1 Chapter 6 Sensitivity Analysis & Duality ( 敏感度分析 & 對偶 )

1

Chapter 6

Sensitivity Analysis & Duality( 敏感度分析 & 對偶 )

2

Max. z = c1x1+ c2x2 +···+ cnxn

s.t. a11x1 + a12x2 +···+ a1nxn ≤ b1

a21x1 + a22x2 +···+ a2nxn ≤ b2

：： am1x1+ am2x2 +···+ amnxn ≤ bm

xi ≥ 0 (i=1,2, ···,n)

3

Different types of changes parameters change the optimal solution

1. Changing the objective function coefficient (ci) NBV (p.276) BV (p.278) Both of NBV and BV --The 100% optimal rule (p.289)

2. Changing the right-hand side of a constraint (bi). Bi (p.282) Many Bi --The 100% feasibility rule (p.292)

3. Changing the column of a (ai). NBV (p.285) BV

4. Adding a new variable or activity. (p.287)5. Adding a new constraint. (Dual Simplex p.300 )

p.262

4

6.1 A Graphical Introduction to Sensitivity Analysis

Sensitivity analysis is concerned with how changes in an LP’s parameters affect the optimal solution.

The optimal solution to the Giapetto problem was z = 180, x1 = 20, x2 = 60 (Point B) and it has x1, x2, and s3 as BV.

How would changes in the problem’s objective function coefficients or the constraint’s right-hand sides change this optimal solution?

p.262

5

Graphical Analysis of the Effect of a Change in an Objective Function Coefficient

If the isoprofit line is flatter than the carpentry constraint, Point A(0,80) is optimal.

Point B(20,60) is optimal if the isoprofit line is steeper than the carpentry constraint but flatter than the finishing constraint.

Point C(40,20) is optimal if the slope of the isoprofit line is steeper than the slope of the finishing constraint.

The isoprofit line is c1x1 + 2x2 = k, the slope of the isoprofit line is just -c1/2.

6

Graphical Analysis of the Effect of a Change in a rhs on the LP’s Optimal Solution

A graphical analysis can also be used to determine whether a change in the rhs of a constraint will make the basis no longer optimal.

In a constraint with a positive slack (or positive excess) in an LPs optimal solution, if we change the rhs of the constraint to a value in the range where the basis remains optimal, the optimal solution to the LP remains the same.

7

Shadow Prices

It is important to determine how a constraint’s rhs changes the optimal z-value.

The shadow price for the ith constraint of an LP is the amount by which the optimal z-value is improved if the rhs of the ith constraint is increased by one. This definition applies only if the change in the rhs of constraint i leaves the current basis optimal.

p.265

8

Importance of Sensitivity Analysis

Values of LP parameters might change. If a parameter changes, sensitivity analysis

shows it is unnecessary to solve the problem again.

Uncertainty about LP parameters. Even if demand is uncertain, the company can

be fairly confident that it can still produce optimal amounts of products.

9

6.2 Some Important Formulas

An LP’s optimal tableau can be expressed in terms of the LP’s parameters.

The formulas are used in the study of sensitivity analysis, duality, and advanced LP topics.

When solving a max problem that has been prepared for solution by the Big M method with the LP having m constraints and n variables. Although some of the variables may be slack, excess, or artificial, they are labeled x1, x2, …,xn.

p.267

10

The LP may then be written as

BV = {BV1, BV2, …, BVn} to be the set of basic variables in the optimal tableau.

NBV = {NBV1, NBV2, …, NBVn} the set of nonbasic variables in the optimal tableau.

cBV is the 1 x m row vector [cBV1 cBV2 ∙∙∙ cBVm].cNBV is the 1 x (n-m) row vector whose elements

are the coefficients of the nonbasic variables (in the order of NBV).

max z = c1x1 + c2x2 + … + cnxn

s.t. a11x1 + a12x2 + … + a1nxn = b1

a21x1 + a22x2 + … + a2nxn = b2

…. …. … … …

am1x1 + am2x2 + … + amnxn = bm

xi ≥ 0 (i = 1, 2, …, n)

11

The m x m matrix B is the matrix whose jth column is the column for BVj in the initial tableau.

Aj is the column (in the constraints) for the variable xj.N is the m x (n-m) matrix whose columns are the

columns for the nonbasic variables (in NBV order) in the initial tableau.

The m x 1 column vector b is the right-hand side of the constraints in the initial tableau.

Matrix algebra can be used to determine how an LP’s optimal tableau (with the set of basic variables BV) is related to the original LP.

12

max z = c1x1 + c2x2 + … + cnxn

s.t. a11x1 + a12x2 + … + a1nxn = b1

a21x1 + a22x2 + … + a2nxn = b2

…. …. … … …

am1x1 + am2x2 + … + amnxn = bm

xi ≥ 0 (i = 1, 2, …, n)

mnn2n1

2n2121

1n1211

a..aa

::::

a..aa

a..aa

..

..

naaaA

bxc

21

n

2

1

n

2

1

n21

b

:

b

b

x

:

x

x

ccc

13

NBA

x

xxccc

BVN

BVNBVBV

mnn2n1

2n2121

1n1211

a..aa

::::

a..aa

a..aa

..

