sensitivity analysis and duality -...

L. Ntaimo (c) 2005 INEN420 TAMU1

Sensitivity Analysis and Duality Part I – Sensitivity Analysis

Based onChapter 6

Introduction to Mathematical Programming: Operations Research, Volume 14th edition, by Wayne L. Winston and Munirpallam Venkataramanan

Lewis Ntaimo


6.1 – Introduction

• Sensitivity analysis (SA) and duality are two of the most important topics in linear programming.

• Sensitivity analysis is concerned with how changes in an LP’s parameters affect the LP’s optimal solution. These parameters are the objective function coefficients, right-hand sides and the technology matrix.



Sensitivity analysis is important for several reasons :

• In many applications, the values of an LP’s parameters may change, e.g. prices, demand. Remember the certainty assumption of linear programming?

• If a parameter changes sensitivity analysis often makes it unnecessary to solve the problem again. Solving an LP with thousands of variables and constraints again would be a chore!

• The knowledge of sensitivity analysis often enables the analyst (you ☺) to determine from the original solution how changes in an LP’s parameters change the optimal solution.



We need the knowledge of matrices to express simplex tableaus inmatrix form before learning how to perform sensitivity analysis on an arbitrary LP.

You already know this!


6.2 – A Graphical Illustration of Sensitivity AnalysisReconsider the Giapetto problem from Chapter 3:

Decision Variables: x1 = number of soldiers produced each week

x2 = number of trains produced each week.

Max z = 3x1 + 2x2

2 x1 + x2 ≤ 100 (finishing constraint)

x1 + x2 ≤ 80 (carpentry constraint)

x1 ≤ 40 (demand constraint)

x1,x2 ≥ 0

Max z = 3x1 + 2x2

2 x1 + x2 + s1 = 100

x1 + x2 + s2 = 80

x1 + s3 = 40

x1, x2 , s1, s2, s3 ≥ 0

Standard Form:


6.2 – A Graphical Illustration of Sensitivity Analysis

X1

X2

20 40 50 60 80

2040

6080

1 00

finishing constraintSlope = -2

carpentry constraintSlope = -1

demand constraint

Feasible RegionA

D

Isoprofit line z = 120 Slope = -3/2

C

B

Giapetto Problem The optimal solution to the Giapetto problem is z = 180, x1 = 20, x2 = 60 (Point B in the figure) with x1, x2, and s3 as basic variables.

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


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.

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

Since a typical isoprofit line is

c1x1 + 2x2 = k,

the slope of the isoprofit line is just -c1/2.X1

X2

20 40 50 60 80

2040

6080

100



demand constraint

Feasible RegionA

D


C

B

Giapetto Problem



In summary:

X1

X2

20 40 50 60 80

2040

6080

100



demand constraint

Feasible RegionA

D


C

B

Giapetto Problem

1. Point A is optimal if -c1/2 ≥ -1 or 0 ≤ c1 ≤ 2 ( -1 is the carpentry constraint slope).

2. Point B is optimal if -2 ≤ -c1/2 ≤ -1 or 2 ≤ c1 ≤ 4 (between the slopes of the carpentry and finishing constraint slopes).

3. Point C is optimal if -c1/2 ≤ -2 or c1≥ 4 ( -2 is the finishing constraint slope).


Let b1 = number of finishing hours in the Giapetto LP.2 x1 + x2 ≤ b1 (finishing constraint)

Then we see that a change in b1

shifts the finishing constraint parallel to its current position.

The current optimal point (Point B) is where the carpentry and finishing constraints are binding.

X1

X2

20 40 50 60 80

2040

6080

10 0

finishing constraint, b1 = 100

carpentry constraint

demand constraint

Feasible Region

A

D

Isoprofit line z = 120

C

B



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.



If we change the value of b1, then as long as the point where the finishing and carpentry constraints intersect are binding remains feasible, the optimal solution will still occur where these constraints intersect.

