Introduction
Preliminaries
Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory games
Sequencing games
Operations Research Games
Henk Norde
Tilburg University, The Netherlands
Sevilla, March 18, 2011
Henk Norde
Operations Research Games
Introduction
Preliminaries
Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory 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
Henk Norde
Operations Research Games
Introduction
Preliminaries
Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory games
Sequencing games
1
Introduction
2
Preliminaries
3
Linear production games
4
Minimum cost spanning tree games
5
Minimum coloring games
6
Inventory games
7
Sequencing games
Henk Norde
Operations Research Games
Introduction
Preliminaries
Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory games
Sequencing games
Introduction
Operations research deals with optimisation problems in which one
decision 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 optimisation
problem is considered where more players have control over part of the
structure underlying the problem.
Basic reference:
Peter Borm, Herbert Hamers, Ruud Hendrickx (2001), Operations
Research Games: A Survey. TOP 9, no. 2, 139-216.
Henk Norde
Operations Research Games
Introduction
Preliminaries
Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory games
Sequencing games
1
Introduction
2
Preliminaries
3
Linear production games
4
Minimum cost spanning tree games
5
Minimum coloring games
6
Inventory games
7
Sequencing games
Henk Norde
Operations Research Games
Introduction
Preliminaries
Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory 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 .
Henk Norde
Operations Research Games
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Notation and notions
The core of a cost game (N, c) is the set
X
X
C (c) := {x ∈ RN :
xi = c(N),
xi ≤ c(S) for all S ⊂ N}
i ∈N
i ∈S
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).
Henk Norde
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Notation and notions
Given a cost game (N, c), the table
x = (xiS )∅6=S∈2N ,i ∈S
is said to be a Population Monotonic Allocation Scheme (PMAS)
(Sprumont (1990)) if
P
(i) efficiency: For all S ⊆ N, S 6= ∅, i ∈S xiS = c(S).
(ii) monotonicity: For all S ⊆ T and for all i ∈ S, xiS ≥ xiT .
P(N, c) denotes the collection of all PMAS-es of (N, c).
Observe that totally balancedness of (N, c) is a necessary condition to
have PMAS, and concavity is a sufficient condition to have a PMAS.
Existence of PMAS provides a form of dynamic stability.
Henk Norde
Operations Research Games
Introduction
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Notation and notions
Example Consider the game (N, 1) where N = {1, 2, 3, 4}. Let
σ = (2, 3, 1, 4) be an order. This order provides a PMAS:
S
{1}
{2}
{3}
{4}
{1, 2}
{1, 3}
{1, 4}
{2, 3}
{2, 4}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
Henk Norde
1
1
−
−
−
0
0
1
−
−
−
0
0
0
−
0
2
−
1
−
−
1
−
−
1
1
−
1
1
−
1
1
3
−
−
1
−
−
1
−
0
−
1
0
−
1
0
0
4
−
−
−
1
−
−
0
−
0
0
−
0
0
0
0
Operations Research Games
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Notation and notions
An order on T ∈ 2N \{∅} is a bijection from T to {1, . . . , |T |}. The
collection of all orders on T is denoted by ΣT .
Theorem
Let T ∈ 2N \{∅}.
σ
(i) Let σ ∈ ΣT . Define the scheme y σ = (yS,i
)S∈2T \{∅},i ∈S by
σ
yS,i
=
1
0
if σ(i) < σ(j) for all j ∈ S\{i}
otherwise
for every S ∈ 2T \{∅} and i ∈ S. Then y σ ∈ P(T , 1T ).
(ii) If y ∈ P(T , 1T ) and y is integer-valued, then there exists an order
σ ∈ ΣT such that y = y σ .
Henk Norde
Operations Research Games
Introduction
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Notation and notions
Example
Consider the game (N, 1) where N = {1, 2, 3, 4}.
S
{1}
{2}
{3}
{4}
{1, 2}
{1, 3}
{1, 4}
{2, 3}
{2, 4}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4}
1
1
−
−
−
0.5
0.5
0.5
−
−
−
0
0
0
−
0
Henk Norde
2
−
1
−
−
0.5
−
−
0.5
0.5
−
0.5
0.5
−
0.5
0.5
3
−
−
1
−
−
0.5
−
0.5
−
0.5
0.5
−
0.5
0.5
0.5
4
−
−
−
1
−
−
0.5
−
0.5
0.5
−
0.5
0.5
0
0
Operations Research Games
Introduction
Preliminaries
Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory games
Sequencing games
1
Introduction
2
Preliminaries
3
Linear production games
4
Minimum cost spanning tree games
5
Minimum coloring games
6
Inventory games
7
Sequencing games
Henk Norde
Operations Research Games
Introduction
Preliminaries
Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory games
Sequencing games
A linear optimization problem
Consider a brewery producing two types of beer, a light beer and a dark
beer.
