3.4 Linear Programming
What is a Linear Program?
 A linear program is a mathematical model
that indicates the goal and requirements of
an allocation problem.
 It has two or more non-negative variables.
 Its objective is expressed as a mathematical
function. The objective function plots as a
line on a two-dimensional graph.
 There are constraints that affect possible
levels of the variables. In two dimensions
these plot as lines and ordinarily define
areas in which the solution must lie.
2
Properties of LP Models
1) Seek to minimize or maximize
2) Include “constraints” or limitations
3) There must be alternatives available
4) All equations are linear
Example LP Model Formulation:
The Product Mix Problem
Decision: How much to make of > 2 products?
Objective: Maximize profit
Constraints: Limited resources
Example: Flair Furniture Co.
Two products: Chairs and Tables
Decision: How many of each to make this
month?
Objective: Maximize profit
Flair Furniture Co. Data
Tables
Chairs
(per table)
(per chair)
Profit
Contribution
$7
$5
Hours
Available
Carpentry
3 hrs
4 hrs
2400
Painting
2 hrs
1 hr
1000
Other Limitations:
• Make no more than 450 chairs
• Make at least 100 tables
Decision Variables:
T = Num. of tables to make
C = Num. of chairs to make
Objective Function: Maximize Profit
Maximize $7 T + $5 C
Constraints:
• Have 2400 hours of carpentry time
available
3 T + 4 C < 2400 (hours)
• Have 1000 hours of painting time available
2 T + 1 C < 1000 (hours)
More Constraints:
• Make no more than 450 chairs
C < 450
(num. chairs)
• Make at least 100 tables
T > 100
(num. tables)
Nonnegativity:
Cannot make a negative number of chairs or tables
T>0
C>0
Model Summary
Max 7T + 5C
(profit)
Subject to the constraints:
3T + 4C < 2400
(carpentry hrs)
2T + 1C < 1000
(painting hrs)
T
C < 450
(max # chairs)
> 100
(min # tables)
T, C > 0
(nonnegativity)
Graphical Solution
• Graphing an LP model helps provide
insight into LP models and their solutions.
• While this can only be done in two
dimensions, the same properties apply to
all LP models and solutions.
Carpentry
Constraint Line
3T + 4C = 2400
C
Infeasible
> 2400 hrs
600
Intercepts
(T = 0, C = 600)
(T = 800, C = 0)
Feasible
< 2400 hrs
0
0
800 T
C
1000
Painting
Constraint Line
2T + 1C = 1000
600
Intercepts
(T = 0, C = 1000)
(T = 500, C = 0)
0
0
500
800 T
Max Chair Line
C
1000
C = 450
Min Table Line
600
450
T = 100
Feasible
0
Region
0 100
500
800 T
C
Objective
Function Line
7T + 5C = Profit
500
Optimal Point
(T = 320, C = 360)
400
300
200
100
0
0
100
200
300
400
500 T
Finding Most Attractive Corner
 The optimal solution will always correspond to a
corner point of the feasible solution region.
 Because there can be many corners, the most
attractive corner is easiest to find visually.
 That is done by plotting two P lines for arbitrary
profit levels.
16
LP Characteristics
• Feasible Region: The set of points that
satisfies all constraints
• Corner Point Property: An optimal
solution must lie at one or more corner
points
• Optimal Solution: The corner point with
the best objective function value is optimal
• Because graphs can be inaccurate due to
human error and often the numbers are
very large or very small it is best to identify
the corner point through a system of
equations. Simply identify which lines
intersect to form the corner point and solve
them simultaneously.
LP Model: Example
RESOURCE REQUIREMENTS
PRODUCT
Bowl
Mug
Labor
(hr/unit)
1
2
Clay
(lb/unit)
4
3
Revenue
($/unit)
40
50
There are 40 hours of labor and 120 pounds of clay
available each day
Decision variables
xb = number of bowls to produce
xm = number of mugs to produce
• We have 240 acres of land to plant corn
and oats. We make a profit of $40 per
acre of corn and $30 per acre of oats.
We have 320 hours of available labor.
Corn takes 2 hours to plant per acre and
oats require 1 hour. How many acres of
each should we plant to maximize our
profits?
•
•
•
•
•
•
•
Xc = acres of corn
Xo = acres of oats
Maximum Profit = 40Xc + 30Xo
Xc+Xo ≤240
2Xc+Xo ≤320
Xc ≥0
Xo ≥0
LP Formulation: Example
Maximize Z = $40 x1 + 50 x2
Subject to
x1 + 2x2 40 hr (labor constraint)
4x1 + 3x2 120 lb (clay constraint)
x1 , x2 0
Solution is x1 = 24 bowls
Revenue = $1,360
x2 = 8 mugs
Graphical Solution: Example
x2
50 –
40 –
4 x1 + 3 x2 120 lb
30 –
20 –
10 –
0–
x1 + 2 x2 40 hr
|
10
|
20
|
30
|
40
|
50
|
60
x1
Extreme Corner Points
x1 = 0 bowls
x2 = 20 mugs
Z = $1,000
x2
40 –
30 –
20 – A
10 –
0–
x1 = 224 bowls
x2 = 8 mugs
x1 = 30 bowls
Z = $1,360
x2 = 0 mugs
Z = $1,200
B
|
10
|
20
| C|
30 40
x1
Mixture
 A rancher is mixing two types of food, Brand X and Brand Y, for






his cattle. If each serving is required to have 60 grams of protein
and 30 grams of fat, where Brand X has 15 grams of protein and 10
grams of fat and costs 80 cents per unit, and Brand Y contains 20
grams of protein and 5 grams of fat, and costs 50 cents per unit,
how much of each type should be used to minimize the cost to the
rancher?
Let X = # of units of Brand X
Let Y = # of units of Brand Y
Minimize Cost = .80X + .50Y
15X + 20Y ≥60
10X + 5Y ≥ 30
Non-negative Constraints