We see that if b1 > 120, x1 will be greater than 40 and will violate the demand constraint. Also, if b1 < 80, x1 will be less than 0 and the nonnegativity constraint for x1 will be violated.

Therefore: 80 ≤ b1 ≤ 120

The current basis remains optimal for 80 ≤ b1 ≤ 120, but the decision variable values and z-value will change.


X1

X2

20 40 50 60 80

2040

6080

10 0


carpentry constraint

demand constraint

Feasible Region

A

D

Isoprofit line z = 120

C

B





In Summary:

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.


6.3 – Shadow Prices

It is important to determine how a constraint’s RHS changes the optimal z-value. We define:

The shadow price for the i th constraint of an LP is the amount by which the optimal z-value is improved (increased in a max problem or decreased in a min problem) if the RHS of the i th constraint is increased by 1. This definition applies only if the change in the RHS of constraint i leaves the current basis optimal.


6.3 – Shadow Prices

Giapetto Example: Max z = 3x1 + 2x2

2 x1 + x2 ≤ 100 (finishing constraint)

x1 + x2 ≤ 80 (carpentry constraint)

x1 ≤ 40 (demand constraint)

x1,x2 ≥ 0

Using the finishing constraint as an example, we know, 100 + ∆ finishing hours are available (assuming the current basis remains optimal).

The LP’s optimal solution is then x1 = 20 + ∆ and x2 = 60 – ∆ with z = 3x1 + 2x2 = 3(20 + ∆) + 2(60 - ∆) = 180 + ∆.

Thus, as long as the current basis remains optimal, a one-unit increase in the number of finishing hours will increase the optimal z-value by $1. So, the shadow price for the first (finishing hours) constraint is $1.


6.4 – Some Important FormulasAn LP’s optimal tableau can be expressed in terms of the LP’s parameters. The formulas we will develop in this section are used in the study of sensitivity analysis, duality, and advanced LP topics.

Assume that we are 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, we will label them x1, x2, …, xn. The LP may then be written as follows:

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)


6.4 – Some Important Formulas• Definitions:

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.

xBV = vector listing the basic variables in the optimal tableau.

xNBV = vector listing the nonbasic variables in the optimal tableau.

cBV = row vector of the initial objective coefficients for the optimal tableau’s basic variables.

cNBV = row vector of the initial objective coefficients for the optimal tableau’s nonbasic variables.


6.4 – Some Important Formulas

• Definitions Cont..

B is an m x m matrix whose j th 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.

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


6.4 – Some Important FormulasAs an example, consider the Dakota Furniture problem without thex2 ≤ 5 constraint (initial tableau):

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

8x1 + 6x2 + x3 + s1 = 48

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

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

Suppose we have found the optimal solution to the LP with the following optimal tableau:

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

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

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

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


6.4 – Some Important FormulasDakota Problem Optimal Tableau:

Since BV = {s1, x3, x1}, cBV = [0 20 60]

Since NBV = {x2, s2, s3}, cNBV = [30 0 0]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

1

3

1

BV

xxs

X

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

2

NBV

ssx

X

Original objective: Max z = 60x1 + 30x2 + 20x3

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

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

- 2x2 + x3 + 2s2 - 4s3 = 8x1 + 1.25x2 - 0.5 s2 + 1.5s3 = 2

BV = {s1, x3, x1}

NBV = {x2, s2, s3


6.4 – Some Important FormulasDakota Problem Initial Tableau:

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

8x1 + 6x2 + x3 + s1 = 48

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

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

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

25.0045.10811

B

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

248

a1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

1.526

a2Since BV = {s1, x3, x1}, , , …

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

82048

bSince NBV = {x2, s2, s3}, ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

105.1012006

N, b1 = 48, b2 = 20, b3 = 8.



We are now in a position to use matrix algebra to determine how an LP’s optimal tableau (with the set of basic variables BV) is related to the original LP.

