ba 452 lesson a.2 solving linear programs 1 1readingsreadings chapter 2 an introduction to linear...

BA 452 Lesson A.2 Solving Linear Programs 11

Readings

Readings

Chapter 2An Introduction to Linear Programming


Overview

Overview


Overview

Graphical Solutions to linear programs arise from graphing the feasible solutions for each constraint and a constant-value line for the objective function to identify the binding constraints to solve.

Slack and Surplus Variables measure the deviation of inequality constraints from binding equalities. Thus they measure how much non-binding constraints can change before they affect an optimum.

Extreme Points are the corners (or vertices) of the feasible region of a linear program. An optimal solution can be found at extreme points. Thus finding extreme points is an alternative to graphing solutions.

Computer Solutions are available to linear programs with many variables and constraints. Computed values include the objective function, decision variables, and slack and surplus variables.

Resource Allocation Problems with Sales Maximums constrain the maximum output D that can be sold at a given price P. The demand curve for output is assumed to be of a special form.


Graphical Solutions

Graphical Solutions


Overview

Graphical Solutions to linear programs arise from graphing the feasible solutions for each constraint and a constant-value line for the objective function to identify which constraints bind (hold with equality) at the optimal solution. Then, solve those constraints to compute the optimal solution.

Graphical Solutions


Graph the first constraint of Example 1 from Lesson I.1, plus non-negativity constraints.

x2

x1

x1 = 6 is the binding edge of the first constraint, where it holds with equality.

The point (6, 0) is on the end of the binding edge of the first constraint plus the non-negativity of x2.

8

7

6

5

4

3

2

1

1 2 3 4 5 6 7 8 9 10

Shaded regioncontains all

feasible pointsfor this constraint

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

Graphical Solutions


2x1 + 3x2 = 19 is the binding edge of the second constraint.

x2

x1

8

7

6

5

4

3

2

1

Shadedregion containsall feasible pointsfor this constraint

Graph the second constraint of Example 1, plus non-negativity constraints.

The point (0, 6 1/3) is on the end of the binding edge of the second constraint plus the non-negativity of x1.

The point (9 1/2, 0) is on the end of the binding edge of the second constraint plus the non-negativity of x2.

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

Graphical Solutions

1 2 3 4 5 6 7 8 9 10


x2

x1

x1 + x2 = 8 is the binding edge of the third constraint

8

7

6

5

4

3

2

1

Shadedregion containsall feasible pointsfor this constraint

Graph the third constraint of Example 1,

plus non-negativity constraints.The point (0, 8) is on the end of the binding edge of the third constraint plus the non- negativity of x1

The point (8, 0) is on the end of the binding edge of the third constraint plus the non-negativity of x2

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

Graphical Solutions

1 2 3 4 5 6 7 8 9 10


x1

x2

8

7

6

5

4

3

2

1

2x1 + 3x2 = 19

x1 + x2 = 8

x1 = 6

Feasible region

Intersect all constraint graphs to define the

feasible region.

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

Graphical Solutions

1 2 3 4 5 6 7 8 9 10


Graph a line with a constant objective-function value. For example, 35 dollars of profit.

x1

(7, 0)

(0, 5)objective function value5x1 + 7x2 = 35objective function value5x1 + 7x2 = 35

8

7

6

5

4

3

2

1

x2

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

Graphical Solutions

1 2 3 4 5 6 7 8 9 10


x1

5x1 + 7x2 = 355x1 + 7x2 = 35

8

7

6

5

4

3

2

1

5x1 + 7x2 = 425x1 + 7x2 = 42

5x1 + 7x2 = 395x1 + 7x2 = 39

Graph alternative constant-value lines.

For example, 35 dollars, 39 dollars, or

42 dollars of profit. x2

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

Graphical Solutions

1 2 3 4 5 6 7 8 9 10


x1

x2

Maximum constant-value line 5x1 + 7x2 = 46Maximum constant-value line 5x1 + 7x2 = 46

Second and third constraints bind at the optimal solutionSecond and third constraints bind at the optimal solution

8

7

6

5

4

3

2

1

Graph the maximum constant-value line,

graph the optimal solution, then determine

the binding constraints.