..

naaaA

bxc

21

n

2

1

n

2

1

n21

b

:

b

b

x

:

x

x

ccc

14

-(1)- x cx c x

x ccZcxZ BVN NBVBV BV

BVN

BVBVN BV

(2)---bB NxB x NxBbB x

Nxb Bx b Nx Bx

bx

x NB bAx

1BVN

1BVBVN

11BV

BVNBVBVNBV

BVN

BV

bBcxcNBcZ

xcNBcbBc

xcNxBcbBc

xcNxBbBc

xcxcZ

1BVBVNNBV

1BV

BVNNBV1

BV1

BV

BVNNBVBVN1

BV1

BV

BVNNBVBVN11

BV

BVNNBVBVBV

)(

)(

)(

15

Simplex Tableau 矩陣形式

BV z XBv XNBV RHS

z 1 0 CBvB-1N-CNBV CBvB-1b

XBv 0 I B-1N B-1b

BV z X RHS

z 1 CBvB-1A-C CBvB-1b

XBv 0 B-1A B-1b

16

Simplex Tableau 矩陣形式

j

jBV

aBaBC

1j

1

j

c

x單一 NBV 之欄表示為

BV z X RHS


XBv 0 B-1A B-1b

建議 :依數字順序計算

17

Max z=x1 + 4x2 + 0s1 + 0s2

s.t. x1 + 2x2 + 1s1 + 0s2 = 6

2x1 + x2 + 0s1 + 1s2 = 8

x1, x2,s1,s2, ≥0

Optimal Basis BV={x2 , s2}

11

02B

04 BV C

1012

0121 A

8

6 b

0041C

121

021

1B

18

121

023

021

121

1012

0121

121

021

AB 1

5

3

8

6

121

021

bB 1

0201

00410242

00411012

0121 02

00411012

0121

121

021

04cABC 1BV

12

8

6 02bBC 1

BV

BV Z X RHS

Z 1 CBvB-1A-C CBvB-1b

XBv 0 B-1A B-1b

19

121

023

021

121

AB 1

5

3 bB 1

0201 CABC 1BV 12 bB C 1

BV

5s s21

x0x21

3s0s21

x1x21

12s0s2x0xz

2121

2121

2121

BV z x1 x2 s1 s2 RHS

z 1 1 0 2 0 12

x2 0 ½ 1 ½ 0 3

s2 0 3/2 0 -½ 1 5

20

Example 1 ： compute the optimal tableau

max. z=x1+4x2

s.t. x1+2x2 ≤6

2x1+x2 ≤8

x1,x2 ≥0

The optimal basis is BV={x2,s2}

p.273

max. z=x1+4x2

s.t. x1+2x2+s1 =6

2x1+x2 +s2=8

x1,x2, s1, s2 ≥0

1

0

11

02

21

21

1BB

21

21

21

21

21

1

2321

21

21

11

11

1

0

1

0

tableau optimal the in sfor column e Th

2

1

1

0aB

tableau optimal the in xfor column The 2

1a)1(

5

3

8

6

1

0bB

tableau optimal the side hand-right The 8

6b)2(

21

21

1

5ss x

3 sxx

2. rowfor BV the iss 1; rowfor BV the is x ,}s,x{BV

2121

123

121

2121

2222

22

5ss x

3 sxx

12 s2 x z

128

6 02bBC

o row stableau' optimal the of side hand-right the )4( 0. tableau optimal of 0 row in s , x tcoefficien )3(

2BC of element first

tableau optimal of 0 row in s tcoefficien )2 (

11-2

1 02caBC

tableau optimal of 0 row in x tcoefficien )1(

021

0 04BC04C

2121

123

121

2121

21

1BV

22

1BV

1

111

BV

1

21

21

1BVBV

23

Six types of changes in an LP’s parameters change the optimal solution

1.Changing the objective function coefficient of a NBV.

2.Changing the objective function coefficient of a BV.

3.Changing the right-hand side of a constraint.

4.Changing the column of a NBV.

5.Adding a new variable or activity.

6.Adding a new constraint. (Dual Simplex )

24

Example ： Dakota problem (p.140)

max z = 60x1 + 30x2 + 20x3

s.t. 8x1+ 6x2 + x3 + s1 = 48

4x1+ 2x2 + 1.5x3 + s2 = 20

2x1+1.5x2 + 0.5x3 + s3 = 8

x1, x2, x3 , s1 , s2 , s3 ≥ 0

z +5x1 + 10s2 + 10s3= 280

-2x2 + s1+2s2 - 8s3 = 24

-2x2 + x3 +2s2 - 4s3 = 8

x1 + 1.25x2 - 0.5s2 +1.5s3= 2

x1, x2, x3 , s1 , s2 , s3 ≥ 0

25

z = 60x1 + 30x2 + 20x3 +0 s1 + 0s2 +0 s3

8x1+ 6x2 + x3 + s1 = 48

4x1+ 2x2 + 1.5x3 + s2 = 20

2x1+1.5x2 + 0.5x3 + s3 = 8

x1, x2, x3 , s1 , s2 , s3 ≥ 0

Optimal Basis BV={s1 , x3 , x1}

20.50

41.50

811

B

60200CBV

10050512

0105124

001168

..

.A

8

20

48

b

000203060C

51500

420

821

B 1

..

26

5150002511

420120

8210201

...

