Linear Programming - Çankaya Üniversitesiie501.cankaya.edu.tr/uploads/files/lp_c.pdf · Linear...

c©Levent Kandiller, Principles of Mathematics in Operations Research, The International Series in

Operations Research and Management Science, Vol. 97, Springer, 2007. ISBN: 0-387-37734-4

Linear ProgrammingLevent Kandiller

Industrial Engineering Department

Cankaya University, Turkey

Linear Programming – p.1/16



Definition

Linear Programming: optimization of a linear objectivefunction subject to finite number (m) of linear constraintswith n unknown and nonnegative decision variables.




Definition

Linear Programming: optimization of a linear objectivefunction subject to finite number (m) of linear constraintswith n unknown and nonnegative decision variables.

Example . The following is an LP:

Min z =2x + 3y

s.t.

2x + y ≥ 6

x + 2y ≥ 6

x, y ≥ 0.




Forms

Standard Form:

Min z =cT x

s.t.

Ax ≥ b

x ≥ θ




Forms

Canonical Form:

Min z =cT x + θT y

s.t.

Ax − y = b

x, y ≥ θ

⇔

Min z =[cT |θT ]

x

y

s.t.

[A| − I]

x

y

= b

x

y

≥ θ.




Geometry

Example . ConsiderMin z =2x + 3y

s.t.

2x + y ≥ 6

x + 2y ≥ 6

x, y ≥ 0.

24 3524 35 2666664 3777775




Geometry


s.t.

2x + y ≥ 6

x + 2y ≥ 6

x, y ≥ 0.

A =

24 1 2 −1 0

2 1 0 −1

35,

b =

24 6

6

35 , c =

2666664 2

3

0

03777775 .




Geometry


s.t.

2x + y ≥ 6

x + 2y ≥ 6

x, y ≥ 0.

A =

24 1 2 −1 0

2 1 0 −1

35,

b =

24 6

6

35 , c =

2666664 2

3

0

03777775 .

0

P

Q

R

x=0

2x+y=6

x+2y=6

y=0

Feasible

set

2x+3y=10

2x+3y=6

2x+3y=0




Basics

Definition . The extreme points of the feasible set areexactly the basic feasible solutions of Ax = b. A solution isbasic when n of its m + n components are zero, and isfeasible when it satisfies x ≥ θ.Phase I of the simplex method finds one basic feasiblesolution, and Phase II moves step by step to the optimalone.




Basics

If we are already at a basic feasible solution x, and for convenience we

reorder its components so that the n zeros correspond to free variables.

24 3524 3524 3524 35




Basics



x =

24 xB

xN = θ

35 , A = [B, N ], cT = (cTB

, cTN

)

24 3524 3524 35




Basics



x =

24 xB

xN = θ

35 , A = [B, N ], cT = (cTB

, cTN

)

Min z =(cTB , cT

N )

24 xB

xN = θ

35s.t.

[B|N ]

24 xB

xN = θ35 = b24 xB

xN = θ

35 ≥ θ.

⇔

Min z =cTBxB + cT

NxN

s.t.

BxB + NxN = b

xB , xN ≥ θ




Simplex Method

Let us take the constraintsBxB + NxN = b ⇔ Bxb = b − NxN ⇔

xB = B−1[b − NxN ] = B−1b − B−1NxN .




Simplex Method


xB = B−1[b − NxN ] = B−1b − B−1NxN .

Now plug xB in the objective functionz = cT

BxB + cT

NxN = cT

B[B−1b − B−1NxN ] + cT

NxN

z = cT

BB−1b + (cT

N− cBB−1N)xN .




Simplex Method


xB = B−1[b − NxN ] = B−1b − B−1NxN .


BxB + cT

NxN = cT


NxN

z = cT

BB−1b + (cT

N− cBB−1N)xN .

If we let xN = θ, then xB = B−1b ≥ θ ⇒ z = cT

BB−1b.




Simplex Method


xB = B−1[b − NxN ] = B−1b − B−1NxN .


BxB + cT

NxN = cT


NxN

z = cT

BB−1b + (cT

N− cBB−1N)xN .

If we let xN = θ, then xB = B−1b ≥ θ ⇒ z = cT

BB−1b.

