CSC 8301– Design & Analysis of Algorithms: Linear...

Villanova University Department of Computing Sciences

CSC 8301– Design &Analysis of Algorithms:Linear Programming

Professor Henry CarterFall 2016


Iterative Improvement

• Start with a feasible solution

• Improve some part of the solution

• Repeat until the solution is optimal

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A Geometric Solution

• Maximize: 3x + 5y

3

x

y

( 0, 2 )

( 0, 0 ) ( 4, 0 )

( 3, 1 )

3x + 5y = 10

3x + 5y = 14

3x + 5y = 20


The Simplex Method

• A convenient means for iterating over extreme points

• Problem must be:

‣ Maximization problem

‣ All constraints must be linear equations with nonnegative right-hand sides

‣ All variables must be nonnegative

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Step 0: Standard Form

• Min-max conversion

• Representing inequalities with slack variables

• Replace negative variables with two nonnegative variables

5


Example Standard Form

• Maximize: 3x + 5ySubject to: x + y ≤ 4

x + 3y ≤ 6x ≥ 0, y ≥ 0

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• Maximize: 3x + 5y + 0u + 0vSubject to: x + y + u = 4

x + 3y + v = 6x, y, u, v ≥ 0


Basic Feasible Solution

• Find an initial extreme point to begin iterative improvement

• For m constraints over n variables, set (n - m) variables to 0

• Common approach is to zero the original variables and set the slack variables accordingly

• For our example, (0, 0, 4, 6)

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Constructing the Simplex Tableau

• Constraint Coefficients

• Objective Row

• Recognize the corresponding solution

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• Maximize: 3x + 5y + 0u + 0v

Subject to: x + y + u = 4x + 3y + v = 6x, y, u, v ≥ 0


Example Tableau

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1 1 1 0 4

1 3 0 1 6

-3 -5 0 0 0

x y u vu

v


Iterating the Extreme Point

• Choose the entering variable (new basic variable)

• Choose the departing variable (new nonbasic variable)

• Pivot Operations

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Example Iteration

11

1 1 1 0 4

1 3 0 1 6

-3 -5 0 0 0

x y u vu

v


Example Iteration

12

1 1 1 0 4

1 3 0 1 6

-3 -5 0 0 0

x y u vu

v

θu=

θv=


Example Iteration

13

1 1 1 0 4

1 3 0 1 6

-3 -5 0 0 0

x y u vu

v


Example Iteration

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1 1 1 0 4

1/3 1 0 1/3 2

-3 -5 0 0 0

x y u vu

v


Example Iteration

15

2/3 0 1 -1/3 2

1/3 1 0 1/3 2

-4/3 0 0 5/3 10

x y u vu

y


Example Iteration

16

2/3 0 1 -1/3 2

1/3 1 0 1/3 2

-4/3 0 0 5/3 10

x y u vu

y


Example Iteration

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1 0 3/2 -1/2 3

0 1 -1/2 1/2 1

0 0 2 1 14

x y u vx

y


Practical Points

• Prevent cycling with Bland’s Rule

• Algorithmically finding an initial feasible solution

• Two-phase Simplex Method

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Analysis

• Number of extreme points grows exponentially

• In practice, between m and 3m iterations are needed

• Operations per iteration: O(mn)

• Other approaches:

‣ Ellipsoid method (polynomial worst-case)

‣ Karmarkar’s algorithm (empirically efficient)

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Practice

• Set up the first simplex tableau for the following linear programming problem:

Maximize: x + 2y

subject to: 4x ≥ yy ≤ 3 + xx ≥ 0, y ≥ 0

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Recap

• Iterative methods start with a solution then iteratively improve it until optimal

• Linear programming optimizes a linear equation with constraints

• The Simplex Method iterates over extreme points in the feasible region

• Very common in practice (even included in Excel add-on)

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Modeling a Transportation Network

• Recall: weighted digraphs

• Represent start and end as a source vertex and a sink vertex

• Represent paths with positive capacity using edge weight

• Flow Network

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Finding a Flow

• Rules:

‣ Flow-conservation

‣ Edge capacity

• Flow value

• Maximize the value

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Ford-Fulkerson

• Initialize a 0-value flow

• Iteratively search for an augmented path

‣ Any path from source to sink with available flow capacity

• When no more augmenting paths can be found, the flow is maximized

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Greedy Approach Example

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1

5

2 33

2 5

1

3 4

4

62


Shortest Augmenting Path

• Uses a BFS to traverse the graph

• Along the path, adjust the incoming flow for each vertex if there is edge capacity available

• If a vertex with no outgoing capacity is encountered, adjust the incoming flow on a back edge

• Augment the current flow each time a new augmenting path reaches the sink

‣ Terminate if no augmenting path to the sink exists

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Example: 1st augmenting path

• Queue:

27

1

5

2 30/3

0/2 0/5

0/1

0/3 0/4

4

60/2

1 ∞ -23456


Example: 2nd augmenting path

• Queue:

28

1

5

2 30/3

2/2 2/5

0/1

0/3 0/4

4

62/2

1 ∞ -23456


Example: no more augmenting paths

• Queue:

29

1

5

2 31/3

2/2 1/5

1/1

1/3 1/4

4

62/2

1 ∞ -23456


Analysis

• Time to find one augmenting path (using adjacency lists): O(|E|)

• Maximum iterations needed: |E||V|/2

• Worst-case: O( )

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Max-flow Min-cut theorem

• A cut is a set of edges that, when removed, eliminate all paths from source to sink

• A min-cut is a cut where the sum of the edge weights is minimized

• The value of a max-flow is equal to the capacity of the min-cut

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Finding the Min-cut

• Queue: 1, 4

32

1

5

2 31/3

2/2 1/5

1/1

1/3 1/4

4

62/2

1 ∞ -234 2 1+56


Practice

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1 52

3 7

6

5

8

4 4

4 6

2


Recap

• Iterative improvement of a max-flow using augmented paths

• Shortest augmenting path algorithm considers both forward and backward edges to find an augmenting path

• The max-flow value of a network is equal to the min-cut capacity

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Next Time...

• Levitin Chapter 10.3-10.4

‣ Remember, you need to read it BEFORE you come to class!

• Homework:

‣ 10.1: 3, 5, 8

‣ 10.2: 1-4, 10

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