logistics. terminology

Operations Research 1 – 06/07 – Chapter 4 1

Operations Research 1

Chapter 4Introduction to Operations Research

Hillier & LiebermannEighth EditionMcGrawHill


Contents• Simplex method: Graphical Algorithm• Simplex method: Algebraic Algorithm• Differences between the graphical and the algebraic

algorithm• Simplex method: Tabular Form• Simplex method: Tie breaking• Simplex method: Other model forms


The Simplex Method – Graphical algorithmx 2

x1

10421 3 5 6 87 9

104

21

35

68

79

Maximize Z = 3x1 + 5x2

subject to

x1 ≤ 4

2x2 ≤ 12

3x1 + 2x2 ≤ 18

and

x1 ≥ 0; x2 ≥ 0

x1 = 4

2x2 = 12

3x1 + 2x2 = 18

Feasibleregion

= Corner-point solution (not feasible)

= Corner-point feasible solution (CPF solution)

For any linear programming problem with n decision variables, two CPF solutions are adjacent to each other if they share n – 1 constraint boundaries.

The two adjacent CPF solutions are connected by a line segment that lies on these same shared constraint boundary. Such a line segment is referred to as an edge of the feasible region.two adjacent

CPF-solutions

= Edge


The Simplex Method – Graphical algorithmx 2

x1

10421 3 5 6 87 9

104

21

35

68

79

x1 = 4

2x2 = 12

3x1 + 2x2 = 18

Feasibleregion

= Corner-point solution (not feasible)

= Corner-point feasible solution (CPF solution)

two adjacent CPF-solutions

= Edge

CPF solution

Its Adjacent CPF solution

(0, 0) (0, 6) and (4, 0)(0, 6) (2, 6) and (0, 0)(2, 6) (4, 3) and (0, 6)(4, 3) (4, 0) and (2, 6)(4, 0) (0, 0) and (4, 3)


x 2

x1

10421 3 5 6 87 9

104

21

35

68

79

Feasibleregion

Initialization: Choose (0,0) as the initial CPF solutionOptimality test: Is (0,0) an optimal CPF solution?

No (Z = 0), adjacent solutions are better.Iteration 1: Move to a better CPF solution in three steps:

1. Choose the direction with the highest rate of increasing Z.

Z increases with a rate of 3 by moving up the x1 axis. Z increases with a rate of 5 by moving along the x2 axis.

We choose to move up the x2 axis. 2. Stop at the first new constraint boundary:

2x2 = 12 (We have to stay in the feasible region).

3. Solve for the intersection of the new set of constraints boundaries: (0, 6).Z = 3 * 0 + 5 * 6 = 30

Optimality test: Is (0, 6) an optimal CPF solution?No, an adjacent solution is better.


The Simplex Method – Graphical algorithm


x 2

x1

10421 3 5 6 87 9

104

21

35

68

79

Feasibleregion

Maximize Z = 3x1 + 5x2Iteration 2: Move to a better CPF solution in three steps:

1. Choose the best direction.Z will increase by moving to the right. Rate = 3.Z will decrease by moving down the x2 axis.We choose to move to the right.

2. Stop at the first new constraint boundary:3x1 + 2x2 = 12 (We have to stay in the feasible region).

3. Solve for the intersection of the new set of constraints boundaries: (2, 6).Z = 3 * 2 + 5 * 6 = 36

Optimality test: Is (2, 6) an optimal CPF solution?Yes, there is no better CPF solution!

The Simplex Method – Graphical algorithm


Graphical algorithm: The Key Solution Concepts

• Solution concept 1– Method focuses solely on CPF-solutions– If there are CPF-solutions, find the best.

• Solution concept 2– Iterative algorithm:

• Initialization: Set up, find a initial CPF-solution• Optimality test: Is the solution optimal? If no, continue. If yes, stop.• Iteration: Perform an iteration to find a better CPF solution. Then go

back to the optimality test.

• Solution concept 3– Whenever possible, choose the origin as the initial CPF-solution

at initialization.– This eliminates the use of algebraic procedures to find and solve

for an initial CPF solution.


