Operations Research Games - IMUS (Instituto de ... · Introduction Preliminaries Linear production...

IntroductionPreliminaries

Linear production gamesMinimum cost spanning tree games

Minimum coloring gamesInventory games

Sequencing games

Operations Research Games

Henk NordeTilburg University, The Netherlands

Sevilla, March 18, 2011

Henk Norde Operations Research Games




Sequencing games

Outline

1 Introduction

2 Preliminaries

3 Linear production games

4 Minimum cost spanning tree games

5 Minimum coloring games

6 Inventory games

7 Sequencing games





Sequencing games

1 Introduction

2 Preliminaries




6 Inventory games

7 Sequencing games





Sequencing games

Introduction

Operations research deals with optimisation problems in which onedecision maker is involved.

Game theory analyses situations in which at least two decision makers(players) are involved.

Operations research games arise in situations where an optimisationproblem is considered where more players have control over part of thestructure underlying the problem.

Basic reference:Peter Borm, Herbert Hamers, Ruud Hendrickx (2001), OperationsResearch Games: A Survey. TOP 9, no. 2, 139-216.





Sequencing games

1 Introduction

2 Preliminaries




6 Inventory games

7 Sequencing games





Sequencing games

Notation and notions

A cooperative cost game is a tuple (N , c) where

• N = {1, 2, ..., n} is the set of players

• c : 2N → R is its characteristic cost function

By convention, c(∅) = 0.

The subgame corresponding to some coalition T ⊆ N is the game

(T , cT )

with cT (S) = c(S) for all S ⊆ T .





Sequencing games


The core of a cost game (N , c) is the set

C (c) := {x ∈ RN :

∑

i∈N

xi = c(N),∑

i∈S

xi ≤ c(S) for all S ⊂ N}

A game (N , c) is called

• balanced if it has a nonempty core;

• totally balanced if the core of every subgame is nonempty;

• concave if c(S ∪ {i})− c(S) ≥ c(T ∪ {i})− c(T ) for all S ⊆ T

and i ∈ N\T . (A concave game is balanced).





Sequencing games


Given a cost game (N , c), the table

x = (xSi )∅6=S∈2N ,i∈S

is said to be a Population Monotonic Allocation Scheme (PMAS)(Sprumont (1990)) if

(i) efficiency: For all S ⊆ N , S 6= ∅,∑

i∈S xSi = c(S).

(ii) monotonicity: For all S ⊆ T and for all i ∈ S , xSi ≥ xTi .

P(N , c) denotes the collection of all PMAS-es of (N , c).

Observe that totally balancedness of (N , c) is a necessary condition tohave PMAS, and concavity is a sufficient condition to have a PMAS.

Existence of PMAS provides a form of dynamic stability.





Sequencing games


Example Consider the game (N , 1) where N = {1, 2, 3, 4}. Letσ = (2, 3, 1, 4) be an order. This order provides a PMAS:

S 1 2 3 4{1} 1 − − −{2} − 1 − −{3} − − 1 −{4} − − − 1

{1, 2} 0 1 − −{1, 3} 0 − 1 −{1, 4} 1 − − 0{2, 3} − 1 0 −{2, 4} − 1 − 0{3, 4} − − 1 0{1, 2, 3} 0 1 0 −{1, 2, 4} 0 1 − 0{1, 3, 4} 0 − 1 0{2, 3, 4} − 1 0 0{1, 2, 3, 4} 0 1 0 0





Sequencing games


An order on T ∈ 2N\{∅} is a bijection from T to {1, . . . , |T |}. Thecollection of all orders on T is denoted by ΣT .

TheoremLet T ∈ 2N\{∅}.

(i) Let σ ∈ ΣT . Define the scheme yσ = (yσS,i )S∈2T\{∅},i∈S by

yσS,i =

{

1 if σ(i) < σ(j) for all j ∈ S\{i}0 otherwise

for every S ∈ 2T\{∅} and i ∈ S . Then yσ ∈ P(T , 1T ).

(ii) If y ∈ P(T , 1T ) and y is integer-valued, then there exists an orderσ ∈ ΣT such that y = yσ.





Sequencing games


ExampleConsider the game (N , 1) where N = {1, 2, 3, 4}.

S 1 2 3 4{1} 1 − − −{2} − 1 − −{3} − − 1 −{4} − − − 1{1, 2} 0.5 0.5 − −{1, 3} 0.5 − 0.5 −{1, 4} 0.5 − − 0.5{2, 3} − 0.5 0.5 −{2, 4} − 0.5 − 0.5{3, 4} − − 0.5 0.5{1, 2, 3} 0 0.5 0.5 −{1, 2, 4} 0 0.5 − 0.5{1, 3, 4} 0 − 0.5 0.5{2, 3, 4} − 0.5 0.5 0

{1, 2, 3, 4} 0 0.5 0.5 0





Sequencing games

1 Introduction

2 Preliminaries




6 Inventory games

7 Sequencing games





Sequencing games

A linear optimization problem

Consider a brewery producing two types of beer, a light beer and a darkbeer.

For the production of one barrel of light beer 4 kg hop is needed and6 kg of corn, for the production of one barrel of dark beer 5 kg ofhop and 2 kg of corn.

The brewery sells light beer with a profit of 68 per barrel and darkbeer with a profit of 42 per barrel.

Daily, the brewery disposes of 70 kg of hop and 72 kg of corn.