AB

10100050

00020306010100203560CABC 1

BV

2

8

24

1bB

280C 1BV bB

BV Z X RHS


XBv 0 B-1A B-1b

z +5x2 + 10s2 + 10s3= 280

-2x2 + s1+2s2 - 8s3 = 24

-2x2 + x3 +2s2 - 4s3 = 8

x1 + 1.25x2 - 0.5s2 +1.5s3= 2

x1, x2, x3 , s1 , s2 , s3 ≥ 0

27

BV Z X RHS


XBv 0 B-1A B-1b

1. Changing the objective functioncoefficient of NBV改變目標函數中“非基本變數”之係數

若 C 改變 , 則改變

Ex. c2=25

The current basis is still optimal.

[ ] [ ][ ] 0 101000100=

000202560 - 10100203560=

C-ABC -1BV

≥

28

Ex. c2=40

[ ] [ ][ ] 1010005 -0=

000204060 10100203560=

C-ABC 1-BV

Ex. For what value of c2 does the current basis remain optimal ?

[ ] [ ][ ]

22

2

2

1-BV

c 35 0 c - 35

O 101000c-350=

00020c60 - 10100203560=

C-ABC

≥⇒≥

≥

0s s x ,6.1x ,2.11x ,2.27s

288z

)278.p( 3.Table

)278.p( 2.Table

tableausimplexbuild

321231 ======

=⇒

⇒

⇒

⇒

The current basis is no longer optimal.

29

Table 2 (p.278)

BV Ratio

z -5x2 +10s2 +10s3 = 280 Z= 280

-2x2 +s1 + 2s2 - 8s3 = 24 s1= 24 None

-2x2 +x3 + 2s2 - 4s3 = 8 x3= 8 None

x1 +1.25x2 -0.5s2 +1.5s3 = 2 x1 = 2 1.6*

“ Final” (Suboptimal) Dakota Tableau ($40/Table)

30

Table 3 (p.278)

BV

z + 4x1 + 8s2 +16s3 = 288 Z= 288

1.6x1 +s1 +1.2s2 -5.6s3 = 27.2 s1= 27.2

1.6x1 +x3 +1.2s2 -1.6s3 = 11.2 x3= 11.2

0.8x1 +x2 -0.4s2 +1.2s3 = 1.6 x2 = 1.6

Optimal Dakota Tableau ($40/Table)

31300bB 1

BVC

BV Z X RHS


XBv 0 B-1A B-1b

2. Changing the objective function coefficient of a BV 改變目標函數中“基本變數”之係數

若 CBv 改變 ,則改變

Ex. c1=70, CBv=[0 20 70]

0 105005.170

000203070 - 1050205.4770C-ABC 1-

BV

≥

32

Ex. For what value of c1 does the current basis remain optimal ?

80 c 560 c5.180c5.04000c25.1700

0002030cc5.180c5.040020c25.140c

0002030c

5.15.00025.11

220120

821020

c200

0002030c

1005.05.12

0105.124

001168

5.15.00

420

821

c200

1

111

11111

11

11

33

CBV

701000550

0002030100701002085100

CAB 1

Ex. c1=100 > 80


0=s =x =x ,4=x ,4=s ,16=s

400=z

)281.p( 5.Table

)281.p( 4.Table

tableausimplexbuild

332121

⇒

⇒

⇒

⇒

34

Table 4 (p.281)

BV Ratio

z +55x2 -10s2 +70s3 = 360 Z= 360

- 2x2 +s1 +2s2 - 8s3 = 24 s1= 24 12

- 2x2 +x3 +2s2 - 4s3 = 8 x3 = 8 4*

x1 +1.25x2 -0.5s2 +1.5s3 = 2 x1 = 2 None

“ Final” (Suboptimal) Tableau if c1=100

35

Table 5 (p.281)

BV

z + 45x2 + 5x3 +50s3 = 400 Z= 400

- x3 + s1 - 4s3 = 16 s1= 16

- x2 +0.5x3 + s2 - 2s3 = 4 s2 = 4

x1 + 0.75x2 +0.25x3 +0.5s3 = 4 x1 = 4

Optimal Dakota Tableau if c1=100

36

BV Z X RHS


XBv 0 B-1A B-1b

3. Changing the right-hand side of a constraint 改變限制式的右邊值

若 b 改變 , 則改變

O

4

12

28

bB 1

Ex. b2=22,

The current basis BV is not changed.

8

22

48

b

300bB 1 BVC

37

Ex. For what value of b2 does the current basis BV remain optimal ?

2416

5012

232

216

4

28

51500

420

821

2

2

2

2

21

b O

b.

b

b

b

..

bB

Ex. b2=30 > 24


11.6chapter

)285.p(6.Table

tableausimplexbuild

3

28

44

bB 1

If the RHS of any constraint is negative, then the current basis is infeasible, a new solution must be found.

38

Table 6 (p.285)

BV

z + 5x2 +10s2 +10s3 = 380 Z= 380

- 2x2 +s1 + 2s2 - 8s3 = 24 s1 = 24

- 2x2 +x3 + 2s2 - 4s3 = 8 x3 = 8

x1 +1.25x2 -0.5s2 +1.5s3 = -3 x1 = - 3

“Final” (infeasible) Dakota Tableau if b2 = 30

39

BV z X RHS


XBv 0 B-1A B-1b

4.Changing the column of a NBV 改變非基本變數之行係數 (ci and ai 均改變 )

若 A 改變 ,則改變

101000100

000203060 10100204060

C

1000.522

0101.524

001158

1.50.50

420

821

60200

CABC 1BV

The current basis is still optimal.

