# EE/Econ 458 The Simplex Method - PowerPoint PPT Presentation

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EE/Econ 458 The Simplex Method. J. McCalley. An approach. Adjacen t corner points are connected by a single line segment on the boundary of the feasible region. One corner point is better than another if it has a higher value of the objective function f. Definitions.

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EE/Econ 458 The Simplex Method

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J. McCalley

## An approach

Adjacent corner points are connected by a single line segment on the boundary of the feasible region.

One corner point is better than another if it has a higher value of the objective function f.

## Definitions

• Pick a corner point at random.

• Move to an adjacent corner point that is better.

• If there are two that are better, move to the one that is best.

• If there are no better adjacent corner points, the current corner point is the solution to the problem.

## An approach

Optimality (stopping) condition: If a corner point feasible solution is equal to or better than all its adjacent corner point feasible solutions, then it is equal to or better than all other corner point feasible solutions, i.e., it is optimal.

Main ideas of proof:

If objective function monotonically increases (decreases) in some direction within the decision-vector space, then each adjacent corner point will become progressively better in the direction of objective function increase (decrease) such that the last corner point must have two adjacent corner points that are worse.

The monotonicity of objective function increase (decrease) is guaranteed by its linearity.

Many problems require non-negativity constraints on decision variables. SCED is like that due to requirement that generation & demand be positive.

## Standard Form – non-negativity constraints

Simplex method requires non-negativity on decision variables to ensure a bounded region.

When decision variable must allow negativity, they can be converted into one, or two decision variables with non-negativity constraints.

Equality constraints:

## Standard Form – eliminating equalities

x3=4-x1

x3 is the slack variable.

It gives “slack” between two sides of the inequality x1<4.

## Slack Variables

We can replace the first inequality x1<4 with

If “slack” is zero, then inequality is satisfied with equality.

Observe that slack variables cannot be negative, because

then the inequality would be violated.

And we can do that with all inequalities, leading to…

Here, we have introduced slack variables within all inequalities

## Equality Form and Augmented Solution

EQUALITY FORM: Has all inequality constraints converted to equality constraints via introduction of slack variables.

AUGMENTED SOLUTION: A solution to the LP that includes values of the decision and slack variables.

Solution: (x1,x2)=(3,2)

Augmented solution:(x1,x2, x3, x4, x5)=(3,2,1,8,5).

Basic solution: An augmented corner-point solution, e.g.,

(x1,x2,x3,x4,x5)=(4,6,0,0,-6). Basic solutions may be feasible or infeasible.

## Basic & basic feasible solutions

Basic Feasible Solution: A feasible augmented corner-point solution, e.g.,

(x1,x2,x3,x4,x5)=(0,6,4,0,6).

We require b ≥ 0

## Positive right-hand-sides

If there is a bi<0, then we can multiply the corresponding constraint by -1 and introduce a slack variable. See end of LPSimplex1 notes for an example.

• Initialization: Start at a corner point solution.

• Iterative step: Move to a better adjacent corner point feasible solution.

• Optimality test: Determine if the current feasible corner point is optimal using our optimality test (if none of its adjacent feasible corner points are better, then the current feasible corner point is optimal).

• If the current feasible corner point is optimal, the solution has been found, and the method terminates.

• If the current feasible corner point is not optimal, then go to 2.

## High-Level Version of Simplex Method

xn+k is the slack variable.

Above equation is satisfied by below basic solution

## Initialization

Because bk ≥0, we are assured xn+k≥0, which satisfies variable non-negativity.

And so an initial BFS is found by letting all decision variables be zero (the origin!).

(x1, x2, x3, x4, x5)=(0,0,4,12,18) is the initial BFS.

Our initial BFS:

(x1, x2, x3, x4, x5)=(0,0,4,12,18)

## Basic & nonbasic variables

For any BFS

Basic variables: not 0.

Nonbasic variables: 0

• n: number of decision variables.

• m: number of constraints.

• We will always have n+m variables in a BFS:

• n are non-basic

• m are basic

• Initialization: Start at a corner point solution.

• Iterative step: Move to a better adjacent corner point feasible solution.

• Optimality test: Determine if the current feasible corner point is optimal using our optimality test (if none of its adjacent feasible corner points are better, then the current feasible corner point is optimal).

• If the current feasible corner point is optimal, the solution has been found, and the method terminates.