The brewery faces the following linear optimization problem:

maximize z = 68x1 + 52x2s.t. 4x1 + 5x2 ≤ 70

6x1 + 2x2 ≤ 72x1, x2 ≥ 0.





Sequencing games

Graphical solution

This linear optimization problem can be solved graphically:

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r

rA = (0, 0)

B = (0, 14)

C = (10, 6)

D = (12, 0)

6x1 + 2x2 = 72

4x1 + 5x2 = 70

x2

x1

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We can ‘read’ the optimal solution from this picture: x1 = 10, x2 = 6and z = 992.





Sequencing games

Dual problem

The corresponding dual optimization problem is:

minimize w = 70y1 + 72y2s.t. 4y1 + 6y2 ≥ 68

5y1 + 2y2 ≥ 52y1, y2 ≥ 0.

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The optimal solution is y1 = 8, y2 = 6 (shadow prices of hop and corn)and w = 992.





Sequencing games

More breweries

Now consider a situation with three breweries, all operating on the samemarket with the same technology.

Brewery 1 disposes daily of 4 kg hop and 33 kg corn.Brewery 2 disposes daily of 6 kg hop and 39 kg corn.Brewery 3 disposes daily of 60 kg hop and 0 kg corn.

Cooperation seems a wise thing to do!





Sequencing games

Many LP problems

The LP problem for brewery 1 reads:

maximize z = 68x1 + 52x2s.t. 4x1 + 5x2 ≤ 4

6x1 + 2x2 ≤ 33x1, x2 ≥ 0.

If, for example, breweries 1 and 3 cooperate, their LP problem becomes:

maximize z = 68x1 + 52x2s.t. 4x1 + 5x2 ≤ 4 + 60

6x1 + 2x2 ≤ 33 + 0x1, x2 ≥ 0.

If all breweries cooperate we get

maximize z = 68x1 + 52x2s.t. 4x1 + 5x2 ≤ 4 + 6 + 60 = 70

6x1 + 2x2 ≤ 33 + 39 + 0 = 72x1, x2 ≥ 0.





Sequencing games

... and many dual problems

The dual problem for brewery 1 is


5y1 + 2y2 ≥ 52y1, y2 ≥ 0,

for breweries 1 and 3


5y1 + 2y2 ≥ 52y1, y2 ≥ 0,

and for all breweries


5y1 + 2y2 ≥ 52y1, y2 ≥ 0.





Sequencing games

Linear production games

Solving all these maximization problems (or, better, all minimizationproblems) we get the following linear production game:

S 1 2 3 12 13 23 123v(S) 68 102 0 170 710 762 992





Sequencing games

Linear production situations

DefinitionA linear production situation is a tuple (N ,R ,P ,A, (bi)i∈N , c) where

N is the finite set of agents;

R is the finite set of resources;

P is the finite set of products;

A ∈ RR×P+ is the nonnegative technology matrix with at least one

positive element in each row and each column; arp is the number ofunits of resource r needed for the production of 1 unit of product p;

bi ∈ RR+ is the nonnegative resource vector of agent i ;

c ∈ RP+ is the vector of selling prices.





Sequencing games

Linear production games, definition

DefinitionLet (N ,R ,P ,A, (bi)i∈N , c) be a linear production situation. Thecorresponding linear production game (N , v) is defined by

v(S) = optimal value of the LP problem

max cT x

s.t. Ax ≤∑

i∈S bix ≥ 0,

for every S ⊆ N .





Sequencing games

Balancedness

Theorem (Owen (1975))Every linear production game is balanced.

ProofStep 1: Determine an optimal solution y∗ ∈ R

R for the dual problem forthe grand coalition N (shadow prices).Step 2: Price the resources of every agent via y∗.Step 3: The resulting vector z ∈ R

N is a core element (this vector iscalled an Owen vector).





Sequencing games

Owen vector, an example

ExampleReconsider the breweries:N = {1, 2, 3}; R = {hop, corn}; P = {light beer, dark beer};

A =( 4 56 2

)

; c = (68 52)T ;

b1 = (4 33)T ;b2 = (6 39)T ;b3 = (60 0)T ;

The optimal solution for the dual problem of N is y∗ = (8 6)T .

The Owen vector isz = (8 · 4 + 6 · 33, 8 · 6 + 6 · 39, 8 · 60 + 6 · 0) = (230, 282, 480).





Sequencing games

Owen vector, an example (continued)

Note that z = (230, 282, 480) is indeed a core element.

S 1 2 3 12 13 23 123v(S) 68 102 0 170 710 762 992

∑

i∈S zi 230 282 480 512 710 762 992

Why is z efficient?

z1 + z2 + z3 = (8 · 4 + 6 · 33) + (8 · 6 + 6 · 39) + (8 · 60 + 6 · 0)

= 8 · (4 + 6 + 60) + 6 · (33 + 39 + 0)

= optimal value dual problem for N

= optimal value primal problem for N

= v(N).





Sequencing games

Owen vector, an example (continued)

Why is, for example, coalition S = {1, 3} ‘happy’?

z1 + z3 = (8 · 4 + 6 · 33) + (8 · 60 + 6 · 0)

= 8 · (4 + 60) + 6 · (33 + 0)

= value objective function of dual problem for {1, 3} in y∗

≥ optimal value dual problem for {1, 3}

= optimal value primal problem for {1, 3}

= v(13).





