Introduction à l'Analyse Numériqueama/courses/math0006-3/course/AN2015-S4.pdf · Quentin Louveaux...

Linear optimization

Quentin Louveaux (ULg - Institut Montefiore) Introduction a l’Analyse Numerique 2013 1 / 28

Production problem

A furniture company produces two types of chairs from beech and oak wood.

The first type (basic) requires 9 boards of beech wood and 2 boards of oak wood. Abasic chair is easy to construct and requires one worker’s hour.

The second type (classic) requires 7 boards of beech wood and 5 boards of oak wood.Due to the more important completion time, the classic chair requires three worker’shours.

The price of a basic chair is 30 euros whereas a classic chair’s price is 70 euros.

The stock of boards of the entreprise is 800 boards of beech wood and 200 boards of oakwood. There are 4 workers working each 40 hours per week.

What is the number of basic and classic chairs that the entreprise must produce for thisweek ?

Production problem

Modeling of the problemChoice of the decision variablesxB = Number of basic chairs to construct this weekxC = Number of classic chairs to construct this week

Objective to optimizemax 30xB + 70xC

ConstraintsBeech : 9xB + 7xC ≤ 800Oak : 2xB + 5xC ≤ 200Work : xB + 3xC ≤ 160Other : xB ≥ 0, xC ≥ 0 (and xB , xC integer).

max 30xB+70xC

s.t. 9xB+ 7xC ≤1000

2xB+ 5xC ≤200

xB+ 3xC ≤160

xB , xC ≥0

max 30xB+70xC

s.t. 9xB+ 7xC ≤1000

2xB+ 5xC ≤200

xB+ 3xC ≤160

xB , xC ≥0

max 30xB+70xC

s.t. 9xB+ 7xC ≤1000

2xB+ 5xC ≤200

xB+ 3xC ≤160

xB , xC ≥0

max 30xB+70xC

s.t. 9xB+ 7xC ≤1000

2xB+ 5xC ≤200

xB+ 3xC ≤160

xB , xC ≥0

Alloy production

The company Steel has received an order for 500 tons of steel to be used in shipbuilding.This steel must have the following characteristics

Chemical element Minimum grade Maximum gradeCarbon (C) 2 3Copper (Cu) 0.4 0.6Manganese (Mn) 1.2 1.65

The company has seven different raw materials in stock that may be used for theproduction of this steel. The following Table lists the grades, available amounts andprices for all raw materials.

Raw material C % Cu % Mn % Availability in t Cost in e/tIron alloy 1 2.5 0 1.3 400 200Iron alloy 2 3 0 0.8 300 250Iron alloy 3 0 0.3 0 600 150Copper alloy 1 0 90 0 500 220Copper alloy 2 0 96 4 200 240Aluminum alloy 1 0 0.4 1.2 300 200Aluminum alloy 2 0 0.6 0 250 165

The objective is to determine the composition of the steel that minimizes the productioncost.

Alloy production

Formulation

Decision Variables

usei : Quantity of alloy i used (i ∈ I )

Objective to optimize

min∑i∈I

pricei usei

Constraints

Carbon : LBC ≤∑

i∈I Ci usei ≤ UBC

Copper : LBCu ≤∑

i∈I Cui usei ≤ UBCu

Manganese :LBMn ≤∑

i∈I Mni usei ≤ UBMn

Availability : 0 ≤ usei ≤ StockiProduction :

