Chapter 4 Sensitivity Analysis 2 [Compatibility Mode]

8/12/2019 Chapter 4 Sensitivity Analysis 2 [Compatibility Mode]

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Summary of formulas:

= -1

RHS of optimal tableaus constraints = B -1b -1 - j BV j j

Slack variable, s i : i th element of cBV

B -1

-

, i - BV Artificial variable, a i : ( i th element of c BV B -1) + M

RHS of optimal Row 0 = c BV B -1b

1


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We must find B -1 to compute all partsof optimal tableau

We must find c BV B-1

to compute theoptimal tableau Row 0.

2


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Exam le 1: Refer to Exam le 4 Cha ter 3For the following LP, the optimal basis isBV = { x ,s }. Compute the optimal tableau.

min z = 2 x 1 - 5 x 2s.t 3x 1+8x 2 12

2x 1+3x 2 6

x1,x2 0 Solution 1: The standard form

min z = 2x 1 - 5x 2s.t 3x 1+8x 2 +s 1 =12

2x 1+3x 2 +s 2 = 6

3

x1,x2 ,s 1,s 2


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Solution 1: The standard formmin z = - 5x

2+s

2+ 2x

1+s

1 [ ] [ ]2 1

5 0 2 0

x x

z

= +

s.t x 2 + x 1+s 1 =3x 2 + s 2 + 2x 1 = 6x1,x2 ,s 1,s 2 0

2 1

2 1

2 1

8 0 3 1 12

3 1 2 0 6

x x

s s

+ =

2 1

2 10, 0

x x

s s

[ ] [ ]

2

2

5 0 , , 2 0 , BV Bv BNV xc x cs

= = =

1

1

, , ,3 1 2 0 6 NBV

x x B N b

s= = = =

Step 1: Find c BV and B -1cBV=[-5 0]

-108 0 8B = so by using Gauss elimination B

3 1 31

=

4

8


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Step 2: Determine the optimal constraints

Row BV z x 2 s 2 x1 s 1 RHS

0 z 1 0 0

1 x 2 0 1 0

2 s 0 0 1

LHS=formula B -1a j

11 1

1 30 38 8

3 2 7 x B a

= = =

RHS=formula B -1b

1

8 81 1

0 18 8

1 30 128 23 6 3

=

5

1 1 3 0 31

8 8

ss a= = =

8 2


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Thus, the constraints of the optimal tableau are

2 18 8 2 x x

=2 17 3 3

8 8 2s s

6


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Step 3: 10

optimal Row 0 [ ]

1 5 0 03 8

18

BV c B = =

1

LHS (coefficient for NBV) RHS

j BV j jc c B a c=

11 0 2 02 68 8 8 8

15 5

BV c c B b= = = =

1 08 8s

Row BV z x s x s RHS

0 z 1 0 0

1 x 0 1 0 3/8 1/8 3/2

72 s 2 0 0 1 7/8 -3/8 3/2


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any constra nt as a negat ve ,BV is now an infeasible basis.

If a variable in Row 0 may have a negative coefficient,

BV is now a suboptimal basis.

Three types of changes in an LP parameter

1 Chan in the ob ective function coefficient of anonbasic variable.

2) Changing the objective function coefficient of abasic variable.

3) Changing the right hand side (RHS) of a

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constraint.


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ROW BV z x 1 x2 x3 s 1 s 2 s 3 RHS

0 z 1 0 5 0 0 10 10 280

1 s 1 0 0 -2 0 1 2 -8 24

2 x 3 0 0 -2 1 0 2 -4 8

3 x 1 0 1 1.25 0 0 -0.5 1.5 2

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1 21

3 2, , B 0 2 4

0 1/ 2 3 / 2 BV NBV x x x s

x s

= = =

order from optimal tableau

or no nee n or er

The onl nonbasic decision variable is x . Coefficient of x 2 of objective function iscurrently c 2 =30

Question: For what value of c 2 would BV remain optimal?

13


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Question:

+

or w a va ue o c 2 wou rema n op ma

so B remain unchange.

nonnegative. Check Formula: B -1b -1 Does b changed? NO

,values. BV is still feasible (remain unchanged)

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Ste 2: Determine the coefficient of NBV inoptimal Row 0Check Formula: =c BVB-1a -c .

BV remain optimal if j 0BV will be subo timal if < 0

The only variable will be change is

12 2 2

6 BV c c B a c

=

[ ] ( ) 0 10 10 2 30 51.5

= + =

15


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BV will remain optimal if5- 0

5 If the coefficient of x 2 is decreased or increased

by 5 or less, (in other word, c 2 30+5=35)

BV will remains optimal. c 2 > , w e su op ma no ongeroptimal). This is because j < 0.

