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TRANSCRIPT
Introduction to Constraint Programming
and Optimization
Tutorial
September 2005
BUAP, Puebla, Mexico
John Hooker
Carnegie Mellon University
Outline
�We will examine 5 example problemsand use them to
illustrate some basic ideas of constraint programming and
optimization.
�Freight transfer
�Traveling salesman problem
�Continuous global optimization
�Product configuration
�Machine scheduling
Outline
�Example: Freight transfer
�Bounds propagation
�Bounds consistency
�Knapsack cuts
�Linear Relaxation
�Branching search
Outline
�Example: Traveling salesman problem
�Filtering for all-differentconstraint
�Relaxation of all-different
�Arc consistency
Outline
�Example: Continuous global optimization
�Nonlinear bounds propagation
�Function factorization
�Interval splitting
�Lagrangean propagation
�Reduced-cost variable fixing
Outline
�Example: Product configuration
�Variable indices
�Filtering for elementconstraint
�Relaxation of element
�Relaxing a disjunction of linear systems
Outline
�Example: Machine scheduling
�Edge finding
�Benders decomposition and nogoods
Example: Freight Transfer
�Bounds propagation
�Bounds consistency
�Knapsack cuts
�Linear Relaxation
�Branching search
Freight Transfer
}3,2,1,0{
8
423
45
7
4050
6090
min
43
21
43
21
43
21
∈≤
++
+≥
++
++
++
jx
xx
xx
xx
xx
xx
xx
�Transport 42 tons of freight using at most 8 trucks.
�Trucks have 4 different sizes: 7,5,4, and 3 tons.
�How many trucks of each size should be used to minimize cost?
�x j= number of trucks of size j.
Cost of
size 4
truck
Freight Transfer
�We will solve the problem by branching search.
�At each node of the search tree:
�Reduce the variable domains using bounds propagation.
�Solve a linear relaxationof the problem to get a bound
on the optimal value.
Bounds Propagation
�The domain of x jis the set of values x jcan have in a some
feasible solution.
�Initially each xjhas domain {0,1,2,3}.
�Bounds propagationreduces the domains.
�Smaller domains result in less branching.
Bounds Propagation
�First reduce the domain of x 1:
423
45
74
32
1≥
++
+x
xx
x
++
−≥
7
34
542
43
21
xx
xx
176
7
33
34
35
421
=
=
⋅−
⋅−
⋅−
≥x
�So domain of x 1is reduced from {0,1,2,3} to {1,2,3}.
�No reductions for x 2, x 3, x 4possible.
Max element
of domain
Bounds Propagation
�In general, let the domain of x ibe {L i, …, Ui}.
�An inequality ax
≥b (with a ≥
0)can be used to raise Lito
−∑ ≠
i
ij
jj
ia
Ua
bL
,m
ax
�An inequality ax
≤b (with a ≥0)can be used to reduce Uito
−∑ ≠
i
ij
jj
ia
La
bU
,m
in
Bounds Propagation
�Now propagate the reduced domains to the other
constraint, perhaps reducing domains further:
84
32
1≤
++
+x
xx
x
−−
−≤
1
84
32
1
xx
xx
717
1
00
18
1=
=
−
−−
≤x
�No further reduction is possible.
Min element of
domain
Bounds Consistency
�Again let {Lj, …, Uj} be the domain of x j
�A constraint set is bounds consistent if for each j:
�x j= Ljin some feasible solution and
�x j= Ujin some feasible solution.
�Bounds consistency ⇒
we will not set x jto any infeasible
values during branching.
�Bounds propagation achieves bounds consistency for a
single inequality.
�7x 1+ 5x 2+ 4x 3+ 3x 4
≥42 is bounds consistent when
the domains are x1
∈{1,2,3} and x2, x 3, x 4
∈{0
,1,2
,3}.
�But not necessarily for a set of inequalities.
Bounds Consistency
�Bounds propagation may not ach
ieve consistency for a set.
�Consider set of inequalities
01
21
21
≥−
≥+
xx
xx
with domains x 1, x 2
∈{0,1},solutions (x1,x2) = (1,0), (1,1).
�Bounds propagation has no effect on the domains.
From x1+ x2
≥1:
01
11
01
11
12
21
=−
=−
≥=
−=
−≥
Ux
Ux
From x1
−x 2
≥1:
10
11
01
11
12
21
=+
=+
≤=
−=
−≥
Lx
Ux
�Constraint set is not bounds consistent because x1 = 0 in no
feasible solution.
