Linear Programming:
Model Formulation
and Graphical
Solution
A Maximization Model Example
The Beaver Creek Pottery Company
Given these limited resources, the company desires to know how
many bowls and mugs to produce each day in order to
maximize profit.
There are 40 hours of labor and 120 kg of clay available each day
for production.
2
Beaver Creek Pottery
Resource Requirements
Product
3
Labor
(hr/unit)
Clay
(kg/unit)
Profit
($/unit)
Bowl
1
4
40
Mug
2
3
50
Modeling
Decision Variables
x1 : number of bowls to produce
x2 : number of mugs to produce
Objective Function
Total Profit = 40 x1 + 50 x2
40 x1 = profit from bowls
50 x2 = profit from mugs
Maximize Z = 40 x1 + 50 x2
4
Model Constraints
Labor Constraint
Total Labor used in production = 1 x1 + 2 x2
1 x1 + 2 x2 ≤ 40 hr
Clay Constraint
Total Clay used in production = 4 x1 + 3 x2
4 x1 + 3 x2 ≤ 120 lb
Non-negativity Constraint
x1 ≥ 0, x2 ≥ 0
5
Linear Programming Model
Maximize Z = 40 x1 + 50 x2
Subject to:
1 x1 + 2 x2 ≤ 40
4 x1 + 3 x2 ≤ 120
x1, x2 ≥ 0
6
Feasible / Infeasible
7
If x1 = 10, x2 = 20
Z = 40 . 10 + 50 . 20 = 1400
If x1 = 20, x2 = 20
Z = 40 . 20 + 50 . 20 = 1800
If x1 = 5, x2 = 10
Z = 40 . 5 + 50 . 10 = 700
If x1 = 2, x2 = 2
Z = 40 . 2 + 50 . 2 = 180
Graphical Solution of
Linear Programming Models
60
50
40
30
Coordinates for
graphical
analysis
8
20
10
0
10
20
30
40
50
60
x2
Graph of the labor constraint line
60
50
40
30
x1 + 2 x2 = 40
20
10
0
9
10
20
30
40
50
60
x1
The labor constraint area
x2
60
M
50
40
L
30
20
x1 + 2 x2 ≤ 40
K
10
0
10
10
20
30
40
50
60
x1
x2
Graph of the labor constraint line
60
50
40
30
4 x1 + 3 x2 = 120
20
10
0
11
10
20
30
40
50
60
x1
The clay constraint area
x2
60
M
50
40
L
30
4 x1 + 3 x2 ≤ 120
20
K
10
0
12
10
20
30
40
50
60
x1
x2
Graph of both model constraints
60
50
40
4 x1 + 3 x2 = 120
30
20
10
x1 + 2 x2 = 40
0
13
10
20
30
40
50
60
x1
The
feasible solution area constraints
x
2
60
50
40
T: Infeasible
4 x1 + 3 x2 = 120
T
30
20
10
0
14
S: Infeasible
R: Feasible
S
R
10
x1 + 2 x2 = 40
20
30
40
50
60
x1
x2
Objective function line for Z = $ 800
60
50
40
30
800 = 40 x1 + 50 x2
20
10
0
15
10
20
30
40
50
60
x1
Alternative
objective function lines for
x2
profits, Z, of $ 800, $ 1200, $ 1600
40
800 = 40 x1 + 50 x2
30
1200 = 40 x1 + 50 x2
20
1600 = 40 x1 + 50 x2
10
0
16
10
20
30
40
x1
Identification
of optimal solution point
x
2
60
50
40
800 = 40 x1 + 50 x2
30
Optimal solution point
20
10
0
17
B
10
20
30
40
50
60
x1
Optimal solution coordinates
x2 40
35
4 x1 + 3 x2 = 120
30
25
20
A
15
10
B
8
x1 + 2 x2 = 40
5
C
18
0
5
10
15
20
25
http://www.baskent.edu.tr/~kilter
24
30
35
40
x1
Solutions at all corners points
x2 40
35
4 x1 + 3 x2 = 120
x1 = 0 bowls
x2 = 20 mugs
Z = $ 1,000
30
25
20
x1 = 24 bowls
x2 = 8 mugs
Z = $ 1,360
A
x1 = 30 bowls
x2 = 0 mugs
Z = $ 1,200
15
10
B
8
5
x1 + 2 x2 = 40
C
19
0
5
10
24
http://www.baskent.edu.tr/~kilter
15
20
25
30
35
40
x1
x2 40The
optimal solution with Z = 70 x1 + 20 x2
35
4 x1 + 3 x2 = 120
30
25
20
A
Optimal solution point
x1 = 30 bowls
x2 = 0 mugs
Z = $ 1,200
15
10
B
5
x1 + 2 x2 = 40
C
20
0
5
10
15
http://www.baskent.edu.tr/~kilter
20
25
30
35
40
x1
21
A Minimization Model Example
The Farmer s Field
The farmer s field requires at least 24 kg. of nitrogen
and 16 kg. of phosphate.
Super-gro costs $6 per bag, and Crop-quick costs $3.
The farmer wants to know how many bags of each brand to
purchase in order to minimize the total cost of fertilizing.
22
The Farmer s Field
Chemical Contribution
Brand
23
Nitrogen
(kg./bag)
Phosphate
(kg./bag)
Super-gro
4
2
Crop-quick
3
4
Linear Programming Model
Minimize Z = 6 x1 + 3 x2
Subject to:
2 x1 + 4 x2 ≥ 16
4 x1 + 3 x2 ≥ 24
x1, x2 ≥ 0
24
Constraint lines for fertilizer model
x2
12
10
8
4 x1 + 3 x2 = 24
6
4
2
2 x1 + 4 x2 = 16
25
0
2
4
http://www.baskent.edu.tr/~kilter
6
8
10
12
14
16
x1
Feasible solution area
x2
12
10
8
4 x1 + 3 x2 = 24
6
Feasible solution area
4
2
2 x1 + 4 x2 = 16
0
26
2
4
6
8
10
12
x1
The optimal solution point
x2
Optimal solution point
12
x1 = 0 bags of Super-gro
x2 = 8 bags of Crop-quick
Z = $ 24
10
8
A
6
Z = 6 x 1 + 3 x2
4
B
2
C
0
27
2
4
6
8
10
12
x1
Irregular Types of
Linear Programming
Problems
Multiple Optimal Solutions
x2 40
35
30
Point B
25
20
Point C
x1 = 24
x2 = 8
Z = $ 1,200
A
x1 = 30 bowls
x2 = 0 mugs
Z = $ 1,200
15
10
B
5
C
29
0
5
10
15
http://www.baskent.edu.tr/~kilter
20
25
30
35
40
x1
An Infeasible Problem
x2
x1 = 4
12
10
C
8
x2 = 6
6
B
4
2
0
30
4 x1 + 2 x2 = 8
A
2
4
6
8
10
12
x1
x2
An Unbounded Problem
x1 = 4
Minimize Z = 4 x1 + 2 x2
12
Subject to:
x1 ≥ 4
x2 ≤ 6
x1, x2 ≥ 0
10
8
x2 = 6
6
4
2
0
31
2
4
6
8
10
12
x1
32