Sequencing games

Assignment 1

TheoremEvery nonnegative and totally balanced cooperative game is a linearproduction game.

Assignment 1Prove this statement for N = {1, 2, 3}.





Sequencing games

1 Introduction

2 Preliminaries




6 Inventory games

7 Sequencing games





Sequencing games

Carpooling

ExampleThree employees of a firm consider the possibility of carpooling in orderto reduce their daily travel cost. Costs of driving a car from one employeeto another or to the firm are given by

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Sequencing games

Carpooling

Plans of carpooling are trees. So, optimal plans of carpooling areminimum cost spanning trees. Here the unique minimum cost spanningtree is {01, 12, 13} with total cost 18.

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Question: How to divide this total cost of 18?

Bird (1976): the vector (7, 5, 6) is a core element of the correspondingmcst game.





Sequencing games

Carpooling

Suppose that a fourth employee is asking whether he can join thecarpoolers 1, 2 and 3. The costs of driving from employee 4 to the otheremployees and the firm are given by

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Sequencing games

Carpooling

A minimum cost spanning tree is now {04, 24, 12, 13} with cost 20. TheBird rule proposes the vector (5, 6, 6, 3) as a way to divide the joint costs.However, player 2 is confronted with an increase in his share of the totalcost, so he will veto the entrance of player 4.

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Question: Is there a possibility of dividing the costs in a monotonic way?In game theoretical terms: Does every minimum cost spanning tree gamehave a PMAS?





Sequencing games

Definitions

A complete weighted graph is a tuple < N ′,w > where

i) N ′ = {0, 1, . . . , n};

ii) w : E → R+, where E = {S : S ⊆ N ′, |S | = 2}.

Elements of N ′ are called nodes. Node 0 is called the source andN = {1, . . . , n} the set of players. Elements of E are called edges and foran l ∈ E the nonnegative number w(l) represents the weight or cost ofedge l .

If w(l) ∈ {0, 1} for every l ∈ E the cost function w is called simple. Thecarrier Ca(w) of w is the set of edges with positive cost, i.e.Ca(w) = {l ∈ E : w(l) > 0}.

A subset Γ of E is called a network. The cost of network Γ isw(Γ) =

∑

l∈Γ w(l).





Sequencing games

Definitions

A path from i to j in Γ is a sequence of nodes i = i0, i1, . . . , ik = j suchthat {is , is+1} ∈ Γ for every s ∈ {0, . . . , k − 1}. A network Γ is aspanning network for S (S ⊆ N) if for every l ∈ Γ we have l ⊆ S ∪ {0}and if for every i ∈ S there is a path in Γ from i to 0.

The minimum cost spanning tree (mcst) game (N , c), corresponding to< N ′,w >, is defined by

c(S) = min{w(Γ) : Γ is a spanning network for S}

for every S ∈ 2N\{∅}.





Sequencing games

Mcst games

ExampleConsider the following complete weighted graph:

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The corresponding mcst game is given by

c(123) = 18;

c(12) = 12; c(13) = 13; c(23) = 15;

c(1) = 7; c(2) = 8; c(3) = 8.Henk Norde Operations Research Games




Sequencing games

Decomposition of mcst games

Note that

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and

c(123) = 18;c(12) = 12; c(13) = 13;c(23) = 15; c(1) = 7;c(2) = 8; c(3) = 8

= 5·

c(123) = 3;c(12) = 2; c(13) = 2;c(23) = 2; c(1) = 1;c(2) = 1; c(3) = 1

+

c(123) = 3;c(12) = 2; c(13) = 3;c(23) = 5; c(1) = 2;c(2) = 3; c(3) = 3.





Sequencing games

Decomposition theorem

TheoremLet w be a cost function with Ca(w) 6= ∅ and let c be the correspondingmcst game. Then there exists a sequence of simple cost functionsw1, . . . ,wk , with

Ca(w) = Ca(w1) ⊃ Ca(w2) ⊃ · · · ⊃ Ca(wk),

and positive numbers α1, . . . , αk such that

w =

k∑

j=1

αjwj .

Moreover, if c1, . . . , ck are the mcst games corresponding to w1, . . . ,wk

respectively, we have

c =

k∑

j=1

αjcj .





Sequencing games

Simple cost functions

Let w be a simple cost function and let S ∈ 2N\{∅} be a coalition. Twonodes i and j in S ∪ {0} are (w , S)-connected if there exists a sequenceof nodes i = i0, . . . , ik = j in S ∪ {0} with w({is , is+1}) = 0 for everys ∈ {0, . . . , k − 1}. A (w , S)-component of S ∪ {0} is a maximal subsetof S ∪ {0} with the property that any two nodes in this subset are(w , S)-connected. The number of (w , S)-components is denoted byn(w , S). Clearly the collection of (w , S)-components form a partition ofS ∪ {0}.

LemmaLet w be a simple cost function and let c be the corresponding mcstgame. Then we have

c(S) = n(w , S)− 1

for every S ∈ 2N\{∅}.





Sequencing games

Example

ExampleConsider the complete weighted graph < N ′,w > with N ′ = {0, . . . , 8}and simple cost function w specified by{l ∈ E : w(l) = 0} = {01, 23, 24, 34, 45, 67}. Let c be the correspondingmcst game. The edges with zero cost are depicted below.