We observe that an LP may be written as follows:

Max z = cBVxBV + cNBVxNBV

s.t. BxBV + NxNBV = b

xBV, xNBV ≥ 0



• Using the format on the left, the Dakota problem can be written as:

.000

ssx

,000

xxs

82048

ssx

.101.5012006

xxs

.20.5041.50811

s.t.

ssx

0]. 0 [30 xxs

60]. 20 [0 zMax

3

2

2

1

3

1

3

2

2

1

3

1

3

2

2

1

3

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡≥

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡≥

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

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⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

Max z = cBVxBV + cNBVxNBV

s.t. BxBV + NxNBV = b

xBV, xNBV ≥ 0



• Multiplying the constraints through by B-1 yields:

B-1BxBV + B-1NxNBV = B-1b =>

xBV + B-1NxNBV = B-1b

Using the Gauss-Jordan method (or your calculator!) for the Dakota problem we know:

Substituting into xBV + B-1NxNBV = B-1b yields:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

=5.15.00420821

B 1-

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2824

.5.15.025.1422822

3

2

2

1

3

1

ssx

xxs


6.4 – Some Important Formulas• Conclusions:

• Column for xj in optimal tableau’s constraints = B-1aj

• Right-hand side of optimal tableau’s constraints xBV = B-1b

e.g. column x2 in the Dakota optimal tableau = B-1a2

e.g. RHS in the Dakota optimal tableau = B-1b

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

8x1 + 6x2 + x3 + s1 = 48

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

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

a2 b

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2824

82048

.1.50.50

420821

RHS

1

3

1

optimal

xxs

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

=25.122

5.126

.5.15.00420821

x2


6.4 – Some Important Formulas• Determining the Optimal Tableau’s Row 0 in Terms of BV and the Initial

LPWe multiple the constraints BxBV + NxNBV = b through by the vector cBVB-1.

cBVxBV + cBVB-1NxNBV = cBVB-1b

We know the original objective function:

z - cBVxBV - cNBVxNBV = 0

Adding the two equations together and eliminating the optimal tableau’s basic variables we obtain:

z + (cBVB-1N – cNBV) xNBV = cBVB-1b


cBVB-1(column of N for xj) – (coefficient of xj in cNBV) = cBVB-1aj - cj

The RHS of row 0 is cBVB-1b and is the optimal objective function value (z-value)

jc = cBVB-1aj - cj is the so called reduced cost for a nonbasic variable xj.

Let be the coefficient of xj in the optimal tableau’s row 0

Thenjc

6.4 – Some Important FormulasFrom z + (cBVB-1N – cNBV) xNBV = cBVB-1b the coefficient of nonbasic variable xj in row 0 is:

Reduced Cost: For any nonbasic variable, the reduced cost is the amount by which its objective function coefficient must be improved before that variable will be a basic variable in some optimal solution to the LP.


6.4 – Summary of Formulas for Computing the Optimal Tableau

• xj column in optimal tableau’s constraints = B-1aj

• RHS of optimal tableau’s constraints = B-1b

• Reduced cost for a nonbasic variable xj: cBVB-1aj – cj

• Coefficient of slack variable si in optimal row 0 = ith element of cBVB-1

• Coefficient of excess variable ei in optimal row 0 =

-(ith element of cBVB-1)

• Coefficient of artificial variable ai in optimal row 0 =

(ith element of cBVB-1) + M (max problem)

• RHS of optimal row 0 = CBVB-1b

=jc


6.4 – Summary of Formulas for Computing the Optimal Tableau

Summary of a Simplex Tableau: This is a MUST know!

AB 1−

bBc 1BV

−cABc 1BV −−

bB 1−

In more detail:

n1

11 aB . . . aB −−

BVxcBVn1 c . . . c

mBVx

1BVx...


