L15 LP Problems

• Homework• Review• Why bother studying LP methods• History• N design variables, m equations• Summary

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H14 part 1

2

x0 11.1993beta 0.2700omega 0.8300

-4

-2

0

2

4

6

8

10

0 0.5 1 1.5 2 2.5 3 3.5

H14 Part 1

3

x0 13.0724beta 0.2362omega 5.6133

-10

-5

0

5

10

15

0 0.5 1 1.5 2 2.5 3 3.5

H14 Part 1

4

x0 12.9713beta 0.2381omega 11.8867

-10

-5

0

5

10

0 0.5 1 1.5 2 2.5 3 3.5

5

Linear

Quadratic

Power

Exponential

a0

4.086

6.335

a1 -0.454 -1.434

a2

0.0775

a3

a4

5.908

a5

-1.113

a6

8.965

a7

-0.463

SSE 9.385 1.914 3.515 0.8304 r2 0.726 0.944 0.897 0.9758

H14

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Curve fitting

7

 

 

 

 

 

)(xf

)(

)(

),(),(),(

),(

444

333

222

111

iii

i

xfyeeerror

xfy

yxPyxPyxPyxP

Curve Fitting

8

xa

a

eaxf

xaxf

xaxaxaaxf

xaxaaxf

xaaxf

7

5

6

4

33

2210

2210

10

)(lexponentia

)(power

)(cubic

)(quadratic

)(linear

Need to find the parameters ai Another way? Especially for non-linear curve fits?

i

iii

i xfyez 22 )]([)( minimize a

Curve Fit example

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Goodness of fit?• R2 = coefficient of determination

0≤ R2 ≤1.• R = correlation coefficien

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Curve Fit example

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Linear Programming Prob.s

12

j

k

jjijKiki

j

k

jjijKiki

j

k

jjijiKiki

kk

bxaorbxaxa

bxaorbxaxa

bxaorbxaxa

tsxcxcxcfMinimize

111

111

111

2211

..)(

x

Why study LP methods

• LP problems are “convex”If there is a solution…it’s global optimum

• Many real problems are LPTransportation, petroleum refining, stock portfolio, airline crew scheduling, communication networks

• Some NL problems can be transformed into LP• Most widely used method in industry

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Std Form LP Problem

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ntojxmtoib

bxaxa

bxaxabxaxa

tsxcxcxcfMin

j

i

mnmnm

nn

nn

nn

1,01,0

..)(

11

22121

11111

2211

x Matrix form

All “≥0” i.e. non-neg.

How do we transform an given LP problem into a Standard LP Prob.?

0x0bbAx

xcx T

..

)(tsfMin

All “=“

Recall LaGrange/KKT method

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0,0

0,0

11

11

11

11

iiiiKiki

Kiki

iiiiKiki

iKiki

sbbsxaxabxaxa

sbbsxaxabxaxa

Add

slack variable

Subtractsurplus variable

Handling negative xi

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0

0

0,0;

jjj

jjj

jjjjj

xmeansxx

xmeansxx

xxxxx

When x is unrestricted in sign:

Transformation example

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Transformation example pg2

18

Trans pg3

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Solving systems of linear equations

n equations in n unknownsProduces a unique solution, for example

20

62242

21

21

xxxx

Elimination methods

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2x42(1)x42x x

row1 usingtutebacksubsti122

220

421 row2to then2 by row1x

622

421

11

21

22

xx

622

421

62242

21

21

xxxx Gaussian Elimination

Elimination methods cont’d

22

622

421

62242

21

21

xxxx

1x1

110

2012- by row2divide row1, torow2

220

421 row2to then2 by row1x

622

421

2

1

x

Gauss-Jordan Elimination

Can we find unique solutions forn unknowns with m equations?

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5 unknowns and 2 equations!

What’s the best you can do?

MUST set 3 xi to zero! Solve for remaining 2. Just like us=0 in LaGrange Method!

m equations= m unknowns

Most we can do is to solve for m unknowns,e.g. we can “solve” for 2 xi

but which 2? 24

Combinations?

2554

53

43

52

42

32

51

41

31

21

54321

m=2, n=5

102

45

)123(12

12345

!25!2

!5!!

!

25

-C

n-mm

nCmn

Combinations from m=2, n=2

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m=2, n=4

43

42

32

41

31

21

4321

62

34

)12(12

1234

!24!2

!4!!

!

24

-C

n-mm

nCmn

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Example 8.2Figure 8.1 Solution to the profit maximization problem. Optimum point = (4, 12). Optimum cost = -8800.

5 unknowns, n=53 equations, m=310 combinations

Example 8.2 cont’d

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Solutions are vertexes (i.e. extreme points, corners) of polyhedron formed by the constraints

Example 8.2 cont’d

• Ten solutions created by setting (n-m) variables to zero, they are called basic solutions

• Some of them were basic feasible solutions• Any solution in polygon is a feasible solution• Variables not set to zero are basic variables• Variables set to zero = non-basic variables

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Canonical form Ex 8.4 & TABLEAU

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124

1

14

1

114

1

28

116

521

421

321

xxx

xxx

xxx

basis

Ex 8.4 cont’d

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0,,124

543

2

1

xxxxx

Pivot row

Pivot column

Method?

1. Set up LP prob in “tableau”2. Select variable to leave basis3. Select variable to enter basis (replace the one

that is leaving)4. Use Gauss-Jordan elimination to form

identity sub-matrix, (i.e. new basis, identity columns)

5. Repeat steps 2-4 until opt sol’n is found!

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Can we be efficient?

• Do we need to calculate all the combinations?• Is there a more efficient way to move from

one vertex to another?• How do we know if we have found the opt

solution, or need to calculate another tableau?

SIMPLEX METHOD! (Next class)

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Summary• Curve fit = min Sum Squared Errors

Min SSE, check R

• Many important LP problems• LP probs are “convex prog probs”• Need to transform into Std LP format

slack, surplus variables, non-negative b and x

• Polygon surrounds infinite # of sol’ns• Opt solution is on a vertex• Must find combinations of basic variables

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