4-optimization linear programming

8/3/2019 4-Optimization Linear Programming

1/71

MAE 550 Optimization inEngineering Design

Linear Programming


2/71

2011 K. English 2

FORMULATION OF LINEARPROGRAMMING PROBLEMS


3/71

2011 K. English 3

Linearization Methods

nTwo classes of optimization theory that are mostthoroughly addressed

Unconstrained problemsCompletely linear problems

n Efficient solution algorithms exist for eachn If we could transform a nonlinear problem into a linear

one, we might be able to significantly speed up solution


4/71

2011 K. English 4

Linearization Methods

nLinearize using Taylor series expansion

n If the function is highly nonlinear, the linear approximationcould result in substantial errors

n Before linearizing nonlinear problems, we will take a sidestep to linear programming (i.e., solving linear optimizationproblems)

!f("x) = f(

"x0 )+!f(

"x0 )(

"x "

"x0 )+O

"x "

"x0( )

2

+#

!f("x) = f(

"x0 )+!f(

"x0 )(

"x "

"x0 )

"x0 is the linearized point


5/71

2011 K. English 5

Linear Programming

n Linear programming is a subset ofoptimization problemsObjective function is linear function of design

variables

Constraints are linear equations or inequalities


6/71

2011 K. English 6

Example

nA company has two grades of inspectors: Grade I andGrade II, who are to be assigned to a quality controlinspection. At least 1800 pieces must be inspected per 8-hr day.

n Grade I inspectors earn $40/hr and can check pieces at arate of 25 per hour, with an accuracy of 98%

n Grade II inspectors earn $30/hr and can check pieces at arate of 15 per hour, with an accuracy of 95%

n Errors cost the company $20 eachn The company has 8 Grade I and 10 Grade 2 inspectorsn Determine the optimal assignment of inspectors that will

minimize the total cost of inspection.


7/71

2011 K. English 7

Example1

2

8 (Grade I)

10 (Grade II)

x

x


8/71

2011 K. English 8

Example1

2

1 2

8 (Grade I)

10 (Grade II)

8(25* ) 8(15* ) 1800 (1800 pieces must be inspected each shift)

x

x

x x

+


9/71

2011 K. English 9

Example

1

2

1 2

1 2

1 2

8 (Grade I)

10 (Grade II)

8(25* ) 8(15* ) 1800 (1800 pieces must be inspected each shift)

200 120 1800

5 3 45

x

x

x x

x x

x x

+

+

+


10/71

2011 K. English 10

Example

1

2

1 2

8 (Grade I)

10 (Grade II)

5 3 45

(Salary) + (Error Charge)*(Parts per hr.)(Error rate) Hourly rate

$40 $20*(25)*(0.02) $50 (Grade I Hourly rate)

$30 $20*(15)*(0.05) $45 (Grade II Hourly rate)

x

x

x x

+

=

+ =

+ =


11/71

2011 K. English 11

Example

1

2

1 2

8 (Grade I)

10 (Grade II)

5 3 45

(Salary) + (Error Charge)*(Parts per hr.)(Error rate) Hourly rate

$40 $20*(25)*(0.02) $50 (Grade I Hourly rate)

$30 $20*(15)*(0.05) $45 (Grade II Hourly rate)

8 5

x

x

x x

Z

+

=

+ =

+ =

= ( )1 2 1 20 45 400 360 x x x x+ = +


12/71

2011 K. English 12

Example

1 2

1 2

1

2

: 400 360

: 5 3 45

0 8

0 10

Minimize Z x x subject to x x

xx

= +

+


13/71

2011 K. English 13

Standard Form

Minimize : Z= c1x1 + c2x2 +!+ cnxn

subject to : a11x

1+ a

12x

2+!+ a

1nx

n= b

1

a21x

1+ a

22x

2+!+ a

2nx

n= b

2

"

am1x

1+ a

m2x

2+!+ a

mnx

n= b

m

x1 ! 0,x2 ! 0 ! xn ! 0

b1! 0,b

2! 0 ! b

m! 0


14/71

2011 K. English 14

Standard Form

Minimize : Z=

cxsubject to : Ax = b

x! 0

b ! 0


15/71

2011 K. English 15

Standard Form

n M constraints, N design variablesn Handling Inequalities Introduce slack variables