∑i∈I usei = Demand

Formulation

Decision Variables

min∑i∈I

pricei usei

Constraints

Carbon : LBC ≤∑

Formulation

Decision Variables

min∑i∈I

pricei usei

Constraints

Carbon : LBC ≤∑

Formulation

Decision Variables

min∑i∈I

pricei usei

Constraints

Carbon : LBC ≤∑

Different forms of linear programming

max 2x1 + 3x2

s.t. 3x1 + x2 ≤ 3

x1 − x2 = 2

x1, x2 ≥ 0

min − 2x1 + 3x2

s.t. 3x1 + 4x2 ≥ 3

x1 − 2x2 ≤ 2

x1 ≥ 0,

− 3 ≤ x2 ≤ 0

min 2x1 − 3x2

s.t. 7x1 − x2 ≤ 3

x1 + 2x2 = 5

x1 ≥ 0, x2 ∈ R

Objective : min ou maxConstraints : ≥,≤,=Bounds : ≥ 0,≤ 0, [l , u],R

max 2x1 + 3x2

s.t. 3x1 + x2 ≤ 3

x1 − x2 = 2

x1, x2 ≥ 0

min − 2x1 + 3x2

s.t. 3x1 + 4x2 ≥ 3

x1 − 2x2 ≤ 2

x1 ≥ 0,

− 3 ≤ x2 ≤ 0

min 2x1 − 3x2

s.t. 7x1 − x2 ≤ 3

x1 + 2x2 = 5

x1 ≥ 0, x2 ∈ R

max 2x1 + 3x2

s.t. 3x1 + x2 ≤ 3

x1 − x2 = 2

x1, x2 ≥ 0

min − 2x1 + 3x2

s.t. 3x1 + 4x2 ≥ 3

x1 − 2x2 ≤ 2

x1 ≥ 0,

− 3 ≤ x2 ≤ 0

min 2x1 − 3x2

s.t. 7x1 − x2 ≤ 3

x1 + 2x2 = 5

x1 ≥ 0, x2 ∈ R

max 2x1 + 3x2

s.t. 3x1 + x2 ≤ 3

x1 − x2 = 2

x1, x2 ≥ 0

min − 2x1 + 3x2

s.t. 3x1 + 4x2 ≥ 3

x1 − 2x2 ≤ 2

x1 ≥ 0,

− 3 ≤ x2 ≤ 0

min 2x1 − 3x2

s.t. 7x1 − x2 ≤ 3

x1 + 2x2 = 5

x1 ≥ 0, x2 ∈ R

max 2x1 + 3x2

s.t. 3x1 + x2 ≤ 3

x1 − x2 = 2

x1, x2 ≥ 0

min − 2x1 + 3x2

s.t. 3x1 + 4x2 ≥ 3

x1 − 2x2 ≤ 2

x1 ≥ 0,

− 3 ≤ x2 ≤ 0

min 2x1 − 3x2

s.t. 7x1 − x2 ≤ 3

x1 + 2x2 = 5

x1 ≥ 0, x2 ∈ R

Different forms of linear programmingWe can go equivalently from one form to the other

Objective :max f (x) ≡ −min −f (x)

max 2x1 − 7x2 ≡ −min−2x1 + 7x2

Constraints :

f (x) ≤ b ≡ −f (x) ≥ −b 2x1 − x2 ≤ 1 ≡ −2x1 + x2 ≥ −1

f (x) = b ≡ f (x) ≤ b et f (x) ≥ b 3x1 − x2 = 3 ≡ 3x1 − x2 ≤ 3 et 3x1 − x2 ≥ 3

f (x) ≤ b ≡ f (x) + s = b, with s ≥ 0 3x1 − 2x2 ≥ 0 ≡ 3x1 − 2x2 − s ≥ 0 with s ≥ 0

Bounds :x ≤ 0 ≡ x := −x et x ≥ 0

y ∈ R → y = y+ − y− and y+, y− ≥ 0 !Not equivalent !

max 2x1 − 7x2 ≡ −min−2x1 + 7x2

Constraints :

f (x) ≤ b ≡ −f (x) ≥ −b 2x1 − x2 ≤ 1 ≡ −2x1 + x2 ≥ −1

Bounds :x ≤ 0 ≡ x := −x et x ≥ 0

max 2x1 − 7x2 ≡ −min−2x1 + 7x2

Constraints :

f (x) ≤ b ≡ −f (x) ≥ −b 2x1 − x2 ≤ 1 ≡ −2x1 + x2 ≥ −1

Bounds :x ≤ 0 ≡ x := −x et x ≥ 0

max 2x1 − 7x2 ≡ −min−2x1 + 7x2

Constraints :

f (x) ≤ b ≡ −f (x) ≥ −b 2x1 − x2 ≤ 1 ≡ −2x1 + x2 ≥ −1

Bounds :x ≤ 0 ≡ x := −x et x ≥ 0

max 2x1 − 7x2 ≡ −min−2x1 + 7x2

Constraints :

f (x) ≤ b ≡ −f (x) ≥ −b 2x1 − x2 ≤ 1 ≡ −2x1 + x2 ≥ −1

Bounds :x ≤ 0 ≡ x := −x et x ≥ 0

max 2x1 − 7x2 ≡ −min−2x1 + 7x2

Constraints :

f (x) ≤ b ≡ −f (x) ≥ −b 2x1 − x2 ≤ 1 ≡ −2x1 + x2 ≥ −1

Bounds :x ≤ 0 ≡ x := −x et x ≥ 0

Standard form

The standard form consists in

Objective : minimization

Constraints : equalities

Bounds : Nonnegative variables

min cT x

s.t. Ax = b

x ∈ Rn+

Exercise : Reduce a given problem into standard form

Standard form

min cT x

s.t. Ax = b

x ∈ Rn+

Standard form

min cT x

s.t. Ax = b

x ∈ Rn+

Standard form

min cT x

s.t. Ax = b

x ∈ Rn+

Graphic representationWe can represent a problem in two dimensions graphically.Example :

max x1+2x2 (1)