Ex.

2

2

5

a2

p.285

40

10100030

000204360 10100204060

C

1005.022

0105.124

001158

5.15.00

420

821

60200

CABC 1BV

Ex. c2=43,

2

2

5

a2


0ssx,1x,12x,31s

283z

)286.p(8.Table

)286.p(7.Table

tableausimplexbuild

322131

41

Table 7 (p.286)

BV Ratio

z - 3x2 +10s2 +10s3 = 280 Z= 280

- 7x2 +s1 + 2s2 - 8s3 = 24 s1= 24 None

- 4x2 +x3 + 2s2 - 4s3 = 8 x3= 8 None

x1 + 2x2 -0.5s2 +1.5s3 = 2 x1 = 2 1*

“ Final” (Suboptimal) Dakota Tableau for New Method of Making Tables

42

Table 8 (p.278)

BV

z +1.5x1 +9.25s2 +12.25s3 = 283 Z= 283

3.5x1 +s1 +0.25s2 -2.75s3 = 31 s1= 31

2x1 +x3 + s2 - s3 = 12 x3= 12

0.5x1 +x2 -0.25s2 +0.75s3 = 1 x2 = 1

Optimal Dakota Tableau for New Method of Making Tables

43

BV Z X RHS


XBv 0 B-1A B-1b

5. Adding a new variable or activity 增加新的活動

若 A, C 改變 ,則改變

If current basis will remain optimal if ≥0 or become nonoptimal if ≤ 0

44

Ex. c4=15,

1

1

1

4a

O 1010050100

00015203060 1010020204060

C

10015.05.12

01015.124

0011168

5.15.00

420

821

60200

CABC 1BV

The current basis is still optimal

45

Exercise: p.288 Dakota problem1. Show that the current basis remains optimal if c3 satisfies

15≤c3≤22.5. If c3=21, find the new optimal solution. Also, if c3=25, find the new optimal solution.

2. If c1=55, show that the new optimal solution does not produce any desks.

3. Show that if b1≥24, the current basis remains optimal. If b1=30, find the new optimal solution.

4.Show that if tables sell for $50 and use 1 board ft of lumber, 3 finishing hours, and 1.5 carpentry hours, the current basis will no longer be optimal. Find the new optimal solution.

5. Dakota is considering manufacturing home computer tables. It sells for $36 and uses 6 board ft of lumber , 2 finishing hours, and 2 carpentry hours. Should the company manufacture any home computer tables?

46

Type 1 BV={s1, x3 , x1}

1.5.50

420

821

1.5

2

6

Change c3 to 20 + Δ

cBVB‑1 = [0 20+Δ 60] = [0 10+2Δ 10-4Δ]

Coefficient of x2 in row 0

=cBVB‑1a2‑c2 = [0 10+2Δ 10-4Δ ] -30 = 5-2Δ

New row 0 is z + (5‑2Δ)x2+(10+2Δ)s2+(10‑4Δ)s3

Current basis remains optimal if (1) 5‑2Δ 0 iff Δ 2.5 (2) 10 + 2Δ 0 iff Δ ‑5 ‑5Δ 2.5 or20‑5 c3 20+2.5.(3)10‑4Δ 0 iff Δ 2.5

47

Thus current basis remains optimal if ‑5 Δ 2.5 or 20‑5 c3 20+2.5.If c3=21 the current solution is still optimal , z increases by 8 to z=288. If c3=25,Δ=5, the tableau for the current optimal basis becomes z ‑5x2 + 20s2 ‑ 10s3 = 320 ‑2x2 + s1 + 2s2 ‑ 8s3 = 24 ‑2x2 + x3 + 2s2 ‑ 4s3 = 8 x1 + 1.25x2 ‑ .5s2 + 1.5s3 = 2Eventually we obtain the new optimal solutionz= 1000/3, x3= 40/3, x1=x2=0.

48

The 100% Rule for Changing objective function coefficient : The 100% Rule

Case 1 – All variables whose objective function coefficients are changed have nonzero reduced costs in the optimal row 0. The current basis remains optimal if and only if

the objective function coefficient for each variable remains within the allowable range .

If the current basis remains optimal, both the values of the decision variables and objective function remain unchanged. If the objective coefficient for any variable is outside the allowable range, the current basis is no longer optimal.

Case 2 – At least one variable whose objective function coefficient is changed has a reduced cost of zero.

p.289

49

The 100% Rule for Changing Right-Hand Sides

Case 1 – All constraints whose right-hand sides are being modified are nonbinding constraints. The current basis remains optimal if and only if

each rhs remains within its allowable range.Then the values of the decision variables and

optimal objective function remain unchanged. If the rhs for any constraint is outside its allowable

range, the current basis is no longer optimal.Case 2 – At least one of the constraints whose rhs is

being modified is a binding constraint (that is, has zero slack or excess).

p.292

50

6.5 Finding the Dual of an LP

Associated with any LP is another LP called the dual. Knowledge of the dual provides interesting economic and sensitivity analysis insights.

When taking the dual of any LP, the given LP is referred to as the primal. If the primal is a max problem, the dual will be a min problem and visa versa.

