Introduction To Operations Research By Frederick Hillier - Test Bank

Introduction To Operations Research By Frederick Hillier – Test Bank

 

Instant Download – Complete Test Bank With Answers

 

 

Sample Questions Are Posted Below

 

Test Bank for Chapter 5

 

Problem 5-1:

Consider the following problem.

 

Minimize   Z =   x₁ + 2 x₂,

subject to

-x₁ + x₂    £ 15

2 x₁  + x₂    £ 90

x₂     ³ 30

and

x₁ ³ 0, x₂ ³ 0.

 

 

(a) Solve this problem graphically

 

(b) Develop a table giving each of the CPF solutions and the corresponding defining equations, BF solution, and nonbasic variables.

 

 

Solution for Problem 5-1:

 

(a)

 

Thus, the optimal solution is (x₁, x₂) = (15, 30) with Z = 75.

 

(b)

The above graphical solution reveals that the problem has three CPF solutions, as listed in the first column of the table below. The corresponding defining equations are the equations of the constraint boundaries (the solid lines in the above graph) that pass through the respective CPF solutions, as listed in the second column of the table below.

To identify the corresponding BF solutions and nonbasic variables, we need to use the augmented form of the problem.

 

Introducing slack variables, x₃ and x₄, and surplus variable x₅, we have

 

Minimize   Z =   x₁ + 2 x₂,

subject to

– x₁ + 2 x₂  + x₃                    = 15

2 x₁ + x₂               + x₄        = 90

x₂                                       – x₅ = 30

and

x₁ ³ 0, x₂ ³ 0, x₃ ³ 0, x₄ ³ 0, x₅ ³ 0.

 

Also introducing an artificial variable x₆ into the last constraint is optional, since this variable would be needed only to initiate the simplex method, which we are not concerned with doing here. If x₆ is introduced, it would become an additional nonbasic variable (so its value would be 0) in the last two columns of the table below.

For each CPF solution, the corresponding BF solution listed in the third column is obtained by calculating  x₃,  x₄  and x₅ in the augmented form of the problem. The nonbasic variables listed in the last column are the variables whose values are 0 in the BF solution.

 

 

CPF Solution (x1, x2)	Defining Equations	BF solution (x1, x2, x3, x4, x5)	Nonbasic variables
(15, 30)	x2   = 30, – x1 + x2  = 15	(15, 30, 0, 30, 0)	x3, x5
(30, 30)	x2   = 30, 2 x1 + x2  = 90	(30, 30, 15, 0, 0)	x4, x5
(25, 40)	– x1 + x2  = 15, 2 x1 + x2  = 90	(25, 40, 0, 0, 10)	x3, x4

 

 

Problem 5-2:

Consider the following problem.

 

Maximize        Z =  2x₁ + 4x₂ + 3x₃,

subject to

x₁ + 3x₂ + 2x₃ ≤ 30

x₁ +   x₂ +   x₃ ≤ 24

3x₁ + 5x₂ + 3x₃ ≤ 60

and

x₁ ≥ 0,   x₂ ≥ 0,   x₃ ≥ 0.

 

You are given the information that  x₁ > 0,  x₂ = 0, and  x₃ >0 in the optimal solution.

 

Using the given information and the theory of the simplex method, analyze the constraints of the problem in order to identify a system of three constraint boundary equations (defining equations) whose simultaneous solution must be the optimal solution (not augmented). Then solve this system of equations to obtain this solution.

 

Solution for Problem 5-2:

 

Since x₁ > 0 and x₃ > 0,  it follows that x₁ = 0 and x₃ = 0 cannot be part of the three defining equations at the optimal solution. Since x₂ = 0, then x₁ + x₃ £ 24 and 3 x₁ + 3 x₃ £ 60 or, equivalently, x₁ + x₃ £ 24 and x₁ + x₃ £ 20. This implies that the second constraint (x₁ + x₂ + x₃ ≤ 24) must have some slack, so its constraint boundary equation cannot be a defining equation at the optimal solution.

We now have ruled out three of the six constraint boundary equations as being part of the defining equations at the optimal solution. For this three-variable problem, each CPF solution including the optimal solution must have three defining equations. Therefore, the three constraint boundary equations not ruled out must be the defining equations for the optimal solution. These defining equations are

 

x₂ = 0,

x₁ + 3 x₂  + 2 x₃ = 30,

3 x₁ + 5 x₂  + 3 x₃ = 60.

 

Solving this system of equations yields the optimal solution as (x₁, x₂, x₃) = (10, 0, 10).

 

 

Problem 5-3:

You are using the simplex method to solve the following linear programming problem.