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Note that c(N) = n(w ,N)− 1 = 4− 1 = 3. If we take for exampleS = {2, 3, 5, 6} we get that {0}, {2, 3}, {5} and {6} are all(w , S)-components. Therefore we also have c(S) = 3.





Sequencing games

Lemma for simple cost functions

LemmaLet w be a simple cost function and let c be the corresponding mcstgame. Then c has a PMAS.

ProofTwo ways to construct a PMAS:

1) Construct the scheme xed by dividing the cost of every(w , S)-component without 0 equally among the players.

2) Let π : N → {1, 2, . . . , n} be an ordering of the players. Define thescheme xπ by assigning the cost of every (w , S)-component without 0 tothe player in this component with the lowest index according to π.





Sequencing games

Example

Consider again the simple cost function

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1

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With π = 48372516 we get

xπN = (0, 0, 0, 1, 0, 0, 1, 1) and xπ2356 = (−, 0, 1,−, 1, 1,−,−).





Sequencing games

Theorem

TheoremEvery mcst game has a PMAS.

Proof

Let c be a mcst game corresponding to cost function w . Decompose w

as a positive combination of simple cost functions with decreasing carrier

w = α1w1 + α2w2 + · · ·+ αkwk

in such a way that the same decomposition is valid for the correspondingmcst games

c = α1c1 + α2c2 + · · ·+ αkck .

Let π : N → {1, . . . , n} be an ordering of the players. Then

xπ,w = α1xπ,w1 + α2x

π,w2 + · · ·+ αkxπ,wk

is a PMAS for c .Henk Norde Operations Research Games




Sequencing games

Two famous algorithms

Two famous algorithms for the determination of a mcst are

Prim’s algorithm (1957): In the first step form an edge of minimal costbetween a node and the source. In every subsequent step form an edge ofa node which is not connected with the source, directly or indirectly, andthe source or a node which is already connected, directly or indirectly,with the source.

Kruskal’s algorithm (1956): In the first step form an edge of minimalcost. In every subsequent step form an edge of minimal cost which doesnot form a cycle with the edges which have already been formed.

Claim

Prim → Bird rule for finding core element

Kruskal → Rule for finding a PMAS





Sequencing games

Example

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PrimFor N = {1, 2, 3}: first form 01, then 13, and finally 23;for S = {2, 3}: first form 02, then 23;etcetera.





Sequencing games

Example

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This yields the following Bird scheme:

S xS,1 xS,2 xS,3123 6 8 1312 6 17 −13 6 − 1323 − 17 81 6 − −2 − 17 −3 − − 18

Note that this scheme is not a PMAS.





Sequencing games

Example

Now we compute PMAS xπ for π = 123 (only for N). Note that

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SoxπN = (6, 6, 6)+?.

Player 1 stops paying!





Sequencing games

Example

Now

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SoxπN = (6, 6, 6) + (0, 2, 2)+?.

Player 3 stops paying!





Sequencing games

Example

... and

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SoxπN = (6, 6, 6) + (0, 2, 2) + (0, 5, 0).

Player 2 stops paying! In all subsequent steps costs have to be allocatedfor simple games where 1, 2 and 3 are all connected to 0.





Sequencing games

Example

So xπN = (6, 13, 8).

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Note that in every step of Kruskal’s algorithm there is one player whogets a connection to the source or to a player with a lower indexaccording to π. That player is paying the cost of that edge.

Moreover note that in this example player 2 is paying the cost of an edge‘far away’.





Sequencing games

Example

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Proceeding in the same way for all coalitions we get:

S xS,1 xS,2 xS,3123 6 13 812 6 17 −13 6 − 1323 − 17 81 6 − −2 − 17 −3 − − 18

Note that this scheme is a PMAS.





Sequencing games

Some references

As xπ is a PMAS for every order π, the average of these n! schemes is aPMAS as well. This value has received a lot of attention in the past:

Equal Remaining Obligation Rule (Feltkamp (1995));

The P-value (Branzei et al. (2004));

The ’folk’ solution for mcst games (Bergantinos, Vidal-Puga (2007),(2008), (2009)).





Sequencing games

The ERO rule

All players pay fractions of the costs of edges;

Each player pays for 1 edge in total;

Connected players have the same remaining obligation.





Sequencing games

Example ERO rule

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Initially, remaining obligations are (1, 1, 1).In step 1 edge 23 is formed, players 2 and 3 pay half of the costs.Remaining obligations are (1, 1

2 ,12 ).

In step 2 edge 13 is formed, player 1 pays 2/3 of the costs, players 2 and3 1/6 of the costs.Remaining obligations are ( 13 ,

13 ,

13).

In step 3 edge 01 is formed. All players pay 1/3 of the costs.So, ERO=( 293 ,

263 ,

263 ).





Sequencing games

Assignment 2

Assignment 2Determine for any b ≥ 0 the ERO solution for the mcst problem below.

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Sequencing games

Monotonic mcst games

Consider the following cost function:

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i

i

i

i

i

0

1 2 3

4 5 6

All depicted edges have cost 1, all other edges have cost 10.Coalitions are allowed to use nodes outside the coalition.Then c(123) = 5, c(12) = c(13) = c(23) = 3.So 2c(123) > c(12) + c(13) + c(23).So (N , c) is not totally balanced (and hence no PMAS exists).





Sequencing games

1 Introduction

2 Preliminaries




6 Inventory games

7 Sequencing games





Sequencing games

Notation and definitions

An (undirected) graph G is a pair G = (N ,E ) where

• N : Vertex set of G

• E : Edge set of G

Edges will be denoted by ij instead of {i , j}.