To find the dual to a max problem in which all the variables are required to be nonnegative and all the constraints are ≤ constraints (called normal max problem) the problem may be written as

p.295

51

Finding the Dual of a Normal Max or Min Problem

Table 14 (p.296)

Primal LP Dual LP

Maxn 個變數

m 個限制式

Minm 個限制式

n 個變數

第 i 個限制式第 i 個變數

第 j 個變數第 j 個限制式

i

i

iiij

b

b

bxa

j

j

jjij

c

c

cya

0

0

i

i

i

y

y

y

不限正負

0

0

j

j

j

x

x

x

不限正負

( 相反 )

( 一致 )

p.295

52

The dual of a normal max problem is called a normal min problem.

max z = c1x1+ c2x2 +…+ cnxn

s.t. a11x1 + a12x2 + … + a1nxn ≤ b1

a21x1 + a22x2 + … + a2nxn ≤ b2

… … … …

am1x1 + am2x2 + … + amnxn ≤ bm

xj ≥ 0 (j = 1, 2, …,n)

min w = b1y1+ b2y2 +…+ bmym

s.t. a11y1 + a21y2 + … + am1ym ≥ c1

a12y1 + a22y2 + … + am2ym ≥ c2

… … … …

a1ny1 + a2ny2 + …+ amnym ≥ cn

yi ≥ 0 (i = 1, 2, …,m)

53

主要問題與對偶問題之關係

Max Min

1 限制式 ≤ ≥ 0 變數

2 ≥ ≤ 0

3 = urs

4 變數 ≥ 0 ≥ 限制式

5 ≤ 0 ≤

6 urs =

54

Finding the Dual of a Normal Max or Min Problem

Table 14 (p.296)Example ： Dakota problem (Table 15)Example ： Diet problem (Table 16)Exercise ： problem 1 (p.301)

p.295

55

Table 14 (p.296)

max z0

min w (x1 ≧0) (x2 ≧0) (xn ≧0)

x1 x2 … xn

(y1 ≧ 0) y1 a11 a12 … a1n ≦ b1

(y2 ≧ 0) y2 a21 a22 … a2n ≦ b2

… … … … … … …

(ym≧ 0) ym am1 am2 … anm ≦ bm

≧ c1 ≧ c2 ≧ cn

56

Table 15 (p.296)

max z

min w (x1 ≧0) (x2 ≧0) (x3 ≧0)

x1 x2 x3

(y1 ≧ 0) y1 8 6 1 ≦ 48

(y2 ≧ 0) y2 4 2 1.5 ≦ 20

(y3≧ 0) y3 2 1.5 0.5 ≦ 8

≧60 ≧30 ≧20

57

Table 16 (p.297)

max z

min w (x1 ≧0) (x2 ≧0) (x3 ≧0) (x4 ≧0)

x1 x2 x3 x4

(y1 ≧ 0) y1 400 3 2 2 ≦ 50

(y2 ≧ 0) y2 200 2 2 4 ≦ 20

(y3 ≧ 0) y3 150 0 4 1 ≦ 30

(y4≧ 0) y4 500 0 4 5 ≦ 80

≧500 ≧6 ≧10 ≧8

58

Transforming into normal formNormal max problem ： step1 ~ 3Normal min problem ： step1 ~ 3

Finding the dual of a nonnormal max problemstep1 ~ 2 (p.299), Example ： table17-18

Finding the dual of a nonnormal min problemstep1 ~ 2 (p.300), Example ： table19-20

Exercise ： problem 3, 4 (p.301)

Finding the Dual of a Nonnormal LPp.298

59

6.6 Economic Interpretation of the Dual Problem

Suppose an entrepreneur wants to purchase all of Dakota’s resources. The entrepreneur must determine the price he or she is willing to pay for a unit of each of Dakota’s resources.

To determine these prices define:y1 = price paid for 1 boards ft of lumber

y2 = price paid for 1 finishing hour

y3 = price paid for 1 carpentry hour

The resource prices y1, y2, and y3 should be determined by solving the Dakota dual.

Example ： Dakota dual

p.302

60

In setting resource prices, the prices must be high enough to induce Dakota to sell.

When the primal is a normal max problem, the dual variables are related to the value of resources available to the decision maker. For this reason, dual variables are often referred to as resource shadow prices.

61

6.7 The Dual Theorem and Its Consequences

The Dual Theorem states that the primal and dual have equal optimal objective function values (if the problems have optimal solutions).

Lemma 1 ： Weak duality implies that if for any feasible solution to the primal and an feasible solution to the dual, the w-value for the feasible dual solution will be at least as large as the z-value for the feasible primal solution. (p.305)

Any feasible solution to the dual can be used to develop a bound on the optimal value of the primal objective function.

p.304

62

Lemma 2 ： Let

be a feasible solution to the primal and

be a feasible solution to the dual. If , then x-bar is optimal for the primal and y-bar is optimal for the dual.

Lemma 3 ： If the primal is unbounded, then the dual problem is infeasible.

Lemma 4 ： If the dual is unbounded, then the primal is infeasible.

nx

x

x

X2

1

myyyy 21byxc

63

The Dual Theorem*** (p.308) Suppose BV is an optimal basis for the primal.

Then cBVB-1 is an optimal solution to the dual. Also .