Graphical algorithm: The Key Solution Concepts• Solution concept 4

– At iteration, only adjacent CPF solutions are considered. – The path to find the optimal solution is along the edges of the feasible

region.• Solution concept 5

– After identification of the current CPF solution, examine all the edges that emanate from that CPF solution.

– The edges lead to other CPF solution.– Choose the edge with the (highest) positive rate of improvement in Z.*

• Solution concept 6– If the rate of improvement in Z is positive, the next adjacent CPF

solution is better than the current CPF solution.*– If the rate of improvement in Z is negative, the next adjacent CPF

solution is worse.*– If there are no positive rates of improvement, the current CPF solution is

optimal.** Only if the objective function is to maximize Z


Notion of the Simplex Method• Besides the graphical algorithm, there is an algebraic

algorithm.• This algorithm is based on solving systems of equations.• We need to translate inequality constraints to equality

constraints.• Therefore we introduce slack variables.

Original Form of the Model


subject tox1 ≤ 4 2x2 ≤ 12

3x1+ 2x2 ≤ 18and x1 ≥ 0; x2 ≥ 0

Augmented Form of the Model


subject to x1 + x3 = 4

2x2 + x4 = 12 3x1 + 2x2 +x5 = 18and xj ≥ 0 for j = 1, 2, 3, 4, 5.


Slack values

FeasibleRegion

InfeasibleRegion

Constraint correspondingwith slack variable

Slack variable > 0 Slack variable < 0Slack variable = 0


Terminology – Graphical & AlgebraicGraphical Algorithm

Algebraic Algorithm

Solution Solution Augmented solution

Solution on a corner point

Corner point solution

Basic solutionAugmented corner point solution

Solution on a feasible corner point

CPF solution Basic Feasible (BF) solutionAugmented CPF solution

Example of a solution (0,6) (0, 6, 4, 0, 6)(Slack included)


Basic and non-basic variables

- =

Thus, we can choose a number of variables (= number of degrees of freedom) to be set equal to any arbitrary value.

These variables are called nonbasic variables. (Together they are the basis.)

The other variables are called basic variables.

Number of variables (in augmented form)

Number of equations Number of degreesof freedom


subject to x1 + x3 = 4

2x2 + x4 = 12 3x1 + 2x2 +x5 = 18and xj ≥ 0 for j = 1, 2, 3, 4, 5.

In this example there are 5 – 3 = 2 degrees of freedom.For example, we choose x1 = 4 and x4 = 2 (x1 and x4 are nonbasic var’s) then this must be true:x3 = 0; x2 = 5; x5 = -4(x3, x2 and x5 are basic var’s)


Properties of a Basic Solution1. Each variable is designated as either a nonbasic

variable or a basic variable.2. The number of basic variables equals the number of

functional constraints.3. The nonbasic variables are set to equal.4. The values of the basic variables are obtained as the

simultaneous solution of the system of equations.5. If the basic variables satisfy the nonnegativity

constraints, the basic solution is a BF solution (feasible).


Solving the Simplex MethodGraphical & Algebraic Interpretations

Method Sequence Graphical Interpretation Algebraic InterpretationInitialization Choose (0, 0) to be the initial

CPF solution.Choose x1 and x2 to be the nonbasic variables (= 0) for the initial BF solution (0, 0, 4, 12, 18)

Optimality test Not optimal, because moving along either edge from (0, 0) increases Z.

Not optimal, because increasing either nonbasic variable (x1 or x2) increases Z.

Iteration 1 step 1 Move up the edge lying on the x2 axis.

Increase x2 while adjusting other variable values to satisfy the system of equations.

Iteration 1 step 2 Stop when the first new constraint boundary (2x2 = 12) is reached.

Stop when the first basic variable (x3, x4 or x5) drops to zero (x4).

Iteration 1 step 3 Find the intersection of the new pair of constraint boundaries: (0, 6) is the new CPF solution.

With x2 now a basic variable and x4 now a nonbasic variable, solve the system of equations: (0, 6, 4, 0, 6) is the new BF solution.