x1+ 2x

2+ 3x

3+ 4x

4! 25

2x1+x

2" 3x

3#12


16/71

2011 K. English 16

Standard Form


1 2 3 4

1 2 3 4 1

1 2 3

1 2 3 2

2 3 4 25

2 3 4 25

2 3 12

2 3 12

x x x x

x x x x S

x x x

x x x S

+ + +

+ + + + =

+

+ =


17/71

2011 K. English 17

Standard Form


n Handling Unrestricted Variables

1 2 3 4

1 2 3 4 1

1 2 3

1 2 3 2

2 3 4 25

2 3 4 25

2 3 12

2 3 12

x x x x

x x x x S

x x x

x x x S

+ + +

+ + + + =

+

+ =

1

1 1 1

1 1

unrestricted

0 0

x

x x x

x x

+

+

=


18/71

2011 K. English 18

Convert into Standard Form1 2 3

1 2 3

1 2 3

1 2 3

1 2

3

: 2 3

: 7

2

3 2 5

, 0unrestricted in sign

Maximize Z x x x

subject to x x x

x x x

x x x

x xx

= +

+ +

+

=


19/71

2011 K. English 19

Convert into Standard Form1 2 3

1 2 3

1 2 3

1 2 3

1 2 4 5

3 4 5

: 2 3

: 7

2

3 2 5

, , , 0=

Maximize Z x x x

subject to x x x

x x x

x x x

x x x x x x x

= +

+ +

+

=

Replace x3 with x4-x5


20/71

2011 K. English 20

Convert into Standard Form( )

( )

( )

( )

1 2 4 5

1 2 4 5

1 2 4 5

1 2 4 5

1 2 4 5

3 4 5

: 2 3

: 7

2

3 2 5

, , , 0

=

Maximize Z x x x x

subject to x x x x

x x x x

x x x x

x x x x

x x x

= +

+ +

+

=

Replace x3 with x4-x5


21/71

2011 K. English 21


( )

( )

( )

1 2 4 5

1 2 4 5

1 2 4 5

1 2 4 5

1 2 4 5

3 4 5

: 2 3

: 7

2

3 2 5

, , , 0

=

Maximize Z x x x x

subject to x x x x

x x x x

x x x x

x x x x

x x x

= +

+ +

+

=

Replace x3 with x4-x5 Multiply both sides of the

third constraint by -1


22/71

2011 K. English 22


( )

( )

( )

1 2 4 5

1 2 4 5

1 2 4 5

1 2 4 5

1 2 4 5

3 4 5

: 2 3

: 7

2

3 2 5

, , , 0

=

Maximize Z x x x x

subject to x x x x

x x x x

x x x x

x x x x

x x x

= +

+ +

+

+ + =




23/71

2011 K. English 23


( )

( )

( )

1 2 4 5

1 2 4 5

1 2 4 5

1 2 4 5

1 2 4 5

3 4 5

: 2 3

: 7

2

3 2 5

, , , 0

=

Maximize Z x x x x

subject to x x x x

x x x x

x x x x

x x x x

x x x

= +

+ +

+

+ + =



Introduce x6 , x7 as slack(and surplus) variables in the

first and second constraint


24/71

2011 K. English 24


( )

( )

( )

1 2 4 5

1 2 4 5 6

1 2 4 5 7

1 2 4 5

1 2 4 5 6 7

3 4 5

: 2 3

: 7

2

3 2 5

, , , , , 0

=

Maximize Z x x x x

subject to x x x x x

x x x x x

x x x x

x x x x x x

x x x

= +

+ + + =

+ =

+ + =






25/71

2011 K. English 25

Convert into Standard Form

1 2 4 5

1 2 4 5 6

1 2 4 5 7

1 2 4 5

1 2 4 5 6 7

: 2 3 3

: 7

2

3 2 2 5

, , , , , 0

Maximize Z x x x x


x x x x x

x x x x

x x x x x x

= +

+ + + =

+ =

+ + =






26/71

2011 K. English 26

Convert into Standard Form

1 2 4 5

1 2 4 5 6

1 2 4 5 7

1 2 4 5

1 2 4 5 6 7

: 2 3 3

: 7

2

3 2 2 5

, , , , , 0

Maximize Z x x x x


x x x x x

x x x x

x x x x x x

= +

+ + + =

+ =

+ + =

1 2 3

1 2 3

1 2 3

1 2 3

1 2

3

: 2 3

: 7

2

3 2 5

, 0

unrestricted in sign

Maximize Z x x x

subject to x x x

x x x

x x x

x x

x

= +

+ +

+

=


27/71

2011 K. English 27

Convert into Standard Form (5 min)