For S ⊆ N ,

• G [S ] = (S ,ES): the subgraph of G induced by S ⊆ N whereES = {ij ∈ E : i ∈ S , j ∈ S}

• G = (N , E): the complement of G whereE = {ij : i , j ∈ N , i 6= j , ij /∈ E}





Sequencing games


A graph is complete if for all i , j ∈ N , ij ∈ E .

A clique in G is an S ⊆ N such that G [S ] is complete.A clique is a maximum clique if it has a maximum size among allcliques.The clique number of a graph is the number of elements of amaximum clique.

An independent set is an S ⊆ N such that G [S ] has no edges.





Sequencing games


A coloring of G = (N ,E ) is a mapping

c : N → N

such that c(i) 6= c(j) for every {i , j} ∈ E .

The chromatic number X (G) of a graph G is the smallest number ofcolors needed to color the vertices of G so that no two adjacentvertices share the same color.

A graph G = (N ,E ) is perfect iff the clique number of every inducedsubgraphs equals the chromatic number of that subgraph.

A graph G = (N ,E ) is complete r-partite iff it can be decomposed in r

pairwise disjoint independent sets and any pair of vertices in twodifferent independent sets are adjacent.





Sequencing games


A directed graph (digraph) is a tuple (V ,D) where

• V : Vertex set of G

• D: Arc set of G

A rooted tree is a digraph for which a special root r ∈ V exists such, thatfor each vertex v ∈ V there is a unique directed path from r to v .





Sequencing games

Minimum coloring game

The minimum coloring game on G is the cost game (N , cG ) where

cG : 2N → R

is defined as cG (S) := X (G [S ]) for all S ⊆ N .





Sequencing games

Minimum coloring game

ExampleLet G be the graph

• • •

1 2 3

Then, the corresponding minimum coloring game is

S ∅ 1 2 3 12 13 23 123

cG (S) 0 1 1 1 2 1 2 2

Observe:

(N , cG ) is totally balanced, since G is perfect (Deng et al. (2000))

C (cG ) = convhull{(1, 1, 0), (0, 1, 1)} since 12 and 23 are maximumcliques (Okamoto (2003a))

(N , cG ) is concave since G is complete r-partite (Okamoto (2003b))





Sequencing games

Graphs and games

Totally balancedness is not enough to obtain ”dynamic stability”.Example

1

2 3

G11

2 3

4 5 G2

Let (N , cG1) be the coloring game w.r.t. G1. Then (1, 1, 0) unique coreelement.Let (N , cG2) be the coloring game w.r.t. G2. Then (0, 0, 1, 1, 1) uniquecore element.So, by including player 4 and 5 player 3 has to pay 1. Hence, thisexample lacks a form of dynamic stability. To be more precise, there doesnot exist a PMAS.





Sequencing games

We have the following relations:

Class of concave games∩

Class of PMAS games∩

Class of totally balanced games.

QuestionWhich graphs characterize the class of coloring games with a PMAS?

Of course, this class of graphs constitutes a subclass of the class ofperfect graphs, and it should contain the class of complete r-partitegraphs.





Sequencing games

Graphs and games

Research in the same spirit in which graphs characterize a class of ORgames with a game theoretical property:

Traveling salesman games (Herer and Penn (1995)):A graph G is 1 sum of K4 and outer planar graphs if and only if thecorresponding Steiner TS game is submodular for any cost function.

Chinese postman games (Granot, Hamers, Tijs (1999)):A graph G is Eulerian (weakly cyclic) if and only if the correspondingChinese postman game is balanced (submodular, totally balanced) forany cost function.

Minimum vertex covering games (Okamoto (2003)):G contains no subgraph isomorphic to K3 or P3 if and only if theminimum vertex covering game is submodular.





Sequencing games

Quasi-threshold graphs

Quasi-threshold graphs are defined in a constructive way: they are thegraphs that can be formed, starting from one-vertex graphs, by thefollowing operations:

1) adding a new vertex that is adjacent to all vertices of aquasi-threshold graph;

2) taking the disjoint union of two quasi-threshold graphs.





Sequencing games


Yan et al. (1996) (in fact, already in Wolk (1965)) characterizedquasi-threshold graphs by the fact that they are the graphs induced by arooted forest. In the induced graph there is an edge between two verticesi and j iff there is a directed path in F between i and j .Example Consider the rooted forest F = (N ,D):

1

2

3

4 5

6

7

8

9

Observe, interchanging position 2 and 3 in F yields the samequasi-threshold graph.





Sequencing games


Solution to non-unicity representation:make a forest in which no vertex has precisely one follower.

Example continued

1

2, 3

4 5

6

7, 8, 9

In the induced graph there is an edge between two vertices i and j iffthere is a directed path in F between i and j or i and j are in the samevertex.





Sequencing games


PropositionLet (N ,E ) be a graph. Then (N ,E ) is a quasi-threshold graph if andonly if a rooted forest F = (V ,D) exists, and for every v ∈ V anonempty subset Mv of N , such that

i) F has no vertices with one follower;

ii) (Mv )v∈V is a partition of N ;

iii) E = {ij : i , j ∈ N , i 6= j , v(i) = v(j) or there is a directed path in F

from v(i) to v(j) or vice versa}.Here v(k) denotes, for every k ∈ N , the unique vertex v ∈ V withk ∈ Mv .

