sensitivity analysis li xiao-lei. a graphical introduction to sensitivity analysis sensitivity...

Sensitivity analysis

LI Xiao-lei

A graphical introduction to sensitivity analysis

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


The Giapetto problemmax z=3x1+2x2

s.t. 2x1 + x2≤100 (finishing constraint)

x1 + x2≤80 (carpentry constraint)

x1 ≤40 (demand constraint)

x1 , x2 ≥0Where

x1=number of soldiers produced per week

x2=number of trains produced per week


0 20 40 60 80 1000

20

40

60

80

100

x

y

3 x+2 y-180 = 0

z=60 z=100

z=180

The optimal solution is z=180, x1=20, x2=60

Graphical analysis of the effect of a change in an objective function coefficient

Let c1 be the contribution to profit by each soldier. For what values of c1 does the current basis remain optimal?

At present, c1=3 and the profit line has the form 3x1+2x2=constant,

or x2=-3x1/2 + constant/2.

Figure 1

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

60

70

80

90

100

x1

x 2

3 x1+2 x

2-120 = 0

finishing constraint, slope=-2

carpentry constraint, slope =-1

demand constraint

profit line z=120, slope=-3/2

B

C

D

A

E


If a change in c1 cause the profit lines to be flatter than the carpentry constraint, the optimal solution will change from point B to a new optimal solution (point A).The slope of each profit line is –c1/2

The profit line will be flatter than the carpentry constraint if –c1/2>-1, or c1<2, and the new optimal solution will be (0,80), point A.


If the profit line s are steeper than the finishing constraint, the optimal solution will change from point B to point C.The slope of the finishing constraint is -

2.

If –c1/2<-2, or c1>4, and the new optimal solution will be (40,20).


In summary, if all other parameters remain unchanged, the current basis remains optimal for 2≤c1≤4. and Giapetto should still manufacture 20 soldiers and 60 trains.

But Giapetto’s profit will change. For instance, if c1=4, Giapetto’s profit will now be 4(20)+2(60)=$200 instead of $180.

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution

Whether a change in the right-hand side of a constraint will make the current basis no longer optimal?

Let b1 the number of available finishing hours. Currently, b1=100.

Figure 2

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

60

70

80

90

100

x1

x 22 x

1+x

2-120 = 0

finishing constraint, slope=-2

carpentry constraint, slope =-1

demand constraint

profit line z=120, slope=-3/2

finishing constraint b1=80finishing constraint b1=120 A

B

C

D


A change in b1 shifts the finishing constraint parallel to its current position.

The current optimal solution is where the carpentry and finishing constraints are binding.

Then as long as the point where the finishing and carpentry constraints are binding remains feasible, the optimal solution will still occur where these constraints intersect.


From figure 2, for 80≤b1≤120, the current basis remains optimal.

With the changing of b1, the values of the decision variables and the objective function value change.

For example, if 80≤b1≤100, the optimal solution will change from point B to some point on the line segment AB. Similarly, if 100≤b1≤120, the optimal solution will change from point B to some point on the line segment BD.


To determine how a change in the right-hand side of a constraint changes the values of the decision variables.Let b1=number of available finishing hours.

If we change b1 to 100+Δ, the current basis remains optimal for -20≤Δ≤20.

Note: as b1 changes, the optimal solution to the LP is still the point where the finishing-hour and carpentry-hour constraints are binding.


Thus, we can find the new values of the decision variables by solving 2x1+x2=100+Δ

and x1+x2=80

This yields x1=20+Δ and x2=60-Δ

Thus, a increase in the number of available finishing hours results in an increase in the number of soldiers produced and a decrease in the number of trains produced.


Let b2=the number of available carpentry hours If we change b2 to 80+Δ, the current ba

sis remains optimal for -20≤Δ≤20.

Note: as b2 changes, the optimal solution to the LP is still the point where the finishing-hour and carpentry-hour constraints are binding.


Thus, if b2=80+ Δ, the optimal solution to the LP is the solution to 2x1+x2=100

and x1+x2=80+ Δ

This yields x1=20-Δ and x2=60+2Δ.

It shows that an increase in the amount of available carpentry hours decrease the number of soldiers produced and increases the number of trains produced.


Let b3=the demand for soldiers.If b3 is changed to 40+Δ, the current basis remains

optimal for Δ≥-20. for Δ in this range, the optimal solution to the LP will still occur where the finishing and carpentry constrains are binding.