Proposition (Optimality Condition ). If the vector (cT

N− cT

BB−1N) is nonnegative,

then no reduction in z can be achieved. Then, the current extreme point

(xB = B−1b, xN = θ) is optimal and the minimum objective function value is

cBB−1b.




Simplex Method

Assume that the optimality condition fails, the usual greedy strat-

egy is to choose the most negative component of cN − cBB−1N ,

known as Dantzig’s rule.




Simplex Method



known as Dantzig’s rule. Thus, we have determined which com-

ponent will move from free to basic, called as entering variable

xe.




Simplex Method



known as Dantzig’s rule. Thus, we have determined which com-

ponent will move from free to basic, called as entering variable

xe. We have to decide which basic component is to become

free, called as leaving variable, xl.




Simplex Method

Let N e be the column of N corresponding to xe. xB = B−1b −

B−1N exe. If we increase xe from 0, some entries of xB may begin

to decrease, and we reach a a neighboring extreme point when a

component of xB reaches 0. It is the component corresponding

to xl.




Simplex Method





to xl. At this extreme point, we have reached a new x which is

both feasible and basic: it is feasible because x ≥ θ, it is basic

since we again have n zero components.




Simplex Method







since we again have n zero components. xe is gone from zero to

α, replaces xl which is dropped to zero.




Simplex Method







since we again have n zero components. xe is gone from zero to

α, replaces xl which is dropped to zero. The other components

of xB might have changed their values, but remain positive.




Simplex Method

Proposition (Min Ratio ). Suppose u = Ne, then the value of xe will be:

α = minxj :basic

(B−1b)j

(B−1u)j= (B−1

b)l

(B−1u)l

and the objective function will decrease to cT

BB−1b − αB−1u.




Simplex Method


α = minxj :basic

(B−1b)j

(B−1u)j= (B−1

b)l

(B−1u)l



Remark (Unboundedness ). The minimum is taken only over positive components of

B−1u, since negative entries will increase xB and zero entries keeps xB as their

previous values. If there are no positive components, then the next extreme point is

infinitely far away, then the cost can be reduced forever; z = −∞! In this case we

term the optimization problem as unbounded.




Simplex Method


α = minxj :basic

(B−1b)j

(B−1u)j= (B−1

b)l

(B−1u)l



Remark (Degeneracy ). Suppose that more than n of the variables are zero or two

different components if the minimum ratio formula give the same minimum ratio. We

can choose either one of them to be made free, but the other will still be in the basis at

zero level. Thus, the new extreme point will have (n + 1) zero components.

Geometrically, there is an extra supporting plane at the extreme point. In degeneracy,

there is the possibility of cycling forever around the same set of extreme points without

moving toward x∗, the optimal solution. In general, one may assume nondegeneracy

hypothesis (xB = B−1b > θ).




Simplex Method

Example . Assume that we are at the extreme point P corresponding to the

following basic feasible solution:

0

P

Q

R

x=0

2x+y=6

x+2y=6

y=0

Feasible

set

2x+3y=10

2x+3y=6

2x+3y=0

24 35 2666664 3777775 2666664 37777752664 3775 0� 1Ah i h i24 35 24 35h i h i24 3524 35h i h i24 35 0� 1A24 35 24 35 24 35 24 3524 35 24 35 24 35 24 3524 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 6

6

0

0

3777775 =

2666664 z1

y

x

z23777775 ,

2664 3775 0� 1Ah i h i24 35 24 35h i h i24 3524 35h i h i24 35 0� 1A24 35 24 35 24 35 24 3524 35 24 35 24 35 24 3524 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 6

6

0

0

3777775 =

2666664 z1

y

x

z23777775 ,

A = [B|N ] =

2664 z1 y x z2

−1 2 1 0

0 1 2 −1

3775 , cT = (cTB|cT

N) =

0� z1 y x z2

0 3 2 0

1A .

h i h i 24 35 24 35h i h i24 3524 35h i h i24 35 0� 1A24 35 24 35 24 35 24 3524 35 24 35 24 35 24 3524 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 6

6

0

0

3777775 =

2666664 z1

y

x

z23777775 ,

A = [B|N ] =

2664 z1 y x z2

−1 2 1 0

0 1 2 −1

3775 , cT = (cTB|cT

N) =

0� z1 y x z2

0 3 2 0

1A .

cTN

− cTB

B−1N =

h2 0

i−h

0 3

i 24 −1 2

0 1

35−1 24 1 0

2 −1

35 .

