University of Colorado at Boulder Yicheng Wang, Phone:303-492-4228, Email:[email protected] Optimization Techniques for Civil and Environmental Engineering

University of Colorado at BoulderYicheng Wang, Phone:303-492-4228, Email:[email protected]

Optimization Techniques forCivil and Environmental Engineering Systems

By

Yicheng Wang


(2) Setting Up the Simplex Method

Original Form of the Model

Augmented Form of the Model


F

G

H

C

D

B

A EH(3,2)

Augmented Form of the Model

For example, H(3,2) is a solution for the original model, which yields the augmented solution ( x1, x2, x3, x4, x5) = (3, 2 ,1 ,8, 5)

For example, G(4,6) is a corner-point infeasible solution, which yields the corresponding basic solution ( x1, x2, x3, x4, x5) = (4, 5 ,0 ,0, -6)

F

G

H

C

D

B

A EH(3,2)


The only difference between basic solutions and corner-point solutions is whether the values of the slack variables are included

Relationship between Corner-Point Solutions and Basic Solutions

In the original model, we have

Corner-point solution

Corner-point feasible (CPF) solution

In the augmented model, we have

Basic solution

Basic Feasible (BF) solution


The corner-point solution (0,0) in the original model corresponds to the basic solution (0, 0, 4,12, 18) in the augmented form, where x1 =0 and x2=0 are the nonbasic variables, and x3=4, x4=12, and x5=18 are the basic variables

F

G

H

C

D

B

A E


Example:

The CPF solution (0,0) in the original model corresponds to the BF solution (0, 0, 4,12, 18) in the augmented form, where x1 =0 and x2=0 are the nonbasic variables, and x3=4, x4=12, and x5=18 are the basic variables

Choose x1 and x4 to be the nonbasic variables that are set equal to 0. The three equations then yield, respectively, x3=4, x2=6 , and x5=6 as the solution for the three basic variables as shown below.


Example:

A(0,0) and B(0,6) are two CPF solutions The corresponding BF solutions are ( x1, x2, x3, x4, x5) = (0, 0 ,4 ,12, 18) and ( x1, x2, x3, x4, x5) = (0, 6 ,4 ,0, 6)

F

G

H

C

D

B

A EH(3,2)

A(0,0) and C(2,6) are two CPF solutions The corresponding BF solutions are ( x1, x2, x3, x4, x5) = (0, 0 ,4 ,12, 18) and ( x1, x2, x3, x4, x5) = (2, 6 ,2 ,0, 0)



(3) The Algebra of the Simplex Method

Use the Wyndor Glass Co. Model to illustrate the algebraic procedure

Initialization

Geometric interpretation Algebraic interpretation

C

D

B

A E


Optimality Test


C

D

B

A E

A(0,0) is not optimal.

Conclusion: The initial BF solution (0,0,4,12,18) is not optimal.

The objective function:

The rate of improvement of Z by the nonbasic variable x1 is 3

The rate of improvement of Z by the nonbasic variable x2 is 5


Iteration1 Step1: Determining the Direction of Movement


C

D

B

A E

Move up from A(0,0) to B(0,6)


Iteration1 Step2: Where to Stop


C

D

B

A E

Stop at B. Otherwise, it will leave the feasible region.

Step 2 determine how far to increase the entering basic variable x2.


Thus x4 is the leaving basic variable for iteration 1 of the example.


Iteration1 Step3: Solving for the New BF Solution


C

D

B

A E

The intersection of the new pair of constraint boundary: B(0,6) Nonbasic variables

Basic variables

Nonbasic variables

Basic variables


(0)

(1)

(2)

(3)

Nonbasic variables: x1= 0

x2= 0Basic variables: x3= 4

x4 =12

x5= 18

Initial BF Solution

Nonbasic variables: x1= 0

x4= 0Basic variables: x3= ？

x2 =6

x5= ？

New BF Solution(0)

(1)

(2)

(3)

Basic variables: x3= 4

x2 =6

x5= 6


Optimality Test


C

D

B

A E

B(0,6) is not optimal, because moving from B to C increases Z.

