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Hillier and Lieberman

Chapter 4


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The Essence of the Simplex

It is an algebraic procedure

Wyndor glass example

Constraint boundaries, figure 4.1


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Wyndor Glass

Maximize Z = 3X1 + 5X2

Subject to:

X1 42X2 12

3X1 + 2X2 18


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Figure 4.1

0, 6

4,0

4,3

2, 6

Corners: (0,0) (4,0) (4.3)

(2,6) and (0,6)


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Corner Point Feasible Solutions

Corner point feasible solutions (CPF solutions)

occur at the intersections of the constraint

boundaries, which are in the feasible region.

The problem has 2 decision variables, so each

corner occurs at the intersection of two constraints.

The dimension of the problem determines the

number of constraints that intersect at each corner.If there are, for example, 5 decision variables the

corner points occur at the intersection of

5 constraints.


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Adjacent feasible solutions

For any linear programming model with n

decision variables, two CPF solutions are

adjacent to each other if they share n-1

constraints. The two adjacent CPF solutions

are connected by a line segment that lies

on the same shared constraint boundaries.

Such a constraint boundary is referred to asan edge of the feasible region.


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Figure 4.1

0, 6

4,0

4,3

2, 6

0,0

(0,0) and (0,6) adjacent





There are 5 edges.

From each CPF solution,

2 edges emanate.

Each CPF solution has 2

adjacent solutions.


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Optimality test

If a CPF solution has no adjacent CPF

solution that is better (in terms of the value

of the objective function) then it must be

an optimal solution.


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Solving the example

Initialization. Choose point (0,0) because

it is convenient (no calculation needed to

show it is feasible)

Optimality test: Conclude that (0,0) is not

optimal because adjacent CPF solutions

are better

Iteration 1: Move to a better CPF solution


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Simplex Iteration

Consider the two edges that emanate from (0,0).Move up the X2 axis because X2 contributesmore to the objective function of Z = 3X1 + 5X2.

Stop at the next constraint boundary, 2X2 = 12

(or X2 = 6). This is point (0,6). You cant movefurther in this direction without leaving thefeasible region.

Find the other adjacent CPF solution for this

point (2,6) from the intersection of constraints.Optimality test: (0,6) is not optimal because

(2,6) is better.

Repeat steps. (2,6) is optimal solution.


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Solution Concepts

Simplex focuses only on CPF solutions, a finiteset.

It is an iterative procedure, meaning a fixed

series of steps is repeated (an iteration) until adesired result is found.

When possible, the initial CPF solution is theorigin because it is convenient. (Possible when

all variables are non-negative.) If origin isinfeasible, special procedures are needed,described in section 4.6 of this chapter.


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Solution Concepts, Continued

At each iteration, we always choose an adjacent

CPF solution. Steps take place along the edges

of the feasible region.

To find the next CPF solution to test, thealgorithm computes the rate of improvement in

that direction. A positive rate of improvement

implies that the adjacent CPF solution is better

than the current one. So we test whether any ofthe edges have a positive rate of improvement.


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Setting up the Simplex Method

We convert the inequalities to equalities by use

of slack variables (and artificial variables for

GE constraints).

Let X3 be a slack for the first constraint

X1 4 becomes X1 + X3 = 4

With X3 0


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Wyndor Glass: Augmented Form

Maximize Z = 3X1 + 5X2

Subject to:

X1 + X3 = 42X2 +X4 = 12

3X1 + 2X2 +X5 = 18

all X variables non-negative


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Augmented Solution

The augmented solution contains values for

the decision variables AND values for the

slacks. So if we augment the solution (3, 2)

in this example by the values of the slacks weget (3, 2, 1, 8, 5).

A basic solution is an augmented corner-point

solution.


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Properties of a basic solution

Each variable is either basic or non-basic

The number of basic variables equals the number offunctional constraints.

The number of non-basic variables equals the total

number of variables minus the number of functionalconstraints.

The non-basic variables are set equal to zero.

The values of the basic variables are found by solving

the simultaneous equation system. If the basic solution satisfies the non-negativity

consideration, it is called a basic feasible solution.


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Example: BF Solution

X1 + X3 = 4

2X2 +X4 = 12

3X1 + 2X2 +X5 = 18

Set X1 and X4 = 0 (non-basic). From

first constraint, X3 =4. From second

constraint, X2 = 6. From third constraint,

X5 =6.

(0, 6, 4, 0, 6)


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Test for adjacent points

Two BF solutions are adjacent if all but

one of their non-basic variables are the same.

All but one of the basic variables would correspond

in the two points, but perhaps with differentvalues.