h i h i 24 3524 35h i h i24 35 0� 1A24 35 24 35 24 35 24 3524 35 24 35 24 35 24 3524 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 6

6

0

0

3777775 =

2666664 z1

y

x

z23777775 ,

A = [B|N ] =

2664 z1 y x z2

−1 2 1 0

0 1 2 −1

3775 , cT = (cTB|cT

N) =

0� z1 y x z2

0 3 2 0

1A .

cTN

− cTB

B−1N =

h2 0

i−h

0 3

i 24 −1 2

0 1

35−1 24 1 0

2 −1

35 .24 −1 2 1 0

0 1 0 1

35→

24 1 −2 −1 0

0 1 0 1

35→

24 1 0 −1 2

0 1 0 1

35⇒ B−1 =

24 −1 2

0 1

35 .

h i h i 24 3524 35h i h i24 35 0� 1A24 35 24 35 24 35 24 3524 35 24 35 24 35 24 3524 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 6

6

0

0

3777775 =

2666664 z1

y

x

z23777775 ,

A = [B|N ] =

2664 z1 y x z2

−1 2 1 0

0 1 2 −1

3775 , cT = (cTB|cT

N) =

0� z1 y x z2

0 3 2 0

1A .

cTN

− cTB

B−1N =

h2 0

i−h

0 3

i 24 −1 2

0 1

35−1 24 1 0

2 −1

35 .

cTN

− cTB

B−1N =

h2 0

i−

h0 3

i 24 −1 2

0 1

3524 1 0

2 −1

35

h i h i24 35 0� 1A24 35 24 35 24 35 24 3524 35 24 35 24 35 24 3524 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

cTN

− cTB

B−1N =

h2 0

i−

h0 3

i 24 −1 2

0 1

3524 1 0

2 −1

35cTN

− cTB

B−1N =

h2 0

i−

h0 3

i 24 3 −2

2 −135 =

0� x z2

−4 3

1A .

24 35 24 35 24 35 24 3524 35 24 35 24 35 24 3524 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

cTN

− cTB

B−1N =

h2 0

i−

h0 3

i 24 −1 2

0 1

3524 1 0

2 −1

35cTN

− cTB

B−1N =

h2 0

i−

h0 3

i 24 3 −2

2 −135 =

0� x z2

−4 3

1A .

Example . Since the first component is negative, P is not optimal; x should enter the basis, i.e.

xe = x, Ne =

24 1

2

35⇒ B−1Ne =

24 3

2

35 , B−1b =

24 6

6

35 =

24 z1

y

35 ,

24 35 24 35 24 35 24 3524 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

xB = B−1b − B−1Nexe =

24 z1

y

35 =

24 6

6

35−

24 3

235x ≥24 0

0

35 .

24 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

xB = B−1b − B−1Nexe =

24 z1

y

35 =

24 6

6

35−

24 3

235x ≥24 0

0

35 .

⇒ α = Min{ 63

= 2, 62

= 3} = 2. Thus, xl = z1, xe = 2, y = 6 − 2α = 2.

24 35 2666664 3777775 2666664 3777775 h i 24 35h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 2

2

0

0

3777775 =

2666664 x

y

z1

z2

3777775 , A =

hB N

i=

24 1 2 −1 0

2 1 0 −1

35 .

h i h i 24 35h i h i24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 2

2

0

0

3777775 =

2666664 x

y

z1

z2

3777775 , A =

hB N

i=

24 1 2 −1 0

2 1 0 −1

35 .

cT =

hcTB

cTN

i=

h2 3 0 0

i, B =

24 1 2

2 1

35 ,

h i h i 24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 2

2

0

0

3777775 =

2666664 x

y

z1

z2

3777775 , A =

hB N

i=

24 1 2 −1 0

2 1 0 −1

35 .

cT =

hcTB

cTN

i=

h2 3 0 0

i, B =

24 1 2

2 1

35 ,h

B I

i

=

24 1 2 1 0

2 1 0 1

35→

24 1 2 1 0

0 −3 −2 1

35→

24 1 0 − 13

23

0 1 23

− 13

35 =

hI B−1

i

.

h i h i 24 3524 35h i h i24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 2

2

0

0

3777775 =

2666664 x

y

z1

z2

3777775 , A =

hB N

i=

24 1 2 −1 0

2 1 0 −1

35 .