−x1+2x2 ≤ 1 (2)

−x1+ x2 ≤ 0 (3)

4x1+3x2 ≤ 12 (4)

x1, x2 ≥ 0 (5)

Graphic representation

max x1+2x2 (1)

−x1+2x2 ≤ 1 (2)

−x1+ x2 ≤ 0 (3)

4x1+3x2 ≤ 12 (4)

x1, x2 ≥ 0 (5)

max x1+2x2 (1)

−x1+2x2 ≤ 1 (2)

−x1+ x2 ≤ 0 (3)

4x1+3x2 ≤ 12 (4)

x1, x2 ≥ 0 (5)

��

1Quentin Louveaux (ULg - Institut Montefiore) Introduction a l’Analyse Numerique 2013 9 / 28

max x1+2x2 (1)

−x1+2x2 ≤ 1 (2)

−x1+ x2 ≤ 0 (3)

4x1+3x2 ≤ 12 (4)

x1, x2 ≥ 0 (5)

��

1+2xx1 =22

max x1+2x2 (1)

−x1+2x2 ≤ 1 (2)

−x1+ x2 ≤ 0 (3)

4x1+3x2 ≤ 12 (4)

x1, x2 ≥ 0 (5)

��

1+2xx1 =42

max x1+2x2 (1)

−x1+2x2 ≤ 1 (2)

−x1+ x2 ≤ 0 (3)

4x1+3x2 ≤ 12 (4)

x1, x2 ≥ 0 (5)

��

1+2xx1 2=4.8

Degenerate cases

In the example we had a unique solution at a vertex of the polyhedron.Some degenerate cases can lead to different solutions.

Degenerate cases

min x1 + x2

s.t. − x1 + x2 ≤ 1

x1, x2 ≥ 0

Degenerate cases

min x1

s.t. − x1 + x2 ≤ 1

x1, x2 ≥ 0

Degenerate cases

max − x1 + x2

s.t. − x1 + x2 ≤ 1

x1, x2 ≥ 0

Degenerate cases

max x1 + x2

s.t. − x1 + x2 ≤ 1

x1, x2 ≥ 0

Degenerate cases

max x1 + 2x2

s.t. − x1 + x2 ≤ 1

− x1 + x2 ≥ 2

x1, x2 ≥ 0x2

Polyhedra

Definition

A polyhedron is a set {x ∈ Rn|Ax ≥ b}

A set of the form Ax ≤ b is also a polyhedron.A set {x ∈ Rn|Ax = b, x ≥ 0} is a polyhedron in standard form.

Definition

Let a ∈ Rn \ {0}.(a) The set {x ∈ Rn|aT x = b} is a hyperplane

(b) The set {x ∈ Rn|aT x ≥ b} is a halfspace

Polyhedra

Definition

A polyhedron is a set {x ∈ Rn|Ax ≥ b}

A set of the form Ax ≤ b is also a polyhedron.A set {x ∈ Rn|Ax = b, x ≥ 0} is a polyhedron in standard form.

Definition

Let a ∈ Rn \ {0}.(a) The set {x ∈ Rn|aT x = b} is a hyperplane

(b) The set {x ∈ Rn|aT x ≥ b} is a halfspace

Convex Sets

Definition

A set S ⊆ Rn is convex if for all x , y ∈ S and all λ ∈ [0, 1], λx + (1− λ)y ∈ S .

Definition

Let x1, . . . , xk be vectors of Rn.

(i) λ1x1 + · · ·+ λkx

k is a conic combination if λ1, . . . , λk ≥ 0

(ii) λ1x1 + · · ·+ λkx

k is a convex combination if λ1, . . . , λk ≥ 0 and λ1 + · · ·+ λk = 1

(iii) The convex hull of x1, . . . , xk is the set of all convex combinations of x1, . . . , xk .

Theorem

(a) The intersection of two convex sets is convex

(b) Every polyhedron is convex

(c) The convex hull of a finite number of points is a polyhedron.