Thus, the optimal solution to the LP is the solution to

2x1+x2=100

and x1+x2=80

This yields x1=20 and x2=60


Note:

In a constraint with positive slack( or positive excess) in an LP’s optimal solution, if we change the right-hand side of the constraint to a value in the range where the current basis remains optimal, the optimal solution to the LP is unchanged.

Shadow prices

We define the shadow price for the ith constraint of an LP to be the amount by which the optimal z-value is improved if the right-hand side of the ith constraint is increased by 1.

This definition applies only if the change in the right-hand side of constraint i leaves the current basis optimal.

Shadow prices

For example, for the Δchanging in the finishing hours, the optimal solution is x1=20+Δ and x2=60-Δ. Then the optimal z-value will equal 3x1+2x2=3(20 +Δ) +2(60-Δ) =180+ Δ. Thus, as long as the current basis remains optimal, a unit increase in the number of available finishing hours will increase the optimal z-value by $1.so the shadow price of the first constraint is $1.

Shadow prices

For the second constraint, if Δchanging in the finishing hours, the optimal solution is x1=20-Δ and x2=60+2Δ. Then the optimal z-value will equal 3x1+2x2=3(20 -Δ) +2(60+2Δ) =180+ Δ. Thus, as long as the current basis remains optimal, a unit increase in the number of available finishing hours will increase the optimal z-value by $1.so the shadow price of the second constraint is $1.

Shadow prices

For the third constraint, the optimal values of the decision variables remain unchanged, as long as the current basis remains optimal. Then the optimal z-value will also remain unchanged, which shows that the shadow price of the third constraint is $0.

It turns out that whenever the slack variable or excess variable for a constraint is positive in an LP’s optimal solution, the constraint will have a zero shadow price.

Shadow prices

Each unit by which constraint i’s right-hand side is increased will increase the optimal z-value( for a max problem) by the shadow price.

Suppose we increase the rhs of the ith constraint of an LP by Δbi , thus, the new optimal z-value is given by(new optimal z-value)=(old optimal z-value) +(constraint i’s shadow price)Δbi

For a minimization problem,(new optimal z-value)=(old optimal z-value) -(constraint i’s shadow price)Δbi

Importance of sensitivity analysis

If a parameter changes, sensitivity analysis often makes it unnecessary to solve the problem again.

A knowledge of sensitivity analysis often enables the analyst to determine from the original solution how changes in an LP’s parameters change the optimal solution.

Some important formulas

Assume a max LP problem with m constraints and n variables. Although some of these variables may be slack, excess, or artificial ones, we choose to label them x1,x2,…,xn. max z=c1x1+c2x2+…+cnxn

s.t. a11x1+a12x2+…+a1nxn=b1

a21x1+a22x2+…+a2nxn=b2 (1) : : : :

am1x1+am2x2+…+amnxn=bm

xi≥0 (i=1,2,…,n)


For the Dakota problem,

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

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

4x1+ 2x2+1.5x3+ s2 =20 (finishing) (1’) 2x1+1.5x2+0.5x3+ s3=8 (carpentry)

x1,x2,x3,s1,s2,s3≥0


Let BVi be the basic variable for row i of the optimal tableau. Also define BV={BV1, BV2 , … , BVm} to be the set of basic variables in the optimal tableau, and define the m × 1 vector xBv=[xBv1, xBv2 , …, xBvm]T

We also defineNBV=the set of nonbasic variables in the opti

mal tableau

xNBv=(n-m) ×1 vector listing the nonbasic variables.


The optimal tableau for the LP 1’ z + 5x2 +10s2+10s3=280

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

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

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

For this optimal tableau, BV1=s1, BV2=x3, and BV3=x1, then xBV=[s1,x3,x1]’

We may choose NBV={x2,s2,s3}, then xNBV=[x

2,s2,s3]’


DEFINATION cBV is the 1×m row vector [cBV1 cBV2 … cB

Vm].

Thus, the elements of cBV are the objective function coefficients for the optimal tableau’s basic variables. For the Dakota problem, BV={s1,x3,x1}, then form (1’) we find that cBV=[0 20 60].


DEFINATION cNBV is the 1×(n-m) row vector whose el

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

If we choose to list the nonbasic variables for the Dakota problem in the order NBV={x2,s2,s3}, then form (1’) we find that cNBV=[30 0 0].


DEFINATION

The m×m matrix B is the matrix whose jth column is the column for BVj in (1).