原題的解 (x-bar)與對偶題的解 (y-bar)存在對應關係，且二者的最佳值相等。

wz

64

6.8 Shadow Prices

The shadow price of the ith constraint is the amount by which the optimal z-value is improved (increased in a max problem and decreased in a min problem) is we increase bi by 1 (from bi to bi+1).

In short, adding points to the feasible region of a max problem cannot decrease the optimal z-value.

p.313

65

6.9 Duality and Sensitivity Analysis

Assuming that a set of basic variables BV is feasible, then BV is optimal if and only if the associated dual solution (cBVB-1) is dual feasible.

This result can be used for an alternative way of doing the following types of sensitivity analysis(6.3, p.276).Change 1 ： Changing the objective function

coefficient of a nonbasic variable.

Change 4 ： Changing the column of a nonbasic variable

Change 5 ： Adding a new activity.

p.323

66

6.10 Complementary Slackness

Theorem 2 ： Let

be a feasible primal solution andbe a feasible dual solution. Then x is primal optimal and y is dual optimal if and only if

siyi = 0 (i=1, 2, …, m)ejyj = 0 (j=1, 2, …, n)

nx

x

x

X2

1

myyyy 21

p.325

67

6.11 The Dual Simplex Method

The dual simplex method ( 對偶單體法 ) maintains a non-negative row 0 (dual feasibility) and eventually obtains a tableau in which each right-hand side is non-negative (primal feasibility).

68

The dual simplex method for a max problemStep 1:Is the right-hand side of each constraint non negative? If not, go to step 2.

Step 2:Choose the most negative basic variable as the variable to leave the basis. The row it is in will be the pivot row. To select the variable that enters the basis, computer the following ratio for each variable xj that has a negative coefficient in the pivot row:

Choose the variable with the smallest ratio as the entering variable. Now use EROs to make the entering variable a basic variable in the pivot row.

Step 3: If there is any constraint in which the right-hand side is negative and each variable has a non-negative coefficient, then the LP has no feasible solution. If no constraint infeasibility is found, return to step 1.

rowpivot in oft Coefficien

0 rowin oft Coefficien

xj

xj

69

Three uses of the dual simplex

1. Finding the new optimal solution after a constraint is added to an LP. (p.330)

2. Finding the new optimal solution after changing a right-hand side of an LP. (p.332)

3. Solving a normal min problem. (p.333)

70

When a constraint is added one of the following

three cases will occur.(p.330)The current optimal solution satisfies the new

constraint.The current optimal solution does not satisfy the

new constraint, but the LP still has a feasible solution.

The additional constraint causes the LP to have no feasible solutions.

Case 1 ： x1+x2+x3 ≦ 11 (p.330)

Case 2 ： x2 ≧ 1 (p.330, table 33-34)

Case 3 ： x1+x2 ≧ 12 (p.331, table 35-38)

71

max z = 60x1 + 30x2 + 20x3

s.t. 8x1+ 6x2 + x3 ≤ 48

4x1+ 2x2 + 1.5x3 ≤ 20

2x1+1.5x2 + 0.5x3 ≤ 8

x1, x2, x3 ≥ 0

BV z x1 X2 x3 s1 s2 s3 RHS

z 1 0 5 0 0 10 10 280

s1 0 0 -2 0 1 2 -8 24

x3 0 0 -2 1 0 2 -4 8

x1 0 1 1.25 0 0 -0.5 1.5 2

A Constraint Is Added to an LPp.331

72

加入限制式 x1+x2+x3 ≤ 11

BV z x1 x2 x3 s1 s2 s3 s4 RHS

z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

s4 0 1 1 1 0 0 0 1 11

BV z x1 x2 x3 s1 s2 s3 RHS

z 1 0 5 0 0 10 10 280

s1 0 0 -2 0 1 2 -8 24

x3 0 0 -2 1 0 2 -4 8

x1 0 1 1.25 0 0 -0.5 1.5 2

Case 1

x1+x2+x3 ≤ 11 ⇒ x1+x2+x3 +s4= 11

73


z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

s4 0 1 1 1 0 0 0 1 11


z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

s4 0 0 0.25 1 0 0.5 -1.5 1 9

ERO

The current optimal solution remains optimal after the constraint x1+x2+x3 ≤ 11 is added.