For the production of one barrel of light beer 4 kg hop is needed and
6 kg of corn, for the production of one barrel of dark beer 5 kg of
hop and 2 kg of corn.
The brewery sells light beer with a profit of 68 per barrel and dark
beer 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
s.t.
z
= 68x1 + 52x2
4x1 + 5x2
6x1 + 2x2
x1 , x2
Henk Norde
≤ 70
≤ 72
≥ 0.
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Graphical solution
This linear optimization problem can be solved graphically:
x2
rB = (0, 14)
c
c
cc p p p
c
pp p p p p pp p B
c
c
c
c
p p pp pp p 6x1 + 2x2 = 72
c
c
c
c
c
c
p p p pp p pBp p
cc
cpc
cc
c
c
pp p Bp p rC = (10, 6)
pppc
c
c
c
c
c
ppp ppp
c
c
c
c
ppc
c
c
c
c
c
pc
pc
c
c
c
c
c
Bp
pc
c
cc
cc
cc
cc
pc
c
c
pp pp p p Bp ppp ppc
c
c
c
c
c
c
p p p pp p p pp p c4x1 + 5x2 = 70
c
c
c
c
c
c
cccc
c
c
p pBp rp p p p p p c
rc
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
p D =p (12, 0)
x1
A = (0, 0)
We can ‘read’ the optimal solution from this picture: x1 = 10, x2 = 6
and z = 992.
Henk Norde
Operations Research Games
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Dual problem
The corresponding dual optimization problem is:
minimize w
s.t.
=
70y1 +
4y1 +
5y1 +
y1 , y2
72y2
6y2
2y2
≥
≥
≥
68
52
0.
p p pp pp p
\
p pp pp pp
p p pp p\
p p pp ppp pp p pp pp pp pp pp pp pp p p pp pp p
p pp\
pp pp p p ppp pp p p pp pp p p pp pp p p pp p
\p pprp p pp pp p pp pp p p pp p p p pp pp
(8, 6) P
p p pp ppP
pp p
P
p p pP
p pp p pp pppp pp
pp
The optimal solution is y1 = 8, y2 = 6 (shadow prices of hop and corn)
and w = 992.
Henk Norde
Operations Research Games
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More breweries
Now consider a situation with three breweries, all operating on the same
market 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!
Henk Norde
Operations Research Games
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Many LP problems
The LP problem for brewery 1 reads:
maximize
s.t.
=
z
68x1 +
4x1 +
6x1 +
x1 , x2
52x2
5x2
2x2
≤
≤
≥
4
33
0.
If, for example, breweries 1 and 3 cooperate, their LP problem becomes:
maximize
s.t.
z
=
68x1 +
4x1 +
6x1 +
x1 , x2
52x2
5x2
2x2
≤
≤
≥
4
33
0.
+
+
60
0
+
+
6
39
+
+
60
0
If all breweries cooperate we get
maximize
s.t.
z
=
68x1 +
4x1 +
6x1 +
x1 , x2
52x2
5x2
2x2
Henk Norde
≤
≤
≥
4
33
0.
Operations Research Games
=
=
70
72
Introduction
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Linear production games
Minimum cost spanning tree games
Minimum coloring games
Inventory games
Sequencing games
... and many dual problems
The dual problem for brewery 1 is
minimize
s.t.
w
=
4y1
4y1
5y1
y1 ,
+
+
+
y2
33y2
6y2
2y2
64y1 +
4y1 +
5y1 +
y1 , y2
33y2
6y2
2y2
70y1 +
4y1 +
5y1 +
y1 , y2
72y2
6y2
2y2
≥
≥
≥
68
52
0,
≥
≥
≥
68
52
0,
≥
≥
≥
68
52
0.
for breweries 1 and 3
minimize
s.t.
w
=
and for all breweries
minimize
s.t.
w
=
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Linear production games
Solving all these maximization problems (or, better, all minimization
problems) we get the following linear production game:
S
1
v (S) 68
2 3
102 0
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12 13 23 123
170 710 762 992
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Linear production situations
Definition
A 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 of
units 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.
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Linear production games, definition
Definition
Let (N, R, P, A, (bi )i ∈N , c) be a linear production situation. The
corresponding linear production game (N, v ) is defined by
v (S)
=
optimal value of the LP problem
max c T x
s.t.
P
Ax ≤ i ∈S bi
x ≥ 0,
for every S ⊆ N.
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Balancedness
Theorem (Owen (1975))
Every linear production game is balanced.
Proof
Step 1: Determine an optimal solution y ∗ ∈ RR for the dual problem for
the grand coalition N (shadow prices).
Step 2: Price the resources of every agent via y ∗ .
Step 3: The resulting vector z ∈ RN is a core element (this vector is
called an Owen vector).