Knapsack Cuts
�Inequality constraints (knapsack constraints) imply cutting
planes.
�Cutting planes make the linear relaxation tighter, and its
solution is a stronger bound.
�For the ≥constraint, each m
axim
al packing corresponds to
a cutting plane (knapsack cut).
423
45
74
32
1≥
++
+x
xx
x
These terms form a maximal packing because:
(a)They alone cannot sum to ≥42, even if each xj
takes the largest value in its domain (3).
(b) No superset of these terms has this property.
Knapsack Cuts
�This maximal packing:
423
45
74
32
1≥
++
+x
xx
x
corresponds to a knapsack cut:
{}
23,4
max
35
37
424
3=
⋅−
⋅−
≥+
xx
Coefficients
of x 3, x 4
Min value of
4x 3+ 3x 4
needed to
satisfy
inequality
Knapsack Cuts
�In general, Jis a packing for ax
≥b (with a ≥0) if
bU
aJ
jj
j<
∑ ∈
∑∑
∉∈
−≥
Jj
Jj
jj
jj
Ua
bx
a
�If Jis a packing for ax
≥b, then the remaining terms must
cover the gap:
�So we have a knapsack cut: ∑
∑
∉∉
∈
−
≥J
jj
Jj
Jj
jj
ja
Ua
b
x}
{m
ax
Knapsack Cuts
�Valid knapsack cuts for
423
45
74
32
1≥
++
+x
xx
x
are
322
32
42
43
≥+
≥+
≥+
xx
xx
xx
Linear Relaxation
jj
jU
xL
xx
xx
xx
xx
xx
xx
xx
xx
xx ≤
≤≥
+≥
+≥
+≤
++
+≥
++
++
++ 322
8
423
45
7
4050
6090
min
32
42
43
43
21
43
21
43
21
�We now have a linear relaxation of the freight transfer
problem:
Branching Search
�At each node of the search tree:
�Reduce domains with bounds propagation.
�Solve a linear relaxation to obtain a lower bound on any
feasible solution in the subtree rooted at current node.
�If relaxation is infeasible, or its optimal value is no better
than the best feasible solution found so far, backtrack.
�Otherwise, if solution of relaxation is feasible, remember it
and backtrack.
�Otherwise, branch by splitting a domain.
3132
31
523
va
lue
)0,2,3,
2(01
23
0123
0123123
==
∈
x
x
rela
xatio
n
infe
asib
le12
3
1232312
∈x
526
va
lue
)0,2,6.2,3(01
23
0123
01233
==
∈
x
x
5.52
7
valu
e
)0,75.2,2,3(
0123
123
0123
==
∈
x
x
solu
tion
fe
asib
le
530
va
lue
)2,0,3,3(012
01233
==
∈
xx
solu
tion
fe
asib
le
530
val
ue)1,2,2,3(
123
1212
3
==
∈
xx
boun
d
todu
eback
trac
k
530
v
alue
)5.0,3,5.1,3(01
23
0123
==
∈
x
x
x 1∈{1,2}
x 1=3
x 2∈{0,1,2}
x 2=3
x 3∈{1,2}
x 3=3
Branching
search
Domains after bounds
propagation
Solution of linear relaxation
Branching Search
This image cannot currently be displayed.
)1,2,2,3()
,,
,(
43
21
=x
xx
x
�Two optimal solutions found (cost = 530): )2,0,3,3(
),
,,
(4
32
1=
xx
xx
Example: Traveling Salesman
�Filtering for all-differentconstraint
�Relaxation of all-different
�Arc consistency
Traveling Salesman
()
{} n
x
xx
c
j
n
ix
xi
i ,,1
,,
diff
eren
t-
all
min
1
1
…
…
∈
∑+
�Salesman visits each city once.
�Distance from city ito city jis cij .
�Minimize distance traveled.
�x i= ith city visited.
Distance from i th
city visited to
next city
(city n+1 = city 1)
�Can be solved by branching + domain reduction +
relaxation.
Filtering for All-different
�The best known filtering algorithm for all-different is based
on m
axim
um cardinality bipartite m
atching and a
theorem of Berge.
�Goal: filter infeasible values from variable domains.
�Filtering reduces the domains and therefore reduces
branching.