• If the current feasible corner point is not optimal, then go to 2.

## High-Level Version of Simplex Method

Move to a better adjacent corner point feasible solution.

• Two feasible corner points are said to be adjacent if

• They satisfy all constraint equations.

• The constraint equations which define them are the same, with exception of just one

• they share n-1 constraint equations, with one constraint equation for one corner point differing from one constraint equation for the other corner point.

## Iterative step

A feasible corner point is the simultaneous solution of a set of n constraint equations that does not violate any constraint equations.

## Iterative step

5 feasible corner points

Move from 1 to 2.

## Iterative step

Start at (0,0).

1. The x2=0 constraint becomes inactive.

2. Move along the x1=0 constraint until you reach the next corner point, which is (0,6).

3. The 2x2=12 constraint becomes active.

Move from 2 to 3.

## Iterative step

1. The x1=0 constraint becomes inactive.

2. Move along the 2x2=12 constraint until you reach the next corner point, which is (2,6).

3. The 3x1+2x2=18 constraint becomes active.

Move from 3 to 4.

## Iterative step

1. The 2x2=12 constraint becomes inactive.

2. Move along the 3x1+2x2=18 constraint until you reach the next corner point, which is (4,3).

3. The x1=4 constraint becomes active.

Move from 4 to 5.

## Iterative step

1. The 3x1+2x2=18 constraint becomes inactive.

2. Move along the x1=4 constraint until you reach the next corner point, which is (4,0).

3. The x2=0 constraint becomes active.

Basic feasible solutions (BFSs)

## Iterative step

Observation 1:

The slack variable, for a particular corner point, is the weighted distance between that corner point and the slack variable’s constraint (weights are the coefficients from constraint equation).

For the constraint corresponding to x1<4 (or x1+x3=4), because the coefficient for x1 is 1, the slack variable is exactly the “distance” between the given corner point and this constraint (from Table 2, this would be 4, 4, 2, 0, 0 for points 1, 2, 3, 4, 5, respectively).

Basic feasible solutions (BFSs)

## Iterative step

Observation 2:

All BFSs have exactly n variables equal to 0 because any BFS is the simultaneous solution of a system of n constraint equations. There are 2 types of such constraint equations that can define a BFS:

• Nonnegativity constraints on decision variablesDV is 0 in the BFS

• Constraints with slack variablesSV is 0 in the BFS

Any BFS has n variables 0. Any BFS has n non-basic variables.

Basic feasible solutions (BFSs)

## Iterative step

Observation 3:

The way we move from one BFS to another is by exchanging exactly one zeroed variable with a non-zeroed variable.

The way we move from one

BFS to another is by exchanging exactly one non-basic variable with a basic variable.

One constraint equation becomes inactive while one constraint equation becomes active.

One variable enters the basis (becomes nonzero) while one variable leaves the basis (becomes zero).

## Iterative step

• Three questions:

• How is the entering basic variable identified?

• How is the leaving basic variable identified?

• How is the new basic feasible solution found?

• How is the entering basic variable identified?

The candidates for the entering basic variable are the n nonbasic variables. In the first step of our example (solution 1 to 2), the candidates are x1 and x2.

Criterion is:

Select the one that improves the objective at the highest rate (i.e., the largest amount of objective per unit change in variable).

## Iterative step

Objective

Increasing either variable, x1 or x2, increases the objective, but x2 increases it the most for a given unit change, since 5>3; so x2 is the entering variable.

2. How is the leaving basic variable identified?

The candidates for the leaving basic variable are the m basic variables. In the first step of our example (solution 1 to 2), the candidates are x3, x4, and x5. The balance between how much the entering variable may increase without pushing any current basic variable negative is controlled by our m constraints.

## Iterative step

Recall initial point, origin, has all n DV zero and all m SV non-zero. There is 1 SV per constraint, so each constraint contains exactly 1 basic variable at initialization. Since iterations always cause each constraint to exchange 1 basic and 1 non-basic variables, each constraint always has just 1 basic variable.

Criterion is: Choose the leaving variable to be the basic variable that hits 0 first as the entering variable is increased, as dictated by one of the m constraint equations.

Recall x2 is our entering variable. We want to identify the constraint, when its basic variable is 0, most limits x2.