Moving to an adjacent point means we swap one

basic variable for a non-basic one. One variableenters; the other leaves.


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Algebra of the Simplex

Initialization: X1 = 0 and X2 = 0

Yields: X3 = 4, X4 = 12, and X5 = 18

BFS (0, 0, 4, 12, 18)

X1(0) + X3 = 4

2X2(0) + X4 =12

3X1(0) + 2X2(0) + X5 = 18

Notice that the BFS can be read immediatelyfrom RHS. The simplex used a procedure

(Gaussian elimination) to convert the equations

to the same convenient form with each iteration.


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Geometric Algebraic

Choose (0,0) as initialCPF solution.

Optimality test: notoptimal because moving

along either edgeincreases Z.

Iteration 1, step 1: Moveup the edge lying on theX2 axis.

Choose X1and X2 to benon-basic for initial BFS(0,0,4,12,18)

Not optimal because

increasing either non-basic variable increasesZ.

Iteration 1, step 1:Increase X2 while

adjusting other variablevalues to satisfy thesystem of equations.


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Geometric Algebraic

Optimality test: Not

optimal because moving

along the edge from (0,6)

to the right increases Z.

Iteration 2: Move along

the edge to the right and

stop when the new

constraint boundary is

reached. Optimality test : (optimal)

Optimality test: Not

optimal because

increasing one non-basic

variable (X1) increases Z.

Iteration 2: Increase X1,

while adjusting other

variables and stop when

the first basic variable

reaches 0. Optimality test: (optimal)


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Direction of Movement

Z = 3X1 + 5X2

If we increase X1 rate of improvement is 3.

If we increase X2 rate of improvement is 5.Choose X2.

X2 is the entering basic variable for iteration 1.


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Where to stop

Increasing X2 increases Z. We want to go asfar as possible without leaving the feasible

region.

Check the constraints.X3 = 4 0 No upper bound on X2

X4 = 122X2 0 X2 6 to keep X4 non-neg

X5 = 182X2 0 X2 9 to keep X5 non-neg

Thus, X2 can be increased to 6, at which point

X4 drops to zero (non-basic). This is the

minimum ratio test.


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Minimum Ratio Test

The objective of the test is to determine which

currently basic variable will drop to zero

and become non-basic. We can rule out the

basic variable in any equation where thecoefficient of the entering variable is zero

or negative since such a basic feasible variable

would not decrease as the entering variable

increases.


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Our Pivot Using Matrix Algebra

We found the largest entry in the Objective

row (greatest increase in Z).

The variable in that column was the

entering variable.

We did the minimum ratio test by dividing

the RHS entries by the positive entries in

the column. The current basic variable is

determined by this equation.


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Solving for the new BFS

Step 3 of the iteration is finding the new BFS.

We know that X1=0, X2=6, and X4=0.

We need to solve for X3 and X5.

Z 3X1 5X2 = 0

X1 +X3 =4

2X2 + X4 = 123X1 +2X2 +X5 = 18

Current coefficients on X4 are (0,0,1,0). We

want to make this the pattern of coefficients for X2.


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Solving for new BF Solution, Continued

To get the desired pattern of coefficients for

X2, we use matrix algebra. Multiple or divide

by a non-zero constant. Add or subtract

multiples of one equation from another.

The current coefficients on X2 are

(-5,0,2,3) and we want them to be(0,0,1,0)


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Transformed Equations

Z - 3X1 + 5/2X4 = 30

X1 +X3 = 4

X2 + X4 = 6

3X1 - X4 + X5 = 6

Since X1 = 0 and X4 = 0, the equations in

this form give the BFS

(0, 6, 4, 0, 6)


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Optimality test for new BF solution

The current objective function gives thevalue of Z in terms of only the current

nonbasic variables.

Z = 30 + 3X1 5/2 X4

If X1 increases, Z will increase.

In iteration 2, we let X1 enter and find a

variable to leave. Or minimum ratio testindicates X5 will leave. We repeat the algebraic

manipulations to reproduce the pattern of coefficients

for X5 (0,0,0,1) as the new coefficients of X1.


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Tabular form

We have done the tabular form already

It provides a summary of important

information

The simplex tableau is a compact display

of the system of equations


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Simplex Tableau

Z X1 X2 X3 X4 X5 RHS

obj 1 -3 -5 0 0 0 0

r1 0 1 0 1 0 0 4r2 0 0 2 0 1 0 12

r3 0 3 2 0 0 1 18

Optimality test: The current BFS is optimal ifevery coefficient in Obj row is non-negative.

Otherwise pivot to obtain a new tableau.