6.5 – Important Observation

The mechanics of sensitivity analysis hinge on the followingimportant observation:

We know that a simplex tableau (for a max problem) for a set of basic variables BV is optimal iff (if and only if) each constraint has a nonnegative RHS and each variable has a nonnegative coefficient in row 0.

• Mathematically, we have two conditions I refer to as follows:

“Optimality condition”:

“Feasibility condition”: 0) xb,B x:(Note 0bB BV1

BV1 ≥=≥ −−

NBVj 0caBcc jj1

BVj ∈∀≥−= −


6.6 – Sensitivity Analysis

Suppose we have solved an LP and have found that BV is an optimal basis. A procedure to determine if any change in the LP will cause the BV to be no longer optimal (to become suboptimal) is as follows:

Step 1 Use the formulas derived in the previous slides to determine how changes in the LP’s parameters change the right hand side of row 0 of the optimal tableau (the tableau having BV as the set of basic variables).

Step 2 If each variable in row 0 has a nonnegative coefficient (optimality condition) and each constraint has a nonnegative RHS (feasibility condition), BV is still optimal. Otherwise, BV is no longer optimal.

If BV is no longer optimal, find the new optimal solution by using the derived formulas to recreate the entire tableau for BV and then continuing the simplex algorithm with the BV tableau as your starting tableau.



There are two reasons why a change in an LP’s parameters cause BV to no longer be optimal:

1. A variable (or variables) in row 0 may have a negative coefficient. In this case, a better (larger z-value) bfs can be obtained by pivoting in a nonbasic variable with a negative coefficient in row 0. If this occurs, the BV is now a suboptimal basis. (This is violation of the optimalitycondition)

2. A constraint (or constraints) may now have a negative RHS. In this case, at least one member of BV will now be negative and BV will no longer yield a bfs. If this occurs, we say they BV is now an infeasible basis. (This is violation of the feasibility condition)



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

1. Changing the objective function coefficient of a nonbasic variable.

2. Changing the objective function coefficient of a basic variable.

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

4. Changing the column of a nonbasic variable.

5. Adding a new variable or activity.

6. Adding a new constraint.



1. Changing the Objective Function Coefficient of a Nonbasic Variable.

• Since B (hence B-1) and b are unchanged, feasibility is not affected.

• However, cbv is unchanged but cj (hence may change). Thus optimalityis affected in this case.

• When is BV still optimal?

jc

If the objective function coefficient of a nonbasic variable xj is changed, the current basis remains optimal if . If , then the current basis is no longer optimal, and xj will be a basic variable in the new optimal solution.

0cj <0cj ≥

Note: The process of computing the new coefficient of a nonbasic variable xjusing is called pricing out xj.0caBcc jj

1BVj ≥−= −



2. Changing the Objective Function Coefficient of a Basic Variable.

• Since B (hence B-1) and b are unchanged, feasibility is not affected.

• However, cbv is changed and hence may change for variable xj. Thus optimality is affected in this case.


jc

If the objective function coefficient of a basic variable xj is changed, the current basis remains optimal if . If for any variable xj, then the current basis is no longer optimal.

0cj <NBV BVand j 0cj ∈∀≥

Note: If the current basis remains optimal, then the values of the decision variables do not change because xBV = B-1b remains unchanged. However, the optimal z-value (z = cBVB-1b) does change.



3. Changing the Right-Hand Side (RHS) of a Constraint.

• Since b does not appear in the optimality condition, changing the RHS of a constraint does not affect optimality but affects feasibility.

• Feasibility requires


0bBx 1BV ≥= −

If the RHS of a constraint is changed, then the current basis (BV) remains optimal if the RHS of each constraint in the tableau remains nonnegative. If the RHS of any constraint becomes negative, then the current basis in infeasible, and a new optimal solution must be found (Use the Dual Simplex Method).

Note: Changing b will change the values of the basic variables and the optimal z-value.