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

Graphical Solutions

1 2 3 4 5 6 7 8 9 10


1 12 3

x1

x2

819 =

x1 = det / det = (8x3-1x19)/(1x3-1x2) = 5

8 119 3

1 12 3

x2 = det / det = (1x19-8x2)/(1x3-1x2) = 3

1 82 19

1 12 3

1x1 + 1x2 = 8 2x1 + 3x2 = 19

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

The optimal solution (x1, x2 ) is where the second and third constraints bind (hold with equality): x1 + x2 = 8 and 2x1 + 3x2 = 19.

Solve those equalities using linear algebra, matrices, determinates (det), and Cramer’s rule:

Graphical Solutions


Summary of a graphical solution procedure Graph the feasible solutions for each constraint. Determine the feasible region that simultaneously

satisfies all the constraints. Draw a constant-value line for the objective function. Move parallel value lines toward larger objective function

values without leaving the feasible region. Any feasible solution on the objective function line with

the largest value is an optimal solution (or optimum). That solution can be found by solving the binding

(equality) constraints.

Graphical Solutions


Slack and Surplus Variables




Overview

Slack and Surplus Variables measure the deviation of inequality constraints from binding equalities. Thus they measure how much non-binding constraints can change before they affect an optimum.


Compute slack variables at the optimum to Example 1.

x1

x2

8

7

6

5

4

3

2

1

Binding secondconstraint:

2x1 + 3x2 = 19

Binding thirdconstraint:x1 + x2 = 8

Binding edge of first constraint:

x1 = 6

Optimalsolution

(x1 = 5, x2 = 3)

Optimalsolution

(x1 = 5, x2 = 3)

s1 = 1

s2 = 0

s3 = 0


Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0

1 2 3 4 5 6 7 8 9 10


Extreme Points

Extreme Points


Extreme Points

Overview

Extreme Points are the corners (or vertices) of the feasible region of a linear program. An optimal solution can be found at extreme points. Thus finding extreme points is an alternative to graphing solutions.


x1

Feasible region

11 22

33

44

55

x2

8

7

6

5

4

3

2

1

1 2 3 4 5 6 7 8 9 10

(0, 6 1/3), where 2x1 + 3x2 = 19 and x1 = 0

(5, 3), where 2x1 + 3x2 = 19 and x1 + x2 = 8

(0, 0)

(6, 2), where x1 + x2 = 8 and x1 = 6

(6, 0), where x2 = 0 and x1 = 6

Compute the extreme points in Example 1 by solving pairs of binding constraints.

Extreme Point Solutions

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0


Evaluate the objective function at each of the extreme points in Example 1.

Point (5,3) thus maximizes the objective function, with value 46. (Likewise, point (0,0) minimizes.)

Extreme point has objective value 5x1 + 7x2 = 44 1/3

Extreme point has objective value 5x1 + 7x2 = 42


22

33

(0, 0)

(6, 0)Extreme point has objective value 5x1 + 7x2 = 30

(6, 2)

44

55

(5, 3)

(0, 6 1/3)


Extreme Point Solutions

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0


Computer Solutions

Computer Solutions


Computer Solutions

Overview

Computer Solutions are available to linear programs with many variables and constraints. Computed values include the objective function, decision variables, and slack and surplus variables.


Computer Solutions

LP problems involving many variables and constraints are now routinely solved with computer packages.

Linear programming solvers are now part of many spreadsheet packages, such as Microsoft Excel.

The Management Scientist program has a convenient LP module. In remainder of this lesson we will interpret the following output:

• objective function value• values of the decision variables• slack and surplus

In a forthcoming lesson, we will interpret the output the shows how an optimal solution is affected by a change in:• a coefficient of the objective function• the right-hand side value of a constraint


To use The Management Scientist 6.0 program, select New under the File menu.

Computer Solutions

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0


Decision Variable Names can be changed, if desired. Enter objective function coefficients in the Objective Function portion

of the input screen. In the Constraints section, enter constraint coefficients, constraint

relationship (<, =,>), where < abbreviates <, and > abbreviates >. And enter the constraint right-hand-side constants.

