TU/e
Algorithms (2IL15) – Lecture 12
Algorithms (2IL15) – Lecture 12
Linear Programming
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Algorithms (2IL15) – Lecture 12
TU/e
Summary of previous lecture
 ρ-approximation algorithm:
algorithm for which computed solution is within factor ρ from OPT
 to prove approximation ratio we usually need lower bound on OPT
(or, for maximization, upper bound)
 PTAS = polynomial-time approximation scheme
= algorithm with two parameters, input instance and ε > 0, such that
– approximation ratio is1+ ε
– running time is polynomial in n for constant ε
 FPTAS = PTAS whose running time is polynomial in 1/ ε
 some problems are even hard to approximate
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TU/e
Algorithms (2IL15) – Lecture 12
Today: linear programming (LP)
 most used and most widely studied optimization method
 can be solved in polynomial time (input size measured in bits)
 can be used to model many problems
 also used in many approximation algorithms (integer LP + rounding)
we will only have a very brief look at LP …
 what is LP? what are integer LP and 0/1-LP ?
 how can we model problems as an LP ?
… and not study algorithms to solve LP’s
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Algorithms (2IL15) – Lecture 12
TU/e
Example problem: running a chocolate factory
Assortment
caca
o
mil
k
hazelnut
s
retail price (100
g)
1
0
0
1.99
Creamy Milk
0.6
0.4
0
1.49
Hazelnut
Delight
0.6
0.2
0.2
1.69
Super Nuts
0.5
0.1
0.4
Cost and availability of ingredients
1.79
Pure Black
cost (kg)
available (kg)
cacao
2.1
50
milk
0.35
50
hazelnuts
1.9
30
How much should we produce of
each product to maximize our profit ?
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Algorithms (2IL15) – Lecture 12
TU/e
Modeling the chocolate-factory problem
cacao
milk
hazelnuts
price (100 g)
1
0
0
1.99
Creamy Milk
0.6
0.4
0
1.49
Hazelnut Delight
0.6
0.2
0.2
Super Nuts
0.5
0.1
0.4
Pure Black
cost (kg)
available (kg)
cacao
2.1
50
1.69
milk
0.35
50
1.79
hazelnuts
1.9
30
variables we want to determine: production (kg) of the products
b = production of Pure Black
m = production of Creamy Milk
h = production of Hazelnut Delight
s = production of Super Nuts
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Algorithms (2IL15) – Lecture 12
TU/e
Modeling the chocolate-factory problem (cont’d)
cacao
milk
hazelnuts
price (100 g)
1
0
0
1.99
Creamy Milk
0.6
0.4
0
1.49
Hazelnut Delight
0.6
0.2
0.2
Super Nuts
0.5
0.1
0.4
Pure Black
cost (kg)
available (kg)
cacao
2.1
50
1.69
milk
0.35
50
1.79
hazelnuts
1.9
30
profits (per kg) of the products
Pure Black
19.9 – 2.1 = 17.8
Creamy Milk
14.9 – 0.6 x 2.1 + 0.4 x 0.35 = 13.5
Hazelnut Delight 15.19
Super Nuts:
16.055
total profit: 17.8 b + 13.5 m + 15.19 h + 16.055 s
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Algorithms (2IL15) – Lecture 12
TU/e
Modeling the chocolate-factory problem
cacao
milk
hazelnuts
price (100 g)
1
0
0
1.99
Creamy Milk
0.6
0.4
0
1.49
Hazelnut Delight
0.6
0.2
0.2
Super Nuts
0.5
0.1
0.4
Pure Black
cost (kg)
available (kg)
cacao
2.1
50
1.69
milk
0.35
50
1.79
hazelnuts
1.9
30
we want to maximize the total profit 17.8 b + 13.5 m + 15.19 h + 16.055 s
under the constraints
b + 0.6 m + 0.6 h + 0.5 s ≤ 50
0.4 m + 0.2 h + 0.1 s ≤ 50
0.2 h + 0.4 s ≤ 30
(cacao availability)
(milk availability)
(hazelnut availability)
This is a linear program: optimize linear function, under set of linear constraints
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Algorithms (2IL15) – Lecture 12
TU/e
Linear programming
n variables; here n=2, in chocolate
example n=4, but often n is large
Find values of real variables x, y such that
y

y≤½ x+2
x − 3 y is maximized
 subject to the constraints
−2x + y ≥ −4
x + y ≥ 3
−½x +y ≤ 2
y ≥ 0
1
−3
y ≥ 2x − 4
x
y≥0
y≥−x+3
m constraints:
objective function; must be
linear function in the variables
goal: maximize (or minimize)