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Owen vector, an example
Example
Reconsider the breweries:
N = {1, 2, 3}; R = {hop, corn}; P = {light beer, dark beer};
4 5 A=
; c = (68 52)T ;
6 2
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 is
z = (8 · 4 + 6 · 33, 8 · 6 + 6 · 39, 8 · 60 + 6 · 0) = (230, 282, 480).
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Owen vector, an example (continued)
Note that z = (230, 282, 480) is indeed a core element.
S
v
(S)
P
i ∈S zi
1
2
3
12 13 23 123
68 102 0 170 710 762 992
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).
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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).
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Assignment 1
Theorem
Every nonnegative and totally balanced cooperative game is a linear
production game.
Assignment 1
Prove this statement for N = {1, 2, 3}.
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Introduction
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1
Introduction
2
Preliminaries
3
Linear production games
4
Minimum cost spanning tree games
5
Minimum coloring games
6
Inventory games
7
Sequencing games
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Carpooling
Example
Three employees of a firm consider the possibility of carpooling in order
to reduce their daily travel cost. Costs of driving a car from one employee
to another or to the firm are given by
7
2i
3i
A@
@ 5
6
A@
A @i 8A
1 8
A
7 A Ai
0
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Carpooling
Plans of carpooling are trees. So, optimal plans of carpooling are
minimum cost spanning trees. Here the unique minimum cost spanning
tree is {01, 12, 13} with total cost 18.
7
2i
3i
A@
@ 5
6
A@
A @i 8A
1 8
A
7 A Ai
0
Question: How to divide this total cost of 18?
Bird (1976): the vector (7, 5, 6) is a core element of the corresponding
mcst game.
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Carpooling
Suppose that a fourth employee is asking whether he can join the
carpoolers 1, 2 and 3. The costs of driving from employee 4 to the other
employees and the firm are given by
7
2i
3i
PP
6
P
@
A@
PP
@ 7
PP
A @5
P 6@
PP@
A @i @i
P
8A
1 4
7
A
7 8
A 3
Ai
0
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Carpooling
A minimum cost spanning tree is now {04, 24, 12, 13} with cost 20. The
Bird 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 total
cost, so he will veto the entrance of player 4.
7
2i
3i
PP
6
P
A@
PP
@ 7
@
PP
A @5
P 6@
PP@
A @i @i
P
8A
1 4
7
A
7 8
A 3
Ai
0
Question: Is there a possibility of dividing the costs in a monotonic way?
In game theoretical terms: Does every minimum cost spanning tree game
have a PMAS?
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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 and
N = {1, . . . , n} the set of players. Elements of E are called edges and for
an l ∈ E the nonnegative number w (l) represents the weight or cost of
edge l.
If w (l) ∈ {0, 1} for every l ∈ E the cost function w is called simple. The
carrier Ca(w ) of w is the set of edges with positive cost, i.e.
Ca(w ) = {l ∈ E : w (l) > 0}.
A subsetP
Γ of E is called a network. The cost of network Γ is
w (Γ) = l∈Γ w (l).
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Definitions
A path from i to j in Γ is a sequence of nodes i = i0 , i1 , . . . , ik = j such
that {is , is+1 } ∈ Γ for every s ∈ {0, . . . , k − 1}. A network Γ is a
spanning 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 \{∅}.
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Mcst games
Example
Consider the following complete weighted graph:
7
2i
3i
A@
@ 5
6
A@
A @i 8A
1 8
A
7 A Ai
0
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.
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Decomposition of mcst games
Note that
7
1
2i
3i
2i
3i
@
A@
A
@ 5
@ 1
6
1
A@
A@
A @i A @i +
8A
1A
1 8
= 5·
1 1
A 7 A
1 A A Ai
Ai
0
0
2
2i
3i
A@
@ 0
1
A@
A @i 3A
1 3
A
2 A Ai
0
and

c(123) = 18;



c(12) = 12; c(13) = 13;
 c(23) = 15; c(1) = 7;


c(2) = 8; c(3) = 8


c(123) = 3;
c(123) = 3;






c(12) = 2; c(13) = 2;
c(12) = 2; c(13) = 3;
= 5·
+
 c(23) = 2; c(1) = 1;
 c(23) = 5; c(1) = 2;




c(2) = 1; c(3) = 1
c(2) = 3; c(3) = 3.