For the Dakota problem, the first column of B is the s1 column in (1’); the second, the x3 column; and the third, the x1 column. Thus,

25.00

45.10

811

B


DEFINATION aj is the column (in the constraints) for

the variable xj in (1).

For example, in the Dakota problem,

5.1

2

6

2a


DEFINATION

N is the m×(n-m) matrix whose columns are the columns for the nonbasic variables (in the NBV order) in (1).

For the Dakota problem, we write NBV={x2,s2,s3}, then

105.1

012

006

N


DEFINATION

The m×1 column vector b is the right-hand side of the constraints in (1).


8

20

48

b

Matrix notation

(1) may be written as

(3)

0,

..

NBVBV

NBVBV

NBVNBVBVBV

xx

bNxBxts

xcxcz

0,

..

NBVBV

NBVBV

NBVNBVBVBV

xx

bNxBxts

xcxcz

Matrix notation

The Dakota problem can be written as

0

0

0

,

0

0

0

8

20

48

101.5

012

006

25.00

45.10

811

..

003060200 max

3

2

2

1

3

1

3

2

2

1

3

1

3

2

2

1

3

1

s

s

x

x

x

s

s

s

x

x

x

s

ts

s

s

x

x

x

s

z

0

0

0

,

0

0

0

8

20

48

101.5

012

006

25.00

45.10

811

..

003060200 max

3

2

2

1

3

1

3

2

2

1

3

1

3

2

2

1

3

1

s

s

x

x

x

s

s

s

x

x

x

s

ts

s

s

x

x

x

s

z

Matrix notation

Multiplying the constraints in (3) through by B-1, we obtain

or (4)

In (4),BVi occurs with a coefficient of 1 in the ith constraint and a zero coefficient in each other constraint. Thus, BV is the set of the basic variables for (4), and (4) yields the constraints for the optimal tableau.

bBNxBBxB NBVBV111

bBNxBx NBVBV11

Matrix notation


Then,

Of course, these are the constraints for the optimal tableau.

5.15.00

420

8211B

2

8

24

5.15.01.25

422-

822-

3

2

2

1

3

1

s

s

x

x

x

s

Matrix notation

From (4), we see that the column of a nonbasic variable xj in the constraints of the optimal tableau is given by

B-1(column for xj in (1) )=B-1aj

The following two equations summarize the preceding discussion:

Column for xj in optimal tableau’s constraints=B-1aj (5)Right-hand side of optimal tableau’s constraints=B-1b (6)

Matrix notation How to express row 0 of the optimal tableau in terms

of BV?We multiply the constraints through by the vector cBVB-1:

(7)

And rewrite the original objective function as,

(8)

Adding (7) to (8), we can eliminate the optimal tableau’s basic variables,

(9)

bBcNxBcxc BVNBVBVBVBV11

0 NBVNBVBVBV xcxcz

bBcxcNBcz BVNBVNBVBV11 )(

Matrix notation

From (9), the coefficient of xj in row 0 is

cBVB-1(column of N for xj)-

(coefficient for xj in cNBV)=cBVB-1aj-cj

the right hand side of row 0 is cBVB-1b

To summarize, we let be the coefficient of xj in the optimal tableau’s row 0. then we have shown that

(10)

and

right-hand side of optimal tableau’s row 0=cBVB-1b (11)

jjBVj caBcc 1

jc

Matrix notation

To illustrate, we determine row 0 of the Dakota problem’s optimal tableau.

Then cBVB-1=[0 10 10], from (10) we find that the coefficients of the nonbasic variables in row 0 are

5.15.00

420

821

]60200[ 1BandcBV

530

5.1

2

6

10100221

2

caBcc BV

Matrix notation

and

Of course, the optimal tableau’s basic variables (x1,x3,and s1) will have zero coefficients in row 0.

