chap5we

(WE) Problem5

Worked Examples for Chapter 5

Example for Section 5.1

Consider the linear programming problem given in Table 6.1 as the dual problem for the Wyndor Glass Co. example. This problem is

Minimize W = 4 y1 + 12 y2 + 18 y3,

subject to y1 + 3 y3 ( 3

2 y2 + 2 y3 ( 5and

y1 ( 0, y2 ( 0, y3 ( 0.(a) Identify the 10 sets of defining equations for this problem. For each one, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or corner-point infeasible solution. For each corner-point solution, give the corresponding basic solution and its set of nonbasic variables.

This problem has five constraints (two functional constraints and three nonnegativity constraints), where the constraint boundary equations for these constraints are obtained by replacing each by =. Since there are three decision variables, each corner-point solution lies at the intersection of three of the constraint boundaries, so the corresponding three constraint boundary equations become the defining equations for that solution. There are 10 different sets of three constraint boundary equations (defining equations).

The basic solutions are obtained from the corner-point solutions by using the augmented form of the model. After introducing surplus variables y4 and y5, the augmented form is

Minimize W = 4 y1 + 12 y2 + 18 y3,

subject to

y1 + 3 y3 - y4 ( 3

2 y2 + 2 y3 - y5( 5and

y1 ( 0, y2 ( 0, y3 ( 0, y4 0, y5 0.

Following are the 10 different ways of choosing three defining equations from the five constraint boundary equations. Solving these three defining equations gives the corner-point solution. The corresponding basic solution is obtained by then using the augmented form of the model to calculate the values of the surplus variables. The three variables that equal zero in the basic solution are the nonbasic variables.

(1) Defining Equations: y1 = 0, y2 = 0, y3 = 0

Corner-point solution: (0, 0, 0)

The corner-point solution (0, 0, 0) is infeasible.

The corresponding basic solution (y1, y2, y3, y4, y5) = (0, 0, 0, -3, -5)

Nonbasic variables: y1, y2, y3(2) Defining Equations: y1 = 0, y2 = 0, y1 + 3 y3 = 3



The corresponding basic solution (y1, y2, y3, y4, y5) = (0, 0, 1, 0, -3)

Nonbasic variables: y1, y2, y4(3) Defining Equations: y1 = 0, y2 = 0, 2 y2 + 2 y3 = 5

Corner-point solution: (0, 0, 2.5)

The corner-point solution (0, 0, 2.5) is a CPF.

The corresponding basic solution (y1, y2, y3, y4, y5) = (0, 0, 2.5, 4.5, 0)

Nonbasic variables: y1, y2, y5(4) Defining Equations: y1 = 0, y1 + 3 y3 = 3, y3 = 0

No solution exists.

(5) Defining Equations: y1 = 0, 2 y2 + 2 y3 = 5, y3 = 0

Corner-point solution: (0, 2.5, 0)

The corner-point solution (0, 2.5, 0) is infeasible.

The corresponding basic solution (y1, y2, y3, y4, y5) = (0, 2.5, 0, -3, 0)

Nonbasic variables: y1, y3, y5(6) Defining Equations: y2 = 0, y1 + 3 y3 = 3, y3 = 0



The corresponding basic solution (y1, y2, y3, y4, y5) = (3, 0, 0, 0, -5)

Nonbasic variables: y2, y3, y4(7) Defining Equations: y2 = 0, 2 y2 + 2 y3 = 5, y3 = 0

No solution exists.

(8) Defining Equations: y2 = 0, y1 + 3 y3 = 3, 2 y2 + 2 y3 = 5

Corner-point solution: (-4.5, 0, 2.5)

The corner-point solution (-4.5, 0, 2.5) is infeasible.

The corresponding basic solution (y1, y2, y3, y4, y5) = (-4.5, 0, 2.5, 0, 0)

Nonbasic variables: y2, y4, y5(9) Defining Equations: y3 = 0, y1 + 3 y3 = 3, 2 y2 + 2 y3 = 5


The corner-point solution (3, 2.5, 0) is a CPF.

The corresponding basic solution (y1, y2, y3, y4, y5) = (3, 2.5, 0, 0, 0)

Nonbasic variables: y3, y4, y5(10) Defining Equations: y1 = 0, y1 + 3 y3 = 3, 2 y2 + 2 y3 = 5


The corner-point solution (0, 1.5, 1) is a CPF.

The corresponding basic solution (y1, y2, y3, y4, y5) = (0, 1.5, 1, 0, 0)

Nonbasic variables: y1, y4, y5Example for Section 5.2

Consider the following model.

Maximize Z = 4 x1 + 3 x2 + 6 x3,

subject to

3 x1 + x2 + 3 x3 30

2 x1 + 2 x2 + 3 x3 40and

x1 0, x2 0, x3 0.