Conclusion: The BF solution (0,6,4,0,6) is not optimal.


The rate of improvement of Z by the nonbasic varialle x1 is 3

The rate of improvement of Z by the nonbasic varialle x4 is -5/2


Iteration2

Step1: Determining the Direction of Movement

Choose x1 to be the entering basic variable

Step2: Where to Stop

The minimum ratio test indicates that x5 is the leaving basic variable

Step3: Solving for the New BF Solution(0)

(1)

(2)

(3)

Nonbasic variables: x1= 0, x4= 0

Basic variables: x3= 2, x2 =6, x1= 2

New BF Solution


Optimality Test


The coefficients of the nonbasic variables x4 and x5 are negative. Increasing either x4 or x5 will decrease Z, so (x1, x2, x3, x4, x5) = (2, 6, 2, 0, 0) must be optimal with Z = 36.

C

D

B

A E

In terms of the original form of the problem (no slack variables), the optimal solution is (x1, x2) = (2, 6) , which yields Z = 3x1+5x2=36.


(4) The Simplex Method in Tabular Form

The tabular form is more convenient form for performing the required calculations. The logic for the tabular form is identical to that for the algebraic form.


Summary of the Simplex Method in Tabular Form

TABLE 4.3b The Initial Simplex Tableau


TABLE 4.3b The Initial Simplex Tableau


Iteration1





Iteration2

Step1: Choose the entering basic variable to be x1

Step2: Choose the leaving basic variable to be x5


Step3: Solve for the new BF solution.

Optimality test: The solution (2,6,2,0,0) is optimal.The new BF solution is (2,6,2,0,0) with Z =36


(5) Tie Breaking in the Simplex Method

Tie for the Entering Basic Variable

The answer is that the selection between these contenders may be made arbitrarily.

The optimal solution will be reached eventually, regardless of the tied variable chosen.

Tie for the Entering Basic Variable


Tie for the Leaving Basci Variable-Degeneracy

If two or more basic variables tie for being the leaving basic variable, choose any one of the tied basic variables to be the leaving basic variable. One or more tied basic variables not chosen to be the leaving basic variable will have a value of zero.

If a basic variable has a value of zero, it is called degenerate.

For a degenerate problem, a perpetual loop in computation is theoretically possible, but it has rarely been known to occur in practical problems. If a loop were to occur, one could always get out of it by changing the choice of the leaving basic variable.


No Leaving Basic Variable – Unbounded Z

If every coefficient in the pivot column of the simplex tableau is either negative or zero, there is no leaving basic variable. This case has an unbound objective function Z

If a problem has an unbounded objective function, the model probably has been misformulated, or a computational mistake may have occurred.


Multiple Optimal Solution

In this example, Points C and D are two CPF Solutions, both of which are optimal. So every point on the line segment CD is optimal.

Fig. 3.5 The Wyndor Glass Co. problem would have multiple optimal solutions if the objective function were changed to Z = 3x1 + 2x2

C (2,6)

E (4,3)

Therefore, all optimal solutions are a weighted average of these two optimal CPF solutions.



Any linear programming problem with multiple optimal solutions has at least two CPF solutions that are optimal. All optimal solutions are a weighted average of these two optimal CPF solutions.

Consequently, in augmented form, any linear programming problem with multiple optimal solutions has at least two BF solutions that are optimal. All optimal solutions are a weighted average of these two optimal BF solutions.





These two are the only BF solutions that are optimal, and all other optimal solutions are a convex combination of these .


(6) Adapting to Other Model Forms

Original Form of the Model Augmented Form of the Model


Artificial-Variable TechniqueThe purpose of artificial-variable technique is to obtain an initial BF solution.

The procedure is to construct an artificial problem that has the same optimal solution as the real problem by making two modifications of the real problem.