cT =

hcTB

cTN

i=

h2 3 0 0

i, B =

24 1 2

2 1

35 ,

cTN

− cTB

B−1N =

h0 0

i−

h2 3

i 24 − 13

23

23

− 13

3524 −1 0

0 −1

35 =

h i h i 24 35 h i




Simplex Method

x =

24 xB

xN

35 =

2666664 2

2

0

0

3777775 =

2666664 x

y

z1

z2

3777775 , A =

hB N

i=

24 1 2 −1 0

2 1 0 −1

35 .

cT =

hcTB

cTN

i=

h2 3 0 0

i, B =

24 1 2

2 1

35 ,

cTN

− cTB

B−1N =

h0 0

i−

h2 3

i 24 − 13

23

23

− 13

3524 −1 0

0 −1

35 =h0 0

i−

h2 3

i 24 13

− 23

− 23

− 13

35 =

h43

13

i> θ.




Simplex Method

Example . Thus, extreme point Q is optimal, cTBB−1b = 10 is the optimal

value of the objective function.

0

P

Q

R

x=0

2x+y=6

x+2y=6

y=0

Feasible

set

2x+3y=10

2x+3y=6

2x+3y=0




Simplex Tableau

We have achieved a transition from the geometry of the simplex

method to algebra so far. In this section, we are going to analyze

a simplex step which can be organized in different ways.

24 35 24 35




Simplex Tableau

The Gauss-Jordan method gives rise to the simplex tableau.

[A‖b] = [B|N‖b] −→ [I|B−1N‖B−1b].

24 35 24 35




Simplex Tableau

The Gauss-Jordan method gives rise to the simplex tableau.

[A‖b] = [B|N‖b] −→ [I|B−1N‖B−1b].

Adding the cost row24 I B−1N B−1b

cTB

cTN

0

35 −→

24 I B−1N B−1b

0 cTN

− cTB

B−1N −cTB

B−1b

35 .




Simplex Tableau


cTB

cTN

0

35 −→

24 I B−1N B−1b

0 cTN

− cTB

B−1N −cTB

B−1b

35 .

The last result is the complete tableau. It contains the solution B−1b, the crucial vector

cNT − cB

T B−1N and the current objective function value cBT B−1b with a superfluous

minus sign indicating that our problem is minimization. The simplex tableau also contains

reduced coefficient matrix B−1N that is used in the minimum ratio. After determining

the entering variable xe, we examine the positive entries in the corresponding column of

B−1N , (v = B−1u = B−1Ne) and α is determined by taking the ratio of(B−1b)j

(B−1Ne)jfor

all positive vj ’s.




Simplex Tableau


cTB

cTN

0

35 −→

24 I B−1N B−1b

0 cTN

− cTB

B−1N −cTB

B−1b

35 .

If the smallest ratio occurs in lth component, then the lth column of B should be replaced

by u. The lth element of (B−1Ne)l = vl is distinguished as pivot element.




Simplex Tableau

It is not necessary to return the starting tableau, exchange twocolumns and start again. Instead we can continue with thecurrent tableau. Without loss of generality, we may assume thatthe first row corresponds to the leaving variable, that is the pivotelement is v1.

26666666666666666666664

3777777777777777777777526666666666666666666664

37777777777777777777775




Simplex Tableau

... 1...0 · · · 0 ∗ · · · ∗

... v1...∗ · · · ∗ (B−1b)1

...

...0...

... ....

... ....

... ....

...0...

... I B−1N...

... v2...

... ....

... ....

... ....

...vm

...

...B−1N B−1b

... 0...0 · · · 0 ∗ · · · ∗

...ce − cTBv

...∗ · · · ∗ −cTBB−1b

26666666666666666666664

3777777777777777777777526666666666666666666664

37777777777777777777775




Simplex Tableau

... 1...0 · · · 0 ∗ · · · ∗

... v1...∗ · · · ∗ (B−1b)1

...

...0...

... ....

... ....

... ....

...0...

... I B−1N...

... v2...

... ....

... ....

... ....

...vm

...