linear function ≤ constant
or ≥ or =, > and < not allowed
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Algorithms (2IL15) – Lecture 12
TU/e
Linear programming
Find values of real variables x, y such that
y=½ x+2

x − 3 y is maximized
 subject to the constraints
−2x + y ≥ −4
x + y ≥ 3
−½x +y ≤ 2
y ≥ 0
1
−3
y=0
y = 2x − 4
y=−x+3
feasible region
= region containing feasible solutions
= region containing solutions satisfying all constraints
feasible region is convex polytope in n-dim space
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TU/e
Algorithms (2IL15) – Lecture 12
Linear programming:
Find values of real variables x1, …, xn such that
 given linear function c1 x1 + c2 x2 + … + cn xn is maximized (or: minimized)
 and given linear constraints on the variables are satisfied
constraints: equalities or inequalities using ≥ or ≤, cannot use < and >
Possible outcomes:
 unique optimal solution: vertex of feasible region
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TU/e
Algorithms (2IL15) – Lecture 12
Linear programming:
Find values of real variables x1, …, xn such that
 given linear function c1 x1 + c2 x2 + … + cn xn is maximized (or: minimized)
 and given linear constraints on the variables are satisfied
constraints: equalities or inequalities using ≥ or ≤, cannot use < and >
Possible outcomes:
 unique optimal solution: vertex of feasible region
 no solution: feasible region in empty
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TU/e
Algorithms (2IL15) – Lecture 12
Linear programming:
Find values of real variables x1, …, xn such that
 given linear function c1 x1 + c2 x2 + … + cn xn is maximized (or: minimized)
 and given linear constraints on the variables are satisfied
constraints: equalities or inequalities using ≥ or ≤, cannot use < and >
Possible outcomes:
 no solution: feasible region in empty
 unique optimal solution: vertex of feasible region
 bounded optimal solution, but not unique
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TU/e
Algorithms (2IL15) – Lecture 12
Linear programming:
Find values of real variables x1, …, xn such that
 given linear function c1 x1 + c2 x2 + … + cn xn is maximized (or: minimized)
 and given linear constraints on the variables are satisfied
constraints: equalities or inequalities using ≥ or ≤, cannot use < and >
Possible outcomes:
 no solution: feasible region in empty
 unique optimal solution: vertex of feasible region
 bounded optimal solution, but not unique
 unbounded optimal solution
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Algorithms (2IL15) – Lecture 12
TU/e
Linear programming: standard form
Maximize
c1 x1 + c2 x2 + … + cn xn
Subject to
a1,1 x1 + a1,2 x2 + … + a1,n xn
a2,1 x1 + a2,2 x2 + … + a2,n xn
....
am,1 x1 + am,2 x2 + … + am,n xn
not: minimize
x1 ≥ 0
x2 ≥ 0
…
xn ≥ 0
≤
≤
b1
b2
≤
bm
only “≤”
(no “=“ and no “≥”)
non-negativity constraints
for each variable
m x n matrix
Maximize c∙x subject to
A x ≤ b and non-negativity constraints on all xi
n-dimensional vectors
c, A, b are input, x must be computed
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Algorithms (2IL15) – Lecture 12
TU/e
Lemma: Any LP with n variables and m constraints can be rewritten as an
equivalent LP in standard form with 2n variables and 2n+2m constraints.
Proof. LP may not be in standard form because
 minimization instead of maximization
− negate objective function:
minimize 2 x1 − x2 + 4x3  maximize −2 x1 + x2 − 4 xn
 some constraints are ≥ or = instead of ≤
− getting rid of =:
replace
by
3 x + x2 − x3 = 5
3 x + x2 − x3 ≤ 5
3 x + x2 − x3 ≥ 5
− changing ≥ to ≤: negate constraint
3 x + x2 − x3 ≥ 5

− 3 x − x2 + x3 ≤ − 5
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Algorithms (2IL15) – Lecture 12
TU/e
Lemma: Any LP with n variables and m constraints can be rewritten as an
equivalent LP in standard form with 2n variables and 2n+2m constraints.