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Pivots

Z X1 X2 X3 X4 X5 RHS ratioobj 1 -3 -5 0 0 0 0

r1 0 1 0 1 0 0 4 **

r2 0 0 2 0 1 0 12 6

r3 0 3 2 0 0 1 18 9


obj 1 -3 0 0 2.5 0 30r1 0 1 0 1 0 0 4 4

r2 0 0 1 0 0.5 0 6 **

r3 0 3 0 0 -1 1 6 2


obj 1 0 0 0 1.5 1 36

r1 0 0 0 1 1/3 -1/3 4

r2 0 0 1 0 0.5 0 6

r3 0 1 0 0 -1/3 1/3 6


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Degeneracy

If one of the degenerate basic variables (basic

variables with a value of zero) retains its zero

value until it is chosen at a subsequent iteration

to be the leaving basic variable, thecorresponding entering variable will be stuck

at zero since it cant be increased without making

the degenerate leaving variable negative, so

the value of Z wont change. Simplex maygo around in a loop, repeating the same sequence

without advancing.


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Unbounded Z

If Z is unbounded, there will be no candidate

for the leaving basic variable.

In these cases, the constraints dont keep thevalue of the objective function from increasing

indefinitely.


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Multiple Optimal Solutions

In this case, the objective function has thesame slope as one of the constraints.

If we change the objective function for the

Wyndor glass problem to:

Z = 3X1 + 2X2

The new objective function has the same

coefficients as the third constraint. The simplex

stops after one optimal solution is found.


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How to tell if there are more optimal

solutions

It may be desirable to identify other optimal

solutions. Some programs tell you they exist.

Whenever a problem has more than one optimalBFS, at least one of the non-basic variables

will have a coefficient of zero in the final

objective function. Other solutions can be

found by pivoting and using the variable with thezero coefficient in the objective row as the

entering variable.


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Multiple Optimal Solutions

Z X1 X2 X3 X4 X5 RHS ratio

obj 1 -3 -2 0 0 0 0

r1 0 1 0 1 0 0 4 4

r2 0 0 2 0 1 0 12 **

r3 0 3 2 0 0 1 18 6


obj 1 0 -2 3 0 0 12

r1 0 1 0 1 0 0 4 4

r2 0 0 2 0 1 0 12 **

r3 0 0 2 -3 0 1 6 2


obj 1 0 0 0 0 1 18

r1 0 1 0 1 0 0 4 4

r2 0 0 0 3 1 -1 6 2

r3 0 0 1 -1.5 0 0.5 3 **


obj 1 0 0 0 0 1 18

r1 0 1 0 0 -1/3 1/3 2

r2 0 0 0 1 1/3 -1/3 2

r3 0 0 1 0 1/2 0 6


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An Equality Constraint

As we saw in McCarl and Spreen an equality

constraint is equivalent to two inequality

constraints. Hillier and Lieberman provide

another way to handle these constraints, withan artificial variable.


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Modified Problem

Change the last constraint so that

3X1 + 2X2 = 18

Z - 3X1 5X2 = 0X1 + X3 = 4

2X2 + X4 = 12

3X1 + 2X2 = 18

There is no obvious BFS because there is

no slack in the third constraint.


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Figure 4.3

4,3

2, 6

0

The BFS is now a

line between the two

points.


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Use an artificial variable

Z - 3X1 5X2 +MX5 = 0X1 + X3 = 4

2X2 + X4 = 12

3X1 + 2X2 + X5 = 18

X5= 18 3x1 2X2

The large M ensures that the artificial variablewill be driven to zero.


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Negative RHS

To eliminate the negative value in theRHS, multiply through by -1. This operation

reverses the sign of the equality.

In doing simplex by hand, all RHS variablesmust be positive. Not required for computer

algorithms.


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Functional Constraints as

We introduce both a surplus variable andan artificial variable so that we can have

a BFS. (We used this technique in solving

our minimization problems earlier.)

Here we will look at an alternative method to

solving minimization problems, the two phase

method.


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Minimization

Minimize Z = jcjXj

Same as

Maximize -Z= j (-cj)Xj


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Radiation Example

Minimize Z = 0.4X1 + 0.5X2 where X1and X2 represent doses of two beams

0.3X1 + 0.1X2 2.7

0.5X1 + 0.5X2 = 6 0.6X1 + 0.4X2 6

2nd and 3rd constraint will require artificial

variables.1st constraint takes a slack and third takes

a surplus


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Problem

-Z + 0.4X1 + 0.5X2 +MX4 + MX6

0.3X1 + 0.1X2 +X3 = 27

0.5X1 + 0.5X2 +X4 = 6

0.6X1 + 0.4X2 -X5 + X6 = 6

X3 is a slack variable

X4 and X6 are artificial variablesX5 is a surplus variable


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Two-Phase Method

Initialization: Add artificial variables to get

an obvious initial solution for the artificial

problem.