Convex Sets

Definition

(i) λ1x1 + · · ·+ λkx

(ii) λ1x1 + · · ·+ λkx

Theorem

Convex Sets

Definition

(i) λ1x1 + · · ·+ λkx

(ii) λ1x1 + · · ·+ λkx

Theorem

Extreme points and vertices

Definition

Let P be a polyhedron. A point x ∈ P is an extreme point of P if there do not exist twopoints y , z ∈ P such that x is a convex combination of y and z .

Definition

Let P be a polyhedron. A point x ∈ P is a vertex of P if there exists c ∈ Rn such thatcT x < cT y for all y ∈ P and y 6= x .

Extreme points and vertices

Definition

Let P be a polyhedron. A point x ∈ P is an extreme point of P if there do not exist twopoints y , z ∈ P such that x is a convex combination of y and z .

Definition

Let P be a polyhedron. A point x ∈ P is a vertex of P if there exists c ∈ Rn such thatcT x < cT y for all y ∈ P and y 6= x .

Bases of a polyhedron

We subdivide the equalities and inequalities into three categories :

aTi x ≥ bi i ∈ M≥

aTi x ≤ bi i ∈ M≤

aTi x = bi i ∈ M=

Definition

Let x be a point satisfying aTi x = bi for some i ∈ M≥,M≤ or M=. The constraint i issaid to be active or tight.

Theorem

Let x ∈ Rn and let I be the set of active constraints for x . The three followingstatements are equivalent.

(i) There exist n linearly independent vectors in {ai |i ∈ I}(ii) span{ai |i ∈ I} = Rn

(iii) The system aTi x = bi , i ∈ I has a unique solution.

We subdivide the equalities and inequalities into three categories :

aTi x ≥ bi i ∈ M≥

aTi x ≤ bi i ∈ M≤

aTi x = bi i ∈ M=

Definition

Let x be a point satisfying aTi x = bi for some i ∈ M≥,M≤ or M=. The constraint i issaid to be active or tight.

Theorem

Let x ∈ Rn and let I be the set of active constraints for x . The three followingstatements are equivalent.

(i) There exist n linearly independent vectors in {ai |i ∈ I}(ii) span{ai |i ∈ I} = Rn

(iii) The system aTi x = bi , i ∈ I has a unique solution.

Definition

Let P be a polyhedron and let x ∈ Rn.

(a) x is a basic solution ifI all equalities (i ∈ M=) are activeI among the active constraints, there are n linearly independent

(b) if x is a basic solution that satisfies all constraints, then x is a feasible basic solution.

Theorem

Let P be a polyhedron and let x ∈ P. The three following statements are equivalent.

(i) x is a vertex

(ii) x is an extreme point

(iii) x is a basic feasible solution

Definition

Let P be a polyhedron and let x ∈ Rn.

(a) x is a basic solution ifI all equalities (i ∈ M=) are activeI among the active constraints, there are n linearly independent

(b) if x is a basic solution that satisfies all constraints, then x is a feasible basic solution.

Theorem

Let P be a polyhedron and let x ∈ P. The three following statements are equivalent.

(i) x is a vertex

(ii) x is an extreme point

(iii) x is a basic feasible solution

Polyhedra in standard form

Consider P = {x ∈ Rn|Ax = b, x ≥ 0}.We assume that the rows of A are linearly independent.

Theorem

A point x is a basic feasible solution if Ax = b and if there exist m indicesB(1), . . . ,B(m) such that

(i) The columns AB(1), . . . ,AB(m) are linearly independent

(ii) If i 6= B(1), . . . ,B(m), then xi = 0

Explanation :

m rowsn rows

)We have n + m constraints and n variables.A basic solution ⇒ n constraints satisfied with equality.The m equalities are automatically satisfied.There are n −m inequalities xi ≥ 0 that are active (the nonbasic variables).There are m inequalities xi ≥ 0 that are possibly not active (basic variables).

Theorem

(ii) If i 6= B(1), . . . ,B(m), then xi = 0

Explanation :

m rowsn rows

Theorem

(ii) If i 6= B(1), . . . ,B(m), then xi = 0

Explanation :

m rowsn rows

Construction of a basis

Procedure ( 6= Algorithm)

(i) Choose m linearly independent columns AB(1), . . . ,AB(m)

(ii) xi = 0 for all i 6= B(1), . . . ,B(m)

(iii) Solve Ax = b for the unknowns xB(1), . . . , xB(m)

If the solution x ≥ 0, then x is a basic feasible solution.We construct the basic matrix as

AB(1) AB(2) · · · AB(m)