Solving the Simplex MethodGraphical & Algebraic Interpretations (continued)

Method Sequence Graphical Interpretation Algebraic InterpretationOptimality test Not optimal, because moving

along the edge from (0, 6) to the right increases Z.

Not optimal, because increasing one nonbasic variable (x1) increases Z.

Iteration 2 step 1 Move along this edge to the right.

Increase x1 while adjusting other variable values to satisfy the system of equations.

Iteration 2 step 2 Stop when the first new constraint boundary (3x1 + 2x2 = 18) is reached.

Stop when the first basic variable (x2, x3, or x5) drops to zero (x5).

Iteration 2 step 3 Find the intersection of the new pair of constraint boundaries: (2, 6) is the new CPF solution.

With x1 now a basic variable and x5 now a nonbasic variable, solve the system of equations: (2, 6, 2, 0, 0) is the new BF solution.

Optimality test (2, 6) is optimal, because moving along either edge from (2, 6) decreases Z.

(2, 6, 2, 0, 0) is optimal, because increasing either nonbasic variable(x4 or x5) decreases Z.


Simplex Method – Tabular Form• Besides the graphical and algebraic forms, there is a tabular form,

which is more convenient to use for solving a problem by hand.

Algebraic Form:

(0) Z - 3x1 - 5x2 = 0(1) x1 + x3 = 4(2) 2x2 + x4 = 12(3) 3x1 + 2x2 +x5 = 18

Tabular FormBasic

variableEq Coefficient of: Right

sideZ x1 x2 x3 x4 x5

Z (0) 1 -3 -5 0 0 0 0x3 (1) 0 1 0 1 0 0 4

x4 (2) 0 0 2 0 1 0 12

x5 (3) 0 3 2 0 0 1 18


Tabular Form: The algorithm• Initialization

– Just like the algebraic algorithm, introduce slack variables.– Select the decision variables to be the initial nonbasic variables

(= 0).– Select the slack variables to be the initial basic variables.

• Optimality Test– The current BF solution is optimal if and only if every coefficent

in row 0 is nonnegative (≥ 0). If it is, stop. Otherwise go to an iteration.

• IterationSee next slides


Tabular Form: The algorithmTabular Form

Basic variable

Eq. Coefficient of: Right sideZ x1 x2 x3 x4 x5

Z (0) 1 -3 -5 0 0 0 0x3 (1) 0 1 0 1 0 0 4

x4 (2) 0 0 2 0 1 0 12

x5 (3) 0 3 2 0 0 1 18

Pivot column• Iteration– Step 1. Determine the entering basic variable by selecting the

variable with the “most negative” coefficient in Eq. (0). Put a box around the column and call this the pivot column.


Tabular Form: The algorithm

62

12 =

92

18 =

Tabular Form Basic

variableEq. Coefficient of: Right

sideRatio

Z x1 x2 x3 x4 x5

Z (0) 1 -3 -5 0 0 0 0 x3 (1) 0 1 0 1 0 0 4 infinite

x4 (2) 0 0 2 0 1 0 12 minimum

x5 (3) 0 3 2 0 0 1 18

• Iteration– Step 2. Determine the leaving basic variable by applying the

minimum ratio test. Put a box around the row with the minimum ratio, this is the pivot row. Call the number in both boxes the pivot number.

Pivot number


Tabular Form: The algorithmIteration: Step 3.Solve for the new BF solution by

changing the column of the entering basic variable (-5, 0, 2, 2) to the column of the leaving basic variable (0, 0, 1, 0) by using elementary row operations:

1. Divide the pivot row by the pivot number. Use this new pivot column in steps 2 and 3.

2. For each other row (including row 0) that has a negative coefficient in the pivot column, add to this row the product of the absolute value of this coefficient and the new pivot row.

3. For each other row that has a negative coefficient in the pivot column, substract from this row the product of this coefficient and the new pivot row.