4. Changing the Column of a Nonbasic Variable.

• Does not affect feasibility ( ) but affects optimality.0bBx 1BV ≥= −

If the column of a nonbasic variable xj is changed, the current basis remains optimal if . If , then the current basis is no longer optimal and xj will be a basic variable in the new optimal tableau.

0cj ≥ 0cj <

Note: If the column of a basic variable is changed, then it is usually difficult to determine whether the current basis is optimal (both the optimality and feasibility conditions are affected). As always, the current basis would remain optimal if the optimality and the feasibility conditions are both satisfied.



5. Adding a New Activity (Addition of a New Variable).

• In many practical situations, opportunities arise to undertake new activities

• Does not affect feasibility ( ) but affects optimality.

• To determine whether a new activity xj will cause the current basis to be no longer optimal, price out the new activity!

0bBx 1BV ≥= −

If a new column (corresponding to a variable xj) is added to an LP, then the current basis remains optimal if . If , then the current basis is no longer optimal and xj will be a basic variable in the new optimal solution.

0cj ≥ 0cj <



6. Adding a New Constraint

Case 1. The current optimal solution satisfies the new constraint. Then the current basis is still optimal.

Case 2. The current optimal solution does not satisfy the new constraint but the LP is still feasible. In this case, use the Dual Simplex Method to determine the new optimal solution.

Case 3. The additional constraint causes the LP to have no feasible solutions. In this case the Dual Simplex Method will show that the LP has become infeasible.


6.7 – Sensitivity Analysis Example

Dakota Furniture Problem (Standard Form):Max z = 60x1 + 30x2 + 20x3 (total revenue)

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

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

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

x1, x2, x3, s1, s2, s3 ≥ 0

x1 = # of desks manufacturedx2 = # of tables manufactured x3 = # of chairs manufactured

Optimal Dakota Tableau:Row z x1 x2 x3 s1 s2 s3 RHS BV

0 1 0 5 0 0 10 10 280 z = 2801 0 0 -2 0 1 2 -8 24 s1 = 242 0 0 -2 1 0 2 -4 8 x3 = 83 0 1 1.25 0 0 -0.5 1.5 2 x1 = 2



• Currently c2 = 30.• For what values of c2 would BV = {s1, x3, x1} remain optimal? (Affects only the optimality

condition)

1. Changing the Objective Function Coefficient of a Nonbasic Variable.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=∴

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

∆+=∆

1.50.5-04-208-21

B ,25.0045.10811

B },x,x,{sBV

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1-131

2

2

10] 10 [0 1.50.5-0

4-208-21

60] 20 [0

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=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

=

Note: For any simplex tableau, B-1 is the m x m matrix consisting of the columns in the current tableau that correspond to the initial tableau’s set of basic variables taken in the same order.

.5or 0,5 if optimal remains BV and holds, 0 c Thus5 c

)30(35

)(301.526

10] 10 [0

cya c-aBccThen

2

2

22

22-1

BV2

≤∆≥∆−≥∆−=∴

∆+−=

∆+−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

−==

Conclusion:

If the price of tables is decreased or increased by $5 or less, BV remains optimal.

Thus, for c2 ≤ 30 + 5 = 35, BV remains optimal.

Also, the z-value remains the same ($280).


6.7 – Sensitivity Analysis Example1. Changing the Objective Function Coefficient of a Nonbasic Variable.

• Consider the case when c2 > 35, e.g. c2 = 40.• We know that BV will now be suboptimal.

method.simplex theofiteration an Performbasis! enter the nowcan xand 0 c

-5

401.526

10] 10 [0

cya c-aBccThen

22

22

22-1

BV2

<∴=

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⎡=

−==


6.7 – Sensitivity Analysis Example1. Changing the Objective Function Coefficient of a Nonbasic Variable.