We introduce x4 and x5 as slack variables for the respective constraints. The augmented form of the model then is Maximize Z = 4 x1 + 3 x2 + 6 x3,

subject to

3 x1 + x2 + 3 x3 + x4 = 30

2 x1 + 2 x2 + 3 x3 + x5 = 40and

x1 0, x2 0, x3 0, x4 0, x5 0.

Using matrix notation, we have

c =, A = , b = , x = .

Now let us apply the revised simplex method step by step to solve this problem.

Iteration 0:Since x4 and x5 are the initial basic variables,

xB =, B = = B-1, so = B-1b =

EMBED Equation.3 = .

cB = Iteration 1:The coefficients in Eq. (0) are

cBB-1A - c = [-4 -3 6] for x1, x2, x3,

cBB-1 = [0 0]

for x4, x5.Since -6 is the most negative coefficient, we choose x3 as the entering basic variable. The coefficients of x3 in Eqs. (1) and (2) are

B-1.

The right-hand side of these equations was identified in Iteration 0 as

B-1b = , where this gives the value of xB = .

Applying the minimum ratio test, since (30/3) < (40/3), we choose x4 as the leaving basic variable. Thus,

xB = .

Since the coefficients of x3 in Eqs. (1) and (2) are and Eq. (1) yielded the leaving basic variable,

= = , soB-1 becomes

B-1 = = .

Therefore,

= B-1b =


cB = .

Iteration 2:The coefficients in Eq. (0) now are

cBB-1A - c = [2 -1 0] for x1, x2, x3,

cBB-1 = [2 0]

for x4, x5.

Since only x2 has a negative coefficient, it becomes the entering basic variable. The coefficients of x2 in Eqs. (1) and (2) are

B-1 = = .

The right-hand side of these equations was identified in Iteration 1 as

B-1b = , where this gives the value of xB = .

Applying the minimum ratio test, since

,

we choose x5 as the leaving basic variable. Thus,

xB = .

Since the coefficients of x2 in Eqs. (1) and (2) are and Eq. (2) yielded the leaving basic variable,

= = ,

so B-1 becomes

B-1 = = .

Therefore,

= B-1b =


cB = .

Optimality Test:The coefficients in Eq. (0) now are

cBB-1A - c = [6 3] - [4 3 6]

= [1 0 0]

for x1, x2, x3,

cBB-1 = [1 1]

for x4, x5.

Since all these coefficients are nonnegative, the current solution is optimal. This solution is

xB = = B-1b = ,

as found in Iteration 2. The other variables are nonbasic variables, so x1 = 0, x4 = 0, and x5 = 0.

Z = 4(0) + 3(10) + 6(20/3) = 70.Example for Section 5.3

Consider the following problem.

MaximizeZ = x1 - x2 + 2 x3,

subject to

x1 + x2 + 3x3 15

2x1 - x2 + x3 2

-x1 + x2 + x3 4

and

x1 0, x2 0, x3 0.

Let x4 , x5, and x6 denote the slack variables for the respective constraints. After the simplex method is applied, a portion of the final simplex tableau is as follows:

Coefficient of:

Basic VariableEqZx1x2x3x4x5x6Right Side

Z(0)10

x4(1)01-1-2

x3(2)00

x2(3)00-

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau.

From the coefficients for (x4, x5, x6) in the final simplex tableau, we observe that

S* = andy* = .

Thus, by the fundamental insight presented in Sec. 5.3, the constraint coefficients for (x1, x2, x3) in the final tableau are

S*A =


The coefficients in the objective function for (x1, x2, x3) in the final tableau are

yA c =

EMBED Equation.3 - = .The final right-hand side is S*b =

EMBED Equation.3 =, and

Z* = y*b = [0 3/2 1/2] = 5. Therefore, the complete final simplex tableau is

Coefficient of:

Basic VariableEqZx1x2x3x4x5x6Right Side

Z(0)13/20003/21/25

x4(1)01001-1-25

x3(2)01/20101/21/23

x2(3)0-3/2100-1/21/21

(b) Identify the defining equations for the CPF solution corresponding to the optimal BF solution in the final simplex tableau.Since the optimal BF solution is (x1, x2, x3, x4, x5, x6) = (0, 1, 3, 5, 0, 0), the corresponding CPF solution is (x1, x2, x3) = (0, 1, 3). The nonbasic variables are x1, x5, x6, and these are the indicating variables that indicate that the following constraints hold with equality.

x1

0

2x1 - x2 + x3 2

-x1 + x2 +x3 4

Therefore, the defining equations for this CPF solution are

x1 = 0,

2x1 - x2 + x3= 2,

-x1 + x2 + x3= 4.

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