Tabular Form

Iteration Basic variable Eq. Coefficient of: Right side

Z x1 x2 x3 x4 x5

0 Z (0) 1 -3 -5 0 0 0 0

x3 (1) 0 1 0 1 0 0 4

x4 (2) 0 0 2 0 1 0 12

x5 (3) 0 3 2 0 0 1 18

1 Z (0)

x3 (1)

x2 (2) 0 0 1 0 1/2 0 6

x5 (3)

1 Z (0) 1 -3 0 0 5/2 0 30

x3 (1) 0 1 0 1 0 0 4

x2 (2) 0 0 1 0 1/2 0 6

x5 (3) 0 3 0 0 -1 1 6

The operations in this example:1. New row (2) = Old row (2) / 22. New row (0) = Old row (0) + 5 * New row (2)3. New row (3) = Old row (3) – 2 * New row (2)


Tabular Form: The algorithmOptimality test: Still not optimal

because there is a negative coefficient in row (0). So at least one more iteration is needed.

Iteration 2:- Determine the pivot row (and

entering basic variable).- Run a minimum ratio test.- Determine the pivot column

(and leaving basic variable).- Perform elementary operations:

Tabular Form

Iteration Basic variable Eq. Coefficient of: Right side

Z x1 x2 x3 x4 x5

0 Z (0) 1 -3 -5 0 0 0 0

x3 (1) 0 1 0 1 0 0 4

x4 (2) 0 0 2 0 1 0 12

x5 (3) 0 3 2 0 0 1 18

1 Z (0) 1 -3 0 0 5/2 0 30

x3 (1) 0 1 0 1 0 0 4

x2 (2) 0 0 1 0 1/2 0 6

x5 (3) 0 3 0 0 -1 1 6

414 =

236 =

2 Z (0) 1 0 0 0 3/2 1 36

x3 (1) 0 0 0 1 1/3 -1/3 2

x2 (2) 0 0 1 0 1/2 0 6

x1 (3) 0 1 0 0 -1/3 1/3 2

- New row (3) = Old row (3) / 3- New row (0) = Old row (0) + 3 * New row (3)- New row (1) = Old row (1) – 1 * New row (3)

Optimality test: There is no negative coefficient in row (0), so the solution is optimal!Solution: Z = 36; x1 = 2; x2 = 6


Tie breaking in the simplex methodYou may have noticed that we never said what to do if the various choice rules of the simplex method do not lead to a clear-cut decision, because of either ties or other similar ambiguities. We discuss these details now.


The entering basic variable, step 1If more than one nonbasic variables are tied for having the largest negative coefficient, then the selection may be made arbitrary.


The leaving basic variable, step 2Definition: Basic Variables (BV’s) with a value of zero are called degenerate. Not very often a choice of a degenerate variable may lead to a loop. R. Bland (Mathematics of Operations Research, 2: 103-107, 1977) has suggested special rules avoiding such loops.


No leaving Basic Variable (BV) – Unbouded ZIf the entering BV could be increased indefinitely without givingnegative values to any of current BV’s, then no variable qualifies to bethe leaving BV.

Basic Variable Eq.

Coefficient of: Right Side RatioZ x1 x2 x3

ZX3

(0)(1)

10

-3 -5 0 1 0 1

0 4 None

With x1= 0 and x2 increasing, x3 = 4 - 1x1 – 0x2 = 4 > 0.

• TABLE 4.9 Initial simplex tableau for the Wyndor Glass Co. Problem without the last two functional constraints.


Multiple Optimal Solutions continuedAny Weighted average of two or more solutions (vectors) where the weights are nonnegative and sum is equal to 1 is called a convex combination of these solutions.

Whenever a problem has more than one optimal BF solution, at least one of the nonbasic variables has a coefficient of zero in the final row 0, so increasing any such variable will not change the value of Z. Therefore, these other optimal BF solutions can be identified (if desired) by performing additional iterations of the simplex method, each time choosing a nonbasic variable with a zero coefficient as the entering basic variable.

If such an iteration has no leaving basic variable, this indicates that the feasible region is unbounded and the entering basic variables can be increased indefinitely without changing the value of Z.


Table 4.10 Basic Iteration Variable Eq.

