kuliah 11 - optimization and linear programming

Optimization and

Linear Programming

Rini Novrianti Sutardjo Tui

The Optimal Decision Problem

Decision which incurs the least cost among the set of permissible decisions

The decisions can be ranked according to whether they incur

greater or lesser cost

The cost of each

decision is known

The set of all permissible decisions is

known

Optimization Methods

Dynamic Programming

Linear Programming

Non-linear Programming

Discrete-time Optimal Control

Linear Programming (LP)

Maximize/minimize : Z = c1x1 + c2x2 + + cnxn

Constraints : a11x1 + a12x2 + + a1nxn = b1

a21x1 + a22x2 + + a2nxn = b2

am1x1 + am2x2 + + amnxn = bm

x1 0 ; x2 0 ; ; xn 0

b1 0 ; b2 0 ; ; bm 0

Standard form of LP problem with m constraints and n variables

Linear Programming in Vector Form

Maximize/minimize : Z = cx

Constraints : Ax = b

x 0

b 0

mnmm

n

n

aaa

aaa

aaa

...

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

...

...

21

22221

11211

Amxn =

n

n

x

x

x

x...

2

1

1

bm

b

b

b m...

2

1

1

nn cccc ...211

Prerequisites of LP Model

All constraints are stated as equations

Constant in right side of constraints

is non-negative

Function of objective is maximize or minimize

All variables are non-negative

Linear Programming Model

Minimize : Z = 40x1 + 36x2

Constraints : x1 8

x2 10

: 5x1 + 3x2 45

x1 0 ; x2 0

x2

x1

5

10

15

5 9

x1 = 8

x2 = 10

A (8, 5/3)

B C

Simplex Method

All constraints have to be equations, if not then slack variable or surplus variable is needed.

Simplex For complex and broad

problem of linear programming

Constant in right side of constraint equations has to be non-negative, if not then it has to be transformed to positive value.

General Model of Simplex Method

Maximize Function of Objective: Maximize

Z C1X1-C2X2- . . . . . CnXn-0S1-0S2-. . .-0Sn = NK

Function of Constraints:

a11X11+a12X12+. . . .+a1nXn+ S1+0S2+. . .+0Sn = b1 a21X21+a22X22+. . . .+a2nXn+ 0S1+1S2+. . .+0Sn = b2

. .. . .. .. . ..=

am1Xm1+am2Xm2+. . . .+amnXn+ S1+0S2+. . .+1Sn = bm Activities Variables Slack Variables

General Model of Simplex Method

Minimize Function of Objective: Minimize

Z C1X1-C2X2- . . . . . CnXn-0S1-0S2-. . .-0Sn = RhV


a11X11+a12X12+. . . .+a1nXn - S1 -0S2-. . . - 0Sn = b1 a21X21+a22X22+. . . .+a2nXn - 0S1-1S2 -. . . - 0Sn = b2

. .. . .. .. . ..=

am1Xm1+am2Xm2+. . . .+amnXn- S1- 0S2 -. . . -1Sn = bm

Activities Variables Surplus Variables

Solving Simplex Model

Formulate LP problem into standard form of LP

Transform LP model into simplex model

Formulate simplex table

Take steps of solving

Simplex Model

Table of Matrix

Column table of basic variables

Row table of Cj - Zj

COLUMN TABLE OF MATRIX OF BASIC VARIABLES

Linear Programming

Example Step 1

Maximize:

Linear Programming Method

Function of Objective:

Maximize: Z=8X1 + 6X2 (in thousand IDR)


Material A : 4X1 + 2X2 60

Material B : 2X1 + 4X2 48

X1, X2 0

Step 2

Simplex Model:

Function of Objective: Maximize

Z 8X16 X20S1- 0S2 = 0


4X1+2X2+ S1+ 0S2 = 60

2X1+4X2+0S1+ 1S2 = 48

X1, X2, S1, S2 0

Step 3 1

Basic Variables

Z X1 X2 S1 S2 RhV

Simplex Table

Step 3 2

Simplex Table

Basic Variables

Z X1 X2 S1 S2 RhV

Z

S1

S2

Step 3 3

Simplex Table

Basic Variables

Z X1 X2 S1 S2 RhV

Z 1 -8 -6 0 0 0

S1

S2

Step 3 4

Simplex Table

Basic Variables

Z X1 X2 S1 S2 RhV

Z 1 -8 -6 0 0 0

S1 0 4 2 1 0 60

S2

Step 3 5

Simplex Table

Basic Variables

Z X1 X2 S1 S2 RhV

Z 1 -8 -6 0 0 0

S1 0 4 2 1 0 60

S2 0 2 4 0 1 48

Step 4 1

First iteration Iteration-0

Basic Variables

Z X1 X2 S1 S2 RhV

Z 1 -8 -6 0 0 0

S1 0 4 2 1 0 60

S2 0 2 4 0 1 48

Step 4 2

Iteration-1: Determining key column

Basic Variables

Z X1 X2 S1 S2 RhV

Z 1 -8 -6 0 0 0

S1 0 4 2 1 0 60

S2 0 2 4 0 1 48

Key column: column with biggest negative value of coefficients of function of constraints.

Step 4 3

Iteration-1: Determining key row

Basic Variables

Z X1 X2 S1 S2 RhV Index

Z 1 -8 -6 0 0 0 -

S1 0 4 2 1 0 60 15

S2 0 2 4 0 1 48 24

Key row: row with smallest index value (positive).