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Decomposition theorem
Theorem
Let w be a cost function with Ca(w ) 6= ∅ and let c be the corresponding
mcst game. Then there exists a sequence of simple cost functions
w1 , . . . , wk , with
Ca(w ) = Ca(w1 ) ⊃ Ca(w2 ) ⊃ · · · ⊃ Ca(wk ),
and positive numbers α1 , . . . , αk such that
w=
k
X
αj wj .
j=1
Moreover, if c1 , . . . , ck are the mcst games corresponding to w1 , . . . , wk
respectively, we have
k
X
c=
αj cj .
j=1
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Simple cost functions
Let w be a simple cost function and let S ∈ 2N \{∅} be a coalition. Two
nodes i and j in S ∪ {0} are (w , S)-connected if there exists a sequence
of nodes i = i0 , . . . , ik = j in S ∪ {0} with w ({is , is+1 }) = 0 for every
s ∈ {0, . . . , k − 1}. A (w , S)-component of S ∪ {0} is a maximal subset
of S ∪ {0} with the property that any two nodes in this subset are
(w , S)-connected. The number of (w , S)-components is denoted by
n(w , S). Clearly the collection of (w , S)-components form a partition of
S ∪ {0}.
Lemma
Let w be a simple cost function and let c be the corresponding mcst
game. Then we have
c(S) = n(w , S) − 1
for every S ∈ 2N \{∅}.
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Example
Example
Consider 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 corresponding
mcst game. The edges with zero cost are depicted below.
2i
6i
H
HH
4i
5i
1i
3i
7i
@
@
8i
@
@i
0
Note that c(N) = n(w , N) − 1 = 4 − 1 = 3. If we take for example
S = {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.
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Lemma for simple cost functions
Lemma
Let w be a simple cost function and let c be the corresponding mcst
game. Then c has a PMAS.
Proof
Two ways to construct a PMAS:
1) Construct the scheme x ed 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 the
scheme x π by assigning the cost of every (w , S)-component without 0 to
the player in this component with the lowest index according to π.
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Example
Consider again the simple cost function
2i
6i
HH
H i
5i
4
7i
1i
3i
@
@
8i
@
@i
0
Then
1 1 1 1 1 1
1 1
ed
xNed = (0, , , , , , , 1) and x2356
= (−, , , −, 1, 1, −, −).
4 4 4 4 2 2
2 2
With π = 48372516 we get
π
xNπ = (0, 0, 0, 1, 0, 0, 1, 1) and x2356
= (−, 0, 1, −, 1, 1, −, −).
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Theorem
Theorem
Every 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 = α1 w1 + α2 w2 + · · · + αk wk
in such a way that the same decomposition is valid for the corresponding
mcst games
c = α1 c1 + α2 c2 + · · · + αk ck .
Let π : N → {1, . . . , n} be an ordering of the players. Then
x π,w = α1 x π,w1 + α2 x π,w2 + · · · + αk x π,wk
is a PMAS for c.
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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 cost
between a node and the source. In every subsequent step form an edge of
a node which is not connected with the source, directly or indirectly, and
the 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 minimal
cost. In every subsequent step form an edge of minimal cost which does
not 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
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Example
8
2i
3i
@
A@
21
13
A@
A @i 17 A
1 18
A
6 A Ai
0
Prim
For N = {1, 2, 3}: first form 01, then 13, and finally 23;
for S = {2, 3}: first form 02, then 23;
etcetera.
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Example
8
2i
3i
A@
@ 21
13
A@
A @i 17 A
1 18
A
6 A Ai
0
This yields the following Bird scheme:
S
123
12
13
23
1
2
3
xS,1
6
6
6
−
6
−
−
xS,2
8
17
−
17
−
17
−
xS,3
13
−
13
8
−
−
18
Note that this scheme is not a PMAS.
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Example
Now we compute PMAS x π for π = 123 (only for N). Note that
8
1
2i
3i
2i
3i
A@
A@
@ 21
@ 1
13
1
A@
A@
A @i A @i +
17 A
1A
1 18
= 6·
1 1
A 6 A
1
A A Ai
Ai
0
0
So
xNπ = (6, 6, 6)+?.
Player 1 stops paying!
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2
2i
3i
A@
@ 15
7
A@
A @i 11 A
1 12
A
0 A Ai
0
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Example
Now
2
1
2i
3i
2i
3i
A@
A@
@ 15
@ 1
7
1
A@
A@
A @i A @i +
11 A
1A
1 12
= 2·
1 1
A 0 A
0
A A Ai
Ai
0
0
So
xNπ = (6, 6, 6) + (0, 2, 2)+?.
Player 3 stops paying!
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0
2i
3i
A@
@ 13
5
A@
A @i 9A
1 10
A
0 A Ai
0
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Example
... and
0
0
2i
3i
2i
3i
@
@
A@
A@
13
5
1
1
A@
A@
A @i A @i +
9A
1A
1 10
= 5·
1 1
A 0 A
0 A A Ai
Ai
0
0
0
2i
3i
@
A@
8
0
A@
A @i 4A
1 5
A
0 A Ai
0
So
xNπ = (6, 6, 6) + (0, 2, 2) + (0, 5, 0).
Player 2 stops paying! In all subsequent steps costs have to be allocated
for simple games where 1, 2 and 3 are all connected to 0.
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Example
So xNπ = (6, 13, 8).