Filtering for All-different
�Consider all-different(x 1,x2,x3,x4,x5) with domains
}6,5,4,3,2,1{
}5,1{
}5,3,2,1{
}5,3,2{
}1{
54321
∈∈∈∈∈
xxxxx
x 1 x 2 x 3 x 4 x 5
1 2 3 4 5 6
Indicate domains with edges
Find maximum cardinality
bipartite matching.
Mark edges in alternating paths
that start at an uncovered vertex.
Mark edges in alternating cycles.
Remove unmarked edges not in
matching.
Indicate domains with edges
Find maximum cardinality
bipartite matching.
x 1 x 2 x 3 x 4 x 5
1 2 3 4 5 6
Mark edges in alternating paths
that start at an uncovered vertex.
Mark edges in alternating cycles.
Remove unmarked edges not in
matching.
Filtering for All-different }6,4{
}6,5,4,3,2,1{
}5{}5,1{
}3,2{}5,3,2,1{
}3,2{}5,3,2{
}1{
55
44
33
221
∈⇒
∈∈
⇒∈
∈⇒
∈∈
⇒∈∈
xx
xx
xx
xxx
Relaxation of All-different
�All-different(x 1,x2,x3,x4) with domains x j
∈{1,2,3,4} has the
convex hull relaxation:
43
21
43
21
++
+=
++
+x
xx
x
32
1
43
2
43
1
42
1
32
1
++
≥
++
++
++
++
xx
x
xx
x
xx
x
xx
x
21
43
42
32
41
31
21
+≥
++++++
xx
xx
xx
xx
xx
xx
1≥
jx
�This is the tightest possible linear relaxation, but it may be
too weak (and have too many inequalities) to be useful.
Arc Consistency
�A constraint set Scontaining variables x 1, …, x nwith
domains D1, …, Dnis arc consistent if the domains contain
no infeasible values.
�That is, for every jand every v
∈Dj, xj= vin some feasible
solution of S.
�This is actually generalized arc consistency(since
there may be more than 2 variables per constraint).
�Arc consistency can be achieved by filtering all infeasible
values from the domains.
Arc Consistency
�The matching algorithm achieves arc consistency for all-
different.
�Practical filtering algorithms often do not achieve full arc
consistency, since it is not worth the time investment.
�The primary tools of constraint programming are
filtering algorithms that achieve or approximate arc
consistency.
�Filtering algorithms are analogous to cutting plane
algorithms in integer programming.
Arc Consistency
�Domains that are reduced by a filtering algorithm for one
constraint can be propagatedto other constraints.
�Similar to bounds propagation.
�But propagation may not achieve arc consistency for a
constraint set, even if it achieves arc consistency for each
individual constraint.
�Again, similar to bounds propagation.
Arc Consistency
�For example, consider the constraint set
),
(di
ffer
ent
-al
l
),
(di
ffer
ent
-al
l
),
(di
ffer
ent
-al
l
32
31
21
xx
xx
xx
}3,2{
}3,2{
}2,1{
321
∈∈∈
xxx
with domains
�No further domain reduction is possible.
�Each constraint is arc consistent for these domains.
�But x 1= 2 in no feasible solution of the constraint set.
�So the constraint set is not arc consistent.
Arc Consistency
�On the other hand, sometimes arc consistency maintenance
alone can solve a problem—without branching.
�Consider a graph coloring problem.
�Color the vertices red, blue or green so that adjacent
vertices have different colors.
�These are binaryconstraints (they contain only 2
variables).
�Arbitrarily color two adjacent vertices red and green.
�Reduce domains to maintain arc consistency for each
constraint.
Graph coloring problem that can be solved by arc consistency
maintenance alone.
Example: Continuous Global Optimization
�Nonlinear bounds propagation
�Function factorization
�Interval splitting
�Lagrangean propagation
�Reduced-cost variable fixing
Continuous Global Optimization
�Today’s constraint programming solvers (CHIP, ILOG
Solver) emphasize propagation.
�Today’s integer programming solvers (CPLEX, Xpress-MP)
emphasize relaxation.
�Current global solvers (BARON, LGO) already combine
propagation and relaxation.
�Perhaps in the near future, all solvers will combine
propagation and relaxation.