100

0

1

0

0 12

BcrowoptimalinsoftCoefficien BV

100

1

0

0

0 13

BcrowoptimalinsoftCoefficien BV

Matrix notation

From (11), the right-hand side of row 0 is

Putting it all together, we see that row 0 is

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

280

8

20

48

101001

bBcBV

Matrix notation

Simplifying formula (10) for slack, excess, and artificial variablesIf xj is the slack variable si, the coefficient of

si in the objective function is 0, and the column for si in the original tableau has 1 in row I and 0 in all other rows. Then (10) yields

Coefficient of si in optimal row 0

= ith element of cBVB-1-0

=ith element of cBVB-1 (10’)

Matrix notation

Similarly, if xj is the excess variable ei, the coefficient of ei in the objective function is 0 and the column for ei in the original tableau has -1 in row i and 0 in all other rows. Then (10) reduces to

Coefficient of ei in optimal row 0

=-( ith element of cBVB-1)-0

=-(ith element of cBVB-1) (10’’)

Matrix notation

Finally, if xj is an artificial variable ai, the objective function coefficient of ai( for a max problem) is –M and the original column for ai has 1 in row i and 0 in all other rows. Then (10) reduces to

Coefficient of ai in optimal row 0

=( ith element of cBVB-1)-(-M)

=(ith element of cBVB-1)+(M) (10’’’)

Summary of formulas for computing the optimal tableau from the initial LP

Column for xj in optimal tableau’s constraints=B-1aj (5)

Right-hand side of optimal tableau’s constraints=B-1b (6)

Coefficient of si in optimal row 0=ith element of cBVB-1 (10’)

Coefficient of ei in optimal row 0=-(ith element of cBVB-1) (10’’)

Coefficient of ai in optimal row 0=-(ith element of cBVB-1)+(M)

(10’’’)

right-hand side of optimal tableau’s row 0=cBVB-1b (11)

We must first find B-1 because it is necessary in order to compute all parts of the optimal tableau.

jjBVj caBcc 1

Example 1

Solution After adding slack variables s1 and s2,

we obtain:

max z=x1+4x2

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

2x1+ x2 +s2=8

Example 1

For the following LP, the optimal basis is BV={x2,s2}, computer the optimal tableau.

max z=x1+4x2

s.t. x1+2x2≤6

2x1+ x2≤8

x1,x2≥0

Example 1

First we compute B-1,

12

1

02

1

and 11

02 1BB

12

1

02

1

and 11

02 1BB

Example 1

Use (5) and (6) to determine the optimal tableau’s constraints.The column for x1 in the optimal tableau is

The other nonbasic variable is s1,

23

21

2

1

121

021

11aB

212

1

0

1

121

021

tableauoptimalin for Column 11

1 aBs

Example 1

From (6)

Since BV is listed as {x2,s2},

5

3

8

6

121

021

tableauoptimal of side hand-Right 1bB

0212

1

021

04c 1BV

B

Example 1

Then (10) yields

From (10’)

112

102

tableauoptimal of 0 rowin oft coefficien

111

1

caBc

x

BV

2 ofelement first

tableauoptimalin oft coefficien1

1

Bc

s

BV

Example 1

(11) yields

Of course, the basic variables x2 and s2 will have zero coefficients in row 0.

Thus, the optimal tableau’s row 0 is

z+x1+2s1=12

126

602

0 row s tableau'optimal theof side hand-right the

1

bBcBV

Example 1

The complete optimal tableau is

z +x1 +2s1 =12

½x1+x2+ ½ s1 =3

3/2x1 - ½ s1 +s2=5


The study of how an LP’s optimal solution depends on its parameters is called sensitivity analysis.

We let BV be the set of basic variables in the optimal tableau. Given a change in an LP, we want to determine whether BV remains optimal.

Important observation:A simplex tableau( for a max problem) for a set of

BV is optimal if and only if each constraint has a nonnegative right-hand side and each variable has a nonnegative coefficient in row 0.


Suppose we have solved an LP and have found that BV is an optimal basis. We can use the following procedure to determine if any change in the LP will cause BV to be no longer optimal.Step 1 Using the preceding formulas, determine

how changes in the LP’s parameters change the right-hand side and row 0 of the optimal tableau.

Step 2 If each variable in row 0 has a nonnegative coefficient and each constraint has a nonnegative right-hand side, BV is still optimal. Otherwise , BV is no longer optimal.


Two reasons why a change in an LP’s parameters causes BV to be no longer optimal. A variable in row 0 may have a negative coefficient.

In this case , a better bfs can be obtained by pivoting in a nonbasic variable with a negative coefficient in row 0. we say that the BV is now a suboptimal basis.

A constraint may now have a negative right-hand side. In this case, at least one member of BV will now be negative and BV will no longer yield a bfs. We say that BV is now an infeasible basis.