8
2i
3i
A@
@ 21
13
A@
A @i 17 A
1 18
A
6 A Ai
0
Note that in every step of Kruskal’s algorithm there is one player who
gets a connection to the source or to a player with a lower index
according 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’.
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Example
8
2i
3i
A@
@ 21
13
A@
A @i 17 A
1 18
A
6 A Ai
0
Proceeding in the same way for all coalitions we get:
S
123
12
13
23
1
2
3
xS,1
6
6
6
−
6
−
−
xS,2
13
17
−
17
−
17
−
xS,3
8
−
13
8
−
−
18
Note that this scheme is a PMAS.
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Some references
As x π is a PMAS for every order π, the average of these n! schemes is a
PMAS 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)).
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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.
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Example ERO rule
6
2i
3i
A@
@ 21
8
A@
A @i 17 A
1 18
A 13 A Ai
0
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, 12 , 12 ).
In step 2 edge 13 is formed, player 1 pays 2/3 of the costs, players 2 and
3 1/6 of the costs.
Remaining obligations are ( 13 , 31 , 13 ).
In step 3 edge 01 is formed. All players pay 1/3 of the costs.
26 26
So, ERO=( 29
3 , 3 , 3 ).
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Assignment 2
Assignment 2
Determine for any b ≥ 0 the ERO solution for the mcst problem below.
7
2i
3i
PP
6
P
A@
PP
@ 7
@
PP
A @5
P 6@
PP@
A @i @i
P
8A
1 7
4
A
7 8
A b
Ai
0
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Monotonic mcst games
Consider the following cost function:
i
1i
3i
H
2H
HH
HH
H
H
H
H
H
HH
H
H i
H i
i
4H
5
6
HH
HH
HH
H i
0
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).
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1
Introduction
2
Preliminaries
3
Linear production games
4
Minimum cost spanning tree games
5
Minimum coloring games
6
Inventory games
7
Sequencing games
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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 where
ES = {ij ∈ E : i ∈ S, j ∈ S}
• Ḡ = (N, Ē ): the complement of G where
Ē = {ij : i, j ∈ N, i 6= j, ij ∈
/ E}
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Notation and definitions
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 all
cliques.
The clique number of a graph is the number of elements of a
maximum clique.
An independent set is an S ⊆ N such that G [S] has no edges.
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Notation and definitions
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 of
colors needed to color the vertices of G so that no two adjacent
vertices share the same color.
A graph G = (N, E ) is perfect iff the clique number of every induced
subgraphs 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 two
different independent sets are adjacent.
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Notation and definitions
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, that
for each vertex v ∈ V there is a unique directed path from r to v .
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Minimum coloring game
The minimum coloring game on G is the cost game (N, c G ) where
c G : 2N → R
is defined as c G (S) := X (G [S]) for all S ⊆ N.
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Minimum coloring game
Example
Let G be the graph
1
2
•
•
3
•
Then, the corresponding minimum coloring game is
S ∅
c G (S) 0
1 2
1 1
3 12 13 23 123
1 2 1 2
2
Observe:
(N, c G ) is totally balanced, since G is perfect (Deng et al. (2000))
C (c G ) = convhull{(1, 1, 0), (0, 1, 1)} since 12 and 23 are maximum
cliques (Okamoto (2003a))
(N, c G ) is concave since G is complete r-partite (Okamoto (2003b))
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Graphs and games
Totally balancedness is not enough to obtain ”dynamic stability”.
Example
2
1
3
G1
2
3
1
4
5 G
2
Let (N, c G1 ) be the coloring game w.r.t. G1 . Then (1, 1, 0) unique core
element.
Let (N, c G2 ) be the coloring game w.r.t. G2 . Then (0, 0, 1, 1, 1) unique
core element.
So, by including player 4 and 5 player 3 has to pay 1. Hence, this
example lacks a form of dynamic stability. To be more precise, there does
not exist a PMAS.
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We have the following relations:
Class of concave games
∩
Class of PMAS games
∩
Class of totally balanced games.
Question
Which graphs characterize the class of coloring games with a PMAS?
Of course, this class of graphs constitutes a subclass of the class of
perfect graphs, and it should contain the class of complete r-partite
graphs.
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Graphs and games
Research in the same spirit in which graphs characterize a class of OR
games 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 the
corresponding 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 corresponding
Chinese postman game is balanced (submodular, totally balanced) for
any cost function.
Minimum vertex covering games (Okamoto (2003)):
G contains no subgraph isomorphic to K3 or P3 if and only if the
minimum vertex covering game is submodular.
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Quasi-threshold graphs
Quasi-threshold graphs are defined in a constructive way: they are the
graphs that can be formed, starting from one-vertex graphs, by the
following operations:
1) adding a new vertex that is adjacent to all vertices of a
quasi-threshold graph;
2) taking the disjoint union of two quasi-threshold graphs.
