4-optimization linear programming
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MAE 550 Optimization inEngineering Design
Linear Programming
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2011 K. English 2
FORMULATION OF LINEARPROGRAMMING PROBLEMS
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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
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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
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Linear Programming
n Linear programming is a subset ofoptimization problemsObjective function is linear function of design
variables
Constraints are linear equations or inequalities
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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.
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Example1
2
8 (Grade I)
10 (Grade II)
x
x
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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
+
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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
+
+
+
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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
+
=
+ =
+ =
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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+ = +
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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
= +
+
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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
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Standard Form
Minimize : Z=
cxsubject to : Ax = b
x! 0
b ! 0
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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
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Standard Form
n M constraints, N design variablesn Handling Inequalities Introduce slack 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
+ + +
+ + + + =
+
+ =
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Standard Form
n M constraints, N design variablesn Handling Inequalities Introduce slack variables
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
+
+
=
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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
= +
+ +
+
=
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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
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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
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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 Multiply both sides of the
third constraint by -1
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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 Multiply both sides of the
third constraint by -1
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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 Multiply both sides of the
third constraint by -1
Introduce x6 , x7 as slack(and surplus) variables in the
first and second constraint
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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
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
= +
+ + + =
+ =
+ + =
Replace x3 with x4-x5 Multiply both sides of the
third constraint by -1
Introduce x6 , x7 as slack(and surplus) variables in the
first and second constraint
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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
subject to x 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
Introduce x6 , x7 as slack(and surplus) variables in the
first and second constraint
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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
subject to x 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
= +
+ +
+
=
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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
= +
+
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Convert into Standard Form (5 min)
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
= +
+ =
+ + =
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Convert into Standard Form (5 min)
1 2
1 2
1
: 2 4
: 2 2
0
Minimize f x x subject to x x
x
= +
+
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Convert into Standard Form (5 min)
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
= +
+
=
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Convert into Standard Form (5 min)
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
= +
+ =
=
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Convert into Standard Form (5 min)
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
= +
+ =
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SOLUTION OF LINEAR
PROGRAMMING PROBLEMS
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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
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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
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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
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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
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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
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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
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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
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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
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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
X5
b
-1 1 -1 0 1 = 5
1 2 0 1 0 = 8
-4 -1 0 0 0 = f-50
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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, 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
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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
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
Add artificial variable, X5Add temporary objective function, w, as a measure of infeasibilityw is the sum of the artificial variables
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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
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
Add artificial variable, X5Add temporary objective function, w, as a measure of infeasibilityw is the sum of the artificial variables
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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 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
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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 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
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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
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
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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
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
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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
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
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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
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
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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
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
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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
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
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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
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
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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
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
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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
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
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Simplex Process for minimization
n Phase I1. Add artificial variables as necessary to achieve canonical form
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Simplex Process for minimization
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
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Simplex Process for minimization
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
3. Minimize using Phase II approaches
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Simplex Process for minimization
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
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
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Simplex Process for minimization
n Phase II
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Simplex Process for minimization
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
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Simplex Process for minimization
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
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
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Simplex Process for minimization
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
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
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
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Simplex Process for minimization
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
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
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
4. Go to step 1 and repeat
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Simplex Process for minimization
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
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
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
4. Go to step 1 and repeat5. The solution is complete
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Simplex Process for minimization
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
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
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
4. Go to step 1 and repeat5. The solution is complete
a) If all associated with non-basic variables are positive, the solution is unique
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Simplex Process for minimization
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
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
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
4. Go to step 1 and repeat5. The solution is complete
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
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Simplex Process for minimization
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
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
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
4. Go to step 1 and repeat5. The solution is complete
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
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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?
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