1 2

1 2

1 2

1 2

: 4 50: 2

2 8

, 0

Minimize Z x x subject to x x

x x

x x

= +

+


28/71

2011 K. English 28


1 2

1 2 3

1 2 4

1 2 3 4

: 4 50

: 2

2 8

, , , 0

Minimize Z x x

subject to x x x

x x x

x x x x

= +

+ =

+ + =

1 2

1 2 3

1 2 4

1 2 3 4

: 4 50

: 22 8

, , , 0

Minimize Z x x

subject to x x x x x x

x x x x

= +

+ =

+ + =


29/71

2011 K. English 29


1 2

1 2

1

: 2 4

: 2 2

0

Minimize f x x subject to x x

x

= +

+


30/71

2011 K. English 30


1 2

1 2

1

: 2 4

: 2 2

0

Minimize f x x

subject to x x

x

= +

+

1 3 4

1 3 4

1 3 4

2 3 4

: 2 4( )

: 2 ( ) 2

, , 0

Minimize f x x x

subject to x x x

x x x

x x x

= +

+

=


31/71

2011 K. English 31


1 2

1 2

1

: 2 4

: 2 2

0

Minimize f x x

subject to x x

x

= +

+

1 3 4

1 3 4 5

1 3 4 5

2 3 4

: 2 4( )

: 2 ( ) 2

, , , 0

Minimize f x x x

subject to x x x x

x x x x

x x x

= +

+ =

=


32/71

2011 K. English 32


1 2

1 2

1

: 2 4

: 2 2

0

Minimize f x x

subject to x x

x

= +

+

1 3 4

1 3 4 5

1 3 4 5

: 2 4 4

: 2 2

, , , 0

Minimize f x x x

subject to x x x x

x x x x

= +

+ =


33/71

2011 K. English 33

SOLUTION OF LINEAR

PROGRAMMING PROBLEMS


34/71

2011 K. English 34

Standard Form

Minimize : Z=

c1x1+

c2x2+!+

cnxn

subject to : a11x

1+ a

12x

2+!+ a

1nx

n= b

1

a21x

1+ a

22x

2+!+ a

2nx

n= b

2

"

am1x

1+ a

m2x

2+!+ a

mnx

n= b

m

x1 ! 0,x2 ! 0 ! xn ! 0

b1! 0,b

2! 0 ! b

m! 0


35/71

2011 K. English 35

Principles of the Simplex Method

n The general form poses the problem as m equations andn variables

n If there are more variables than equations (i.e., m


36/71

2011 K. English 36


37/71

2011 K. English 37

Example Problem

X1

X2

X3

X4

b

1 -1 1 0 = -5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Min f (!

x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1 + 2x2 " 8x

1# 0 x

2# 0


38/71

2011 K. English 38

Example Problem

X1

X2

X3

X4

b

1 -1 1 0 = -5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Min f (!

x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1 + 2x2 " 8x

1# 0 x

2# 0Not Canonical


39/71

2011 K. English 39

Example Problem

X1

X2

X3

X4

b

1 -1 1 0 = -5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Min f (!

x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1 + 2x2 " 8x

1# 0 x

2# 0Not Canonical

Multiply by -1


40/71

2011 K. English 40

Example Problem

X1

X2

X3

X4

b

1 -1 1 0 = -5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Min f (!

x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1 + 2x2 " 8x

1# 0 x

2# 0Not Canonical

X1 X2 X3 X4 b

-1 1 -1 0 = 5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Multiply by -1


41/71

2011 K. English 41

Example Problem

X1

X2

X3

X4

b

1 -1 1 0 = -5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Min f (!