Illustrate in example

The Dakota furniture example x1=number of desks manufactured

x2=number of tables manufactured

x3=number of chairs manufactured

The objective function was

max z=60x1+30x2+20x3


The initial tableau was z -60x1 -30x2 - 20x3 =0

8x1+ 6x2+ x3 +s1 =48 (lumber constraint)

4x1+ 2x2+1.5x3 +s2 =20 (finishing constraint)

2x1+1.5x2+0.5x3 +s3=8 (carpentry constraint)

The optimal tableau was z +5x2 +10s2 +10s3 =280

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

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

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

BV={s1,x3,x1} and NBV={x2,s2,s3}. The optimal bfs is z=280,s1=24,x3=8,x1=2,x2=0,s2=0,s3=0


Six types of changes in an LP’s parameters: Change 1 changing the objective function coeffici

ent of a nonbasic variable Change 2 changing the objective function coeffici

ent of a basic variable Change 3 changing the right-hand side of a const

raint Change 4 changing the column of a nonbasic var

iable Change 5 adding a new variable or activity Change 6 adding a new constraint


Changing the objective function coefficient of a nonbasic variable

The only nonbasic decision variable is x2. the objective function coefficient of x2 is c2=30.

How would a change in c2 affect the optimal solution?


Suppose we change c2 from 30 to 30+Δ. From (6), since the B-1 and b are unchanged, the ri

ght-hand side of BV’s tableau (B-1b) has not changed, so BV is still feasible.

From (10), BV will remain optimal if c2≥0, and BV will be suboptimal if c2<0.

From preceding, we know that cBVB-1=[0 10 10].

5)30(

5.1

2

6

101002c


Thus, c2≥0 holds if 5-Δ≥0,or Δ≤5.This means that if the price of tables is de

creased or increased by $5 or less, BV remains optimal.

If BV remains optimal after a change in a nonbasic variable’s objective function coefficient, the values of the decision variables and the optimal z-value remain unchanged.

Illustrate in example For example, if the price of tables increase to

$33(c2=33), the optimal solution to the Dakota problem remains unchanged.

On the other hand, if c2>35, BV will no longer be optimal, because c2<0.

For example, if c2=40,

540

5.1

2

6

101002

c


Now the BV “final” tableau is, Basic

variables 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*

This is not an optimal tableau, and we can increase z by making x2 a basic variable in row 3.


Changing the objective function coefficient of a basic variable

In the Dakota problem, x1 and x3 are basic variables. Since we are not changing B or b, (6) shows

that the right-hand side of each constraint will remain unchanged, and BV will remain feasible.

Since we are changing cBV. From (10), this may change more than one coefficient in row 0.


The Dakota problem,Suppose that c1 is changed from 60 to 60+Δ.

5.1105.0100

5.15.00

420

821

60200

5.15.00

420

821

25.00

45.10

811

1

1

Bc

BB

BV

Illustrate in example Noting that,

The coefficient of each nonbasic variable in the new row 0 is,

20,30,60,

5.0

5.1

1

,

5.1

2

6

,

2

4

8

321321

cccaaa

5.110 ofelement second0 rowin oft coefficien

5.010 ofelement second0 rowin oft coefficien

25.1530

5.1

2

6

5.1105.0100

13

12

221

2

Bcs

Bcs

caBcc

BV

BV

BV


Thus, row 0 of the optimal tableau is now z+(5+1.25Δ)x2+(10-0.5Δ)s2+(10+1.5Δ)s3=?

BV will remain optimal if anf only if the following hold:

5+1.25Δ≥0 Δ≥-4

10-0.5Δ≥0 Δ≤20

10+1.5Δ≥0 Δ≥-20/3

the current basis will remain optimal if and only if -4≤Δ≤20.


Thus, as long as 56=60-4≤c1≤60+20=80, the current basis remains optimal.

Suppose c1=70, z=70x1+30x2+20x3.

revenue should increase by 2(10)=$20, and new revenue=280+20=$300.

Illustrate in example If c1<56 or c1>80, the current basis is no longer op

timal. Suppose c1=100(Δ=40), how can we determine th

e new optimal solution?

360

8

20

48

701000 row of side handright

705.1100 rowin t coefficien

105.0100 rowin t coefficien

00 rowin t coefficien

0,5525.15,0

1

3

2

1

321

bBc

s

s

s

ccc

BV


The BV tableau is , Basic

variables 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


Changing the right-hand side of a constraint Effect on the current basis

Since b does not appear in (10), changing the right-hand side of a constraint will leave row 0 of the optimal tableau unchanged;

changing a right-hand side cannot cause the current basis to become suboptimal.

Will affect the right-hand side of the constraints in the optimal tableau.


