L17 LP part3

• Homework• Review• Multiple Solutions• Degeneracy• Unbounded problems• Summary

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H16 8.35

2

1 2

1 2 3 4

1 2 3 4

( ) 2. .1 2 1 0 103 2 0 1 18

0i

Min f x xs tx x x xx x x x

x

x

1 2

1 2

1 2

( ) 2. .

2 103 2 18

0i

Max f x xs tx xx x

x

x

Canonical…thereforeFeasible!

8.35 cont’d

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Simplex Tableau

row basic x1 x2 x3 x4 bb/

a_pivota x3 -1 2 1 0 10 -10 n/ab x4 3 2 0 1 18 6 minc c' -2 -1 0 0 0

First Tableau

row basic x1 x2 x3 x4 bb/

a_pivot+Re to Ra d x3 0 2.66667 1 0.33333 16/Rb by 3 e x1 1 0.66667 0 0.33333 6 2*Re+Rb f c' 0 0.33333 0 0.66667 12

f+12=0f= - 12

8.39

4

1 2

1 2 3 4

1 2 3 4

( ) 2. .2 1 0 50 0 1 2

0i

Min f x xs tx x x xx x x x

x

x

Canonical…thereforeFeasible!

8.39 cont’d

5

1 2

1 2 3 4

1 2 3 4

( ) 2. .2 1 0 50 0 1 2

0i

Min f x xs tx x x xx x x x

x

x

8.39 cont’d

6

8.44

7

1 2

1 2

1 2

1 2

( ). .4 3 9

2 62 6

0i

Max z x xs tx xx xx xx

x

1 2

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

( ). .4 3 1 0 0 91 2 0 1 0 62 1 0 0 1 6

0i

Min f x xs tx x x x xx x x x xx x x x x

x

x

Canonical…thereforeFeasible!

8.44 cont’d

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Simplex Tableaurow basic x1 x2 x3 x4 x5 b b/a_pivota x3 4 3 1 0 0 9 2.25 minb x4 1 2 0 1 0 6 6c x5 2 1 0 0 1 6 3d c' -1 -1 0 0 0

First Tableaurow basic x1 x2 x3 x4 x5 b b/a_pivot

e x1 1 0.75 0.25 0 0 2.25 3f x4 0 1.25 -0.25 1 0 3.75 3 ming x5 0 -0.5 -0.5 0 1 1.5 -3 n/ah c' 0 -0.25 0.25 0 0 2.25

f=-2.25

8.44

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First Tableaurow basic x1 x2 x3 x4 x5 b b/a_pivot

e x1 1 0.75 0.25 0 0 2.25 3f x4 0 1.25 -0.25 1 0 3.75 3 ming x5 0 -0.5 -0.5 0 1 1.5 -3 n/ah c' 0 -0.25 0.25 0 0 2.25

f=-2.25

Second Tableaurow basic x1 x2 x3 x4 x5 b b/a_pivot

h x1 1 0 0.4 -0.6 0 0i x2 0 1 -0.2 0.8 0 3j x5 0 0 -0.6 0.4 1 3k c' 0 0 0.2 0.2 0 3

f = - 3

1

2

5

3 4

033

, 0( ) 3

3

xxxx xfz f

x

Transforming LP to Std Form LP

1. If Max, then f(x) = - F(x)2. If x is unrestricted, split into x+ and x-, and

substitute into f(x) and all gi(x) and renumber all xi

3. If bi < 0, then multiply constraint by (-1)

4. If constraint is ≤, then add slack si5. If constraint is ≥, then subtract surplus si10

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.

0x0bbAx

xcx T

..

)(tsfMin

All “=“

Canonical form Ex 8.4 & TABLEAU

12

124

1

14

1

114

1

28

116

521

421

321

xxx

xxx

xxx

basisall +1

Simplex Method – Part 1 of 2Single Phase Simplex Method

When the Standard form LP Problem has only≤ inequalties…. i.e. only slack variables, we can solve using the Single-Phase Simplex Method!(i.e. canonical form!)

If surplus variables exist… we need the Two-Phase Simplex Method –with artificial variables… Sec 8.6-7 (after Spring Break)

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Single-Phase Simplex Method1. Set up LP prob in a SIMPLEX tableau

add row for reduced cost, cj’ and column for min-ratio, b/a label the rows (using letters) of each tableau

2. Check if optimum, all non-basic c’≥0? 3. Select variable to enter basis(from non-basic)

Largest negative reduced cost coefficient/ pivot column

4. Select variable to leave basis Use min ratio column / pivot row

5. Use Gauss-Jordan elimination on rows to form new basis, i.e. identity columns

6.Repeat steps 2-5 until opt solution is found!14

Special cases?

• Multiple solutions• Unbounded problems• Degenerate solutions

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Multiple Solutions

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Non-basic ci’=0

Non-unique global solutions, ∆f = 0

Unbounded problem

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Pivot column coefficients aij < 0

1 2 3

2 3 1

2 1 1 00 2

x x xx x x

Degenerate solution

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First Tableaurow basic x1 x2 x3 x4 x5 b b/a_pivot

e x1 1 2 1 0 0 3 3/1f x4 0 3 2 1 0 0 0/2 ming x5 0 4 -1 0 1 0 neg n/ah c' 0 -2 -4 0 0 2.25

f=-2.25

Want to bring in x3 for x4… but the min ratio rule says no amount of x3!...

Therefore no change in f either.

Simplex method will move to a solution, slowlySometimes it will “cycle” forever.

More Terms• Degererate basic solution - one or more basic

variables has a zero value in a basic solution (i.e. b=0)• Degererate basic feasible solution - one or more basic

variables has a zero value in a basic feasible solution (i.e. b=0)

• Optimum basic solution – basic feasible solution with minimum f.

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Test 3• T/F 15 pts• M/choice (terms????) 10 pts• Excel Curve Fitting -set up Excel equations, for

one of five analytical equations, using cell labels only e.g. B4, C6 (i.e. no naming of variables) (25 pts)

• Transform prob to Standard LP Form (25 pts)• Solve LP problem using Simplex (25 pts)

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Summary• Simplex Method moves efficiently from one

feasible combination of basic variables to another.• Use Single-Phase Simplex Method when only

“slack” type constraints.• Multiple solutions• Unbounded solutions/problems• Degenerate Basic Solution• Degenerate Basic Feasible Solution

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