1

Fair Cost-Sharing Method for the

Minimum Spanning Tree Game

邱冠凱 胡啟政 劉宗灝 文國煒

2

Outline

• Coalition Game• Dissatisfaction

• MWD

• MAD

3

Coalition Game

邱冠凱

4

Example

• Players = {1,2,3}

• All nonempty subset (named as coalition)– {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}

• A cost function c related to all coalitions.– c({1}) = v1, c({2}) = v2, ..., c({1,2,3}) = v7

• c(S) is the amount that the players in the coalition S have to pay collectively in order to have access to a service.

5

Core

• The problem is to find the core of this coalition game.

• Core is a cost distribution of the grand coalition such that no other coalition can obtain an outcome better for all its members than the current assignment.

• There may not exist any core.– Emptynness of the core.

• There may exist many cores.– Some players would unhappy with the cost allocation.

6

Example

• We want to find the cost allocation {x1, x2, x3} such that – x1+x2+x3 = c({1,2,3})– x1 c({1})≦– x2 c({2})≦– x3 c({3})≦– x1+x2 c({1, 2})≦– x1+x3 c({1, 3})≦– x2+x3 c({2, 3})≦

7

Cooperative Game

• Given a solution in the core, there is no incentive for a player to leave the grand coalition.

• Coalition game is also named as cooperative game.

• The minimum spanning tree game is a classical cooperative game problem.

8

Minimum Spanning Tree Game

For any coalition S µ V, c(S) =minfPe2TS

cejTS is a spanning treeof G[S [ f rg]g, whereG[S [ f rg] denotes thesubgraph of G inducedby theset of vertices S [ f rg

r

a

bc

d e f

3 3

3 3

3

3

2 1

2 1

G: connected networkr: service providerV = {a, b, c, d, e, f}: clients

c({a, b, d, e}) = 10

c({b, d, e, f}) = 9

9

The core of MST game

• Bird proposed a cost allocation rule known as Bird’s rule.

• Bird’s rule ensures that no coalition has incentive to be formed.– We don’t prove it here.

r

a

bc

d e f

3 3

3 3

3

3

2 1

2 1

xa = 2, xb = 3, xc = 1, xd = 2, xe = 3, xf = 1,xa + xb + xc+ xd+ xe+ xf = 12 = c({a, b, c, d, e, f}).

10

Outline

• Coalition Game

• Dissatisfaction• MWD

• MAD

11

Dissatisfaction

胡啟政

12

Dissatisfaction

• Definition: For the cost a player have to pay in one solution, dissatisfaction is the ratio of the current cost to the best cost (the smallest one) of all solutions.

• Dissatisfaction of an agent i in a solution is:

•

iCy

ii y

xCx

min

),(

Viix )(

1),( Cxi

13

Cost Sharing with Fairness

• Two optimization problems:– MWD: minimize the worst dissatisfaction

• i.e. Find an allocation which

minimizes

– MAD: minimize the average dissatisfaction• i.e. Find an allocation which

minimizes

Cxx Vii )(

),(max CxiVi

Cxx Vii )(

),( CxiVi

14

With Bird’s Rule…

• The dissatisfaction of a vertex v to a mcst T is:

• Two optimization problems become– SPANNING TREE-MWD: Find a mcst that minimizes t

he worst dissatisfaction– SPANNING TREE-MAD: Find a mcst that minimizes t

he average dissatisfaction

→ Polynomial time algorithms for them.

)},'({min

),(),(

' vT

vTCT

GTTBirdv

15

Work With Trees…

• The fee of a vertex depends on the path between the root and it. β(T1, a) = 41

β(T1, b) = 11

β(T1, c) = 12

β(T1, d) = 39

β(T1, e) = 10T1

T2

β(T2, a) = 11

β(T2, b) = 12

β(T2, c) = 39

β(T2, d) = 10

β(T2, e) = 41

16

FEE

• To determine the dissatisfaction of a vertex v ∈V, it’s necessary to know the set of fees it can have to pay, which is F(v).

• From Bird’s rule, β(B, v) can only take values in L(v) = { | [x, v] ∈ E }],[ vxc

β(B, v) ],[ vxc

17

FEE

• Example:

G: :GTL(x) = {1, 2, 3}

L(y) = {1, 4}

L(z) = {2, 4}

F(x) = {3}

F(y) = {1}

F(z) = {2}

18

To compute F(v)…

Find all mcsts and then trace them?