Note As long as the right-hand side of each

constraint in the optimal tableau remains nonnegative, the current basis remains feasible and optimal.

If at least one right-hand side in the optimal tableau becomes negative, the current basis is no longer feasible and therefore no longer optimal.


Suppose we change the amount of finishing hours (b2) in the Dakota problem.

Currently,b2=20, we change b2 to 20+Δ, from (6)

The current basis will remain optimal if and only if the following hold:

24+2Δ≥0 Δ≥-12

8+20Δ≥0 Δ≥-4

2-0.50Δ≥0 Δ≤4

5.02

28

224

8

20

48

5.15.00

420

821

8

20

481B


For -4≤Δ≤4, 20-4≤b2≤20+4, or 16≤b2≤24, the current basis remains feasible and therefore optimal.


Effect on decision variables and zEven if the current basis remains optimal, the values of

the decision variables and z change.

The new values of the basic variables are as follows

From (11),

1

12

28

8

22

48

5.15.00

420

8211

1

3

1

bB

x

x

s

300

8

22

48

10100) new( New 1

bBcvaluez BV


When the current basis is no longer optimalSuppose we change b2 to 30. from (6), the right-

hand side of the constraints in the tableau for BV={s1,x3,x1} is

From (11), the right-hand side of row 0 is

3

28

44

8

30

48

5.15.00

420

8211bB

380

8

30

48

101001

bBcBV


The tableau is, Basic

variables

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

-2x2 +s1+ 2s2 -8s3=44 s1=44

-2x2+x3 +2s2 -4s3=28 x3=28

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

Since x1=-3, BV is no longer feasible or optimal.


Changing the column of a nonbasic variable x2 is a nonbasic variable in the optimal solution,

this means that Dakota should not manufacture any tables at present.

Suppose, the price of tables increased to $43 and, because of changes in production technology, a table required 5 board feet of lumber, 2 finishing hours and 2 carpentry hours.


Here we are changing elements of the column for x2 in the original problem. The change leaves B and b unchanged. Thus, the right-hand side of the optimal tableau remains unchanged.

From (10),the only part of row 0 that is changed is c2; the current basis will remain optimal if and only if c2≥0 holds.


Use (10) to compute the new coefficient of x2 in row 0.

Since c2<0, the current basis is no longer optimal. c2=-3 means that each table now increases revenues by $3.

0343

2

2

5

10100

2

2

5

and 43

10100

2

22

1

c

ac

BcBV


From (5),

The tableau is now, Basic

variables

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

-7x2 +s1+ 2s2 -8s3=24 s1=24

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

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

2

4

7

2

2

5

5.15.00

420

821

21aB


Adding a new activity Suppose that Dakota is considering

making footstools. A stool sells for $15 and requires 1 board foot of lumber, 1 finishing hour, and 1 carpentry hour. Should the company manufacture any stools?


Our new initial tableau is z -60x1 -30x2 -20x3-15x4 =0

8x1 +6x2 +x3 +x4+s1 =48

4x1 +2x2+1.5x3 +x4 +s2 =20

2x1+1.5x2+0.5x3 +x4 +s3=8

We call the addition of the x4 column to the problem adding a new activity.


From (6), we see that the right-hand sides of all constraints in the optimal tableau will remain unchanged.

From (10), we see that the coefficient of each of the old variables in row 0 will remain unchanged.

Compute c4, the current basis will remain optimal if c4≥0.


Since

The current basis is still optimal.This means that each stool manufactured will

decrease revenues by $5. for this reason, we choose not to manufacture any stools.

515

1

1

1

10100

1

1

1

and 15

4

44

c

ac

Summary of sensitivity analysis

Change in initial problem

Effect on optimal tableau

Current basis is still optimal if

Changing nonbasic objective function coefficient cj

Coefficient of xj in optimal row 0 is changed

Coefficient of cj in row 0 for current basis is still nonnegative

Changing basic objective function coefficient cj

Entire row 0 may change

Each variable still has a nonnegative coefficient in row 0

Changing right-hand side of a constraint

Right-hand side of constraints and row 0 are changed

Right-hand side of each constraint is still nonnegative

Changing the column of a nonbasic variable xj or adding a new variable xj

Changes the coefficient for xj in row 0 and xj’s constraint column in optimal tableau

The coefficient of xj in row 0 is still nonnegative

sensitivity analysis li xiao-lei. a graphical introduction to sensitivity analysis sensitivity...

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