Do not enter non-negativity constraints. They are assumed.

Computer Solutions

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0


Maximized objective function value = 46

Optimal x1 = 5 and x2 = 3

Under the Solution menu, select Solve for the Optimal Solution.

Computer Solutions

Example 1:

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 > 0 and x2 > 0


Resource Allocation with Sales Maximums



Overview

Resource Allocation Problems with Sales Maximums constrain the maximum output D that can be sold at a given price P. The demand curve for output is assumed to be of the following special form: At price P, demand quantity is D At any price less than P, demand quantity does not increase, but remains D At any price greater than P, demand quantity drops from D to 0



Question: The Monet Company produces four type of picture frames: 1, 2, 3, 4. The four types differ in size, shape and materials used. Each type requires a certain amount of skilled labor,

metal, and glass:• Each frame of type 1 uses 2 hours of labor, 4 ounces

of metal, 6 ounces of glass, and sells for $28.50.• Each frame of type 2 uses 1 hour of labor, 2 ounces

of metal, 2 ounces of glass, and sells for $12.50.• Each frame of type 3 uses 3 hours of labor, 1 ounces

of metal, 1 ounces of glass, and sells for $29.25.• Each frame of type 4 uses 2 hours of labor, 2 ounces

of metal, 2 ounces of glass, and sells for $21.50.



Each week, Monet can buy up to 4000 hours of skilled labor and 10,000 ounces of glass.

The unit costs are $8.00 per labor hour, $0.50 per ounce of metal, and $0.75 per ounce of glass.

Market constraints are such that it is impossible to sell more than:• 1000 type 1 frames, • 2000 type 2 frames, • 500 type 3 frames, • 1000 type 4 frames.

How can Monet maximize its weekly profit?



Answer: First, compute unit profit for each type of frame: Frame 1: Unit profit = Sales price – input costs = 28.5 – 2x8 – 4x.5 – 6x.75 = 28.5 – 16 – 2 – 4.5 = $6.00 Frame 2: Unit profit = 12.5 – 1x8 – 2x.5 – 2x.75 = 12.5 – 8 – 1 – 1.5 = $2.00 Frame 3: Unit profit = 29.25 – 3x8 – 1x.5 – 1x.75 = 29.25 – 24 – .5 – .75 = $4.00 Frame 4: Unit profit = 21.5 – 2x8 – 2x.5 – 2x.75 = 21.5 – 16 – 1 – 1.5 = $3.00



Let xi be weekly sales of frames of type i (i = 1, 2, 3, 4).

Maximize 6x1 + 2x2 + 4x3 + 3x4 (profit objective)

subject to 2x1 + x2 + 3x3 + 2x4 4000 (labor constraint)

6x1 + 2x2 + x3 + 2x4 10,000 (glass constraint)

x1 1000 (frame 1 sales constraint)




x1, x2, x3, x4 0 (nonnegativity constraints) 6x1 = Profit from frames of type 1

2x1 = Labor used in frames of type 1

4x1 = Glass used in frames of type 1



Maximize 6x1 + 2x2 + 4x3 + 3x4

subject to 2x1 + x2 + 3x3 + 2x4 4000

6x1 + 2x2 + x3 + 2x4 10,000

x1 1000

x2 2000

x3 500

x4 1000

x1, x2, x3, x4 0





subject to 2x1 + x2 + 3x3 + 2x4 4000

6x1 + 2x2 + x3 + 2x4 10,000

x1 1000

x2 2000

x3 500

x4 1000

x1, x2, x3, x4 0


Maximum weekly profit of $10,000.

Xi is optimal weekly sales of frames of type i (i = 1, 2, 3, 4).



subject to 2x1 + x2 + 3x3 + 2x4 4000

6x1 + 2x2 + x3 + 2x4 10,000

x1 1000

x2 2000

x3 500

x4 1000

x1, x2, x3, x4 0


End of Lesson A.2

BA 452 Quantitative Analysis

ba 452 lesson a.2 solving linear programs 1 1readingsreadings chapter 2 an introduction to linear...

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