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Quasi-threshold graphs
Yan et al. (1996) (in fact, already in Wolk (1965)) characterized
quasi-threshold graphs by the fact that they are the graphs induced by a
rooted forest. In the induced graph there is an edge between two vertices
i and j iff there is a directed path in F between i and j.
Example Consider the rooted forest F = (N, D):
4
5
3
9
2
6
1
8
7
Observe, interchanging position 2 and 3 in F yields the same
quasi-threshold graph.
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Quasi-threshold graphs
Solution to non-unicity representation:
make a forest in which no vertex has precisely one follower.
Example continued
4
5
2, 3
6
1
7, 8, 9
In the induced graph there is an edge between two vertices i and j iff
there is a directed path in F between i and j or i and j are in the same
vertex.
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Quasi-threshold graphs
Proposition
Let (N, E ) be a graph. Then (N, E ) is a quasi-threshold graph if and
only if a rooted forest F = (V , D) exists, and for every v ∈ V a
nonempty 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 with
k ∈ Mv .
We will refer to the triple (V , D, (Mv )v∈V ) as the as the rooted forest
representation of quasi-threshold graph (N, E ).
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(P4 , 2K2)-free graphs
Yan et al. (1996) show that the class of quasi-threshold graphs coincides
with the class of (P4 , C4 )-free graphs, i.e. the graphs which do not have
subgraphs isomorphic to P4 or C4 .
2
1
P4
3
2
4
1
3
2
4
1
3
4
2K2
C4
The complements of P4 and C4 are isomorphic to P4 and 2K2
respectively.
The complements of quasi-threshold graphs, i.e. the (P4 , 2K2 )-free
graphs, are used to characterize coloring games with a PMAS.
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(P4 , 2K2)-free graphs
Obviously, the rooted forest representation of a quasi-threshold graph can
also 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 vertex
or to different vertices between which a directed path exists.
Example
I2
I1
4
5
2, 3
I3
6
I4
7, 8, 9
1
The maximal independent sets are: I1 = {1, 2, 3, 4}, I2 = {1, 2, 3, 5},
I3 = {1, 6}, and I4 = {7, 8, 9}.
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(P4 , 2K2)-free graphs
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 the
union of disjoint vertices.
Example
The rooted tree representation of the 3-partite graph with independent
sets I1 = {1}, I2 = {2, 3} and I3 = {4, 5, 6}.
1
2, 3
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PMAS existence
Theorem
Let G = (N, E ) be a graph and let (N, c G ) be the corresponding
minimum coloring game. If P(N, c G ) 6= ∅ then G is (P4 , 2K2 )-free.
Proof
Let x ∈ P(N, c G ) 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 of
generality we can assume again that S = {1, 2, 3, 4}, 12 ∈ E , 34 ∈ E ,
13 ∈
/ E , 14 ∈
/ E , and 24 ∈
/ E . Note that c G ({1, 2, 4}) = c G ({1, 3, 4}) = 2
G
G
and c ({1, 3}) = c ({1, 4}) = c G ({2, 4}) = 1. By efficiency and
monotonicity of x we get
4
=
=
≤
=
=
c G ({1, 2, 4}) + c G ({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
c G ({1, 3}) + c G ({1, 4}) + c G ({2, 4})
3, which yields a contradiction.
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PMAS existence
Proposition
Let G = (N, E ) be a (P4 , 2K2 )-free graph with rooted forest
representation (V , D, (Mv )v∈V ), and let (N, c G ) be the corresponding
minimum coloring game. For every S ∈ 2N \{∅} we have
c G (S) = | max(VS )|, where VS := {v ∈ V : Mv ∩ S 6= ∅}.
Example
4
5
2, 3
6
1
7, 8, 9
Then c G {1, 2, 3, 7, 8} = 2, c G ({1, 6}) = 1 and c G (N) = 4.
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PMAS existence
Theorem
Let G = (N, E ) be a graph and let (N, c G ) be the corresponding
minimum coloring game. Then, P(N, c G ) 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 forest
representation.
4 a
5 b
2, 3
c
6 d
1 e
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PMAS existence
In each vertex chose an order of the players. Take, for example
σa = (4) 4 a
σc = (3, 2)
b 5 σb = (5)
2, 3
c
d 6 σd = (6)
σe = (1) 1 e
f
7, 8, 9
σf = (8, 7, 9)
Define the {0, 1}-valued scheme x = (xS,i )S∈2N \{∅},i ∈S by

 1 if v (i) ∈ max(VS ) and
xS,i =
σv(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 and
xS,i = 0 otherwise.
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Structure of PMAS-es
Proposition
Let G = (N, E ) be a (P4 , 2K2 )-free graph with rooted forest
representation (V , D, (Mv )v∈V ), let (N, c G ) be the corresponding
minimum coloring game. Then the restricted game to a coalition of
players that are all contained in the same vertex of the rooted forest
representation, yields a unit game, i.e.,
c G (S) = 1 for every S ∈ 2Mv \{∅}.
