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
Chairman 35 20
CEO 15
Layoffs

6. 2. Each item is given to party that assigned it more points
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
Chairman 35 20
CEO 15
Layoffs
Glaxo
SmithKline
HQ (25)
Name (10)
Chairman (35)
CEO (35)
Layoffs (25)
Points=60
Points=70

7. 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
Glaxo
SmithKline
HQ (25)
Name (10)
Chairman (35)
CEO (35)
Layoffs (25)
Points=60
Points=70
Layoffs Point Ratio: 25/20=1.25
Name Point Ratio: 10/5=2
CEO Point Ratio: 2.33

8. So we transfers layoffs, but not the whole value
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
(1-X) = X
X = 7/9
So, (2/9) = (7/9) = 64
Glaxo
SmithKline
HQ (25)
Name (10)
Chairman (35)
CEO (35)
2/9 Layoffs (4)
7/9 Layoffs (19)
Points=64

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
BOB
CAROL
TED
ALICE
Boat-$20,875
$7,625
$6,625

12. 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?

13. Dividing up possessions for Divorce
Bob’s Ranking
Pension
House
Investments
Vehicles
1: Pension
3: Investments
1: House
3: Pension
Carol’s Ranking
2: House
4: Vehicles
2: Investments

14. Bottom-Up Strategy
Bob
Carol A C B E D
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 C A B
Carol D E

15. 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

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 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

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
Player 1 cuts cake into 3 pieces of same size. He hands 3 pieces to player 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.
Player 3 chooses a piece
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)
Player 1 receives remaining piece
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 Apportionment method
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
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
Kentucky
68,705
1.995
Rhode Island
68,446
1.988
Delaware
55,540
1.613
Totals
3,615,920
105
Original Plan for Congressional Apportionment (1790)

25. 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

26. 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

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
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
Kentucky
68,705
1.995
Rhode Island
68,446
1.988
Delaware
55,540
1.613
Totals
3,615,920
105

28. 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
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. Apportioning 30/31 Teaching Assistants:
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. 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

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
52/3=17.333
Pre-Calc 33 1
33/2=16.5
Calculus 15
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

33. Hamilton vs. Jefferson Method
1820 Census: NY Population-1,368, US Population-8,969,878
House size 213
Standard divisor = 42,112 NY’s quota =
Hamilton method→33 seats
Jefferson method→ with d=39,900
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 state’s quota
2. The tentative apportionment ni is the rounded quota
3. Calculate sum of tentative apportionments

35. 4. If tentative apportionments don’t 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/√ ni(ni±1)
Exercise 2

38. 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%

39. 77th Congress
Michigan was given 17 seats with a population of million (rep share=3.234)
Arkansas was given 7 seats with a population of million (rep share=3.591)
So the relative difference was
x100% = 11.04%
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