Linear Programming: Model Formulation and Graphic Solution Chapter 2 Lecture 2
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
Objective Function Total Profit = 40 x1 + 50 x2
x2 : number of mugs to produce 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
If x1 = 5, x2 = 10 If x1 = 2, x2 = 2
Z = 40 . 5 + 50 . 10 = 700 Z = 40 . 2 + 50 . 2 = 180
If x1 = 10, x2 = 20
Z = 40 . 10 + 50 . 20 = 1400
If x1 = 20, x2 = 20
Z = 40 . 20 + 50 . 20 = 1800
7
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 x profits, Z, of $ 800, $ 1200, $ 1600 2
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
15Prof. M.A.Shouman 20 25
24
30
35
40
x1
Solutions at all corners points
x2 40 35
4 x1 + 3 x2 = 120 x1 = 0 bowls
30
x2 = 20 mugs Z = $ 1,000
25 20
x1 = 24 bowls
A
x2 = 8 mugs Z = $ 1,360 x1 = 30 bowls
15
x2 = 0 mugs Z = $ 1,200
10
B
8 5
x1 + 2 x2 = 40
C 19
0
5
10
15Prof. M.A.Shouman 20 24 25
30
35
40
x1
x2 40 The
optimal solution with Z = 70 x1 + 20 x2
35
4 x1 + 3 x2 = 120 30 25 20
A Optimal solution point x1 = 30 bowls
15
x2 = 0 mugs Z = $ 1,200
10
B
5
x1 + 2 x2 = 40
C 200
5
10
15
Prof. M.A.Shouman 20 25
30
35
40
x1
21
A Minimization Model Example The Farmer’s Field The farmer’s field requires at least 16 kg. of nitrogen and 24 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
6
Prof. 8 M.A.Shouman10
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
10 8
Z = $ 24
A
6
Z = 6 x1 + 3 x 2 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
A
Point C
x1 = 24
x1 = 30 bowls
x2 = 8
x2 = 0 mugs
Z = $ 1,200
Z = $ 1,200
15 10
B
5
C 290
5
10
15
Prof. M.A.Shouman 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
Z=
Subject to: x1 ≥ 4
4 x1
10
+2
x2 ≤ 6
x2
8
x1, x2 ≥ 0 x2 = 6
6 4 2
0 31
2
4
6
8
10
12
x1
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