.can increase z by making x 2 as basic variables

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For example: c 2=40

1

6

0 10 10 2 40 5c c B a c = = =

1.5 Final (suboptimal) Tablaeu


-

1 s 1 0 0 -2 0 1 2 -8 24

2 x 3 0 0 -2 1 0 2 -4 8

3 x 0 1 1.25 0 0 -0.5 1.5 2

17Pivot element


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Optimal Tablaeu


1 s 1 0 1.6 0 0 1 1.2 -5.6 27.2

2 x 3 0 1.6 0 1 0 1.2 -1.6 11.2

3 x 2 0 0.8 1 0 0 -0.4 1.2 1.6

18

1 1 , 2 . , 3 .


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4.3.2 Changing the Objective Function

+

c BV will be change to [0 20 60+ ]

nonnegative.Check Formula: B -1bDoes B -1 changed? NODoes b changed? NO

Therefore, RHS of constraints have nonnegativevalues. BV is still feasible (remain unchanged)

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Step 2 : Determine the coefficient of NBV in optimalow

Check Formula: j=c BVB-1a j-c j.BV remain optimal if 0

BV will be suboptimal if j < 0

1 2 8

[ ] [ ]1 0 20 60 0 2 4 0 10 0.5 10 0.5

0 0.5 1.5

BV c B = + = +

1 3 1

12 2 2

, , are , t e r coe c ent n opt ma ta aeu ow must st e

6

0 10 0.5 10 0.5 2 30 5 1.25 BV

s x x

c c B a c = = + = +

2

1.5

sc c

= 1 1

2 2 2th element of 10 0.5 BV s s BV B a c c B = =

20

3 3 3 3th element of 10 0.5s BV s s BV c c B a c c B = = = +


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2

0, 5 1.25 0 4

0, 10 0.5 0 20

c

c

+

3 0, 10 0.5 0 -20/3sc +

Therefore, current basi

1

s remain optimal for -4 20

or 56=60-4 60+20=80c

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Let c 1=100(or =40)

1 3 1, , are BV, their coefficient in optimaltablaeu Row 0 must still be 0s x x

1 3 1

2

0

5 1.25 55

sc c c

c

= = =

= + =2

3

10 0.5 10

10 0.5 70s

s

c

c

= = = + =

[ ]1RHS of row 0, 0 10 70 20 360 BV c B b = =

22


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Final (suboptimal tableau)

0 z 1 0 55 0 0 -10 70 360

1 s 1 0 0 -2 0 1 2 -8 24

2 x 3 0 0 -2 1 0 2 -4 8

3 x 1 0 1 1.25 0 0 -0.5 1.5 2

23

Pivot element


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Optimal tableau

0 z 1 0 45 5 0 0 50 400

1 s 1 0 0 0 -1 1 0 -4 16

2 s 3 0 0 -1 0.5 0 1 -2 4

3 x 1 0 1 0.75 0.25 0 0 0.5 4

z=400, when x 1=4, x 2=0, x 3=0

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4.3.3 Changing the RHS of a Constraint

Changing the RHS cannot cause the currentbasic to become suboptimal. As long as the

,remains feasible and optimal.

Su ose we chan e b from 20 to 20+ Step 1: Determine whether RHS of constraint is

nonnegative. -

Does B -1 changed? NO1 0 2 4 20

0 0.5 1.5 8

B b

= +

Therefore,RHS of constraints

24 2 8 2

+ = +

25

.


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Current basis will remain optimal iff

24 2 0 12+

8 2 0 4

2 0.5 0 4

+

-4 4 or 16 =20-4 b2 20+4 =24

26


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If 16 b 24 the current basis remain o timal

(s 1,x3,x1) but the values of decision variables and zchange. uppose 2= w ere + , = so

11

1 2 8 48 28s

= = =

1 0 0.5 1.5 8 1 x

[ ]148

0 10 10 20 2 300 BV z c B b

= = + =

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When the current basis is no longer optimal:

Suppose b 2=30 (or =10) RHS of optimal constraints:

11

31 2 8 48 440 2 4 20 10 22

s x B b

= = + =

1

1

. .

Since 3, BV is no longer feasible

of row 0

x

x

RHS

=

[ ]148

0 10 10 20 10 380 BV z c B b

= = + =

We will use dual simplex algorithm to solve LPs whenthe initial tableau has one or more ve RHS and each

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variable in Row 0 has nonnegative coefficient.

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Chapter 4 Sensitivity Analysis 2 [Compatibility Mode]