We will refer to the triple (V ,D, (Mv )v∈V ) as the as the rooted forestrepresentation of quasi-threshold graph (N ,E ).





Sequencing games

(P4, 2K2)-free graphs

Yan et al. (1996) show that the class of quasi-threshold graphs coincideswith the class of (P4,C4)-free graphs, i.e. the graphs which do not havesubgraphs isomorphic to P4 or C4.

1

2 3

4

P4

1

2 3

4

C4

1

2 3

4

2K2

The complements of P4 and C4 are isomorphic to P4 and 2K2

respectively.

The complements of quasi-threshold graphs, i.e. the (P4, 2K2)-freegraphs, are used to characterize coloring games with a PMAS.





Sequencing games


Obviously, the rooted forest representation of a quasi-threshold graph canalso be used to describe the complementary (P4, 2K2)-free graph G . In G

there are no edges between i and j iff i and j belong to the same vertexor to different vertices between which a directed path exists.Example

1

2, 3

4 5

6

7, 8, 9

I1 I2

I3

I4

The maximal independent sets are: I1 = {1, 2, 3, 4}, I2 = {1, 2, 3, 5},I3 = {1, 6}, and I4 = {7, 8, 9}.





Sequencing games


Observe that the class of r -partite graphs is a subclass of the class of(P4, 2K2)-free graphs.

The rooted forest representation of a r -partite graph consists of theunion of disjoint vertices.

ExampleThe rooted tree representation of the 3-partite graph with independentsets I1 = {1}, I2 = {2, 3} and I3 = {4, 5, 6}.

1 4, 5, 62, 3





Sequencing games

PMAS existence

TheoremLet G = (N ,E ) be a graph and let (N , cG ) be the correspondingminimum coloring game. If P(N , cG ) 6= ∅ then G is (P4, 2K2)-free.

ProofLet x ∈ P(N , cG ) and suppose that G is not (P4, 2K2)-free. Let S ∈ 2N

be such that |S | = 4 and G [S ] is either P4 or 2K2. Without loss ofgenerality we can assume again that S = {1, 2, 3, 4}, 12 ∈ E , 34 ∈ E ,13 /∈ E , 14 /∈ E , and 24 /∈ E . Note that cG ({1, 2, 4}) = cG ({1, 3, 4}) = 2and cG ({1, 3}) = cG ({1, 4}) = cG ({2, 4}) = 1. By efficiency andmonotonicity of x we get

4 = cG ({1, 2, 4}) + cG ({1, 3, 4})= x{1,2,4},1 + x{1,2,4},2 + x{1,2,4},4 + x{1,3,4},1 + x{1,3,4},3 + x{1,3,4},4≤ x{1,4},1 + x{2,4},2 + x{2,4},4 + x{1,3},1 + x{1,3},3 + x{1,4},4= cG ({1, 3}) + cG ({1, 4}) + cG ({2, 4})= 3, which yields a contradiction.





Sequencing games

PMAS existence

PropositionLet G = (N ,E ) be a (P4, 2K2)-free graph with rooted forestrepresentation (V ,D, (Mv )v∈V ), and let (N , cG ) be the correspondingminimum coloring game. For every S ∈ 2N\{∅} we havecG (S) = |max(VS )|, where VS := {v ∈ V : Mv ∩ S 6= ∅}.

Example

1

2, 3

4 5

6

7, 8, 9

Then cG{1, 2, 3, 7, 8} = 2, cG ({1, 6}) = 1 and cG (N) = 4.Henk Norde Operations Research Games




Sequencing games

PMAS existence

TheoremLet G = (N ,E ) be a graph and let (N , cG ) be the correspondingminimum coloring game. Then, P(N , cG ) 6= ∅ if and only if G is a(P4, 2K2)-free graph.

Proof:Proof of ”only if”-part has already been provided.Assume that G is (P4, 2K2)-free and consider its rooted forestrepresentation.

1 e

2, 3 c

4 a 5 b

6 d

7, 8, 9 f





Sequencing games

PMAS existence

In each vertex chose an order of the players. Take, for example

1 eσe = (1)

2, 3 cσc = (3, 2)

4 aσa = (4) 5b σb = (5)

6d σd = (6)

7, 8, 9f σf = (8, 7, 9)

Define the {0, 1}-valued scheme x = (xS,i )S∈2N\{∅},i∈S by

xS,i =

1 if v(i) ∈ max(VS ) andσv(i)(i) > σv(i)(j) for every j ∈ (S ∩Mv(i))\{i}

0 otherwise,

For example, take S = {1, 2, 3, 7, 8}. Then xS,i = 1 if i = 3, 8 andxS,i = 0 otherwise.





Sequencing games

Structure of PMAS-es

PropositionLet G = (N ,E ) be a (P4, 2K2)-free graph with rooted forestrepresentation (V ,D, (Mv )v∈V ), let (N , cG ) be the correspondingminimum coloring game. Then the restricted game to a coalition ofplayers that are all contained in the same vertex of the rooted forestrepresentation, yields a unit game, i.e.,

cG (S) = 1 for every S ∈ 2Mv \{∅}.