Continuous Global Optimization
�Consider a continuous global optimization problem:
]2,0[],1,0[
14
22m
ax
21
21
21
21
∈∈
=≤
++
xx
xx
xx
xx
�The feasible set is nonconvex, and there is more than one
local optimum.
�Nonlinear programming techniques try to find a local
optimum.
�Global optimization techniques try to find a global optimum.
]2,0[],1,0[
22
14m
ax
21
21
21
21
∈∈
≤+
=+
xx
xx
xx
xx
Feasible set
Global optimum
Local optimum
x 1
x 2
Continuous Global Optimization
�Branch by splitting continuous interval domains.
�Reduce domains with:
�Nonlinear bounds propagation.
�Lagrangean propagation (reduced cost variable fixing)..
�Relaxthe problem by factoring functions.
Continuous Global Optimization
Nonlinear Bounds Propagation
81
241
41
21
=⋅
=≥
Ux
�To propagate 4x 1x 2 = 1, solve for x 1= 1/4x 2 and x2= 1/4x 1.
Then:
41
141
41
12
=⋅
=≥
Ux
�This yields domains x 1
∈[0.125,1], x2
∈[0.25,2].
�Further propagation of 2x 1+ x2
≤2 yields
x 1∈
[0.1
25,0
.875
],
x 2∈
[0.2
5,1.
75].
�Cycling through the 2 constraints converges to the fixed point
x 1 ∈
[0.1
46,0
.854
], x
2∈
[0.2
93,1
.707
].
�In practice, this process is terminated early, due to
decreasing returns for computation time,
Propagate intervals
[0,1], [0,2]
through constraints
to obtain
[1/8,7/8], [1/4,7/4]
x 1
x 2
Nonlinear Bounds Propagation
�Factor complex functions into elementary functions that have
known linear relaxations.
�Write 4x 1x 2= 1 as 4y= 1 where y= x1x 2.
�This factors 4x 1x 2into linear function 4yand bilinear
function x1x 2.
�Linear function 4yis its own linear relaxation.
�Bilinear function y= x1x 2has relaxation:
21
21
12
21
21
12
21
21
12
21
21
12
UL
xL
xU
yU
Ux
Ux
U
LU
xU
xL
yL
Lx
Lx
L
−+
≤≤
−+
−+
≤≤
−+Function Factorization
�We now have a linear relaxation of the nonlinear problem:
jj
jU
xL
UL
xL
xU
yU
Ux
Ux
U
LU
xU
xL
yL
Lx
Lx
L
xxy
xx
≤≤
−+
≤≤
−+
−+
≤≤
−+
≤+=
+
21
21
12
21
21
12
21
21
12
21
21
12
21
21
22
14m
ax
Function Factorization
Sol
ve li
near
rel
axat
ion.
x 1
x 2
Interval Splitting
Sol
ve li
near
rel
axat
ion.
x 1
x 2
Sin
ce s
olut
ion
is in
feas
ible
, sp
lit a
n in
terv
al a
nd b
ranc
h.
]1,25.0[
2∈
x
]75.1,1[
2∈
x
Interval Splitting
x 1
x 2
x 1
x 2
]75.1,1[
2∈
x]1,
25.0[2∈
x
]75.1,1[
2∈
x]1,
25.0[2∈
x
Solution of
relaxation is
feasible,
value = 1.25
This becomes
best feasible
solution
x 1
x 2
x 1
x 2
]75.1,1[
2∈
x]1,
25.0[2∈
x
Solution of
relaxation is
feasible,
value = 1.25
This becomes
best feasible
solution
x 1
x 2
x 1
x 2Solution of
relaxation is
not quite
feasible,
value = 1.854
jj
jU
xL
UL
xL
xU
yU
Ux
Ux
U
LU
xU
xL
yL
Lx
Lx
L
xxy
xx
≤≤
−+
≤≤
−+
−+
≤≤
−+
≤+=
+
21
21
12
21
21
12
21
21
12
21
21
12
21
21
22
14m
ax
Lagrange multiplier (dual variable)
in solution of relaxation is 1.1
Lagrangean Propagation
Optimal value of relaxation is 1.854
�Any reduction ∆in right-hand side of 2x 1+ x2
≤2 reduces
the optimal value of relaxation by 1.1 ∆.
�So any reduction ∆in the left-hand side (now 2) has the
same effect.
Lagrangean Propagation
�Any reduction ∆in left-hand side of 2x 1+ x2
≤2 reduces the
optimal value of the relaxation (now 1.854) by 1.1 ∆.
