post optimality

7/27/2019 Post Optimality

1/19

Duality and SensitivityAnalysis

Merton TrucksModel

101Model 102 Availability

Contribution $3000 $5000

Eng. Assy. 1 2 4000

Metal Stmp. 2 2 6000

101 Assy. 2 5000

102 Assy 3 4500

Optimal Product Mix: 2000 Model 101s and 1000 Model 102sOptimal Contribution: $11,000,000

How much is Engine Assembly capacity worth to Merton Trucks?

Increase the Engine capacity availability by 1, and resolve.

The difference in the total contribution = worth of capacity/unit= $2000/hr

2

Merton Trucks (Scaled)

Model 101 Model 102 Availability


Eng. Assy. 1/4000 unit 2/4000 unit 1 unit

Metal Stmp. 2/6000 unit 2/6000 unit 1 unit

unit = 4000 hr

unit = 6000 hr

101 Assy. 2/5000 unit 1 unit

102 Assy 3/4500 unit 1 unit



unit = 5000 hr

unit = 4500 hr

Increase the Engine capacity availability by 1, and resolve.The difference in the total contribution = worth of capacity/unit

= 1 million/unit = 1 million/4000 hr= $250/hr

3


2/19

Worth of Engine Capacity

% Increase Contribution

Worth

(per hour

Capacity)Ori inal 11 000 000, ,

0.5% increase $11,040,000 $2000.00

1% increase $11,080,000 $2000.00

5% increase $11,400,000 $2000.00

10% increase $11,800,000 $2000.00

4

Merton Trucks

Model 101 Model 102 Availability


Eng. Assy. 1 2 4400

Engine Assembly Capacity is now 4400 hours

Metal Stmp. 2 2 6000

101 Assy. 2 5000

102 Assy 3 4500



5



Worth

(per hourCapacity)

Ori inal 11 800 000


, ,

6


3/19



Worth

(per hour

Capacity)Ori inal 11 800 000


, ,

0.5% increase $11,844,000 $2000.00

7



Worth

(per hour

Capacity)


, ,

0.5% increase $11,844,000 $2000.00

1% increase $11,888,000 $2000.00

8



Worth

(per hourCapacity)

Ori inal 11 800 000


, ,

0.5% increase $11,844,000 $2000.00

1% increase $11,888,000 $2000.00

5% increase $12,000,000 $909.09

9


4/19



Worth

(per hour

Capacity)Ori inal 11 800 000


, ,

0.5% increase $11,844,000 $2000.00

1% increase $11,888,000 $2000.00

5% increase $12,000,000 $909.09

10% increase $12,000,000 $454.55

10

Merton Trucks

Engine Assembly capacity = 4000 hrs11

Merton Trucks

Engine Assembly capacity by 1%12


5/19

Merton Trucks


Merton Trucks


Merton Trucks (new)

Engine Assembly capacity = 4400 hrs15


6/19

Merton Trucks (new)


Merton Trucks (new)


Merton Trucks (new)



7/19

Forming the Dual

Primal in standard form Dual in standard form

... +++ bxaxaxan

x

n

cxcxc +++ ...

2211

Max

..ts 11221111 +++ mm cyayaya L

Min

..tsm

y

m

bybyb +++ ...

2211

0,...,,

...

............

............

...

...

21

2211

22222121

+++

+++

n

mnmnmm

nn

nn

xxx

bxaxaxa

bxaxaxa

0,,,

............

............

21

2211

22222112

+++

+++

m

nmmnnn

mm

yyy

cyayaya

cyayaya

K

L

L

19

Input (Primal in standard form):{ Maximization objective

{ Non-negative decision variables

{ Less than or equal to () type constraints

Forming the Dual

Output (Dual):{ Minimization objective

{ One dual variable for each primal constraint

{ Non-negative dual variables

{ Greater than or equal to () type constraints

{ One constraint for each primal variable20

Input (Primal in standard form):{ Minimization objective

{ Non-negative decision variables

{ Greater than or equal to () type constraints

Forming the Dual

Output (Dual):{ Maximization objective

{ One dual variable for each primal constraint

{ Non-negative dual variables

{ Less than or equal to () type constraints

{ One constraint for each primal variable

21


8/19

Primal in non-standard form

22

Primal-Dual Relationship

Primal ProblemObjective: Max

Constraint i :

== form

>= form

Variable j:

xj >= 0

xj urs

xj


9/19

Dual for non-standard Primal

yyy

yyy

321

321

20295

151025

++

++Min

..ts

freeyyy

yyy

yyy

yyy

321

321

321

321

,0,0

40543

25685

30743

+

=+

25

Primal-Dual Example

Primal Dual

04520302 4321 +++ xxxxMax yyy 321 151025 ++Min

0,0,

155672

104849

253535

4311

4321

4321

4321

=+

+

++

xfree,xxx

xxxx

xxxx

xxxx..ts

freeyyy

yyy

yyy

yyy

yyy

321

321

321

321

321

,0,0

40543

25685

30743

20295

+

=+

++..ts

26

maxx

Z x x= +5 41 2

Primal in non-standard form

+ =

=

6 5

8 10

0

1 2

1 2

1

2 1

x x

x

x x; urs27


10/19

Dual for non-standard Primal

miny

w y y y= +6 5 101 2 3

=

+

8 6 4

0

1 2 3

1 2

1 2 3

y y y

y y

y y y; , urs

28

Shadow Price

Shadow Price of a Resource: Price of selling an infinitesimalquantity of that resource.

Also

quantity of that resource.

Generally, both are equal, except when the optimal solution isdegenerate.

Read the documentation to interpret the meaning in case of adegenerate optimal solution.