Proof (cont’d). LP may not be in standard form because
 minimization instead of maximization
 some constraints are ≥ or = instead of ≤
 variables without non-negativity constraint
− for each such variable xi introduce two new variables ui and vi
− replace each occurrence of xi by (ui − vi)
− add non-negativity constraints ui ≥ 0 and vi ≥ 0
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Algorithms (2IL15) – Lecture 12
TU/e
Lemma: Any LP with n variables and m constraints can be rewritten as an
equivalent LP in standard form with 2n variables and 2n+2m constraints.
Proof (cont’d).
 variables without non-negativity constraint
− for each such variable xi introduce two new variables ui and vi
− replace each occurrence of xi by (ui − vi)
− add non-negativity constraints ui ≥ 0 and vi ≥ 0
new problem is equivalent to original problem :
for any original solution there is new solution with same value
− if xi ≥ 0 then ui = xi and vi = 0, otherwise ui = 0 and vi = − xi
and vice versa
− set xi = (ui − vi)
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TU/e
Algorithms (2IL15) – Lecture 12
Lemma: Any LP with n variables and m constraints can be rewritten as an
equivalent LP in standard form with 2n variables and 2n+2m constraints.
Instead of standard form, we can also get so-called slack form:
– non-negativity constraint for each variable
– all other constraints are =, not ≥ or ≤
 Standard form (or slack form): convenient for developing LP algorithms
 When modeling a problem: just use general form
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TU/e
Algorithms (2IL15) – Lecture 12
Algorithms for solving LP’s
simplex method
− worst-case running time is exponential
− fast in practice
interior-point methods
− worst-case running time is polynomial in input size in bits
− some are slow in practice, others are competitive with simplex method
LP when dimension (=number of variables) is constant
− can be solved in linear time (see course Advanced Algorithms)
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TU/e
Algorithms (2IL15) – Lecture 12
Modeling a problem as an LP
 decide what the variables are (what are the choices to be made?)
 write the objective function to be optimized (should be linear)
 write the constraints on the variables (should be linear)
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Algorithms (2IL15) – Lecture 12
TU/e
Example: Max Flow
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Algorithms (2IL15) – Lecture 12
TU/e
2 /3
4 / 10
2 /2
2 /2
3 /3
source s
2 /2
t
sink
1/5
1 /1
1 /5
0/3
flow = 1, capacity = 5
Flow: function f : V x V → R satisfying
 capacity constraint:
0 ≤ f (u,v) ≤ c(u,v ) for all nodes u,v
 flow conservation: for all nodes u ≠ s, t we have flow in = flow out:
∑v in V f (v,u) = ∑v in V f (u,v)
value of flow: |f | = ∑v in V f (s,v) − ∑v in V f (v,s)
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Algorithms (2IL15) – Lecture 12
TU/e
Modeling Max Flow as an LP
 decide what the variables are (what are the choices to be made?)
for each edge (u,v) introduce variable xuv
( xuv represents f(u,v) )
 write the objective function to be optimized (should be linear)
maximize ∑v in V xsv − ∑v in V xvs
(note: linear function)
 write the constraints on the variables (should be linear)
xuv ≥ 0
for all pairs of nodes u,v
xuv ≤ c(u,v)
for all pairs of nodes u,v
∑v in V xvu − ∑v in V xuv = 0
for all nodes u ≠ s, t
(note: linear functions)
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Algorithms (2IL15) – Lecture 12
TU/e
Modeling Max Flow as an LP
Now write it down nicely
maximize
∑v in V xsv − ∑v in V xv,s
subject to
xuv ≥ 0
for all pairs of nodes u,v
xuv ≤ c (u,v)
for all pairs of nodes u,v
∑v in V xvu − ∑v in V xuv = 0
for all nodes u ≠ s, t
Conclusion: Max Flow can trivially be written as an LP
(but dedicated max-flow algorithm are faster than using LP algorithms)
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TU/e
Algorithms (2IL15) – Lecture 12
Example: Shortest Paths
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TU/e
Algorithms (2IL15) – Lecture 12
Shortest paths
weighted, directed graph G = (V,E )



weight (or: length) of a path
= sum of edge weights
δ (u,v)
= distance from u to v
= min weight of any path from u to v
shortest path from u to v
= any path from u to v of weight δ (u,v)
weighted, directed graph
v3
v1
v4
−2
1
2
1
4
v5
2
v6
2.3
v2
weight = 2
v7
δ(v1,v5) = 2
Is δ (u,v) always well defined?
No, not if there are negative-weight cycles.