Example

1 e

2, 3 c

4 a 5b

6d

7, 8, 9f

GHenk Norde Operations Research Games




Sequencing games


PropositionLet G = (N ,E ) be a (P4, 2K2)-free graph with rooted forestrepresentation (V ,D, (Mv )v∈V ), let (N , cG ) be the correspondingminimum coloring game and let x ∈ P(N , cG ). Theny = (yS,i )S∈2Mv \{∅},i∈S , defined by yS,i = xS,i for every S ∈ 2Mv\{∅} andi ∈ S , is such that y ∈ P(Mv , 1Mv

).

So, the restriction of a PMAS of the coloring game to a coalition whichplayers coincide with the set of players of one vertex, generates a PMASof the unit game on this coalition.





Sequencing games


TheoremLet G = (N ,E ) be a (P4, 2K2)-free graph with rooted forestrepresentation (V ,D, (Mv )v∈V ) and let (N , cG ) be the correspondingminimum coloring game. Let y v ∈ P(Mv , 1Mv

) for every v ∈ V . Thenthere is a unique x ∈ P(N , cG ) such that xS,i = y v

S,i for every v ∈ V ,

S ∈ 2Mv \{∅} and i ∈ S , and this scheme x is given by

xS,i =

{

yv(i)S∩Mv(i),i

if v(i) ∈ max(VS )

0 if v(i) /∈ max(VS )

for every S ∈ 2N\{∅} and i ∈ S .





Sequencing games


Example

1 e

2, 3 c

4 a 5b

6d

7, 8, 9f

The only coalitions that have a non-trivial PMAS are {2, 3} and {7, 8, 9}.





Sequencing games


Take as PMAS-es of {2, 3} and {7, 8, 9}:

S 2 3 7 8 9{2} 1 −{3} − 1{2, 3} 0.5 0.5{7} 1 − −{8} − 1 −{9} − − 1{7, 8} 1 0 −{7, 9} 1 − 0{8, 9} − 1 0{7, 8, 9} 1 0 0

Then for coalition {1, 2, 5, 7, 8} the scheme x is generated as follows:





Sequencing games


Observe that cG ({1, 2, 5, 7, 8}) = 2. Further, player 1 and 2 are not invertices that are in maxVS . Hence, they receive both 0. The otherplayers are in maxVS , hence their payoff has to be read from theirPMAS-es of their unit game.

S 5 7 8 9{5} 1{7, 8} 1 0 −

Hence, xS is:

S 1 2 3 4 5 6 7 8 9{1, 2, 5, 7, 8} 0 0 − − 1 − 1 0 −





Sequencing games


TheoremLet G = (N ,E ) be a (P4, 2K2)-free graph with rooted forestrepresentation (V ,D, (Mv )v∈V ), let (N , cG ) be the correspondingminimum coloring game, and let x ∈ P(N , cG ).Then x is integer-valued if and only if there is a collection of ordersσ = (σv )v∈V such that x = xσ.





Sequencing games

Assignment 3

Assignment 3Construct a (P4, 2K2)-free graph G such that the correspondingminimum coloring game (N , cG ) admits precisely 72 integer-valuedPMAS-es. What is the minimum number of players needed?





Sequencing games

1 Introduction

2 Preliminaries




6 Inventory games

7 Sequencing games





Sequencing games

One firm ordering problem

Consider a firm that faces a (deterministic) demand of d units of somegood per unit of time. It is not allowed to run out of stock and the leadtime (time between placement of an order and arrival of the goods) isassumed to be zero. The firm faces two kinds of costs:

ordering costs: for each order the firm places it has to pay a fixedcost of a, independent of the quantity ordered;

holding costs: the costs of keeping one unit of good in stock for oneunit of time is h.

Question:What is the optimal ordering quantity of the firm?





Sequencing games

Optimal ordering quantity

Let Q denote the ordering quantity. Then time between two successiveorders is Q/d , or, equivalently, there are on average d/Q orders per unitof time. The average inventory is Q/2, so the average holding costs pertime unit equal hQ/2. So the average inventory costs per unit of timeare:

AC (Q) = a ·d

Q+ h ·

Q

2.

The optimal ordering size and the optimal number of orders per time unitare

Q∗ =

√

2ad

hand m∗ =

d

Q∗=

√

dh

2a.

The minimal average costs are

AC (Q∗) = 2am∗.





Sequencing games

n-firm inventory situation

DefinitionAn n-firm inventory situation is a tuple (N , d , h, a), where

N is the finite set of firms;

d ∈ RN++ is the vector of demand levels;

h ∈ RN++ is the vector of holding cost parameters;

a > 0 is the cost per order.





Sequencing games

Inventory game

What is the optimal ordering decision for coalition S ⊆ N?First note that in optimum firms always synchronize their cycle lengths,i.e. there is a constant c such that

Qi

di= c

for every i ∈ S . Here Qi refers to the ordering quantity for firm i inoptimum. Now average cost per unit of time is a function of c :

AC (c) = a ·1

c+∑

i∈S

hi ·Qi

2

= a ·1

c+

c

2

∑

i∈S

hi · di .





Sequencing games

Inventory game

Optimal value of c is

c∗ =

√

2a∑

i∈S hidi.

Optimal ordering quantities for i ∈ S are given by

Q∗i =

√

2ad2i

∑

i∈S hidi.

The optimal number of orders per time unit is

mS =1

c∗=

√

∑

i∈S hidi

2a=

√

∑

i∈S

m2i .

Minimal average costs are given by

c(S) = AC (c∗) = 2amS = 2a

√

∑

i∈S

m2i .





