advanced operations research models instructor: dr. a. seifi teaching assistant: golbarg kazemi...

Advanced Operations Research Models

Instructor: Dr. A. Seifi

Teaching Assistant: Golbarg Kazemi

Kazemi_Golbarg@yahoo.com

Kazemi_Golbarg@aut.ac.ir

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Today’s lecture

•Course Outline

•Administrivia

•A little LP review

•Economic Interpretation

•Duality theory

•Shadow price

•Assignments2

What you will learn

•Mathematical Modeling– learn a variety of ways of modeling real-

world problems as structured mathematical problems

•Solution Methods– learn to use powerful optimization tools

to solve the problems arising in your mathematical models

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Course Outline

•Linear Optimization Models– Formulations, Sensitivity analysis

•Data Envelopment Analysis and Resource Allocation– Applications and Methods

• Integer Programming– Formulations, Solution methods

•Dynamic Programming– Formulation and Applications

•Column Generation technique

•Quadratic programming– Duality Theorem and QP Solution Methods

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Administrivia•Prerequisite: – Operation Research 1

•Reading material: Wayne L. Winston “ Operation Research: Application and Algorithms”; highly recommended

•Softwares: LINGO- LINDO- NEOS on-line optimizer- XPRESS-MP

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Administrivia (contd.)•Grading

– Assignments: 10%– Projects: 10%-15% 2 projects– Midterm: 30%, Final: 50%

•The assignments would be from the imparted subjects each session will be given to you at the end of the TA session and, you should send them in a week to this address:

Kazemi_ Golbarg@yahoo.com

• Feedback, comments, suggestions, questions are always most welcome. 6

LP review: Definitions

Linear programming problem: – problem of maximizing or minimizing a linear

function of a finite number of variables – subject to a finite number of linear constraints:

, or = constraints

max/min f(x) = c1x1 + c2x2 + … + cnxn

subject to ai1x1 + ai2x2 + … + ainxn bi i=1,…,mFeasible point: xRn s.t. x satisfies all constraints

Feasible region: set of all feasible pointsP = {xRn: x satisfies all

constraints}

=

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LP review: more definitions

max/min f(x) = c1x1 + c2x2 + … + cnxn

subject to a11x1 + a12x2 + … + a1nxn b1

a21x1 + a22x2 + … + a2nxn

b2

a31x1 + a32x2 + … + a3nxn = b3

Decision variables: should completely describe all decisions to be made

Objective function

Constraints

Optimal solution: feasible solution with best (max/min) objective-function valueOptimal value: objective-f’n. value at an optimal solution 8

Economic Interpretation

Suppose that we have to produce three types of materials 1, 2 and 3 by means two different resources human being and wood:

Product 1 profit :5

Product 2 profit: 2

Product 3 profit: 3

Resources 1 2 3 Max. amount

Human resource

1 2 2

8

wood 1 4 3

7

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Continue….

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Simplex Tableau….

Basic Var.

Row No.

Z X1 X2 X3 S1 S2 R.H.S

Z 0 1 -5 -2 -3 0 0 0

S1 1 0 1 2 2 1 0 8

S2 2 0 3 4 1 0 1 7

Basic Var.

Row No.

Z X1 X2 X3 S1

S2 R.H.S

Z 0 1 0 14/3 -4/3 0 5/3 35/3

S1 1 0 0 2/3 5/3 1 -1/3 17/3

X1 2 0 1 4/3 1/3 0 1/3 7/3

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Optimal Tableau…

Basic Var. Row No.

Z X1 X2 X3 S1 S2 R.H.S

Z 0 1 0 26/5 0 4/5 7/5 81/5

X3 1 0 0 2/5 1 3/5 -1/5 17/5

X1 2 0 1 6/5 0 -1/5 2/5 6/5

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Economic Interpretation

Necessary Source for Product 2

How can we produce it?

4/3 Decrease of product 1

Available resource

2 units of S1

4 units of S2

4/3*1=4/3

4/3*3=4

2/3

-

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Interpretation…

Lost Profit:

1.4/3 decrease of product 1: 4/3*5=20/3

2.2/3 consumption of resource 1: 2/3*0=0

Total lost: 0 + 20/3=20/3

Gained profit by producing 1 unite of product 2: 2

Result: 20/3 – 2= 14/3

So we lost!

Note: What’s shadow price?!

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Duality…

00 , x x

18 2xx 3

12 2x

4 s.t x

5x 3xMax z

21

21

2

1

21

dual

0,y,y y

52y2y

33y ys.t.

18y12y4yMin y

321

32

31

3210

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Hints….

Primal problem(Max) Dual problem(Min)

constraint i objective function

r.h.s. constraint

variable =

unrestricted

variable i r.h.s.

objective function variable

constraint unrestricted

=

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Duality theorem….

•Lemma 1: Given a primal linear program, the dual problem of the dual linear program becomes the original primal problem.

•Weak Duality theorem: If x0 is a primal feasible solution and y0 is dual feasible, then .0T0T ybxc

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Corollaries……

Corollary 1 If x0 is primal feasible , y0 is a dual

feasible, and , then x0 and y0 are optimal solutions to the respective problem

Corollary 2

If the primal problem is unbounded above, then the dual problem is infeasible.