�The optimal value should not be reduced below the lower
bound 1.25 (value of best feasible solution so far).
�So we should have 1.1 ∆
≤1.854 −1.25, or
25.185
4.1
)]2(
2[1.12
1−
≤+
−x
x
451
.11.1
25.185
4.1
22
21
=−
−≥
+x
xor
�This inequality can be propagated to reduce domains.
�Reduces domain of x 2from [1,1.714] to [1.166,1.714].
Reduced-cost Variable Fixing
λL
vb
xg
−−
≥)(
Lower bound on
optimal value of
original problem
Optimal value of
relaxation
�In general, given a constraint g(x)
≤bwith Lagrange
multiplier
λ, the following constraint is valid:
�If g(x)
≤b is a nonnegativity constraint
−xj
≤0 with
Lagrange multiplier
λ(i.e, reduced cost of x jis −
λ), then
λL
vx
j
−−
≥−
λL
vx
j
−≤
or
�If xjis a 0-1 variable and (v
−L)/
λ< 1, then we can fix
x jto 0. This is reduced cost variable fixing.
Example: Product Configuration
�Variable indices
�Filtering for elementconstraint
�Relaxation of element
�Relaxing a disjunction of linear systems
Product Configuration
�We want to configure a computer by choosing the type of
power supply, the type of disk drive, and the type of memory
chip.
�We also choose the number of disk drives and memory chips.
�Use only 1 type of disk drive and 1 type of memory.
�Constraints:
�Generate enough power for the components.
�Disk space at least 700.
�Memory at least 850.
�Minimize weight subject to these constraints.
Mem
ory
Mem
ory
Mem
ory
Mem
ory
Mem
ory
Mem
ory
Pow
ersu
pply
Pow
ersu
pply
Pow
ersu
pply
Pow
ersu
pply
Dis
k dr
ive
Dis
k dr
ive
Dis
k dr
ive
Dis
k dr
ive
Dis
k dr
ive
Per
sona
l com
pute
r
Product Configuration
Product Configuration
Product Configuration
�Let t i= type of component iinstalled.
qi = quantity of component iinstalled.
�These are the problem variables.
�This problem will illustrate the elementconstraint.
jU
vL
jA
qv
vc
jj
j
iij
ti
j
jj
j
i
al
l ,
al
l ,
min
≤≤=∑∑
Am
ount
of a
ttrib
ute
jpr
oduc
ed
(< 0
if c
onsu
med
):
mem
ory,
hea
t, po
wer
, w
eigh
t, et
c.
Qua
ntity
of
com
pone
nt i
inst
alle
d
Product Configuration
jU
vL
jA
qv
vc
jj
j
iij
ti
j
jj
j
i
al
l ,
al
l ,
min
≤≤=∑∑
Am
ount
of a
ttrib
ute
jpr
oduc
ed
(< 0
if c
onsu
med
):
mem
ory,
hea
t, po
wer
, w
eigh
t, et
c.
Qua
ntity
of
com
pone
nt i
inst
alle
d
Am
ount
of a
ttrib
ute
jpr
oduc
ed b
y ty
pe t
iof
com
pone
nt i
Product Configuration
jU
vL
jA
qv
vc
jj
j
iij
ti
j
jj
j
i
al
l ,
al
l ,
min
≤≤=∑∑
Am
ount
of a
ttrib
ute
jpr
oduc
ed
(< 0
if c
onsu
med
):
mem
ory,
hea
t, po
wer
, w
eigh
t, et
c.
Qua
ntity
of
com
pone
nt i
inst
alle
d
t i is
a v
aria
ble
in
dex
Am
ount
of a
ttrib
ute
jpr
oduc
ed b
y ty
pe t
iof
com
pone
nt i
Product Configuration
jU
vL
jA
qv
vc
jj
j
iij
ti
j
jj
j
i
al
l ,
al
l ,
min
≤≤=∑∑
Am
ount
of a
ttrib
ute
jpr
oduc
ed
(< 0
if c
onsu
med
):
mem
ory,
hea
t, po
wer
, w
eigh
t, et
c.
Uni
t cos
t of p
rodu
cing
at
trib
ute
j
Qua
ntity
of
com
pone
nt i
inst
alle
d
t i is
a v
aria
ble
in
dex
Am
ount
of a
ttrib
ute
jpr
oduc
ed b
y ty
pe t
iof
com
pone
nt i
Product Configuration
�Variable indices are implemented with the elementconstraint.