2. How is the leaving basic variable identified?

## Iterative step

• The above table tells us the following:

• We can increase x2 up to 6, but if we go further, x4 will become negative.

• We can increase x2up to 9, but if we go further, x5 will become negative.

• So we cannot increase x2 beyond 6, and x4becomes the leaving variable

• x4 becomes 0 in the new BFS.

• (x4becomes non-basic in the new BFS)

• and of course, x2 becomes non-zero

• (x2 becomes basic).

3. How is the new basic feasible solution found?

## Iterative step

• The new BFS could be found from:

• we know x1 is still non-basic and therefore x1=0

• x4 just became non-basic and therefore x4=0

• the most limiting constraint in the above table, x2=6

• the other constraints:

• Row 1 of table: x3=4-x1=4-0=4

• Row 3 of table: x5=18-3x1-2x2=18-3(0)-2(6)=6

• (x1,x2,x3,x4,x5)=(0,6,4,0,6), and we are done with this iteration.

• Then check for optimality.

• 3. How is the new basic feasible solution found?

But we need a more structured way (that we can teach to a computer).

Write the objective function like this:

## Iterative step

For the initial solution, we can write F together with constraint equations:

The above is set up for our initial solution: (x1, x2, x3, x4, x5)=(0,0,4,12,18). We observe values of the basic variables are the right-hand-sides of (8), (9), (10).

Because x1 and x2 are non-basic (0), the right-hand-side of (7) is the objective value at this solution.

3. How is the new basic feasible solution found?

## Iterative step

Goal: Have x2 enter the basis & x4 leave the basis so that, at the new solution, our system of equations is in the same form as above where:

• the values of the basic variables at the solution can be directly read off as the right-hand-sides of those equations and

• the value of the objective at that solution can be directly read off as the right-hand-side of the objective equation.

To accomplish this…

• Because x2 is going to be non-zero, it must show up in only one constraint equation, with a 1 as its coefficient (so it can be directly read off), and it must not appear in the objective equation (since it will not be 0).

• Because x4 is going to be 0, it must appear in the objective equation (so the right-hand-side of the objective equation will directly give F).

3. How is the new basic feasible solution found?

## Iterative step

Entering variable: x2

Leaving variable: x4

We want to eliminate (make 0) the coefficient of x2 (-5) in row 1.

We will use row 3 as our “pivot row” (and a33 our pivot) so that the leaving variable (x4) appears in row 1 after eliminating x2.

3. How is the new basic feasible solution found?

Join the matrix with the right-hand-side vector, like this

Pivot

## Iterative step

Pivot row

Use Gaussian Elimination! For an n × n matrix:

Choose the row from which the leaving variable was identified to be the pivot row. Let this be row k so that akk is the pivot.

Divide row k by akk.

Eliminate all ajk, j=1,…,n except j=k. This means to make all elements directly above and beneath the pivot equal to 0 by adding an appropriate multiple of the pivot row to each row above or beneath the pivot.

3. How is the new basic feasible solution found?

## Iterative step

Recalling (x2, x3, x5) are basic variables (with non-zero values) and (x1, x4) are non-basic variables (with zero values), one can immediately read off

x3=4 (from second row)

x2=6 (from third row)

x5=6 (from fourth row)

3. How is the new basic feasible solution found?

## Iterative step

Recalling (x2, x3, x5) are basic variables (with non-zero values) and (x1, x4) are non-basic variables (with zero values), one can immediately read off

x3=4 (from second row)

x2=6 (from third row)

x5=6 (from fourth row)

How to identify the entering variable?

Select the variable that improves the objective

at the highest rate (i.e., the largest amount

of objective per unit change in variable).

How to identify the leaving variable?

Choose the leaving variable to be the one that

hits 0 first as the entering variable is increased,

as dictated by one of the m constraint equations.

How is the new BFS found?

Using the equation used to identify the leaving variable as the pivot row, eliminate the entering variable from all other equations.

## Iterative step – Key ideas

If there are any remaining variables with positive coefficients in current objective function expression, then the current solution may still be improved; take another iteration.

If not, the current solution may not be further improved, and it is therefore optimal.

## Optimality test

From example:

Since x1 has a positive coefficient, F may still be improved. Therefore this solution is not optimal.

In the right-hand-form, optimality occurs when all coefficients are negative.

In the left-hand-form, optimality occurs when all coefficients are negative.