Example
4 a
b 5
2, 3
c
d 6
1 e
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Structure of PMAS-es
Proposition
Let G = (N, E ) be a (P4 , 2K2 )-free graph with rooted forest
representation (V , D, (Mv )v∈V ), let (N, c G ) be the corresponding
minimum coloring game and let x ∈ P(N, c G ). Then
y = (yS,i )S∈2Mv \{∅},i ∈S , defined by yS,i = xS,i for every S ∈ 2Mv \{∅} and
i ∈ S, is such that y ∈ P(Mv , 1Mv ).
So, the restriction of a PMAS of the coloring game to a coalition which
players coincide with the set of players of one vertex, generates a PMAS
of the unit game on this coalition.
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Theorem
Let G = (N, E ) be a (P4 , 2K2 )-free graph with rooted forest
representation (V , D, (Mv )v∈V ) and let (N, c G ) be the corresponding
minimum coloring game. Let y v ∈ P(Mv , 1Mv ) for every v ∈ V . Then
v
there is a unique x ∈ P(N, c G ) such that xS,i = yS,i
for every v ∈ V ,
Mv
S ∈ 2 \{∅} and i ∈ S, and this scheme x is given by
(
v(i )
yS∩Mv (i ) ,i if v (i) ∈ max(VS )
xS,i =
0
if v (i) ∈
/ max(VS )
for every S ∈ 2N \{∅} and i ∈ S.
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Structure of PMAS-es
Example
4 a
b 5
2, 3
c
d 6
1 e
f
7, 8, 9
The only coalitions that have a non-trivial PMAS are {2, 3} and {7, 8, 9}.
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Structure of PMAS-es
Take as PMAS-es of {2, 3} and {7, 8, 9}:
S
{2}
{3}
{2, 3}
{7}
{8}
{9}
{7, 8}
{7, 9}
{8, 9}
{7, 8, 9}
2
3
1
−
−
1
0.5 0.5
7
8
9
1 −
− 1
− −
1 0
1 −
− 1
1 0
−
−
1
−
0
0
0
Then for coalition {1, 2, 5, 7, 8} the scheme x is generated as follows:
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Structure of PMAS-es
Observe that c G ({1, 2, 5, 7, 8}) = 2. Further, player 1 and 2 are not in
vertices that are in max VS . Hence, they receive both 0. The other
players are in max VS , hence their payoff has to be read from their
PMAS-es of their unit game.
S
{5}
{7, 8}
5
1
7 8
9
1 0
−
4
−
5 6
1 −
Hence, xS is:
S
1
{1, 2, 5, 7, 8} 0
2 3
0 −
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Structure of PMAS-es
Theorem
Let G = (N, E ) be a (P4 , 2K2 )-free graph with rooted forest
representation (V , D, (Mv )v∈V ), let (N, c G ) be the corresponding
minimum coloring game, and let x ∈ P(N, c G ).
Then x is integer-valued if and only if there is a collection of orders
σ = (σv )v∈V such that x = x σ .
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Assignment 3
Assignment 3
Construct a (P4 , 2K2 )-free graph G such that the corresponding
minimum coloring game (N, c G ) admits precisely 72 integer-valued
PMAS-es. What is the minimum number of players needed?
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1
Introduction
2
Preliminaries
3
Linear production games
4
Minimum cost spanning tree games
5
Minimum coloring games
6
Inventory games
7
Sequencing games
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One firm ordering problem
Consider a firm that faces a (deterministic) demand of d units of some
good per unit of time. It is not allowed to run out of stock and the lead
time (time between placement of an order and arrival of the goods) is
assumed to be zero. The firm faces two kinds of costs:
ordering costs: for each order the firm places it has to pay a fixed
cost of a, independent of the quantity ordered;
holding costs: the costs of keeping one unit of good in stock for one
unit of time is h.
Question:
What is the optimal ordering quantity of the firm?
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Optimal ordering quantity
Let Q denote the ordering quantity. Then time between two successive
orders is Q/d, or, equivalently, there are on average d/Q orders per unit
of time. The average inventory is Q/2, so the average holding costs per
time unit equal hQ/2. So the average inventory costs per unit of time
are:
Q
d
+h· .
AC (Q) = a ·
Q
2
The optimal ordering size and the optimal number of orders per time unit
are
r
r
2ad
d
dh
∗
∗
Q =
and m = ∗ =
.
h
Q
2a
The minimal average costs are
AC (Q ∗ ) = 2am∗ .
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n-firm inventory situation
Definition
An 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;
N
h ∈ R++ is the vector of holding cost parameters;
a > 0 is the cost per order.