�The yin xyis a variable index.
�Constraint programming solvers implement x yby replacing
it with a new variable zand adding the constraint
element(y,(x1,…,xn),z).
�This constraint sets zequal to the y th element of the list
(x1,…,xn).
�So is replaced by and the new constraint
Variable Indices
This image cannot currently be displayed.
()
ijij
ni
iji
iz
Aq
Aq
t),
,,
(,el
emen
t1…
∑ iij
ti
iA
q
�The element constraint can be propagated and relaxed.
∑ iijz
for each i.
Filtering for Element
)),
,,
(,(
elem
ent
1z
xx
yn
…
can be processed with a domain reduction algorithm that
maintains arc consistency.
othe
rwis
e}{
if
}|
{
{j
j
j
y
j
x
yz
x
xx
yy
Dj
xz
z
D
jD
DD
DD
jD
D
DD
D
=←
∅≠
∩∩
←
∩←
∈∪
Domain of z
Example...
)),
,,
,(,
(el
emen
t4
32
1z
xx
xx
y
The initial domains are:
}70,
50,40{
}90,
80,50,
40{
}20,
10{
}50,
10{
}4,3,1{
}90,
80,60,
30,20{
4321
======
xxxxyz
DDDDDD
The reduced domains are:
}70,
50,40{
}90,
80{
}20,
10{
}50,
10{
}3{
}90,
80{
4321
======
xxxxyz
DDDDDD
Filtering for Element
Relaxation of Element
�element(y,(x1,…,xn),z) can be given a continuous relaxation.
�It implies the disjunction
)(
kk
xz
=∨
�In general, a disjunction of linear systems can be given a
convex hullrelaxation.
�This provides a way to relax the element constraint.
Relaxing a Disjunction of Linear Systems
�Consider the disjunction
)(
kk
kb
xA
≤∨
�It describes a union of polyhedra defined by Ak x
≤bk .
�Every point xin the convex hull of this union is a convex
combination of points
in the polyhedra.
11
bx
A≤
22
bx
A≤
33
bx
A≤
x
1x
2x
3x
Convex hull
kx
Relaxing a Disjunction of Linear Systems
�So we have
∑
∑
≥=≤
=
kk
k
kk
kk
kk
kb
xA
xx
0,1
al
l ,
αα
α
Using the change of variable
we get the convex hull relaxation
kk
kx
xα
=
∑
∑
≥=≤
=
kk
k
kk
kk
k
k
kb
xA
xx
0,1
al
l , α
αα
Relaxation of Element
�The convex hull relaxation of element(y,(x1, …, x n), z) is the
convex hull relaxation of the disjunction , which
simplifies to
ix
x
xz
kik
i
kkk
al
l ,
∑∑ ==
)(
kk
xz
=∨
()
ji
zA
At
jU
vL
zv
vc
ijij
mij
i
ij
jj
ijj
jj
j
, al
l ,
),,
,(,
elem
ent
al
l,
,
min
1…
∑∑≤
≤=
Product Configuration
�Returning to the product configuration problem, the model
with the element constraint is
with domains {
}{
}{
}
{}{
}3,2,1,
,1
al
l],
,[
,,
,,
,,
,
32
1
32
1
∈∈∈
∈∈
∈
q
jU
Lv
CB
At
BA
tC
BA
t
jj
j
Type of power supply
Type of disk
Type of memory
Number of
power
supplies
Number of disks,
memory chips
Product Configuration
�We can use knapsack cuts to help filter domains. Since
, the constraint implies
where each zijhas one of the values
],
[j
jj
UL
v∈
∑=
jij
jz
v∑
≤≤
j
U jij
L jv
zv
ijm
iij
iA
qA
q,
,1…
�This implies the knapsack inequalities
∑≤
iijk
ki
jA
qL
}{
max
ji
ijkk
iU
Aq
≤∑
}{
min
�These can be used to reduce the domain of qi.
