Readings
Readings
Chapter 2
An Introduction to Linear Programming
BA 452 Lesson A.2 Solving Linear Programs
1
Overview
Overview
BA 452 Lesson A.2 Solving Linear Programs
2
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
3
Graphical Solutions
Graphical Solutions
BA 452 Lesson A.2 Solving Linear Programs
4
Graphical Solutions
Overview
Graphical Solutions to linear programs arise from graphing
the feasible solutions for each constraint and a constantvalue 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.
BA 452 Lesson A.2 Solving Linear Programs
5
Graphical Solutions
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Graph the first constraint of Example 1 from
Lesson I.1, plus non-negativity constraints.
x2
8
7
6
5
4
x1 = 6 is the binding edge of
the first constraint, where it
holds with equality.
Shaded region
contains all
feasible points
for this constraint
The point (6, 0) is on the end
of the binding edge of the first
constraint plus the nonnegativity of x2.
3
2
1
1
2
3
4
5
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
x1
6
Graphical Solutions
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Graph the second constraint of Example
1, plus non-negativity constraints.
The point (0, 6 1/3) is on the
x2
end of the binding edge of the
second constraint plus the
non-negativity of x1.
8
7
6
2x1 + 3x2 = 19 is the binding
edge of the second constraint.
5
4
3
2
1
The point (9 1/2, 0) is on the end of
the binding edge of the second
constraint plus the
non-negativity of x2.
Shaded
region contains
all feasible points
for this constraint
1
2
3
4
5
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
x1
7
Graphical Solutions
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Graph the third constraint of Example 1,
plus non-negativity constraints.
The point (0, 8) is on the end
x2
of the binding edge of the
third constraint plus the nonnegativity of x1
8
7
6
x1 + x2 = 8 is the binding edge
of the third constraint
5
4
3
2
1
Shaded
region contains
all feasible points
for this constraint
1
2
3
4
5
The point (8, 0) is on the end of the
binding edge of the
third constraint plus the
non-negativity of x2
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
x1
8
Graphical Solutions
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Intersect all constraint graphs to define the
feasible region.
x2
x1 + x2 = 8
8
7
x1 = 6
6
5
4
3
2x1 + 3x2 =
19
Feasible
region
2
1
1
2
3
4
5
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
x1
9
Graphical Solutions
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Graph a line with a constant objective-function
value. For example, 35 dollars of profit.
x2
8
7
(0, 5)
6
objective function value
5x1 + 7x2 = 35
5
4
3
2
(7, 0)
1
1
2
3
4
5
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
x1
10
Graphical Solutions
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Graph alternative constant-value lines.
For example, 35 dollars, 39 dollars, or
x2
42 dollars of profit.
8
7
5x1 + 7x2 = 35
6
5x1 + 7x2 = 39
5
4
5x1 + 7x2 = 42
3
2
1
1
2
3
4
5
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
x1
11
Graphical Solutions
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Graph the maximum constant-value line,
graph the optimal solution, then determine
x2 the binding constraints.
8
Maximum constant-value
line 5x1 + 7x2 = 46
Second and third constraints
bind at the optimal solution
7
6
5
4
3
2
1
1
2
3
4
5
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
x1
12
Graphical Solutions


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:
1x1 + 1x2 = 8
2x1 + 3x2 = 19
1
2
1
3
x1
x2
=
8
19
x1 = det
8 1
19 3
/ det
1
2
1
3
= (8x3-1x19)/(1x3-1x2) = 5
x2 = det
1 8
2 19
/ det
1
2
1
3
= (1x19-8x2)/(1x3-1x2) = 3
BA 452 Lesson A.2 Solving Linear Programs
13
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.
BA 452 Lesson A.2 Solving Linear Programs
14
Slack and Surplus Variables
Slack and Surplus Variables
BA 452 Lesson A.2 Solving Linear Programs
15
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
16
Slack and Surplus Variables

