# Chapter 19 - Linear Programming ```BUAD306
Chapter 19 – Linear Programming
Optimization
QUESTION: Have you ever been limited to
what you can get done because you don’t
have enough ________?



Examples of how we optimize in daily
decisions?
Why do we strive for optimization?
Factors that limit optimization?
What is Linear Programming
A model consisting of linear
relationships that represent a firm’s
objective and resource constraints
 Problems are typically referred to as
constrained optimization problems

LP Objectives
Maximize profits
 Maximize outputs
 Minimize costs
 Determine combinations of outputs to
meet/surpass goals

Product Planning - Landscaping
 Portfolio Selection - Investments
 Route Planning / Pricing - Airlines

LP Assumptions
Linearity - The impact of decision
variables is linear in constraints and
objective function
 Divisibility – non-integer values of
decision variables are acceptable
 Certainty – values of parameters are
known and constant
 Non-Negativity – negative values of
decision variables are unacceptable

LP Concepts
Objective - The goal of the LP model
 Decision Variables - Amounts of either
inputs or outputs
 Constraints - Limitations of available
alternatives: Equal to, Greater than,
Less than
 Feasible Solution Space - The set of
all feasible combinations of decision
variables as defined by the constraints

Constraints

Restrictions on the company’s resources






Time
Labor
Energy
Materials
Money
Restriction guidelines


recipe for making food products
engineering specifications
LP Steps
Step 1: Set up LP model
Step 2: Plot the constraints
Step 3: Identify the feasible solution
space
Step 4: Plot the objective function
Step 5: Determine the optimal solution
LP Model Example
X1 = Quantity of Product 1 to produce
X2 = Quantity of Product 2 to produce
Maximize: 5x1 + 8x2
(profit)
Decision Variables
Objective Function
Subject to:
Labor:
Material:
Product 1:
x2 &gt;= 0
2x1 + 4x2 &lt;= 250 hours
7x1 + 6X2 &lt;= 100 pounds
X1
&gt;=
10 units
Non-negativity constraint
Constraints
LP Example
A pottery company employs artisans
to produce clay bowls and mugs. The
two primary resources used by the
company are special pottery clay and
skilled labor. Given these limited
resources, the company would like to
determine how many bowls and mugs
to produce each day to maximize
profit.
The two products have the following resource
requirements for production and selling price per
item produced:
Labor
Clay Revenue
Product (h/unit) (lb/unit) (\$/unit)
Bowl
1
3
25
Mug
2
2
20
There are 40 hours of labor and 60 pounds of clay
available each day.
Formulate this problem as an LP model.
Plot Constraints
Step 1: For each constraint, set x1 = 0, get value for x2
Step 2: For each constraint, set x2 = 0, get value for x1
Step 3: Plot these as coordinates and then draw
constraint lines
Labor Constraint:
x1 + 2x2 &lt;= 40 hrs
Clay Constraint:
3x1 + 2x2 &lt;= 60 lbs.
Plot Constraints
50
45
40
35
Clay
x2
30
25
20
15
10
5
Labor
0
0
10
20
30
x1
40
50
Identify Feasible Solution
Space
This is the area that satisfies all of the
LP model’s constraints simultaneously.
Includes all of the points on the borders
as well.
50
45
40
35
x2
30
25
20
15
10
5
0
0
10
20
30
40
50
Plot the Objective Function

In order to locate the point in the feasible
solution space that will maximize revenue,
we now need to plot the Objective Function:
• Maximize Z = \$25x1 + 20x2