Suboptimal TableauRow z x1 x2 x3 s1 s2 s3 RHS BV MRT

0 1 0 -5 0 0 10 10 280 z = 2801 0 0 -2 0 1 2 -8 24 s1 = 24 none2 0 0 -2 1 0 2 -4 8 x3 = 8 none3 0 1 1.25 0 0 -0.5 1.5 2 x1 = 2 1.6*

Optimal TableauRow z x1 x2 x3 s1 s2 s3 RHS BV

0 1 4 0 0 0 8 16 288 z = 2881 0 1.6 0 0 1 1.2 -5.6 27.2 s1 = 27.22 0 1.6 0 1 0 1.2 -1.6 11.2 x3 = 11.23 0 0.8 1 0 0 -0.4 1.2 1.6 x2 = 1.6

• Thus, if c2 = 40, the optimal solution to the Dakota Problem changes to x1 = 0, x2 = 1.6, x3 = 11.2, and z = 288.

• The increase in the price of tables causes Dakota Furniture to manufacture tables instead of desks!


6.7 – Sensitivity Analysis Example2. Changing the Objective Function Coefficient of a Basic Variable.

• Consider changing c1 from its current value of c1 = 60.• For what values of c1 would BV = {s1, x3, x1} remain optimal? (Affects both the

optimality and feasibility conditions)

zero. be stillmust 0 rowin tscoefficien their , variablesbasic are xand , x,s Since

:0 row new thecompute nowcan We]1.510 0.5-10 [0

B ]60 20 [0

Bcy Let

},x,x,{sBV Since.60 cThen

.c changed have eby which wamount thedenote Let

131

1-

1-BV

131

1

1

∆+∆=∆+=

=

=∆+=

∆


6.7 – Sensitivity Analysis Example2. Changing the Objective Function Coefficient of a Basic Variable.

∆+=

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡∆+∆=

−==

25.15

301.526

]1.510 0.5-10 [0

cyac-aBcc :x:have we variablesnonbasicfor However,

22221-

BV22

-20/3). iff (true 05.110 0 c 20) iff (true 05.010 0 c -4) iff (true 025.15 0 c

holdfollowingtheiffoptimal RemainsBV

6

5

2

≥∆≥∆+⇒≥≤∆≥∆−⇒≥≥∆≥∆+⇒≥

.300$ 20(8) 30(0)70(2)

20x 30x70x z 70 c e.g. :changemay value-z optimal theHowever,

optimal. remains BV800756 70, c e.g. unchanged. are sbv' then of values theoptimal, remains BV If

.80c56 :Range. 204-if optimal remainsBV

3211

1

1

=++=++=⇒=

⇒≤≤=

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∆−=

−⎥⎥⎥

⎦

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⎢⎢⎢

⎣

⎡∆+∆=

−==

5.001

0010

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cyac-aBcc :s 5555-1

BV52

∆+=

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5.101

0100

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cyac-aBcc :s 6666-1

BV63



• When the current BV is no longer optimal.2. Changing the Objective Function Coefficient of a Basic Variable.

70)40(5.1105.110c :s 10)40(5.0105.001c :s

(bv) 0c :s (bv) 0c : x

55)40(25.1525.15c : x(bv) 0c : x

:0 row is that costs, reduced theCompute20. 40 60-100 then 100,cConsider

63

52

41

33

22

11

1

=+=∆+=−=−=∆−=

==

=+=∆+==

>==∆=

solution! optimal newget tobasis theinto sEnter .suboptimal now is }x,x,{sBV 100,c If

.360 82048

]70 01- 0[

ybbBc 0 row of RHS]70 01- 0[

]5.101 5.001 0[Bcy Let

2

1311

1BV

1BV

===

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

==

=∆+∆−==

−

−



• When the current BV is no longer optimal.2. Changing the Objective Function Coefficient of a Basic Variable.

c1 = 100Suboptimal Tableau

Row z x1 x2 x3 s1 s2 s3 RHS BV MRT0 1 0 55 0 0 -10 70 360 z = 3601 0 0 -2 0 1 2 -8 24 s1 = 24 122 0 0 -2 1 0 2 -4 8 x3 = 8 4*3 0 1 1.25 0 0 -0.5 1.5 2 x1 = 2 none

