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Lecture 25

Two-Phase Simple Upper Bounded Simplex Algorithm

Example:

Minimize –2x1 – x2

Subject to x1 + x2 < 6

0 < x1 < 5

0 < x2 < 5

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The Graph

x1

x2

optimum

x1 bound

x2 bound

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Convert To An Equality Constraint

Minimize –2x1 – x2

Subject to x1 + x2 +x3 = 6

0 < x1 < 5

0 < x2 < 5

0 < x3 < Note Bound on the slack variable

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Add An Artificial Variable

Minimize –2x1 – x2

Subject to x1 + x2 +x3 + A= 6

0 < x1 < 5

0 < x2 < 5

0 < x3 <

0 < A <

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Phase I Problem

Minimize ASubject to x1 + x2 +x3 + A= 6

0 < x1 < 5

0 < x2 < 5

0 < x3 <

0 < A <

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Step 0. Initialization

BASIC = {A at 6}

NONBASIC = {x1 at 0, x2 at 0, x3 at 0}

B = 1

B-1 = 1

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Step 1. Pricing

v = cBB-1 = 1(1) = 1

cbar1 = c1 – va1 = 0 – 1(1) = -1

cbar2 = c2 – va2 = 0 – 1(1) = -1

cbar3 = c3 – va3 = 0 – 1(1) = -1

All price favorably!

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Step 2. Optimality

Not optimal

Let p = 3 - that is x3 is allowed to increase from 0

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Step 3. Direction

y = B-1a3 = 1(1) = 1

= 1 Why?

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Step 4 Step Size

- = min{, (xj-ℓj)/(yj) : yj > 0}

+ = min{, (uj-xj)/(-yj) : yj < 0}

= min{- , + , up - ℓp}

- = min{, (6-0)/1} = 6+ = min{} = = min{6, , -0} = 6

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Step 5. New Point

XB = xB - y = 6 – 6(1)(1) = 0

A = 0

xp = x3 = ℓ3 + = 0 + 6 = 6

Current Point: [x1,x2,x3,A] = [0,0,6,0]

BASIC = {x3 at 6}

NONBASIC = {x1 at 0, x2 at 0, A at 0}B = 1 B-1 = 1

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Step 1. Pricing

v = cBB-1 = 0(1) = 0

cbar1 = c1 – va1 = 0 – 0(1) = 0

cbar2 = c2 – va2 = 0 – 0(1) = 0

All price unfavorably!

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Step 2. Optimality

Optimal For Phase I

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Phase II Problem

Minimize -2x1 – x2

Subject to x1 + x2 +x3 = 6

0 < x1 < 5

0 < x2 < 5

0 < x3 <

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Step 1. Pricing

Current Point: [x1,x2,x3] = [0,0,6]

BASIC = {x3 at 6}

NONBASIC = {x1 at 0, x2 at 0}B = 1 B-1 = 1

v = cBB-1 = 0(1) = 0

cbar1 = c1 – va1 = -2 – 0(1) = -2

cbar2 = c2 – va2 = -1 – 0(1) = -1

All price favorably!

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Step 2. Optimality

Not Optimal

Let p = 1

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Step 3. Direction

y = B-1a1 = 1(1) = 1

= 1 Why?

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Step 4 Step Size

- = min{, (xj-ℓj)/(yj) : yj > 0}

+ = min{, (uj-xj)/(-yj) : yj < 0}

= min{- , + , up - ℓp}

- = min{, (6-0)/1} = 6+ = min{} = = min{6, , 5-0} = 5

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Step 5. New Point

XB = xB - y = 6 – 5(1)(1) = 1

x3 = 1

xp = x1 = ℓ1 + = 0 + 5 = 5

Current Point: [x1,x2,x3] = [5,0,1]

BASIC = {x3 at 1}

NONBASIC = {x1 at 5, x2 at 0}B = 1 B-1 = 1 Note: The basis stayed the same.

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Step 1. Pricing

v = cBB-1 = 0(1) = 0

cbar1 = c1 – va1 = -2 – 0(1) = -2 Unfavorable Why?

cbar2 = c2 – va2 = -1 – 0(1) = -1 Favorable

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The Graph

x1

x2

current point

x1 bound

x2 bound

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Step 2. Optimality

Not Optimal

Let p = 2

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Step 3. Direction

y = B-1a1 = 1(1) = 1

= 1 Why?

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Step 4 Step Size

- = min{, (xj-ℓj)/(yj) : yj > 0}

+ = min{, (uj-xj)/(-yj) : yj < 0}

= min{- , + , up - ℓp}

- = min{, (1-0)/1} = 1+ = min{} = = min{1, , 5-0} = 1

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Step 5. New Point

XB = xB - y = 1 – 1(1)(1) = 0

x3 = 0

xp = x2 = ℓ2 + = 0 + 1 = 1

Current Point: [x1,x2,x3] = [5,1,0]

BASIC = {x2 at 1}

NONBASIC = {x1 at 5, x3 at 0}B = 1 B-1 = 1

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Step 1. Pricing

v = cBB-1 = -1(1) = -1

cbar1 = c1 – va1 = -2 – (-1)(1) = -1 Unfavorable Why?

cbar3 = c3 – va3 = 0 – (-1)(1) = 1 Unfavorable Why?

Optimality Obtained

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The Graph

x1

x2

current point

x1 bound

x2 bound