We will select and arbitrary level of revenue,
such as \$300, in order to plot the objective
function.
• 300 = \$25x1 + 20x2
• if x1 = 0, then x2 = 12 (0, 15)
• if x2 = 0, then x1 = 12 (12, 0)
Identify Feasible Solution
Observation: Every point on this
Space
Line in the feasible solution area
Will result in a revenue of \$300.
(Every combination of X1 and X2
on this line in this space will yield
an \$300 revenue total.)
50
45
40
CAN WE DO BETTER
THAN \$300 GIVEN
THE CONSTRAINTS???
35
x2
30
25
20
15
10
5
0
0
10
20
30
40
50
Identify Feasible Solution
Space
Shift the objective function line to locate
the optimal solution point.
50
45
40
35
x2
30
25
20
15
10
5
0
0
10
20
30
40
50
Identify Feasible Solution
Space
Shift the objective function line to locate
the optimal solution point.
50
45
40
35
x2
30
25
20
15
10
5
0
0
10
20
30
40
50
Identify Feasible Solution
Space
Shift the objective function line to locate
the optimal solution point.
50
45
40
35
x2
30
25
20
15
10
5
0
0
10
20
30
40
50
Identify Feasible Solution
Space
Shift the objective function line to locate
the optimal solution point.
50
45
40
This is where the objective function line is
TANGENT to the feasible solution space!
35
x2
30
25
20
15
10
5
0
0
10
20
30
40
50
What if???
What if objective function was:
100x1 + 20x2?? How does that change
the look of the objective function line?
50
What if STEEP
obj function
slope?
45
40
35
x2
30
What if FLAT
obj function
slope?
25
20
15
10
5
0
0
10
20
30
40
50
Solving for the Optimal
Solution
Given our constraint equations:
• x1 + 2x2 = 40 (Labor)
• 3x1 + 2x2 = 60 (Clay)
We set them equal to each other and solve for the
optimal quantities of X1 and X2…
Solving for the Optimal
Solution
Thus, the optimal solution is to produce:
And to maximize the objective function:
Maximize Z: 25x1 + 20x2
Z=
50
45
(0,0) =
40
35
(0,20) =
x2
30
(20,0) =
25
20
(10,15) =
15
10
5
0
0
10
20
30
x1
40
50
Example #1
Delstate Jewelers manufacturers two types of semi-precious gems for
sale to its wholesale customers, Gem 1 and Gem 2. Delstate realizes
revenue of \$4.00 for each Gem 1 sold and revenue of \$3 per gem for
each Gem 2 sold. To produce the gems, the company must cut and
polish each gem. The company is limited to a certain number of
minutes for each machine that is used in the cutting and polishing
process as shown in the table below. Given this information,
determine the optimal quantities of Gem 1 and Gem 2 that maximize
revenue.
Product
Cutting
Polishing
Gem 1
14
7
Gem 2
6
12
Total Minutes
Available
84
84
Example #2
A leather shop makes custom, hand-tooled briefcases and luggage.
The shop makes a \$400 profit from each briefcase and \$200 profit from
each piece of luggage. The shop has a contract to provide a local store
with exactly 30 items each month. A tannery supplies the shop with at
least 80 square yards of leather per month. The shop must at least use
this amount, but can use more if needed. Each briefcase requires 2
square yards of leather, each piece of luggage requires 8 square yards
of leather. From past performance, the shop owner knows that they
cannot make more than 20 briefcases per month.
Determine the number of briefcases and luggage to produce each
month in order to maximize revenue. Formulate an LP model and solve
graphically. Determine the optimal quantities of briefcases and luggage
and the maximized revenue amount. Also determine if there is any
slack or surplus.
In Class Example
max 5x1 + 6x2
st
A 2x1+3x2&lt;=60
B 4x1+x2&lt;=80
C x1+x2&lt;=50
N/N
x1, x2 &gt;=0
Will email
to you!!
NEW In Class Example
max 2x1 + 3x2
st
A x1+3x2&lt;=60
B 2x1+2x2&lt;=80
C x1+x2&lt;=35
D 3x1+2x2&lt;=120
N/N
x1, x2 &gt;=0
Will email
to you!!
Binding
 Slack/Surplus
 Redundant Constraints
 Multiple Optimal Solutions

Binding Constraint

When a constraint forms the optimal
corner point of the feasible solution
space, it is BINDING.
Moo!
Slack and Surplus
A slack variable is a variable
to a &lt;= constraint to make it an
equality.
 A surplus variable is a variable
representing an excess above a
resource requirement that is
subtracted from a &gt;= constraint to
make it an equality.

Redundant Constraint
A constraint that does not form a
unique boundary of the feasible
solution space.
 Its removal would not alter the
feasible solution space.

Multiple Optimal Solutions
In some instances, you may get an
objective function which is parallel
to a constraint line
 This ends up NOT being tangent to
the feasible solution space!

Sensitivity Analysis

Assessing the impact of
potential changes to the
numerical values of an LP
model
• Objective function
coefficients
• Right-hand values of
constraints
• Constraint coefficients
Max 25x1 + 20x2
S.T.
L: 1x1 + 2x2 &lt;= 40
C: 3x1 + 2x2 &lt;= 60
N/N: x1, x2 &gt;=0
LINDO
Computer-based modeling system for
solving linear programming problems
 All PCs in the Purnell computer lab
have LINDO installed

LINDO Steps


Designate
functions within
LINDO
 Max/Min
 Constraints
Solve with
Sensitivity Analysis
Notice that the results
we obtained in our graphical
approach.
Objective Function Ranges
With this report, we can determine the range in which the objective coefficients can
exist before the optimal solution changes:
Revenue coefficient for Bowls can range between 10 and 30 without impacting
solution
Revenue coefficient for Mugs can range between 16.67 and 50 without impacting
solution
RHS Ranges
With this report, we can determine the
range in which the RHS of the
constraints can change while yielding
the same savings/cost indicated by the
RHS for Labor constraint can range
between 20 and 60
RHS for Clay constraint can range
between 40 and 120

The amount by which the value of the
objective function would change if
there is a one unit change in the RHS
of that constraint
If constraint is binding, shadow price =
the amount objective function will
change for each one unit change in the
RHS
 If constraint has slack, shadow price = 0
and increasing or decreasing RHS has
no effect - would only increase
slack/surplus!

Reminder for Range
Analysis:
When you change the objective
function values and test, you are
verifying whether the optimal
combination changes…
 When you change the constraint RHS
of constraints, the optimal combination
WILL change– what you are verifying is
whether the shadow/dual price is in
effect and if, thus, the change is worth
it!

Review of
LP Handout #3
Homework #5 – PIES!
Max Z:
1.5x1 + 1.2X2
Subject to:
Flour:
Sugar:
Time:
N/N:
3x1 + 3x2 &lt;= 2100
1.5x1 + 2x2 &lt;= 1200
6x1 + 3x2 &lt;= 3600
X1, X2 &gt;= 0
Homework #5
```