Optimal TableauRow z x1 x2 x3 s1 s2 s3 RHS BV

0 1 0 45 5 0 0 50 400 z = 4001 0 0 0 -1 1 0 -4 16 s1 = 162 0 0 -1 0.5 0 1 -2 4 s2 = 43 0 1 0.75 0.25 0 0 0.5 4 x1 = 4

• Thus, if c1 = 100, the optimal solution to the Dakota Problem changes to x1 = 4, x2 = x3 = 0, and z = 400.

• The increase in the price of desks causes Dakota Furniture to stop making chairs but make desks only!


6.7 – Sensitivity Analysis Example3. Changing the RHS of a constraint.

• Consider changing b2 from its current value of b2 = 20.• For what values of b2 would BV = {s1, x3, x1} remain optimal? (Affects feasibility condition only)

)4 iff (true 00.5 2 -4) iff (true 02 8 -12) iff (true 02 24

0bB optimalremain toBVFor

5.0228224

820

48

1.50.5-04-208-21

bB

:become sconstraint theof RHS then the, 20bLet

1-

1-

2

≤∆≥∆−≥∆≥∆+≥∆≥∆+≥

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆−∆+∆+

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡∆+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

∆+=

.300 82248

10] 10 [0

b) (newBcz

11228

82248

B x22, b e.g.

b) new(Bxchange.may z and variablesbasic

theof values theoptimal, remains BV ifEven 42 b 16

44-:Range

1BV

1

1

3

1

BV2

1BV

2

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

=

≤≤⇒≤∆≤⇒

−

−

−

xxs


6.7 – Sensitivity Analysis Example3. Changing the RHS of a constraint.

• Remember that shadow price/dual multipliers are y = cBVB-1?

• The important concept of how dual multipliers (shadow prices) can be used to determine how changes in the RHS change the optimal z-value should be apparent by now.

• What if BV is no longer optimal?

solution. optimal new get the toAlgorithmSimplex Dual theuse tableau,infeasiblean have weSince

.38083048

]10 10 [0 yb bBc 0 Row of RHS

32844

83048

1.50.5-0

4-208-21

bB

then ,24 30bLet

1-BV

1-

2

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡===

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

>=


6.7 – Sensitivity Analysis Example4. Changing the Column of a Nonbasic Variable.

• Suppose the price of tables increased from $30 to $43 and that due to changes in technology the table now requires 5 board ft of lumber, 2 finishing hours, and 2 carpentry hours.

• Would this change the optimal solution to the Dakota Furniture problem?

286). pageon 8 and 7 Table (Seealgorithmsimplex apply the then and }x,x,{s BVfor tableau Recreate

optimal.longer no is }x,x,{s BV 0c Since3

43225

10] 10 [0

cya c.out x pricing 0 rowin xoft coefficien

new thecompute tocaBcc useSimply

.1.526

a 30, c Before .225

a 43, c

131

1312

222

22

221-

BV2

2222

==<

−=

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

−=⇒−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==


6.7 – Sensitivity Analysis Example5. Adding a New Activity.

• Suppose Dakota Furniture decides to start making stools. The price of a stool is $15 and requires 1 board ft of lumber, 1 finishing hours, and 1 carpentry hours.

• Should Dakota Furniture manufacture stools?

• Price out the new activity!

stool.any emanufacturnot should Furniture Dakota Therefore, $5.by revenue drecrease

wouldstooleach that means This $5. iscost reduced Theoptimal. still is }x,x,{s BV 0c Since

5

15111

10] 10 [0

caBc cThen

15 c and 111

aThen

ed.manufactur stools of # Let x

1312

441

BV4

44

4

=≥=

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

−=

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

=

−

sensitivity analysis and duality -...

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