...B−1N B−1b

... 0...0 · · · 0 ∗ · · · ∗

...ce − cTBv

...∗ · · · ∗ −cTBB−1b

The first step in the pivot operation is to divide the leavingvariable’s row by the pivot element to create 1 in the pivot entry.Then, we have

26666666666666666666664

3777777777777777777777526666666666666666666664

37777777777777777777775




Simplex Tableau

The first step in the pivot operation is to divide the leavingvariable’s row by the pivot element to create 1 in the pivot entry.Then, we have26666666666666666666664

... 1v1

...0 · · · 0 ∗ · · · ∗... 1

... ∗ · · · ∗ α

...

...0...

... ....

... ....

... ....

...0...

... I B−1N...

... v2

...... .

...... .

...... .

......vm

...

...B−1N B−1b

... 0...0 · · · 0 ∗ · · · ∗

...ce − cTB

v... ∗ · · · ∗ −cT

BB−1b

37777777777777777777775 .

26666666666666666666664

37777777777777777777775




Simplex Tableau26666666666666666666664

... 1v1

...0 · · · 0 ∗ · · · ∗... 1

... ∗ · · · ∗ α

...

...0...

... ....

... ....

... ....

...0...

... I B−1N...

... v2

...... .

...... .

...... .

......vm

...

...B−1N B−1b

... 0...0 · · · 0 ∗ · · · ∗

...ce − cTB

v... ∗ · · · ∗ −cT

BB−1b

37777777777777777777775 .

For all the rows except the objective function row, do:For row i, multiply v1*(the updated first row) and subtract fromrow i. For the objective function row, multiply the first row by(ce − cB

T v) and subtract from the objective function row.

26666666666666666666664

37777777777777777777775




Simplex Tableau

What we have at the end is another simplex tableau.26666666666666666666664

... 1v1

...0 · · · 0 ∗ · · · ∗... 1

...∗ · · · ∗ α

...

... −v2

v1

...... .

...... .

...... .

......−vm

v1

...

... I ∗...

...0...

... ....

... ....

... ....

...0...

... ∗

+

.

.

.

+

...−ce−cTBv

v1

...0 · · · 0 ∗ · · · ∗... 0

...∗ · · · ∗ −cTB

B−1b − α(ce − cTB

v)

37777777777777777777775 .




Simplex Tableau

Example . The starting tableau at point P

0

P

Q

R

x=0

2x+y=6

x+2y=6

y=0

Feasible

set

2x+3y=10

2x+3y=6

2x+3y=0

2666664 3777775 24 35

2664 3775 266666664 377777775




Simplex Tableau

Example . The starting tableau at point P is

A b

cT 0

=

B N b

cTB cT

N 0

=

−1 2 1 0 6

0 1 2 −1 6

0 3 2 0 0

2666664 3777775 24 35

2664 3775 266666664 377777775




Simplex Tableau

Example . The starting tableau at point P is

A b

cT 0

=

B N b

cTB cT

N 0

=

−1 2 1 0 6

0 1 2 −1 6

0 3 2 0 0

The final tableau after Gauss-Jordan iterations is2666664 z1 y x z2 RHS

z1 1 0 3 −2 6

y 0 1 2 −1 6

z 0 0 −4 3 −18

3777775 =

24 I B−1N B−1b

0 cTN

− cTB

B−1N −cost

35

2664 3775 266666664 377777775




Simplex Tableau

Example . The final tableau after Gauss-Jordan iterations is2666664 z1 y x z2 RHS

z1 1 0 3 −2 6

y 0 1 2 −1 6

z 0 0 −4 3 −18

3777775 =

24 I B−1N B−1b

0 cTN

− cTB

B−1N −cost

35Since the reduced cost for x is −4 < 0, x should enter the basis. The

minimum ratio α = Min{62 , 6

3} = 2 due to z1, thus z1 should leave the

basis.

2664 3775 266666664 377777775




Simplex Tableau

Example . The final tableau after Gauss-Jordan iterations is2666664 z1 y x z2 RHS

z1 1 0 3 −2 6

y 0 1 2 −1 6

z 0 0 −4 3 −18

3777775 =

24 I B−1N B−1b

0 cTN

− cTB

B−1N −cost

35Since the reduced cost for x is −4 < 0, x should enter the basis. The

minimum ratio α = Min{62 , 6

3} = 2 due to z1, thus z1 should leave the

basis.2664 1 0 3 −2 6

0 1 2 −1 6

0 0 −4 3 −18

3775 −→

266666664 z1 y x z2 RHS

x 13

0 1 − 23

2

y − 23

1 0 13

2

−z 43

0 0 13

−10

377777775




Simplex Tableau2664 1 0 3 −2 6

0 1 2 −1 6

0 0 −4 3 −18

3775 −→

266666664 z1 y x z2 RHS

x 13

0 1 − 23

2

y − 23

1 0 13

2

−z 43

0 0 13

−10

377777775Example . Thus, x∗ = 2 = y∗ ⇒ z∗ = 10.