Augmented Form of the Artificial Problem

Initial Form of the Artificial Problem The Real Problem


The feasible region of the Real Problem The feasible region of the Artificial Problem


Converting Equation (0) to Proper FormThe system of equations after the artificial problem is augmented is

To algebraically eliminate from Eq. (0), we need to subtract from Eq. (0) the product, M times Eq. (3)


Application of the Simplex MethodThe new Eq. (0) gives Z in terms of just the nonbasic variables (x1, x2)

The coefficient can be expressed as a linear function aM+b, where a is called multiplicative factor and b is called additive term.

When multiplicative factors a’s are not equal, use just multiplicative factors to conduct the optimality test and choose the entering basic variable.

When multiplicative factors are equal, use the additive term to conduct the optimality test and choose the entering basic variable.

M only appears in Eq. (0), so there’s no need to take into account M when conducting the minimum ratio test for the leaving basic variable.


Solution to the Artificial Problem


Functional Constraints in ≥ Form


The Big M method is applied to solve the following artificial problem (in augmented form)

The minimization problem is converted to the maximization problem by


Solving the Example

The simplex method is applied to solve the following example.

The following operation shows how Row 0 in the simplex tableau is obtained.



The Real Problem The Artificial Problem


The Two-Phase Method

Since the first two coefficients are negligible compared to M, the two-phase method is able to drop M by using the following two objectives.

The optimal solution of Phase 1 is a BF solution for the real problem, which is used as the initial BF solution.


Summary of the Two-Phase Method


Phase 1 Problem (The above example)

Example:

Phase2 Problem

Example:


Solving Phase 1 Problem


Preparing to Begin Phase 2


Solving Phase 2 Problem


How to identify the problem with no feasible solutons

The artificial-variable technique and two-phase method are used to find the initial BF solution for the real problem.

If a problem has no feasible solutions, there is no way to find an initial BF solution.

The artificial-variable technique or two-phrase method can provide the information to identify the problems with no feasible solutions.


To illustrate, let us change the first constraint in the last example as follows.

The solution to the revised example is shown as follows.


(7) Shadow Prices(0)

(1)

(2)

(3)

Resource bi = production time available in Plant i for the new products.

How will the objective function value change if any bi is increased by 1 ?


b2: from 12 to 13

Z: from 36 to 37.5

△Z=3/2

b1: from 4 to 5

Z: from 36 to 36

△Z=0

b3: from 18 to 19

Z: from 36 to 37

△Z=1



indicates that adding 1 more hour of production time in Plant 2 for the twonew products would increase the total profit by $1,500.

The constraint on resource 1 is not binding on the optimal solution, so there is a surplus of this resource. Such resources are called free goods

The constraints on resources 2 and 3 are binding constraints. Such resources are called scarce goods.

H(0,9)


(8) Sensitivity Analysis

Maximize

b, c, and a are parameters whose values will not be known exactly until the alternative given by linear programming is implemented in the future.

The main purpose of sensitivity analysis is to identify the sensitive parameters.

A parameter is called a sensitive parameter if the optimal solution changes with the parameter.


How are the sensitive parameters identified?

In the case of bi , the shadow price is used to determine if a parameter is a sensitive one.

For example, if > 0 , the optimal solution changes with the bi.

However, if = 0 , the optimal solution is not sensitive to at least small changes in bi.

For c2 =5, we have

c1 =3 can be changed to any other value from 0 to 7.5 without affecting the optimal solution (2,6)


(8) Parametric Linear Programming

Sensitivity analysis involves changing one parameter at a time in the original model to check its effect on the optimal solution.

By contrast, parametric linear programming involves the systematic study of how the optimal solution changes as many of the parameters change simultaneously over some range.

Documents

University of Colorado at Boulder Yicheng Wang, Phone:303-492-4228, Email:[email protected] Optimization Techniques for Civil and Environmental Engineering