19

Arborescences

• Assume arborescences are oriented from the root to the leaves. For convenience to deal with the problem, we can build from G=( ,E,c) a weighted digraph H=( ,A,c).

• For each edge [x,y] ∈ E, there are two arcs (x,y) and (y,x) in A with cost

rV rV

],[),(),( yxxyyx ccc

20

Arborescences

• Example:

G: H:

21

Arborescences

• Thus, given a mcst T of an undirected connected graph G, there exists in H a corresponding mcsa B, and C(T) = C(B).r

22

FEE

FEEInput: A digraph H = ( , A, c), a vertex v ∈ VStep 1: Compute any mcsa of H and let be its total costStep 2: L := { | (x, v) ∈ A }Step 3: F(v) := Step 4: For each l of L do

A’ := A \ {(x,v) | ≠ l }H’ := ( , A, c )Compute a mcsa B’ in H’If B’ exists and its total cost is Then F(v) := F(v) U { l }

End ForOutput: F(v)

rVr

optC

),( vxc

rV '|Ar

optC

),( vxc

23

FEE

• Example: To compute F(x)…

C = 6, L(x) = { 2, 3 }

opt

r

x y z

3 3

2

2

1

1

l = 2

3

F(x) = { }2

l = 3

, 3

24

FEE

• Computing a mcsa : O(mn) ( m = |A|, n = | | )

Total time : O(mn )

• After F(v) is computed, F (v) and F (v) can be determined. Also, dissatisfactions are determined.

And we can start to solve optimization problems now!!

r rV2

max min

25

Outline

• Coalition Game

• Dissatisfaction

• MWD• MAD

26

The Minimum Worst Dissatisfaction Problem

劉宗灝

27

MWD – Minimum Worst Dissatisfaction Problem

'

Among all spanning arborescences of minimum cost,

find one the minimizes the worst dissatisfaction over

all vertices.

argmin max{ ( , )}

( , ) ( , )

min {

H

H

B T v Birdv V

v BirdB T

B C

B vB C

( ', )}B v

The set of allocations from Bird’s rule

v在 B上的 cost

v在所有 mcsar中最小的 cost

28

MWD – Minimum Worst Dissatisfaction Problem

• The maximum Fmax(v) / Fmin(v) among all vertices is an upper bound on the worst dissatisfaction.

• The idea of the algorithm is decrease the upper bound until it reached the optimal.

29

How to decrease the upper bound?

• Removing some arcs each time.

• Which arc?– Arcs that cause worst dissatisfaction.

30

Algorithm

1. Input: A digraph H = (Vr, A, c)

2. 用 FEE 算每個點的 F(v)

3. 算出一棵 H 的 mcsar ，其 cost 為 C1

4. 在 H 上挑一個最大不滿意度 (Fmax(v) / Fmin(v)) 最高的點 v

5. 把造成 Fmax(v) 的邊從 H 上移除6. 在 H 上重算一棵 mcsar

1. 若不存在，或其 cost > C1，則回傳 C1 對應之 mcsar。2. 否則 goto step 4

31

Example

r

x y z

3 3 3

2 1

32

Digraph

r

x y z

3 3 3

2 1

12

F(x) = {2,3}F(y) = {1,2,3}F(z) = {1,3}

mcsars of this digraph have cost of 6

33

Pick node z

r

x y z

3 3 3

2 1

12

F(x) = {2,3}F(y) = {1,2,3}F(z) = {1,3}

34

(r,z) removed.

r

x y z

3 3

2 1

12

F(x) = {2,3}F(y) = {1,2,3}F(z) = {1}

35

mcsar still exists

r

x y z

3 3

2 1

12

F(x) = {2,3}F(y) = {1,2,3}F(z) = {1}

36

Pick node y

r

x y z

3 3

2 1

12

F(x) = {2,3}F(y) = {1,2,3}F(z) = {1}

37

(r,y) removed.

r

x y z

3

2 1

12

F(x) = {2,3}F(y) = {1,2}F(z) = {1}

38

mcsar still exists

r

x y z

3

2 1

12

F(x) = {2,3}F(y) = {1,2}F(z) = {1}

39

Pick node y

r

x y z

3

2 1

12

F(x) = {2,3}F(y) = {1,2}F(z) = {1}

40

(x,y) removed.