Pivot

Remark . All the pivot operation can be handled by multiplying the inverse of

the following elementary matrix.

h i h i




Pivot

E =

26666666666666666666666666666666664

1... v1

...0 . . . 0

. 0... .

... . .

.... .

... . .

0 .... .

... . .

1... vl

...0 . . . 0

0 . . . 0... .

...1

. .... .

... 1 0

. .... .

... .

. .... .

... 0 .

0 . . . 0...vm

... 1

37777777777777777777777777777777775⇔ E−1 =

266666666666666666666666666666666641

... −v1

vl

...

.... .

...

.... .

... 0

.... .

...... 1

vl

......

...1...

... .

0... .

... .

... .... .

...−vm

vl

... 1

37777777777777777777777777777777775

h i h i




Pivot

Thus, the pivot operation ish

I B−1N B−1b

i−→

hE−1I E−1B−1N E−1B−1b

i.




Pivot


I B−1N B−1b

i−→


i.

New basis is BE (B except the lth column is replaced by

u = N e) and basis inverse is (BE)−1 = E−1B−1. This is called

product form of the inverse.




Pivot


I B−1N B−1b

i−→


i.

New basis is BE (B except the lth column is replaced by

u = N e) and basis inverse is (BE)−1 = E−1B−1. This is called

product form of the inverse.

Thus, if we store E−1’s then we can implement the simplex

method on a simplex tableau.




Revised Simplex Method

Let us investigate what calculations are really necessary in the simplex

method. Each iteration exchanges a column of N with a column of B,

and one has to decide which columns to choose, beginning with a

basis matrix B and the current solution xB = B−1b.




Revised Simplex MethodS1. Compute row vector λ = cT

BB−1 and then cTN − λN .






S2. If cTN − λN ≥ θ, stop; the current solution is optimal.

Otherwise, if the most negative component is eth, choose eth

column of N (u) to enter the basis.









S3. Compute v = B−1u.










S4. Calculate ratios of B−1b to v = B−1u, admitting only positive

components of v. If there exists none, the minimal cost is

−∞; if the smallest ratio occurs at component l, then lth

column of current B will be replaced with u.










S4. Calculate ratios of B−1b to v = B−1u, admitting only positive

components of v. If there exists none, the minimal cost is

−∞; if the smallest ratio occurs at component l, then lth

column of current B will be replaced with u.

S5. Update B (or B−1) and the solution is xB = B−1b. Return to

S1.




Revised Simplex MethodRemark . We need to compute λ = c−1

BB−1, v = B−1u, and xB = B−1b. A

popular way is to work only on B−1. We can update B−1’s by premultiplying E−1’s.







The excessive computing (multiplying with E−1’s) could be avoided by directly

reinverting the current B at a time and deleting the current E−1’s that contain the

history.








reinverting the current B at a time and deleting the current E−1’s that contain the history.

Remark . The alternative way of computing λ, v and xB is λB = cT

B, Bv = u, and

BxB = b. Then, the standard decompositions (B = QR or PB = LU ) lead

directly to these solutions.








reinverting the current B at a time and deleting the current E−1’s that contain the history.

Remark . The alternative way of computing λ, v and xB is λB = cT

B, Bv = u, and

BxB = b. Then, the standard decompositions (B = QR or PB = LU ) lead

directly to these solutions.

Remark . How many simplex iterations do we have to take? There are at most(

n

m

)

extreme points. In the worst case, the simplex method may travel almost all of the

vertices. Thus, the complexity of the simplex method is exponential. However,

experience supports the following average behavior. The simplex method travels about

m extreme points, which means an operation count of about m2n, which is

comparable to ordinary elimination to solve Ax = b, and that is the reason of its

success.


Linear Programming - Çankaya Üniversitesiie501.cankaya.edu.tr/uploads/files/lp_c.pdf · Linear...

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