x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1 + 2x2 " 8x

1# 0 x

2# 0Not Canonical

X1 X2 X3 X4 b

-1 1 -1 0 = 5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Multiply by -1Not Canonical


42/71

2011 K. English 42

Example Problem Min f (!

x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X1

X2

X3

X4

b

-1 1 -1 0 = 5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Not Canonical

Add artificial variable, X5

X1

X2

X3

X4 X

5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50


43/71

2011 K. English 43


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X1

X2

X3

X4

b

-1 1 -1 0 = 5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Not Canonical

Add artificial variable, X5

X1

X2

X3

X4

X5

b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50


44/71

2011 K. English 44


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X1

X2

X3

X4

b

-1 1 -1 0 = 5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Not Canonical

Add artificial variable, X5Add temporary objective function, w, as a measure of infeasibilityw is the sum of the artificial variables

X1

X2

X3

X4

X5

b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

0 0 0 0 1 = w


45/71

2011 K. English 45


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X1

X2

X3

X4

b

-1 1 -1 0 = 5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Not Canonical

X1

X2

X3

X4

X5

b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

0 0 0 0 1 = w



46/71

2011 K. English 46


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X1

X2

X3

X4

b

-1 1 -1 0 = 5

1 2 0 1 = 8

-4 -1 0 0 = f-50

Not Canonical

X1

X2

X3

X4

X5

b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

0 0 0 0 1 = w

Not Canonical



47/71

2011 K. English 47


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

0 0 0 0 1 = w

Not Canonical

Subtract row 1 from row 4 to get canonical formX

1X

2X

3X

4X

5b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

0-(-1) 0-1 0-(-1) 0-0 1-1 = w-5


48/71

2011 K. English 48


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

0 0 0 0 1 = w

Not Canonical

Subtract row 1 from row 4 to get canonical formX

1X

2X

3X

4X

5b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5


49/71

2011 K. English 49


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X2 will enter the basis (only negative coefficient and we are minimizing the objective function)

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

X1

X2

X3

X4

X5

b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5


50/71

2011 K. English 50


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

X2 will enter the basis (only negative coefficient and we are minimizing the objective function)Calculating the b/a ratios shows that row 2 has the minimum value and X4 will leave the basis

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

X1

X2

X3

X4

X5

b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

5/1=5

8/2=4


51/71

2011 K. English 51


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

Divide row 2 by 2 to get the coefficient for X2 to be 1

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

5/1=5

8/2=4

X1

X2

X3

X4

X5

b

-1 1 -1 0 1 = 5

=0.5 2/2=1 0/2=0 =0.5 0/2=0 = 8/2=4

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5


52/71

2011 K. English 52


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

Divide row 2 by 2 to get the coefficient for X2 to be 1

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

5/1=5

8/2=4

X1

X2

X3

X4

X5

b

-1 1 -1 0 1 = 5

0.5 1 2=0 0.5 0 = 4

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5


53/71

2011 K. English 53


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

Divide row 2 by 2 to get the coefficient for X2 to be 1Subtract the new row 2 from row 1

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

5/1=5

8/2=4

X1

X2

X3

X4

X5

b

-1-0.5=-1.5 1-1=0 -1-0=-1 0-0.5=-0.5 1-0=1 = 5-4=1

0.5 1 0 0.5 0 = 4

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5


54/71

2011 K. English 54


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

Divide row 2 by 2 to get the coefficient for X2 to be 1Subtract the new row 2 from row 1Add the new row 2 to row 3

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

5/1=5

8/2=4

X1

X2

X3

X4

X5

b

-1.5 0 -1 -0.5 1 = 1

0.5 1 0 0.5 0 = 4

-4+0.5=-3.5 -1+1=0 0+0=0 0+0.5=0.5 0+0=0 = f-50+4=f-46

1 -1 1 0 0 = w-5


55/71

2011 K. English 55


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

Divide row 2 by 2 to get the coefficient for X2 to be 1Subtract the new row 2 from row 1Add the new row 2 to row 3Add the new row 2 to row 4