r

x y z

3

2 1

1

F(x) = {2,3}F(y) = {1}F(z) = {1}

41

Any more mcsar ?

r

x y z

3

2 1

1

F(x) = {2,3}F(y) = {1}F(z) = {1}

NO

42

Recover (x,y), find mcsar.

r

x y z

3

2 1

12

F(x) = {2,3}F(y) = {1,2}F(z) = {1}

43

Solution

r

x y z

3

2 1

12

F(x) = {2,3}F(y) = {1,2}F(z) = {1}

44

Worst dissatisfaction = 2.

r

x y z

3

2 1

F(x) = {2,3}F(y) = {1,2}F(z) = {1}

45

Outline

• Coalition Game

• Dissatisfaction

• MWD

• MAD

46

The Minimum Average Dissatisfaction Problem

文國煒

47

MAD Problem

• Give a graph, among all spanning arborescences of minimum cost, find one that minimizes the average dissatisfaction.

• Minimize the average dissatisfaction also minimize the worst dissatisfaction.

However, this is not always true.

48

Example 1

r

a e

b c d

41 41

11 10

12 39

T T`Fmax(a)/Fmin(a)=41/11

Fmax(b)/Fmin(b)=1

Fmax(c)/Fmin(c)=1

Fmax(d)/Fmin(d)=39/10

Fmax(e)/Fmin(e)=1

The Worst:3.9

Average:2.12

Fmax(a)/Fmin(a)=1

Fmax(b)/Fmin(b)=12/11

Fmax(c)/Fmin(c)=39/12

Fmax(d)/Fmin(d)=1

Fmax(e)/Fmin(e)=41/10

The Worst:4.1

Average:2.08

49

MAD Problem

• Minimize the sum or the average dissatisfaction are equivalent.

• The SPANNING TREE-MAD problem can be described as follows:

Vv

BirdvTB CBH ),(minarg

50

Solving Procedure

• First– The initial graph G=(Vr, E, c) is turned into a di

agraph H=(Vr, A, c).– For each edge[x, y] belongs to E, A has corre

sponding arcs (x, y) and (y, x)

– c(x, y)=c(y, x)=c[x, y]

51

Solving Procedure

• Second– Remove each arc that c(x, y)<Fmin(y)

• Third– For each arc (x, y) belongs to A, has a weight

w(x, y) = c(x, y)/Fmin(y).

– The weight w(x, y) captures the dissatisfaction of y if the arc (x, y) is chosen.

52

Solving Procedure

• Forth– Computing a spanning arborescence rooted

in r which minimizes the total weight can provide the optimal solution.

53

Example 2

r

h

a b c d

efg

4

4

3 2 1

3 2 1

3

T T`T``

54

Example 2

r

h

a b c d

efg

4,4/3

4,4/3

3,3/2

3,1

3,1

3,3/2

2,2

2,1

2,1

2,2

1,1

1,1

1,1

1,1

3,33,3

T Average dissatisfaction: (4/3+3/2+2+1+3+1+1+1)/8≈1.47T` Average dissatisfaction: (4/3+3/2+2+1+3+1+1+1)/8≈1.47

Average dissatisfaction: (4/3+3/2+2+1+1+2+3/2+4/3)/8≈1.45T``

T`` is not a minimum cost spanning tree!!

cost, weight

55

Solving Procedure

• Forth– For each arc (x, y), combines c(x, y) and w(x, y) in to a co

mposite cost and generates the new H*.

– Computing a spanning arborescence rooted in r with which minimizes the total cost can provide the optimal solution.

10,)1( ),(),(),(

~

yxyxyx wcc~

c

56

Algorithm

57

Example 3

h

a b c d

efg

4

4

3 2 1

3 2 1

3

G

4,4/3

4,4/3

3,3/2

3,3/2

3,1

3,1

2,2

2,1

2,2

2,1

1,1

1,1 3,3

3,3

1,1

1,1HA mcsar B is determined and its sum of dissatisfactions is D=4/3+3/2+1+3+1+1+1=71/6

λ= 71/77

4λ+(1- λ)4/3

4λ+(1- λ)4/3

3λ+(1- λ)3/2

3λ+(1- λ)

3λ+(1- λ)3/2

3λ+(1- λ)

2λ+(1- λ)

2λ+(1- λ)

2

2

1

1

1

1

3

3

H*B*Optimal solution

58

Thank you!