Product Configuration
�Using the lower bounds L j, for power, disk space and memory,
we have the knapsack inequalities
}40
0,
300
,25
0m
ax{
}0,0m
ax{
}0,0,0m
ax{
850
}0,0,0m
ax{
}80
0,
500
max
{}0,0,0
max
{70
0
}30
,25
,20
max
{}
50,
30m
ax{
}15
0,
100
,70
max
{0
32
1
32
1
32
1
q
q
q
++
≤+
+≤
−−
−+
−−
+≤
which simplify to
32
32
1
400
850
800
700
2030
150
0
q
≤≤−
−≤
�Propagation of these reduces domain of q3from {1,2,3} to {3}.
Product Configuration
�Propagation of all constraints reduces domains to
]10
40,
565
[]
1200
,90
0[
]16
00,
700
[]
45,0[
]90,
75[]
600
,14
0[
]35
0,
350
[
]12
0,
900
[]0,0[
]0,0[
]0,0[]
1600
,70
0[
]0,0[
]75
,90
[]
30,
75[
]15
0,
150
[
},
{}
,{
}{
}3{}2,1{
}1{
43
21
3424
14
3323
13
3222
12
3121
11
32
1
32
1
∈∈
∈∈
∈∈
∈∈
∈∈
∈∈
∈−
−∈
−−
∈∈
∈∈
∈∈
∈∈
vv
vv
zz
z
zz
z
zz
z
zz
z
CB
tB
At
Ct
q
Product Configuration
�Using the convex hull relaxation of the element constraints, we
have a linear relaxation of the problem:
ki
q
jv
qA
jq
Av
iq
jv
vv
iq
q
jq
Av
vc
iki
U ji
ijk
Dk
ii
ijk
Dk
L j
U ii
L i
U jj
L j
Dk
iki
iD
kik
ijk
j
jj
j
it
it
it
it
, al
l ,0
al
l ,
}{
min
al
l
},{
max
al
l ,
al
l ,
al
l ,
al
l ,
min ≥
≤
≤≤
≤≤
≤==
∑
∑∑∑∑
∑ ∈
∈
∈
∈
Current bounds on vj
Current bounds on qi
qik> 0 when type k
of component iis
chosen, and qikis the
quantity installed
Product Configuration
�The solution of the linear relaxation at the root node is
)70
5,
900
,10
00,
15()
,,
(
0 s
othe
r
321
41
33
22
11
====
==
==
vv
qqq
ijBAC
…
Use 1 type C power supply (t 1 = C)
Use 2 type A disk drives (t2 = A)
Use 3 type B memory chips (t3 = B)
�Since only one qik> 0 for each i, there is no need to
branch on the ti s.
�This is a feasible and therefore optimal solution.
�The problem is solved at the root node.
Min weight is 705.
Example: Machine Scheduling
�Edge finding
�Benders decomposition and nogoods
Machine Scheduling
�We want to schedule 5 jobs on 2 machines.
�Each job has a release time and deadline.
�Processing times and costs differ on the two
machines.
�We want to minimize total cost while observing
the time widows.
�We will first study propagation for the 1-m
achine
problem (edge finding).
�We will then solve the 2-machine problem using
Benders decompositionand propagation on the
1-machine problem
Machine Scheduling
Processing time on
machine A
Processing time on
machine B
�Jobs must run one at a time on each machine (disjunctive
scheduling). For this we use the constraint
Machine Scheduling
),
(edi
sjun
ctiv
pt
t= (t 1,…t n)
are start times
(variables)
p= (pi1,…pin)
are processing
times
(constants)
�The best known filtering algorithm for the disjunctive
constraint is edge finding.
�Edge finding does not achieve full arc or bounds
consistency.
�The problem can be written
Machine Scheduling
() )
|(
),|
(e
disj
unct
iv
al
l ,
min
iy
pi
yt
jd
pt
f
jij
jj
jj
yj
jj
y
j
j
==
≤+
∑
Start time of job j
Machine assigned
to job j
�We will first look at a scheduling problem on 1 machine.
Edge Finding
�Let’s try to schedule jobs 2, 3, 5 on machine A.
�Initially, the Earliest Start Time Ejand Latest End Time Lj
are the release dates r jand deadlines dj :
]7,3[]
,[
]7,2[]
,[
]8,0[]
,[
55
33
22
===
EL
EL
EL
05
10
Job 2
Job 3
Job 5
E3
L 3
= time window
Edge Finding
�Job 2 must precede jobs 3 and 5, because:
�Jobs 2, 3, 5 will not fit between the earliest start time of
3, 5 and the latest end time of 2, 3, 5.