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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
=c
di
for every i ∈ S. Here Qi refers to the ordering quantity for firm i in
optimum. Now average cost per unit of time is a function of c:
AC (c) =
=
1 X
Qi
+
hi ·
c
2
i ∈S
1 cX
hi · di .
a· +
c
2
a·
i ∈S
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Optimal value of c is
∗
c =
s
2a
.
i ∈S hi di
P
Optimal ordering quantities for i ∈ S are given by
s
2adi2
∗
Qi = P
.
i ∈S hi di
The optimal number of orders per time unit is
rP
sX
1
i ∈S hi di
mS = ∗ =
=
mi2 .
c
2a
i ∈S
Minimal average costs are given by
∗
c(S) = AC (c ) = 2amS = 2a
sX
mi2 .
i ∈S
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Theorem
Definition
Let (N, d, h, a) be an inventory situation. The corresponding inventory
game (N, c) is defined by
sX
c(S) = 2a
mi2 ,
i ∈S
where
mi =
r
hi di
2a
for every i ∈ N.
Theorem (Meca et al. (1999))
Every inventory game is concave.
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3
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4
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5
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6
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7
Sequencing games
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sequencing situations
In a 1-machine sequencing situation there is a queue of players
N = {1, 2, . . . , n}, each with one job, in front of a machine. Each job
must 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} and
for 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) with
p, α ∈ RN
++ .
Theorem (Smith (1956))
A processing order that minimizes total costs is obtained by ordering the
jobs in decreasing order with respect to the urgency index ui = αpii .
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Example
Consider the following sequencing situation:
M
1
2
3
α1 = 6
α2 = 8
α3 = 4
p1 = 3
p2 = 2
p3 = 4
u1 = 2
u2 = 4
u3 = 1
Total costs are 6 · 3 + 8 · 5 + 4 · 9 + 3 · 10 = 124.
Optimal processing order is
M
2
4
1
4
α4 = 3
p4 = 1
u4 = 3
3
Optimal costs are 8 · 2 + 3 · 3 + 6 · 6 + 4 · 10 = 101.
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Terminology
Let σ : N → {1, . . . , n} be an order and i ∈ N. Then
X
C (σ, i) =
pj
j∈N:σ(j)≤σ(i )
is the completion time of job i under order σ.
For coalition S ⊆ N and order σ let
X
CS (σ) =
αi C (σ, i)
i ∈S
denote the total costs of coalition S.
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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 denoted
by A(S).
The corresponding sequencing game (N, v ) is defined by
v (S) = max (CS (σ0 ) − CS (σ))
σ∈A(S)
for every S ⊆ N.
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Example
Consider coalition S = {1, 2, 4}.
M
1
2
3
4
α1 = 6
α2 = 8
α3 = 4
α4 = 3
p1 = 3
p2 = 2
p3 = 4
p4 = 1
u1 = 2
u2 = 4
u3 = 1
u4 = 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.
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Example
M
1
2
3
4
α1 = 6
α2 = 8
α3 = 4
α4 = 3
p1 = 3
p2 = 2
p3 = 4
p4 = 1
u1 = 2
u2 = 4
u3 = 1
u4 = 3
Proceeding in the same way for all coalitions we get
S
1
v (S) 0
2 3
0 0
4
0
12 13 14 23 24 34 123 124 134 234 1234
12 0 0 0 0 8
12 12
8
8
23
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Neighbor jobs
Consider the interchange of two neighboring jobs:
i
→
j
j
i
The change of total costs equals
αj pi − αi pj .
This change is positive if
uj > ui .
The cost savings, attainable by players i and j, equal
gij = max{0, αj pi − αi pj }.
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connected coalitions
A coalition S is connected with respect to order σ if for all i, j ∈ S and
for all k ∈ N with σ(i) < σ(k) < σ(j) we have k ∈ S.
Theorem
Let (N, σ0 , p, α) be a sequencing situation and let (N, v ) be the
corresponding sequencing game. Then for any coalition S that is
connected with respect to σ0 we have
X
v (S) =
gij .
i ,j∈S:σ0 (i )<σ0 (j)
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Theorem
Any coalition T can be partitioned into maximally connected subsets
(components) with respect to σ. This collection of components is
denoted by T \σ.
Theorem
Let (N, σ0 , p, α) be a sequencing situation and let (N, v ) be the
corresponding sequencing game. Then for any coalition T we have
X
v (T ) =
v (S).
S∈T \σ0
Theorem
Every sequencing game is convex.
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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 Splitting
rule: go from the initial order to the optimal order for N by stepwise
interchanging two misplaced neighboring jobs and divide the profits of
such an interchange evenly among the two jobs.
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Example
M
1
2
α1 = 6
α2 = 8
p1 = 3
p2 = 2
u1 = 2
u2 = 4
Cost savings 12, to be divided by players
M
2
1
3
α3 = 4
p3 = 4
u3 = 1
1 and 2.
4
α4 = 3
p4 = 1
u4 = 3
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).
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Thanks
Thank you for your attention
Henk Norde
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