3.591-3.234 3.234
40
Discussion

• In what instances can you think of that have used
  proportional procedures? Were they envy-free?
• Can you think of other uses for apportionment?
  Which method would be best?
• Homework (7th Edition)
• Chapter 13 3
• Chapter 14 19