The objective of this phase is to find a

BFS to the real problem.

We minimize the artificial problem.


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The artificial problem

Minimize Z = X4 + X6

0.3X1 + 0.1X2 +X3 = 27

0.5X1 + 0.5X2 +X4 = 6

0.6X1 + 0.4X2 -X5 + X6 = 6


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The Real Problem

Minimize Z = 0.4X1 + 0.5X2

0.3X1 + 0.1X2 +X3 = 27

0.5X1 + 0.5X2 = 6

0.6X1 + 0.4X2 -X5 = 6


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Phase 1: Obtain the starting objective

Min Z = X4 + X6 becomes Max -Z = X4 X6

Use elementary row operations to zero these out.

(E.g. add or subtract multiples of the other rows

until the coefficients on X4 and X6 are zero.

That gives us

Z -1.1X1 - 0.9X2 +X5 = -12

We perform the usual simplex maximization and

find a BFS (6,6,0,3,0,0,0)


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Z X1 X2 X3 X4 X5 X6

-1 -1.1 -0.9 0 0 1 0 -12

0 0.3 0.1 1 0 0 0 2.7

0 0.5 0.5 0 1 0 0 6

0 0.6 0.4 0 0 -1 1 6

Z X1 X2 X3 X4 X5 X6

-1 0 -16/30 11/3 0 1 0 -2.10 1 1/3 10/3 0 0 0 9

0 0 1/3 -5/3 1 0 0 1.5

0 0 1/5 -2 0 -1 1 0.6

Z X1 X2 X3 X4 X5 X6

-1 0 0 -5/3 0 -5/3 8/3 -0.5

0 1 0 20/3 0 5/3 -5/3 8

0 0 0 5/3 1 5/3 -5/3 0.5

0 0 1 -10 0 5 5 3

Z X1 X2 X3 X4 X5 X6

-1 0 0 0 1 0 1 0

0 1 0 0 -4 -5 5 6

0 0 0 1 3/5 1 -1 0.3

0 0 1 0 6 5 -5 6

Phase 1


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Phase 2

Start from the final tableau of phase 1

Drop the artificial variables

Substitute the phase 2 objective function

Restore proper form by eliminating the

basic variables X1 and X2 from the

objective row (use row operations)

Solve the phase 2 problem


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Phase 2

Z X1 X2 X3 X5 RHS

-1 0 0 0 -0.5 -5.4

0 1 0 0 -5 6

0 0 0 1 1 0.3

0 0 1 0 5 6

Z X1 X2 X3 X5

-1 0 0 0.5 0 -5.25

0 1 0 5 0 7.5

0 0 0 1 1 0.3

0 0 1 -5 0 4.5

Optimal Solution: X1 = 7.5, X2 = 4.5, X3 (slack)=0,

X5 (surplus) = 0.3


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Variables allowed to be negative

Weve dealt with this in McCarl and Spreen.

Two types, one is bounded, one unbounded.

X1 -10 construct X1* = X1 + 10 and substitute

X1 can take any value. X1 = X1(+) and X1(-)

adds an extra column to the problem.


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Section 4.7 Post-Optimality Analysis

Re-optimization

Shadow prices (weve covered)

Sensitivity analysis

Parametric programming


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Re-optimization

If we have found an optimal solution for oneversion of an LP and then want to run a

slightly different version, it is inefficient to

start from scratch. Re-optimization involves

deducing how changes in the original model

get carried into the final solution (our original

optimal solution). The final tableau is then

revised in light of the new information and theiterations start from there. (If the tableau that

emerges is not feasible, a dual method can

be applied. )


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Sensitivity Analysis

Sensitivity parameters are those that cant be

changed without changing the optimal solution.

If a shadow price on a constraint is positive, then

an increase in the RHS of that constraint willchange the optimal solution. If it is zero, there

is a range over which it can be changed with

no change in the optimal solution.

Also, if a reduced cost is non-zero, there will

be a range over which the related Cj can be

changed without affecting the solution.


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Parametric Programming

Related to sensitivity analysis

Involves the systematic study of how the

optimal solution changes as many

parameters change simultaneously oversome range

Trade-offs in parameter values can be

studied, e.g. some resources (bi) can beincreased if others are reduced


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Section 4.9 Interior Point Solution

Starts from inside the feasible region

Moves along a path from the interior to the

boundary

Large problems can be solved more

efficiently

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