Compute slack variables at the optimum to
Example 1.
x2
Binding third
constraint:
x1 + x2 = 8
8
7
Binding edge of
first constraint:
x1 = 6
s3 = 0
6
s1 = 1
5
Binding second
constraint:
2x1 + 3x2 = 19
4
3
2
1
Optimal
solution
(x1 = 5, x2 = 3)
1
2
3
4
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
s2 = 0
5
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
x1
17
Extreme Points
Extreme Points
BA 452 Lesson A.2 Solving Linear Programs
18
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
19
Extreme Point Solutions

x2
Compute the extreme points in Example 1
by solving pairs of binding constraints.
8
7
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
5 (0, 6 1/3), where 2x1 + 3x2 = 19 and x1 = 0
6
5
4
3
Feasible
region
2
1
4 (5, 3), where 2x1 + 3x2 = 19 and x1 + x2 =
8
3 (6, 2), where x1 + x2 = 8 and x1 = 6
2 (6, 0), where x2 = 0 and x1 = 6
1 (0, 0)
1
2
x1
3
4
5
6
7
8
9
10
BA 452 Lesson A.2 Solving Linear Programs
20
Extreme Point Solutions

Evaluate the objective function at each of the
extreme points in Example 1.
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
1
Extreme point (0, 0) has objective value 5x1 + 7x2 = 0
2
Extreme point (6, 0) has objective value 5x1 + 7x2 = 30
3 Extreme point (6, 2) has objective value 5x1 + 7x2 = 42
4
Extreme point (5, 3) has objective value 5x1 + 7x2 = 46
5 Extreme point (0, 6 1/3) has objective value 5x1 + 7x2 = 44 1/3

Point (5,3) thus maximizes the objective function, with
value 46. (Likewise, point (0,0) minimizes.)
BA 452 Lesson A.2 Solving Linear Programs
21
Computer Solutions
Computer Solutions
BA 452 Lesson A.2 Solving Linear Programs
22
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
23
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
24
Computer Solutions

To use The Management Scientist 6.0 program,
select New under the File menu.
BA 452 Lesson A.2 Solving Linear Programs
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
25
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.
BA 452 Lesson A.2 Solving Linear Programs
26
Computer Solutions
Example 1:
Max
5x1 + 7x2
s.t.
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0

Under the Solution menu, select
Solve for the Optimal Solution.
Maximized objective function value = 46
Optimal x1 = 5 and x2 = 3
BA 452 Lesson A.2 Solving Linear Programs
27
Resource Allocation with Sales Maximums
Resource Allocation with Sales
Maximums
BA 452 Lesson A.2 Solving Linear Programs
28
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
BA 452 Lesson A.2 Solving Linear Programs
29
Resource Allocation with Sales Maximums
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.
BA 452 Lesson A.2 Solving Linear Programs
30
Resource Allocation with Sales Maximums



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?
BA 452 Lesson A.2 Solving Linear Programs
31
Resource Allocation with Sales Maximums
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
BA 452 Lesson A.2 Solving Linear Programs
32
Resource Allocation with Sales Maximums

Let xi be weekly sales of frames of type i (i = 1, 2, 3, 4).
Maximize
subject to



6x1 + 2x2 + 4x3 + 3x4 (profit objective)
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
BA 452 Lesson A.2 Solving Linear Programs
33
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
x 1, x 2, x 3, x 4  0
BA 452 Lesson A.2 Solving Linear Programs
34
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
x 1, x 2, x 3, x 4  0
BA 452 Lesson A.2 Solving Linear Programs
35
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
x 1, x 2, x 3, x 4  0


Maximum weekly profit of
$10,000.
Xi is optimal weekly sales of
frames of type i (i = 1, 2, 3,
4).
BA 452 Lesson A.2 Solving Linear Programs
36
BA 452
Quantitative Analysis
End of Lesson A.2
BA 452 Lesson A.2 Solving Linear Programs
37