Coefficient of: Right OptimalSide solutionZ x1 x2 x3 x4 x5

Z0 x3

x4

x5

(0)(1)(2)(3)

1000

-3 -2 0 0 0 1 0 1 0 0 0 2 0 1 0 3 2 0 0 0

0 no 4 12 18

Z1 x1

x4

x5

(0)(1)(2)(3)

1000

0 -2 3 0 0 1 0 1 0 0 0 2 0 1 0 0 2 -3 0 1

12 no 4 12 6

Z2 x1

x4

x2

(0)(1)(2)(3)

1000

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

18 Yes 4 6 3

Z3 x1

x3

x2

(0)(1)(2)(3)

1000

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

18 Yes 2 2 6


Equality ConstraintsAny equality constraint

ai1x1 + ai2x2 + ……+ ainxn = bi

Actually is equivalent to a pair of inequality constraints:ai1x1 + ai2x2 + ……+ ainxn ≤ bi

ai1x1 + ai2x2 + ……+ ainxn ≥ bi

Example: Suppose that the Wyndor Glass Co. Problem is modified to require that Plant 3 be used at full capacity. The only resulting change in the LP model is that the third constraint, 3x1 + 2x2 ≤ 18, instead becomes an equality constraint

3x1 + 2x2 = 18,so that the complete model becomes the one shown in the upper right hand corner of figure 4.3. This figure als shows in darker ink the feasible region which now consists of just the line segment connecting (2,6) and (4,3).

After the slack variables still needed for the inequality constraints are introduced, the system of equations for the augmented form of the problem becomes

(0) Z – 3x1 – 5x2 = 0(1) x1 + x3 = 4(2) 2x2 + x4 =12(3) 3x1 + 2x2 = 18


Figure 4.3

(2,6)

(4,3)

Maximize Z = 3x1 + 5x2,Subject to X1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 = 18And x1 ≥ 0, x2 ≥ 0

• Figure 4.3

When the third constraint becomes an equality constraint, the feasible region for the Wyndor Glass Co. problem becomes the line segment between (2,6) and (4,3).


MinimizationMinimization problems can be converted to equivalent

maximization problems:

Minimizing

is equivalent to

Maximizing

∑=

=n

1jjjxcZ

∑=

−=−n

1jjj x)c(Z


Variables Allowed to be negativeSuppose that the Wyndor Glass Co. Problem is changed so that product 1 already is in production, and the first decision variable x1 represents the increase in its production rate. Therefore, a negative value of x1 would indicate that product 1 is to be cut back by that amount. Such reductions might be desirable to allow a larger production rate for the new, more profitable product 2, so negative values should be allowed for x1 in the model.


Variables with Bound on the Negative Values Allowed. (sheet 1 of 2)

Consider any decision variable xj that is allowed to have negative values which satisfy a constraint of the form

xj ≥ Lj , Where Lj is some negative constant. This constraint can

be converted to a nonnegativity constraint by making the change of variables

x’j = xj – Lj, so x’j ≥ 0. Thus, x’j + Lj would be substituted for xj throughout the

model, so that the redefined decision variable x’j cannot be negative. (This same technique can be used when Lj is positive to convert a functional constraint xj ≥ Lj to a nonnegativity constraint x’j ≥ 0.)


Variables with Bound on the Negative Values Allowed. (continued)

To illustrate, suppose that the current production rate for product 1 in the Wyndor Glass Co. problem is 10. With the definition of x1 just given, the complete model at this point is the same as that given in Sec.3.1 except that the nonnegativity constraint x1 ≥ 0 is replaced byx1 ≥ -10To obtain the equivalent model needed for the simplex method, this decision variable would be redefined as the total production rate of product 1x’j = x1 + 10,which yields the changes in the objective function and constraints as shown:

Z = 3x1 + 5x2

x1 ≤ 4 2x2 ≤ 123x1 + 2x2 ≤ 18x1 ≥ -10, x2 ≥ 0

Z = 3(x’1-10) + 5x2

x’1 - 10 ≤ 4 2x2 ≤ 123(x’1-10)+ 2x2 ≤ 18x’1 - 10 ≥ -10, x2 ≥ 0

Z = -30 + 3x’1 + 5x2

x’1 ≤ 14 2x2 ≤ 123x’1 + 2x2 ≤ 48x’1 ≥ 0, x2 ≥ 0

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