)The nonbasic matrix N corresponds to nonbasic indices.The basic vector is xB = (xB(1), . . . , xB(m)) and the nonbasic vector corresponds to theother indices.We have

ABxB = b

xN = 0

ABxB + ANxN = b

Example

(ii) xi = 0 for all i 6= B(1), . . . ,B(m)

AB(1) AB(2) · · · AB(m)

ABxB = b

xN = 0

ABxB + ANxN = b

Example

(ii) xi = 0 for all i 6= B(1), . . . ,B(m)

AB(1) AB(2) · · · AB(m)

ABxB = b

xN = 0

ABxB + ANxN = b

Example

(ii) xi = 0 for all i 6= B(1), . . . ,B(m)

AB(1) AB(2) · · · AB(m)

ABxB = b

xN = 0

ABxB + ANxN = b

Example

(ii) xi = 0 for all i 6= B(1), . . . ,B(m)

AB(1) AB(2) · · · AB(m)

ABxB = b

xN = 0

ABxB + ANxN = b

Example

Some important remarks

Correspondence between the base and the basic solutionTwo different bases could lead to the same solution x .

Adjacent BasesTwo bases are adjacent they differ by only one index.Differently stated they have n − 1 indices in common !

Degenerescence

Definition

A basic solution x ∈ Rn is degenerate if more than n constraints are active at the solution.

Degenerescence for a standard form

Let P = {x ∈ Rn|Ax = b, x ≥ 0}, with A ∈ Rm×n. A basic solution x is degenerate if xhas more than n −m zero elements.

Remark : Degenerescence may be representation-dependent.A non degenerate basis can be degenerate in another representation of the problem andconversely.Example :

Degenerescence

Definition

Degenerescence

Definition

The simplex algorithm

Principle

Start at a feasible basic solution

Check whether the current basis is optimal

If not, find a direction of improvement

The direction of improvement leads to either . . . a better feasible basic solution

or . . . proving that the problem is unbounded

Fundamental question

Let x be a feasible basic solution.Find a direction d such that x + θd is feasible for some θ ≥ 0

Principle

Finding a feasible direction from a basic feasible solution

Given : x = (xB xN) ∈ PFind d such that x + dθ ∈ P

We want to change at least one nonbasic solution.Select one index j ∈ N

dj = 1 di = 0 for all i 6= j

We have Ax = b and A(x + θd) = b ⇒ Ad = 0

0 = Ad =n∑

AB(i)dB(i) + Aj

BdB + Aj

dB = −B−1Aj is the j th basic direction

0 = Ad =n∑

AB(i)dB(i) + Aj

BdB + Aj

0 = Ad =n∑

AB(i)dB(i) + Aj

BdB + Aj

0 = Ad =n∑

AB(i)dB(i) + Aj

BdB + Aj

Taking care of the nonnegativity constraints

Nonbasic variables : xi (i 6= j) : = 0 : ⇒ OK !xj : goes in the positive direction ⇒ OK !

Basic VariablesNondegenerate case : xB > 0Therefore xB + θdB ≥ 0 for θ sufficiently small

Degenerate case : xB(i) = 0 for some i2 cases(1) (dB)i = (−B−1Aj)i ≥ 0 ⇒ feasible for θ small enough(2) (dB)i = (−B−1Aj)i < 0 ⇒ not feasible for any θ

Checking optimality

Essential question : Is the current basis optimal or can we find a basic direction thatimproves the objective function ?Definition : cB = (cB(1), . . . , cB(m))Let us compute the rate of objective change for the j th basic direction.d = (dB dN) with dB = −B−1Aj dN = ej

cTd = cTdB + cTdN

= −cTB B−1Aj + cj

Definition

The reduced cost of the j th nonbasic variable is

cj = cj − cTB B−1Aj .

Checking optimality

Essential question : Is the current basis optimal or can we find a basic direction thatimproves the objective function ?Definition : cB = (cB(1), . . . , cB(m))Let us compute the rate of objective change for the j th basic direction.d = (dB dN) with dB = −B−1Aj dN = ej

cTd = cTdB + cTdN

= −cTB B−1Aj + cj

Definition

The reduced cost of the j th nonbasic variable is

cj = cj − cTB B−1Aj .

Optimality Criterion

Theorem

Consider a basic feasible solution x

(i) If c ≥ 0 then x is optimal

(ii) If x is optimal and nondegenerate then c ≥ 0.