Index value: RV/key column

Key number

Step 4 4

Iteration-1: Changes of row value

Basic Variables

Z X1 X2 S1 S2 RhV

Z

X1 0 1 0 15

S2

New key row: (previous key row)/key number

Other row value: (previous row value) ( (new key row) x key column of the row itself).

Step 4 5


Basic Variables

Z X1 X2 S1 S2 RhV

Z

X1 0 1 0 15

S2 0 0 3 - 1 18


Other row value: (previous row value) (new key row) x key column of the row itself

Step 4 6


Basic Variables

Z X1 X2 S1 S2 RhV

Z 1 0 - 2 2 0 120

X1 0 1 0 15

S2 0 0 3 - 1 18


Other row value: (previous row value) (new key row) x key column of the row itself

Step 4 7

Basic Variables

Z X1 X2 S1 S2 RhV

Z 1 0 - 2 2 0 120

X1 0 1 0 15

S2 0 0 3 - 1 18

Iteration-2: Determining key column

Step 4 8

Basic Variables


Z 1 0 - 2 2 0 120 -

X1 0 1 0 15 30

S2 0 0 3 - 1 18 6

Iteration-2: Determining key row

Step 4 9

Basic Variables


Z

X1

X2 0 0 1 - 1/6 1/3 6 -


Step 4 10

Basic Variables


Z 1 0 0 5/3 2/3 132 -

X1 0 1 0 1/3 - 1/6 12 -

X2 0 0 1 - 1/6 1/3 6 -


Solution

Optimal

X1 = 12

X2 = 6 Zmax = IDR

132,000

ROW TABLE OF MATRIX OF CJ - ZJ

Linear Programming

Example Step 1

Maximize:

Linear Programming Method

Function of Objective:

Maximize: Z=3X + 5Y


2X 8

3Y 15

6X + 5Y 30

X, Y 0

Step 2

Simplex Model:

Function of Objective: Maximize

Z=3X+5Y+0S1+0S2+0S3


2X+0Y+ 1S1+ 0S2+0S3 = 8

0X+3Y+ 0S1+ 1S2+0S3 = 15

6X+5Y+ 0S1+ 0S2+1S3 = 30

X,Y, S1, S2 , S3 0

Step 3 Enter each coefficient into simplex table

Basic Var. Obj. Cj 3 5 0 0 0 Qty. Ratio

X Y S1 S2 S3

S1 0 8 2 0 1 0 0

S2 0 15 0 3 0 1 0

S3 0 30 6 5 0 0 1

Zj 0 0 0 0 0

Cj-Zj 3 5 0 0 0

Step 4 Determine key column


X Y S1 S2 S3

S1 0 8 2 0 1 0 0

S2 0 15 0 3 0 1 0

S3 0 30 6 5 0 0 1

Zj 0 0 0 0 0

Cj-Zj 3 5 0 0 0

Key column: column with biggest positive value of Cj Zj (maximize), or column with biggest negative value of Cj Zj (minimize).

Step 5 Determine key row


X Y S1 S2 S3

S1 0 8 2 0 1 0 0 8/0 = ~

S2 0 15 0 3 0 1 0 15/3 = 5

S3 0 30 6 5 0 0 1 30/5 = 6

Zj 0 0 0 0 0

Cj-Zj 3 5 0 0 0

Key row: row with the smallest value of quantity ratio.

rowrespondingofcolumnkeyofiablesoftcoefficien

valueQratioQuantity

______var__

__

Step 6 Determine pivot point


X Y S1 S2 S3

S1 0 8 2 0 1 0 0 8/0 = ~

S2 0 15 0 3 0 1 0 15/3 = 5

S3 0 30 6 5 0 0 1 30/5 = 6

Zj 0 0 0 0 0

Cj-Zj 3 5 0 0 0

Step 7 Change value of key row

New value : previous key row value/pivot point

New value for row S2:

15/3 0/3 3/3 0/3 1/3 0/3

5 0 1 0 1/3 0

Step 8 Change value of non-key row

New value = previous value (FR x previous value of key row) FR (fixed ratio) is coefficient of variables of key column of corresponding row/pivot point

Row S1:

FR = 0/3 = 0,

Since value of FR is 0, therefore for row S1, new value = previous value

Row S2:

FR = 5/3,

30 6 5 0 0 1

(5/3) x 15 0 3 0 1 0

5 6 0 0 -5/3 1

Step 9 Change table in step 3


X Y S1 S2 S3

S1 0 8 2 0 1 0 0

Y 5 5 0 1 0 1/3 0

S3 0 5 6 0 0 -5/3 1

Zj 0 5 0 5/3 0

Cj-Zj 3 0 0 -5/3 0

Change in basic variable: variables of key column (Y) replace variable of key row (S2)

Step 10 Optimality test

Simplex table is considered optimal if value of Cj- Zj 0 for case of maximize and value of Cj Zj 0 for case of minimize. Table in step 9 indicates that there is still positive value of Cj-Zj, therefore optimal condition has not been reached yet. Repeat step 3 9 until optimized solution is achieved.

Step 11 Repeat step 3 9

Basic Var.

Obj. Cj 3 5 0 0 0 Qty. Ratio

X Y S1 S2 S3

S1 0 38/6 0 0 1 5/9 -2/6

Y 5 5 0 1 0 1/3 0

X 3 5/6 1 0 0 -5/18 1/6

Zj 3 5 0 15/18 3/6

Cj-Zj 0 0 0 -15/18 -3/6

Solution

Optimal

X = 5/6

Y = 5 Z = (3 x

5/6) + (5 x 5) = 27.5

M i n i n g M a n a g e m e n t

kuliah 11 - optimization and linear programming

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