 

Maximize        Z = 6x₁ + 5x₂ – x₃ + 4x₄,

subject to

3x₁ + 2x₂ – 3x₃ +   x₄  ≤ 120

3x₁ + 3x₂ +  x₃ + 3x₄  ≤ 180

and

x₁ ≥ 0,  x₂ ≥ 0,  x₃ ≥ 0,   x₄ ≥ 0.

 

You have obtained the following final simplex tableau where x₅ and x₆ are the slack variables for the respective constraints.

 

		Coefficient of:
Basic Variable	Eq.	Z	x1	x2	x3	x4	x5	x6	Right Side
Z	(0)	1	0		0				Z*
x1	(1)	0	1		0
x3	(2)	0	0		1		–

 

Use the fundamental insight presented in Sec. 5.3 of the textbook to identify Z*, , and .

 

Solution for Problem 5-3:

 

The fundamental insight presented in Sec. 5.3 of the textbook states in part that

 

Z* =y*b,

 

= S*b,

 

where y* =  is the vector of coefficients of the slack variables in Eq. (0),

 

S* =

 

is the matrix of coefficients of the slack variables in the remaining equations, and b is the vector of the original right-hand sides. (Since S = B^-1, where B is the basis matrix, and y* = c_BB^-1, where c_B is the vector of the objective function coefficients of the basic variables, this alternative notation may be used throughout this solution.)

 

Therefore,  the final right-hand side is

= S* b = =, and

 

Z^* = y*b =  = 315.

 

 

 

Problem 5-4:

Consider the following problem.

 

Maximize        Z = 2x₁ + 4x₂ + 3x₃,

subject to

x₁ + 3x₂ + 2x₃ = 20

x₁ + 5x₂          ≥ 10

and

x₁ ≥ 0,  x₂ ≥ 0, x₃ ≥ 0.

 

Let   be the artificial variable for the first constraint. Let x₅ and  be the surplus variable and artificial variable, respectively, for the second constraint.

You are now given the information that a portion of the final simplex tableau is as follows:

 

		Coefficient of:
Basic Variable	Eq.	Z	x1	x2	x3		x5		Right Side
Z	(0)	1				M+2	0	M
x1	(1)	0				1	0	0
x5	(2)	0				1	1	-1

 

 

(a) Extend the fundamental insight presented in Sec. 5.3 of the textbook to identify the missing numbers in the final simplex tableau. Show your calculations.

 

(b) Identify the defining equations of the CPF solution corresponding to the optimal solution in the final simplex tableau.

 

Solution for Problem 5-4:

 

(a)

If  and  had been placed in adjacent columns in the initial tableau, their coefficients in Eqs. (1) and (2) would form an identity matrix. Therefore, by the logic of the fundamental insight presented in Sec. 5.3 of the textbook, it is the coefficients of these variables in the final tableau that play the key role. In particular, using the notation of the fundamental insight, y* = [2   0] is the vector of coefficients of the artificial variables in Eq. (0) after subtracting their values [M   M] in the initial tableau (before restoring proper form from Gaussian elimination). Similarly,

 

S* =

 

is the matrix of coefficients of the artificial variables in the remaining equations. (The notation of Sec. 5.2, replacing S* by B^-1 and replacing y* by c_BB^-1, also may be used instead.) Therefore, using the other notation employed in Sec. 5.3 as well, we can obtain the missing elements in the final tableau from the formulas for the fundamental insight as follows.

 

The final constraint columns for (x₁, x₂, x₃) are

 

            S*A = =.

 

The final coefficients in Eq. (0) for (x₁, x₂, x₃) are

 

            y*A – c =  –  = .

The final right-hand side is

S*b = =, and

Z^* = y*b =  = 40.

 

Therefore, after inserting the missing values, the complete final simplex tableau is

 

Basic		Coefficient of:		Right
Variable	Eq.	Z	x1         x2         x3                 x5	Side
Z	(0)	1	0          2         1          M+2    0         M	40
x1	(1)	0	1          3         2          1          0         1	20
x5	(2)	0	0          -2         2          1          1         -1	10

 

 

(b)

The nonbasic variables in the above final tableau yielding the optimal solution are x₂, x₃, ,  and . x₂ and x₃ are the indicating variables for the x₂ ≥ 0 and x₃ ≥ 0 constraints, and  is the indicating variable for the x₁ +3x₂ + 2x₃ = 20 constraint. (  is not an indicating variable except when both x₅ and  are nonbasic variables.) Therefore, the defining equations are

 

x₁ + 3 x₂  + 2 x₃   = 20,

x₂              = 0,

x₃   = 0.