Definition

A basis matrix B is said to be optimal if

(i) B−1b ≥ 0 and

(ii) c = −cTB B−1A ≥ 0

Optimality Criterion

Theorem

Consider a basic feasible solution x

(i) If c ≥ 0 then x is optimal

(ii) If x is optimal and nondegenerate then c ≥ 0.

Definition

A basis matrix B is said to be optimal if

(i) B−1b ≥ 0 and

(ii) c = −cTB B−1A ≥ 0

Simplex algorithm

Iterative step : From a basis B doIf cj ≥ 0 for all j

then the current basis is optimalelse select a nonbasic variable j with cj < 0

We bring j into the basisWe now look for θ maximal such that x + θd ∈ P

If di ≥ 0 for all i ∈ Bthen x + θd ≥ 0 for all θ ≥ 0

The problem is unbounded and OPT = −∞If di < for some i ∈ B

then θ∗ = min{i∈B|dB(i)<0}

(− xB(i)

Variable i achieving the minimum goes out of the basisWe move to the next vertex (basic feasible solution)

B ← B ∪ {j} \ {i}New point:= x + θ∗d

Simplex algorithm

(− xB(i)

Simplex algorithm

(− xB(i)

Simplex algorithm

(− xB(i)

Simplex algorithm

(− xB(i)

Simplex algorithm

(− xB(i)

Simplex algorithm

(− xB(i)

The operation of moving to another basis

In an iteration of the simplex algorithm, in the basis, we replace the variable i such that

i = arg min{−xB(i)

dB(i)| i ∈ B with dB(i) < 0}

by the variable j entering the basis (that was chosen with cj < 0).

Theorem

(i) The columns AB(k), k 6= i and Aj are linearly independent

(ii) The vector y = x + θ∗d is a basic feasible solution associated with the new basismatrix.

The operation of moving to another basis

In an iteration of the simplex algorithm, in the basis, we replace the variable i such that

i = arg min{−xB(i)

dB(i)| i ∈ B with dB(i) < 0}

by the variable j entering the basis (that was chosen with cj < 0).

Theorem

(i) The columns AB(k), k 6= i and Aj are linearly independent

(ii) The vector y = x + θ∗d is a basic feasible solution associated with the new basismatrix.

Finding an intial basis

2 issues : finding a feasible solution and finding the corresponding basis.

One easy case

Consider a problem of the type Ax ≤ b where b ≥ 0.The standard form, if we add the slack variables

a11x1 + · · ·+ a1nxn+s1 = b1

.... . .

am1x1 + · · ·+ amnxn +sm= bm

The point (x1 = 0, · · · , xn = 0) is a feasible solution and (s1 · · · sm) is the correspondingbasis.

In general

It is not easy to find a feasible basis.One can try any basis (choice of m variables) but there is no guarantee that it is feasible.In some cases, the problem is infeasible → it might be impossible to find a basic feasiblesolution.

Finding an intial basis

2 issues : finding a feasible solution and finding the corresponding basis.

One easy case

Consider a problem of the type Ax ≤ b where b ≥ 0.The standard form, if we add the slack variables

a11x1 + · · ·+ a1nxn+s1 = b1

.... . .

am1x1 + · · ·+ amnxn +sm= bm

The point (x1 = 0, · · · , xn = 0) is a feasible solution and (s1 · · · sm) is the correspondingbasis.

In general

It is not easy to find a feasible basis.One can try any basis (choice of m variables) but there is no guarantee that it is feasible.In some cases, the problem is infeasible → it might be impossible to find a basic feasiblesolution.

Using a linear program and the simplex algorithm to find an initial basis

Phase I

Consider a problem of the type Ax = b, x ∈ Rn+ where we assume (after suitable

multiplication of rows by −1) that b ≥ 0.

min ξ1 + · · ·+ ξm

a11x1 + · · ·+ a1nxn + ξ1 = b1

.... . .

am1x1 + · · ·+ amnxn + ξm = bm

If the optimal solution of the phase 1 problem is > 0 ⇒ the initial problem isinfeasible.

If the initial problem is feasible ⇒ the phase 1 problem has 0 as optimal solution

One more issue : how to obtain an initial basis for the original problem without ξivariables in it

Phase I

min ξ1 + · · ·+ ξm

a11x1 + · · ·+ a1nxn + ξ1 = b1

.... . .

Phase I

min ξ1 + · · ·+ ξm

a11x1 + · · ·+ a1nxn + ξ1 = b1

.... . .

Introduction à l'Analyse Numériqueama/courses/math0006-3/course/AN2015-S4.pdf · Quentin Louveaux...

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