Sequencing games

Theorem

DefinitionLet (N , d , h, a) be an inventory situation. The corresponding inventorygame (N , c) is defined by

c(S) = 2a

√

∑

i∈S

m2i ,

where

mi =

√

hidi

2a

for every i ∈ N .

Theorem (Meca et al. (1999))Every inventory game is concave.





Sequencing games

1 Introduction

2 Preliminaries




6 Inventory games

7 Sequencing games





Sequencing games

sequencing situations

In a 1-machine sequencing situation there is a queue of playersN = {1, 2, . . . , n}, each with one job, in front of a machine. Each jobmust be processed by the machine, and the processing time of job i is pi .The players are initially ordered according to σ0 : N → {1, 2, . . . , n} andfor every player there is a cost of αi per unit of time spent in the system.The corresponding sequencing situation is denoted by (N , σ0, p, α) withp, α ∈ R

N++.

Theorem (Smith (1956))A processing order that minimizes total costs is obtained by ordering thejobs in decreasing order with respect to the urgency index ui =

αi

pi.





Sequencing games

Example

Consider the following sequencing situation:

M 1α1 = 6p1 = 3u1 = 2

2α2 = 8p2 = 2u2 = 4

3α3 = 4p3 = 4u3 = 1

4α4 = 3p4 = 1u4 = 3

Total costs are 6 · 3 + 8 · 5 + 4 · 9 + 3 · 10 = 124.Optimal processing order is

M 2 4 1 3

Optimal costs are 8 · 2 + 3 · 3 + 6 · 6 + 4 · 10 = 101.





Sequencing games

Terminology

Let σ : N → {1, . . . , n} be an order and i ∈ N . Then

C (σ, i) =∑

j∈N :σ(j)≤σ(i)

pj

is the completion time of job i under order σ.For coalition S ⊆ N and order σ let

CS(σ) =∑

i∈S

αiC (σ, i)

denote the total costs of coalition S .





Sequencing games

Sequencing game

Let S ⊆ N be a coalition and i ∈ N . Then

P(σ, i) = {j ∈ N |σ(j) < σ(i)}

is the set of predecessors of player i with respect to σ.An order σ is called admissible for S if

P(σ, j) = P(σ0, j)

for every j ∈ N\S . The collection of all admissible orders for S is denotedby A(S).

The corresponding sequencing game (N , v) is defined by

v(S) = maxσ∈A(S)

(CS (σ0)− CS(σ))

for every S ⊆ N .Henk Norde Operations Research Games




Sequencing games

Example

Consider coalition S = {1, 2, 4}.

M 1α1 = 6p1 = 3u1 = 2

2α2 = 8p2 = 2u2 = 4

3α3 = 4p3 = 4u3 = 1

4α4 = 3p4 = 1u4 = 3

Admissible orders for S are 1234 and 2134. So optimal order for S is

M 2 1 3 4

Hence v(124) = C124(1234)− C124(2134) = 88− 76 = 12.





Sequencing games

Example

M 1α1 = 6p1 = 3u1 = 2

2α2 = 8p2 = 2u2 = 4

3α3 = 4p3 = 4u3 = 1

4α4 = 3p4 = 1u4 = 3

Proceeding in the same way for all coalitions we get

S 1 2 3 4 12 13 14 23 24 34 123 124 134 234 1234v(S) 0 0 0 0 12 0 0 0 0 8 12 12 8 8 23





Sequencing games

Neighbor jobs

Consider the interchange of two neighboring jobs:

i j j i→

The change of total costs equals

αjpi − αipj .

This change is positive ifuj > ui .

The cost savings, attainable by players i and j , equal

gij = max{0, αjpi − αipj}.





Sequencing games

connected coalitions

A coalition S is connected with respect to order σ if for all i , j ∈ S andfor all k ∈ N with σ(i) < σ(k) < σ(j) we have k ∈ S .

TheoremLet (N , σ0, p, α) be a sequencing situation and let (N , v) be thecorresponding sequencing game. Then for any coalition S that isconnected with respect to σ0 we have

v(S) =∑

i ,j∈S:σ0(i)<σ0(j)

gij .





Sequencing games

Theorem

Any coalition T can be partitioned into maximally connected subsets(components) with respect to σ. This collection of components isdenoted by T\σ.

TheoremLet (N , σ0, p, α) be a sequencing situation and let (N , v) be thecorresponding sequencing game. Then for any coalition T we have

v(T ) =∑

S∈T\σ0

v(S).

TheoremEvery sequencing game is convex.





Sequencing games

Equal Gain Splitting rule

As a sequencing game is convex there is an abundance of core elements.One ‘nice’ core element is obtained by applying the Equal Gain Splittingrule: go from the initial order to the optimal order for N by stepwiseinterchanging two misplaced neighboring jobs and divide the profits ofsuch an interchange evenly among the two jobs.





Sequencing games

Example

M 1α1 = 6p1 = 3u1 = 2

2α2 = 8p2 = 2u2 = 4

3α3 = 4p3 = 4u3 = 1

4α4 = 3p4 = 1u4 = 3

Cost savings 12, to be divided by players 1 and 2.

M 2 1 3 4

Cost savings 8, to be divided by players 3 and 4.

M 2 1 4 3

Cost savings 3, to be divided by players 1 and 4.EGS rule provides core element (7.5, 6, 4, 5.5).





Sequencing games

Thanks

Thank you for your attention


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