29

Max

..ts11

2

1

21

+

xx

xx

Shadow Price

X1 =1

Optimal

0,

2

11

21

+

xx

xx

What is the price of buying an infinitesimal quantity of theresource represented by X1 + X2 2?

What is special about this optimal solution?

=

X1 + X2 =2

30


11/19

Max

..ts

1

1

2

1

21

+

x

x

xx

Shadow Price

X1 =1

Optimal

0,

2

11

21

+

xx

xx=

X1 + X2 =3

What is the price of buying an infinitesimal quantity of theresource represented by X1 + X2 2?

31

Max

..ts

1

1

2

1

21

+

x

x

xx

Shadow Price

X1 =1

Optimal

0,

2

21

21

+

xx

xx2 =

X1 + X2=1.5

What is the price of selling an infinitesimal quantity of theresource represented by X1 + X2 2?

32

Using MS Excel Solver

Advertised:

Shadow prices denote the rate of increase in objectivefunction values when the right hand side of the constraint isincreased by a small amount

At degenerate solutions, this is not the full answer.

Do not trust Excel Sensitivity Report if your solution

happens to be degenerate! Use your judgment to interpret

the values you get according to the context.

33


12/19

Laws of Duality Strong Law of Duality:

o If theprimal problem hasa finite optimum, then at optimal solution:

Objective value of Primal = Objective value of Dual

o

Primal Unbounded

Dual Infeasibleo Primal Infeasible Dual Unbounded or Infeasible

Primal Obj at optimality = Dual Obj at Optimality

Obj value of any non-optimal feasible solution for

the Maximization problem

Obj value of any non-optimal feasible solution for

the Minimization problem

Weak Law of Duality: Each feasible solution for the primal (maximization)problem hasan objectivevaluethat is less than or equal to theobjective valueofevery feasible solution to the dual (minimization) problem. 34

Complementary Slackness

Consider an optimal solution to the primal problem.

{ If a constraint is non-binding at the solution, i.e., has a strictlypositive slack, then the dual variable (shadow price)

solution to the dual.

{ If the dual variable (shadow price) corresponding to a particularconstraint has a strictly positive value in an optimal solution tothe dual, then the constraint is binding at an optimal solution tothe primal problem.

Slack Shadow Price = 0

35


Slack in Primal

Constraint

Corresponding

shadow priceAllowed?

Positive Positive Not Allowed

Positive Zero Allowed

Zero Positive Allowed

Zero Zero Allowed

36


13/19


Slack in Primal

Constraint

Corresponding

shadow priceAllowed?

Positive Positive Not Allowed

Positive Zero Allowed

Zero Positive Allowed

Zero Zero Allowed

Notice that complementary slackness is validONLY at an optimal solution.

37

Reduced Costs

The reduced cost of a coefficient of a decision variable in theobjective function is the minimum amount by which thecoefficient should be reduced in order that the decisionvariable achieves a non-zero level in an optimal solution.

What is the reduced cost for a decision variable already atnon-zero value in an optimal solution?

For a minimization problem, reduced costs are either ZERO or?

For a maximization problem, reduced costs are either ZEROor ?

Can a decision variable at zero level have a reduced cost ofzero?

38

Reduced Costs

Min

..ts 12121

+

+

xx

xxX

1+ X

2=1

0, 11 xx

An optimal solution: (x1, x2) = (1, 0)

What is the reduced cost of x2?

p ma X1 + X2

39


14/19

Sensitivity Analysis

Sensitivity analysis tells us the maximum amount by which wecan change any of the coefficients in a linear program such

that the set of constraints that determine an optimal solutiondoes not change.

{ We are concerned with changing only one coefficient and keeping allothers fixed.

{ We are bothered only about the set of constraints that define theoptimal solution they should not change. But, the optimal solutioncan change, the objective function value can change.

40

Sensitivity AnalysisOptimumValue = 11 MillionModel_101 = 2000Model_102 = 1000

Original Model

41

Sensitivity AnalysisOptimumValue = 13 MillionModel_101 = 2000

Model_102 = 1000

Objective function coefficient increasesZ = 3000 Model_101 + 5000 Model_102 to Z = 4000 Model_101 + 5000 Model_102

42


15/19

Sensitivity AnalysisOptimumValue = ???Model_101 = 2000Model_102 = 1000


43

Sensitivity AnalysisOptimumValue = 17.5 MillionModel_101 = 2000Model_102 = 1000


44

Sensitivity AnalysisOptimumValue = 11 MillionModel_101 = 2000

Model_102 = 1000

Original Model

45


16/19

Sensitivity AnalysisOptimumValue = 10.5 MillionModel_101 = 2000Model_102 = 1000

Objective function coefficient decreasesZ = 3000 Model_101 + 5000 Model_102 to Z = 2750 Model_101 + 5000 Model_102

46



47

Sensitivity AnalysisOptimumValue = 9.5 MillionModel_101 = 1000

Model_102 = 1500


48


17/19


Objective function coefficient change

49


Original Model

50


Model_102 = 1200

Engine Assy. RHS increasesModel_101 + 2 Model_102 4000 to Model_101 + 2 Model_102 4200

51


18/19


Engine Assy. RHS increasesModel_101 + 2 Model_102 4000 to Model_101 + 2 Model_102 4600

52


Original Model

53


Model_102 = 800

Engine Assy. RHS decreasesModel_101 + 2 Model_102 4000 to Model_101 + 2 Model_102 3800

54


19/19


Engine Assy. RHS decreasesModel_101 + 2 Model_102 4000 to Model_101 + 2 Model_102 3100

55


Engine Assy. RHS change

56

post optimality

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