0T0T ybxc

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(Contd.)

Corollary 3

If the dual problem is unbounded below, then the primal problem is infeasible.

Note： The converse of the above two is not true .

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Strong Duality Theorem….

If either the primal or the dual problem has a finite optimal solution, then so does the other and they achieve the same optimal objective value.

If either problem has an unbounded objective value, then the other has no feasible solution.

If either problem has no feasible solution, then the other has either no feasible solution or unbounded.

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Complementary slackness

Primal Problem Dual Problem

Max cTx Min bTy

s.t. s.t.

For the primal problem, define ： primal slackness vector ( )

Dual problem, define ： dual slackness vector ( )

bxA

0x

cy TA

0y

0xbs A

0cyr TA

mRs

nRr

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Continue……

For any primal feasible solution x and dual feasible solution y, we know

i.e. is the duality gap between the primal feasible solution x and dual feasible solution y . This duality gap vanishes , if and only if and

• .

ysxr0 TT

ysxr TT

0xr T

0ys T

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Complementary slackness

Let x be a primal feasible solution and y be a dual feasible solution to a symmetric pair of linear problem. Then x and y become an optimal solution pair if and only if the complementary slackness conditions:

either

or xj = 0

either

or yi = 0

are satisfied.

0)(r jT

j A cy

n1,2j

0)(s ii Axb

m1,2i

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Karush-Kuhn-Tucker (K-K-T) Optimality Conditions Theorem

Given a linear programming problem in its standard form, vector x is an optimal solution to the problem if and only if then exist vector y and r, s such that1. Ax+s=b, (primal feasibility)2. (dual feasibility)3. (complementary

slackness)

( ) for canonical form

0s0x ,0rcry ,TA0ys0xr TT ,

0ysxr TT

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Revised simplex

•What?! This method is a modified version of the Primal Simplex Method.

•Why?! It is designed to exploit the fact that in many practical applications the coefficient matrix {aij} is very sparse, namely most of its elements are equal to zero.

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Continue….

•Bottom line:

Don’t update all the columns of the simplex tableau: update only those columns that you need!

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Example…

BASIC VAR.

ROW NO.

Z X1 X2 X3 S1 S2 R.H.S

Z 0 1 -4 -3 -6 0 0 0

S1 1 0 3 1 3 1 0 30

S2 2 0 2 2 3 0 1 40

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Basic Var.

Row NO.

Z X1 X2 X3 S1 S2 R.H.S

Z 0 1 1 0 0 1 1 70

X3 1 0 4/3 0 1 2/3 -1/3 20/3

X2 2 0 -1 1 0 -1 1 10

Continue….

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Continue….

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Economic Interpretation of the dual

shadow price, marginal price, equilibrium price.

Given the standard L.P. model, the primal problem can be viewed as a process of providing different services ( ) to meet a set of customer demands (Ax=b) in a maximum profit manner (max cTx).

0x

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Continue……..

For a non-degenerate optimal solution x* obtain from revised simplex method,

we have with an optimal value

since for a small enough in demand, we know B-1(b+ b)＞ 0 and

is an optimal B.F.S. to the following problem:

Max cTx s.t. Ax=b+ b

0

b

0

xx

1* B* B 0bc 1* Bz T

B

0bx 1* BB

0

bb

0

xx

)(*

1* BB

0x

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Continue…..

The optimal value of above problem is

where y is the dual variables .The above equation says the increment cost of satisfying an incremental demand b is equal

to y*Tb.yi* can be though of the marginal cost , of the

providing one unit of the i-th demand at optimum.

)(1* bbc Bz TB

)let ( 1* BTB

T cy

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Continue…..

i.e. it indicates the maximum unit profit one has received for satisfying additional demands when an optimal is achieved.

Called shadow price or marginal cost.

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Assignments!

1. Use revised simplex method to solve the following LP:

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Continue…

2. Interpret the Simplex tableaus (first and second and the last table are given) below as a production problem (Write a scenario for it, and a model then interpret it)

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Basic Var.

Row No.

Z X1 X2 S1 S2 S3 R.H.S

Z 0 1 -3 -5 0 0 0 0

S1 1 0 1 0 1 0 0 4

S2 2 0 0 2 0 1 0 12

S3 3 0 0 2 0 0 1 18

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Basic Var.

Row No.

Z X1 X2 S1 S2 S3 R.H.S

Z 0 1 -3 0 0 5/2 0 30

S1 1 0 1 0 1 0 0 4

X2 2 0 0 1 0 1/2 0 6

S3 3 0 3 0 0 -1 1 6

Basic Var.

Row No.

Z X1 X2 S1 S2 S3 R.H.S

Z 0 1 0 0 0 3/2 1 36

S1 1 0 0 0 0 1/3 -1/3 2

X2 2 0 0 0 1 1/2 0 6

X1 3 0 1 1 0 -1/3 1/3 2

Continue….

3. In the previous question determine the shadow prices, and if the sources vary from 4,12 and 18 to 5,14,18 what happens to the profit?!

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I will be waiting for your assignments until next Sunday 12:00 PM

Don’t forget ….

Thanks a lot

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