74

x2 ≥ 1 ⇒ -x2 +e4=- 1

TABLE 33

BV z x1 x2 x3 s1 s2 s3 e4 RHS

z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

e4 0 0 -1 0 0 0 0 1 -1


z 1 0 5 0 0 10 10 280

s1 0 0 -2 0 1 2 -8 24

x3 0 0 -2 1 0 2 -4 8

x1 0 1 1.25 0 0 -0.5 1.5 2

Case 2

加入限制式 x2 ≥ 1

75

TABLE 33 EV

BV z x1 x2 x3 s1 s2 s3 e4 RHS r

z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

e4 0 0 -1 0 0 0 0 1 -1 LV

TABLE 34


z 1 0 0 0 0 10 10 5 275

s1 0 0 0 0 1 2 -8 -2 26

x3 0 0 0 1 0 2 -4 -2 10

x1 0 1 0 0 0 -1/2 3/4 5/4 3/4

x2 0 0 1 0 0 0 0 -1 1

ERO

76

Case 3

加入限制式 x1 +x2 ≥ 12


z 1 0 5 0 0 10 10 280

s1 0 0 -2 0 1 2 -8 24

x3 0 0 -2 1 0 2 -4 8

x1 0 1 1.25 0 0 -0.5 1.5 2

x1 +x2 ≥ 12 ⇒ -x1 -x2 +e4 = - 12

TABLE 35


z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

e4 0 -1 -1 0 0 0 0 1 -12

77

TABLE 36 EV


z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

e4 0 0 0.25 0 0 -0.5 1.5 1 -10 LV

TABLE 35 EV


z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

e4 0 -1 -1 0 0 0 0 1 -12 LV

78

TABLE 37 EV


z 1 0 10 0 0 0 40 20 80

s1 0 0 -1 0 1 0 -2 4 -16

x3 0 0 -1 1 0 0 2 4 -32 LV

x1 0 1 1 0 0 0 0 -1 12

s2 0 0 -0.5 0 0 1 -3 -2 20

TABLE 36 EV


z 1 0 5 0 0 10 10 0 280

s1 0 0 -2 0 1 2 -8 0 24

x3 0 0 -2 1 0 2 -4 0 8

x1 0 1 1.25 0 0 -0.5 1.5 0 2

e4 0 0 0.25 0 0 -0.5 1.5 1 -10 LV

79

TABLE 38


z 1 0 10 0 0 0 60 60 -240

s1 0 0 0 -1 1 0 -4 0 16

x2 0 0 1 -1 0 0 -2 -4 32

x1 0 1 0 1 0 0 2 3 -20

s2 0 0 0 -0.5 0 1 -4 -4 36

TABLE 37 EV


z 1 0 10 0 0 0 40 20 80

s1 0 0 -1 0 1 0 -2 4 -16

x3 0 0 -1 1 0 0 2 4 -32 LV

x1 0 1 1 0 0 0 0 -1 12

s2 0 0 -0.5 0 0 1 -3 -2 20

80

If the right-hand side of a constraint is changed and the current basis becomes infeasible, the dual simplex can be used to find the new optimal solution.(p.332)

-Example ： finishing hours=30, table 39-40

81

max z = 60x1 + 30x2 + 20x3

s.t. 8x1+ 6x2 + x3 ≤ 48

4x1+ 2x2 + 1.5x3 ≤ 20

2x1+1.5x2 + 0.5x3 ≤ 8

x1, x2, x3 ≥ 0BV z x1 X2 x3 s1 s2 s3 RHS

z 1 0 5 0 0 10 10 280

s1 0 0 -2 0 1 2 -8 24

x3 0 0 -2 1 0 2 -4 8

x1 0 1 1.25 0 0 -0.5 1.5 2

Changing a RHS of an LPp.332

82

1.50.5-0

4-20

821

B ,

20.50

41.50

811

B

8

20

48

b ,

1005.05.12

0105.184

001168

A

60200C , 000203060C

}x,x,s{BV

1

BV

131

b2=20→30TABLE 39


z 1 0 5 0 0 10 10 0 380

s1 0 0 -2 0 1 2 -8 0 44

x3 0 0 -2 1 0 2 -4 0 28

x1 0 1 1.25 0 0 -0.5 1.5 0 -3

83

TABLE 39 EV


z 1 0 5 0 0 10 10 380

s1 0 0 -2 0 1 2 -8 44

x3 0 0 -2 1 0 2 -4 28

x1 0 1 1.25 0 0 -0.5 1.5 -3 LV

TABLE 40


z 1 20 30 0 0 0 40 320

s1 0 4 3 0 1 0 -2 32

x3 0 4 3 1 0 0 2 16

s2 0 -2 -2.5 0 0 1 -3 6

84

Solving a Normal Min Problem

Min z=x1+2x2

s.t. x1 -2x2 + x3 ≥ 4

2x1 + x2 - x3 ≥ 6

x1 , x2, x3 ≥ 0

p.333

Let z’=-z, z’= -x1 - 2x2

z’ + x1 + 2x2 = 0

x1 -2x2 + x3 - e1 = 4

2x1 + x2 - x3 - e2 = 6

85

1.RHS ( 負值 ) 最小離開 , e2 leaving2.ratio 最小進入 , x1entering

Solving a Normal Min Problem

Table 42

BV z x1 x2 x3 e1 e2 RHS

z 1 1 2 0 0 0 0e1 0 -1 2 -1 1 0 -4e2 0 -2 -1 1 0 1 -6

ratio |1/(-2) | |1/(-1) | min

Table 41

z’ + x1 + 2x2 = 0-x1+2x2 - x3 + e1 = -4-2x1 - x2 + x3 + e2 = -6

p.334

86

1.RHS ( 負值 ) 最小離開 , e1 leaving2.ratio 最小進入 , x3 entering

Table 43


z 1 0 3/2 1/2 0 1/2 -3e1 0 0 5/2 -3/2 1 -1/2 -1x1 0 1 1/2 -1/2 0 -1/2 3

ratio min31

2/32/1

12/1

2/1

Table 44


z 1 0 7/3 0 1/3 1/3 -10/3x3 0 0 -5/3 1 -2/3 1/3 2/3x1 0 1 -1/3 0 -1/3 - 1/3 10/3

ratio min

z’ = -10/3, z = 10/3, x1 = 10/3, x3 = 2/3, x2 = 0

87

z + 2x1 + x3 = 0

- x1 - x2 + x3 + e1 = ‑5

- x1 + 2x2 - 4x3 + e2 = -8

Exercise : p.335 #1

Use the dual simplex method to solve the following LP:

