Post on 15-Jan-2016
BA 452 Lesson A.2 Solving Linear Programs 11
Readings
Readings
Chapter 2An Introduction to Linear Programming
BA 452 Lesson A.2 Solving Linear Programs 22
Overview
Overview
BA 452 Lesson A.2 Solving Linear Programs 33
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.
BA 452 Lesson A.2 Solving Linear Programs 44
Graphical Solutions
Graphical Solutions
BA 452 Lesson A.2 Solving Linear Programs 55
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
BA 452 Lesson A.2 Solving Linear Programs 66
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
BA 452 Lesson A.2 Solving Linear Programs 77
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
BA 452 Lesson A.2 Solving Linear Programs 88
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
BA 452 Lesson A.2 Solving Linear Programs 99
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
BA 452 Lesson A.2 Solving Linear Programs 1010
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
BA 452 Lesson A.2 Solving Linear Programs 1111
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
BA 452 Lesson A.2 Solving Linear Programs 1212
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
BA 452 Lesson A.2 Solving Linear Programs 1313
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
BA 452 Lesson A.2 Solving Linear Programs 1414
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
BA 452 Lesson A.2 Solving Linear Programs 1515
Slack and Surplus Variables
Slack and Surplus Variables
BA 452 Lesson A.2 Solving Linear Programs 1616
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.
BA 452 Lesson A.2 Solving Linear Programs 1717
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
Slack and Surplus Variables
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
BA 452 Lesson A.2 Solving Linear Programs 1818
Extreme Points
Extreme Points
BA 452 Lesson A.2 Solving Linear Programs 1919
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.
BA 452 Lesson A.2 Solving Linear Programs 2020
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
BA 452 Lesson A.2 Solving Linear Programs 2121
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
Extreme point has objective value 5x1 + 7x2 = 011
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 has objective value 5x1 + 7x2 = 46
Extreme Point Solutions
Example 1:
Max 5x1 + 7x2
s.t. x1 < 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
BA 452 Lesson A.2 Solving Linear Programs 2222
Computer Solutions
Computer Solutions
BA 452 Lesson A.2 Solving Linear Programs 2323
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.
BA 452 Lesson A.2 Solving Linear Programs 2424
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
BA 452 Lesson A.2 Solving Linear Programs 2525
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
BA 452 Lesson A.2 Solving Linear Programs 2626
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
BA 452 Lesson A.2 Solving Linear Programs 2727
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
BA 452 Lesson A.2 Solving Linear Programs 2828
Resource Allocation with Sales Maximums
Resource Allocation with Sales Maximums
BA 452 Lesson A.2 Solving Linear Programs 2929
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
Resource Allocation with Sales Maximums
BA 452 Lesson A.2 Solving Linear Programs 3030
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.
Resource Allocation with Sales Maximums
BA 452 Lesson A.2 Solving Linear Programs 3131
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?
Resource Allocation with Sales Maximums
BA 452 Lesson A.2 Solving Linear Programs 3232
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
Resource Allocation with Sales Maximums
BA 452 Lesson A.2 Solving Linear Programs 3333
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)
x2 2000 (frame 2 sales constraint)
x3 500 (frame 3 sales constraint)
x4 1000 (frame 4 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
Resource Allocation with Sales Maximums
BA 452 Lesson A.2 Solving Linear Programs 3434
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
Resource Allocation with Sales Maximums
BA 452 Lesson A.2 Solving Linear Programs 3535
Resource Allocation with Sales Maximums
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
BA 452 Lesson A.2 Solving Linear Programs 3636
Maximum weekly profit of $10,000.
Xi is optimal weekly sales of frames of type i (i = 1, 2, 3, 4).
Resource Allocation with Sales Maximums
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
BA 452 Lesson A.2 Solving Linear Programs 3737
End of Lesson A.2
BA 452 Quantitative Analysis