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

5/1=5

8/2=4

X1

X2

X3

X4

X5

b

-1.5 0 -1 -0.5 1 = 1

0.5 1 0 0.5 0 = 4

-3.5 0 0 0.5 0 = f-46

1+0.5=1.5 -1+1=0 1+0=1 0+0.5=0.5 0+0=0 = w-5+4=w-1


56/71

2011 K. English 56


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

Divide row 2 by 2 to get the coefficient for X2 to be 1Subtract the new row 2 from row 1Add the new row 2 to row 3Add the new row 2 to row 4

X1 X2 X3 X4 X5 b

-1 1 -1 0 1 = 5

1 2 0 1 0 = 8

-4 -1 0 0 0 = f-50

1 -1 1 0 0 = w-5

5/1=5

8/2=4

X1

X2

X3

X4

X5

b

-1.5 0 -1 -0.5 1 = 1

0.5 1 0 0.5 0 = 4

-3.5 0 0 0.5 0 = f-46

1.5 0 1 0.5 0 = w-1


57/71

2011 K. English 57


x) = !4x1! x

2+50

s.t. g1(!

x) = x1! x

2" !5

g2(!

x) = x1+ 2x

2" 8

x1# 0 x

2# 0

All coefficients in row 4 are positive, so w cannot be decreased furtherX5 still has a valueThis tells us that a basic feasible solution to the original problem (without the artificial variable)cannot be found

X1

X2

X3

X4

X5

b

-1.5 0 -1 -0.5 1 = 1

0.5 1 0 0.5 0 = 4

-3.5 0 0 0.5 0 = f-46

1.5 0 1 0.5 0 = w-1


58/71

2011 K. English 58

Simplex Process for minimization

n Phase I1. Add artificial variables as necessary to achieve canonical form


59/71

2011 K. English 59


n Phase I1. Add artificial variables as necessary to achieve canonical form2. Create a new objective function (w) equal to the sum of the

artificial variables


60/71

2011 K. English 60




3. Minimize using Phase II approaches


61/71

2011 K. English 61




3. Minimize using Phase II approachesn If w cannot be reduced to zero, the solution to

the original problem does not exist.

n If w has been reduced to zero, delete theartificial variables and the artificial objective (w)from the system and proceed with Phase II


62/71

2011 K. English 62


n Phase II


63/71

2011 K. English 63


n Phase II1. Find the minimum ci, i=1,n.

Denote this variable with a subscript k (ck). If ck is non-negative, go to step 5


64/71

2011 K. English 64




2. If ck is negative Find the minimum (bj/ ajk) where j=1,m for all positive ajk If all ajk are non-positive, the solution is unbounded. Otherwise, let the subscript r denote the row corresponding to this minimum ark is the pivot element


65/71

2011 K. English 65





3. Pivot on ark. Divide row r by ark Subtract ajk times the new row r from row j, where j=1, m+1 and jr


66/71

2011 K. English 66






4. Go to step 1 and repeat


67/71

2011 K. English 67






4. Go to step 1 and repeat5. The solution is complete


68/71

2011 K. English 68







a) If all associated with non-basic variables are positive, the solution is unique


69/71

2011 K. English 69







a) If all associated with non-basic variables are positive, the solution is uniqueb) If some associated with a non-basic variable is zero, the solution is not unique


70/71

2011 K. English 70







a) If all associated with non-basic variables are positive, the solution is uniqueb) If some associated with a non-basic variable is zero, the solution is not uniquec) If some is negative, the solution is unbounded


71/71

Group Problem (In-Class)A company has 2 grades of inspectors, who are to be assigned to a

quality control inspection process. It is required that at least 1800 piecesbe inspected per 8 hour day.

n Grade 1 inspectors can check pieces at the rate of 25 per hour, with anaccuracy of 98%.

n Grade 2 inspectors can check pieces at the rate of 15 per hour with anaccuracy of 95%.

n The wage rate of a Grade 1 inspector is $40 per hour, while a Grade2inspector earns $30 per hour.

n Each error made by an inspector costs the company $20.n The company has available 8 Grade 1 inspectors and 10 Grade 2

inspectors.

What is the optimal assignment of inspectors that will minimize the totalcost of inspection?

4-optimization linear programming

Documents