Max z=-2x1-x3

s.t. x1 + x2 - x3 ≥ 5

x1 - 2x2 + 4x3 ≥ 8

x1 , x2, x3 ≥ 0

z + 2x1 + x3 = 0

x1 +x2 - x3 - e1 = 5

x1 - 2x2 + 4x3 - e2 = 8

⇒

88

z + 2x1 + x3 = 0

- x1 - x2 + x3 + e1 = ‑5

- x1 + 2x2 - 4x3 + e2 = -8

1.RHS ( 負值 ) 最小離開 , e2 leaving2.ratio 最小進入 , x3entering

Exercise : p.335 #1


z 1 2 0 1 0 0 0e1 0 -1 -1 1 1 0 -5e2 0 -1 2 -4 0 1 -8

ratio |2/(-1) | |1/(-4) | min

89


z 1 2 0 1 0 0 0e1 0 -1 -1 1 1 0 -5e2 0 -1 2 -4 0 1 -8

ratio |2/(-1) | |1/(-4) | min


z 1 7/4 1/2 0 0 1/4 -2e1 0 -5/4 -1/2 0 1 1/4 -7x3 0 1/4 -1/2 1 0 -1/4 2

ratio min5

7

4/5

4/7

1

2/1

2/1

1.RHS ( 負值 ) 最小離開 , e1 leaving2.ratio 最小進入 , x2 entering

90


z 1 1/2 0 0 1 1/2 -9x2 0 5/2 1 0 -2 -1/2 14x3 0 3/2 0 1 -1 -1/2 9

z = ‑9, x2 = 14, x3 = 9

91

主要單體法與對偶單體法知差異(Max problem)

差異項目主要單體法對偶單體法

1 0 列不受正負限制 * 均為非負值

2 RHS 均為非負值不受正負限制 *

3 判斷最佳解 0 列均為非負值 RHS 均為非負值

4 選擇進入變數最負 0 列最負 RHS

5 RatioMin{RHS/ 正係數 }

Min{|0 列係數 / 負係數 |}

* 最佳解除外

92

對偶問題

Max. Z=x1 - 2x2 + 3x3

s.t. 2x1 + x2 + 4x3 - x4 ≦ - 4

-x1 + 2x2 - x3 + 4x4 ≧ 5

x1 - 3x2 + 2x3 + x4 = 6

x1≧0, x2≧0, x3 urs, x4≦0

--- y1

--- y2

--- y3

Min. w= -4y1 +5y2 + 6y3

s.t. 2y1 - y2 + y3 ≦ 1

y1 + 2y2 - 3y3 ≧ -2

4y1 - y2 + 2y3 = 3

-y1 + 4y2 + y3 ≦ 0

y1≧0, y2 ≦ 0, y3 urs

93

求線性規劃之對偶問題

0x,x

3x3x2

5x2x

3xxs.t.

x4x3Zmin

21

21

21

21

21

94

0s,u,u,x,x

3sx3x2

5ux2x

3uxxs.t.

x4x3ZZmax

32121

321

221

121

21

0s,u,u,x,x

3sx3x2

5ux2x

3uxxs.t.

0x4x3Zmax

32121

321

221

121

21

95

故 x1=1,x2=2,u1=0,u2=0,s3=7 Z 有最大值 11 ，即 Z 有最小值 11

96

利用對偶單體法求解下 LP 問題

0x,x,x

5x3x2x

5xxx

3xx3x2s.t.

x3x2xZmax

321

321

321

321

321

97

0x,x,x

5x3x2x

5xxx

5xxx

3xx3x2s.t.

x3x2xZmax

321

321

321

321

321

321

0u,u,s,s,x,x,x

5ux3x2x

5uxxx

5sxxx

3sxx3x2s.t.

0x3x2xZmax

4321321

4321

3321

2321

1321

321

98

Z x1 x2 x3 s1 s2 u3 u4 RHS

Z 1 1 2 3 0 0 0 0 0

s1 0 2 3 1 1 0 0 0 3

s2 0 1 1 1 0 1 0 0 5

u3 0 1 1 1 0 0 1 0 5

u4 0 1 2 3 0 0 0 1 5

Z 1 0 3 2 0 0 1 0 5

s1 0 0 1 1 1 0 2 0 7

s2 0 0 0 0 0 1 1 0 0

x1 0 1 1 1 0 0 1 0 5

u4 0 0 3 4 0 0 1 1 0

Z 1 0 1 0 2 0 5 0 19

x3 0 0 1 1 1 0 2 0 7

s2 0 0 0 0 0 1 1 0 0

x1 0 1 2 0 1 0 1 0 2

u4 0 0 7 0 4 0 7 1 28

Z 1 0 0 0 18

7 0 6

1

7 23

x3 0 0 0 1 3

7 0 1

1

7 3

s2 0 0 0 0 0 1 1 0 0

x1 0 1 0 0 1

7 0 1

2

7 6

x2 0 0 1 0 4

7 0 1

1

7 4

故 x1 = 6, x2 = 4, x3 = 3,

s1 = s2 = u3 = u4 =0 時，

Z 有最大值 23

Documents

1 Chapter 6 Sensitivity Analysis & Duality ( 敏感度分析 & 對偶 )