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Algorithms (2IL15) – Lecture 12
TU/e
Modeling single-source single-target shortest path as an LP
Problem: compute distance δ(s,t) from given source s to given target t
 decide what the variables are (what are the choices to be made?)
for each vertex v introduce variable xv
( xv represents δ(s,v) )
 write the objective function to be optimized (should be linear)
minimize xt maximize xt
 write the constraints on the variables (should be linear)
xv ≤ xu + w(u,v)
xs = 0
for all edges (u,v) in E
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Algorithms (2IL15) – Lecture 12
TU/e
Modeling single-source single-target shortest path as an LP
variables: for each vertex v we have a variable xv
LP:
maximize xt
subject to xv ≤ xu + w(u,v)
xs = 0
for all edges (u,v) in E
Lemma: optimal solution to LP = δ(s,t).
Proof. (assume for simplicity that δ(s,t) is bounded)
≥ : consider solution where we set xv = δ(s,v) for all v
− solution is feasible and has value dist(s,t)  opt solution ≥ δ(s,t)
≤ : consider opt solution, and shortest path s = v0,v1,…,vk,vk+1 = t
− prove by induction that xi ≤ δ(s,vi)  opt solution ≤ δ(s,t)
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TU/e
Algorithms (2IL15) – Lecture 12
Example: Vertex Cover
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Algorithms (2IL15) – Lecture 12
TU/e
∩
G = (V,E) is undirected graph
vertex cover in G: subset C V such that for each edge (u,v) in E
we have u in C or v in C (or both)
Vertex Cover (optimization version)
Input: undirected graph G = (V,E)
Problem: compute vertex cover for G with minimum number of vertices
 Vertex Cover is NP-hard.
 there is a 2-approximation algorithm running in linear time.
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Algorithms (2IL15) – Lecture 12
TU/e
Modeling Vertex Cover as an LP
 decide what the variables are (what are the choices to be made?)
for vertex v introduce variable xv
( idea: xv = 1 if v in cover, xv = 0 if v not in cover)
 write the objective function to be optimized (should be linear)
minimize ∑v in V xv
(note: linear function)
 write the constraints on the variables (should be linear)
− for each edge (u,v) write constraint xu + xv ≥ 1 (NB: linear function)
− for each vertex v write constraint
xu in {0,1}
not a linear constraint
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TU/e
Algorithms (2IL15) – Lecture 12
integrality constraint: “xi must be integral”
0/1-constraint: “xi must 0 or 1”
integer LP: LP where all variables have integrality constraint
0/1-LP: LP where all variables have 0/1-constraint
(of course there are also mixed versions)
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Algorithms (2IL15) – Lecture 12
TU/e
Theorem: 0/1-LP is NP-hard.
Proof. Consider decision problem: is there feasible solution to given 0/1-LP?
Which problem do we use in reduction? already saw reduction from Vertex Cover;
let’s do another one: 3-SAT
Need to transform 3-SAT formula into instance of 0/1-LP
( x1 V x2 V ¬x3)
Λ (x2 V ¬x4 V ¬x5) Λ (¬x2 V x3 V x5 )
variable yi for each Boolean xi
yi = 1 if xi = TRUE and yi = 0 if xi = FALSE
maximize y1 (not relevant for decision problem, pick arbitrary function)
y1 + y2 + (1−y3) ≥ 1
y2 + (1−y4) + (1−y5 ) ≥ 1
(1 − y2) + y3 +
y5
≥ 1
yi in {0,1} for all i
subject to
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TU/e
Algorithms (2IL15) – Lecture 12
Theorem: 0/1-LP is NP-hard.
problem can be modeled as “normal” LP
 problem can be solved using LP algorithms
 problem can be solved efficiently
problem can be modeled as integer LP (or 0/1-LP)
 problem can be solved using integer LP (or 0/1-LP) algorithms
 does not mean that problem can be solved efficiently
(sometimes can get approximation algorithms by relaxation and rounding
see course Advanced Algorithms)
 there are solvers (software) for integer LPs that in practice are quite
efficient
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Algorithms (2IL15) – Lecture 12
TU/e
Summary
 what is an LP? what are integer LP and 0/1-LP?
 any LP can be written in standard form (or in slack form)
 normal (that is, not integer) LP can be solved in polynomial time
(with input size measured in bits)
 integer LP and 0/1-LP are NP-hard
 when modeling a problem as an LP
−
−
−
−
define variables and how they relate to the problem
describe objective function (should be linear)
describe constraints (should be linear, < and > not allowed)
no need to use standard or slack form, just use general form
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