�Therefore job 2 must end before jobs 3, 5 start.
05
10
Job 2
Job 3
Job 5
min{E
3,E5}
max{L2,L3,L5}
pA2
pA3
pA5
Edge Finding
�Jobs 2 precedes jobs 3, 5 because:
05
10
Job 2
Job 3
Job 5
min{E
3,E5}
max{L2,L3,L5}
pA2
pA3
pA5
53
25
35
32
},
min
{}
,,
max
{A
AA
pp
pE
EL
LL
++
<−
�This is called edge finding because it finds an edge in the
precedence graph for the jobs.
Edge Finding
�In general, job kmust precede set Sof jobs on machine iwhen
05
10
Job 2
Job 3
Job 5
min{E
3,E5}
max{L2,L3,L5}
pA2
pA3
pA5
∑ ∪∈
∈∪
∈<
−}
{}
{}
{m
in}
{m
axk
Sj
ijj
Sj
jk
Sj
pE
L
Edge Finding
�Since job 2 must precede 3 and 4, the earliest start times of jobs
3 and 4 can be updated.
�They cannot start until job 2 is finished.
05
10
Job 2
Job 3
Job 5
pA2
[E3,L3] updated to [3,7]
Edge Finding
This image cannot currently be displayed.
�Edge finding also determines that job 3 must precede job 5, since
05
10
Job 2
Job 3
Job 5
pA2
[E5,L5] updated to [5,7]
53
55
33
25
4}3
min
{}7,7
max
{}
min
{}
,m
ax{
AA
pp
EL
L+
=+
=<
=−
=−
pA3
pA5
�This updates [E5,L5] to [5,7], and there is too little time
to run job 5. The schedule is infeasible.
Benders Decomposition
�We will solve the 2-machine scheduling problem with logic-
based Benders decomposition.
�First solve an assignment problem that allocates jobs to machines
(master problem).
�Given this allocation, schedule the jobs assigned to each machine
(subproblem).
�If a machine has no feasible schedule:
�Determine which jobs cause the infeasibility.
�Generate a nogood(Benders cut) that excludes assigning
these jobs to that machine again
�Add the Benders cut to the master problem and re-solve it.
�Repeat until the subproblem is feasible.
�We will solve it as an integer programming problem.
�Let binary variable xij= 1 when yj= i.
Benders Decomposition
cuts
B
ende
rs
min∑ j
jy
jf
�The m
aster problem(assigns jobs to machines) is:
}1,0{
cuts
B
ende
rs
min ∈∑
ij
ijij
ij
x
xf
Benders Decomposition
�We will add a relaxation of the subproblemto the master
problem, to speed up solution.
8
5
3
10
8
5
min
55
44
33
22
55
44
33
54
55
44
33
22
11
55
44
33
22
55
44
33
≤+
++
≤+
+≤≤
++
++
≤+
++
≤+
+
∑
BB
BB
BB
BB
AA
AA
AA
AA
AA
AA
AA
AA
AA
AA
AA
AA
AA
AA
AA
AA
ijij
ij
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xp
xf
If jobs 3,4,5 are all
assigned to machine
A, their processing
time on that
machine must fit
between their
earliest release time
(2) and latest
deadline (7).
Benders Decomposition
�The solution of the master problem is
0 s
othe
r
14
53
21
==
==
==
ij
BA
AA
A
x
xx
xx
x
�Assign jobs 1,2,3,5 to machine A, job 4 to machine B.
�The subproblem is to schedule these jobs on the 2 machines.
�Machine B is trivial to schedule (only 1 job).
�We found that jobs 2, 3, 5 cannot be scheduled on machine A.
�So jobs 1, 2, 3, 5 cannot be scheduled on machine A.
Benders Decomposition 1
)1(
)1(
)1(
53
2≥
−+
−+
−A
AA
xx
x
�So we create a Benders cut(nogood) that excludes assigning
jobs 2, 3, 5 to machine A.
�These are the jobs that caused the infeasibility.
�The Benders cut is:
�Add this constraint to the master problem.
�The solution of the master problem now is:
0 s
othe
r
14
35
21
==
==
==
ij
BB
AA
A
x
xx
xx
x
Benders Decomposition
�So we schedule jobs 1, 2, 5 on machine A, jobs 3, 4 on
machine B.
�These schedules are feasible.
�The resulting solution is optimal, with min cost 130.