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Operations Research IChapter 02 (continued)
Modeling with Linear Programming
Dr. Ayham Jaaron
First semester 2013/2014August 2013
Model formulationExample for practice
• Farmer Jones decided to supplement his income by baking and selling two types of cakes, chocolate and vanilla. Each chocolate cake sold gives a profit of $3, and the profit on each vanilla cake sold is $5. each chocolate cake requires 20 minutes of baking time and uses 4 eggs and 4 pounds of flour, while each vanilla cake requires 40 minutes of baking time and uses 2 eggs and 5 pounds of flour. If farmer Jones has available only 260 minutes of baking time, 32 eggs, and 40 pounds of flour, how many of each type of cake should be baked in order to maximize farmer Jones’ profit?
Problem No.4 (book page 26)
Problem 14 (Book page 21)Construct the LP model that represents the problem
below? Maximize steam generated!!!
Operations Research IChapter 02 (continued)Graphical LP Solution
Dr. Ayham Jaaron
Graphical LP Solution (2 variables)
The graphical procedure includes two steps:
Determination of the feasible solution space
Determination of the Optimum solution from among all the feasible points in the solution space.
Graphical solution types
• Maximization Problems• Minimization problems
• We shall try with Maximization problems first
Example 1: The Reddy Mikks Company
Maximize Z= 5 X1 + 4 X2Let’s try now !
Example 1: The Reddy Mikks Company
Maximize Z= 5 X1 + 4 X2
Example 1: The corner points technique
Another Technique: using arbitrary values
Example (2) on Graphical Method
Resolve using the Graphical Method for the following problem:Maximize Z = 3x + 2y subject to: 2x + y ≤ 18 2x + 3y ≤ 42 3x + y ≤ 24 x ≥ 0 , y ≥ 0
Example (2)...Continued
• Initially we draw the coordinate system correlating to an axis the variable x, and the other axis to variable y, as we can see in the following figures.
Example (2)...continued
Example (2)...continued
• we proceed to determine the extreme points in the feasible region, candidates to optimal solutions, that are the O-F-H-G-C points figure's. Finally, we evaluate the objective function ( 3x + 2y ) at those points, which result is picked up in the following board. As G point provides the bigger value to the objective Z, such point constitutes the optimal solution, we will indicate x = 3; y = 12, with optimal value Z = 33.
Example (2)...continued
Example (3): Reddy Mikks Company problemGraphical Solution of Maximization Model (1 of 12)
Figure 2.2 Coordinates for Graphical Analysis
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
X1 is bowls
X2 is mugs
Labor ConstraintGraphical Solution of Maximization Model (2 of 12)
Figure 2.3 Graph of Labor Constraint
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Labor Constraint AreaGraphical Solution of Maximization Model (3 of 12)
Figure 2.4 Labor Constraint Area
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Clay Constraint AreaGraphical Solution of Maximization Model (4 of 12)
Figure 2.5 Clay Constraint Area
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Both ConstraintsGraphical Solution of Maximization Model (5 of 12)
Figure 2.6 Graph of Both Model Constraints
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Feasible Solution AreaGraphical Solution of Maximization Model (6 of 12)
Figure 2.7 Feasible Solution Area
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Objective Function Solution = $800Graphical Solution of Maximization Model (7 of 12)
Figure 2.8 Objection Function Line for Z = $800
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Alternative Objective Function Solution LinesGraphical Solution of Maximization Model (8 of 12)
Figure 2.9 Alternative Objective Function Lines
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Optimal SolutionGraphical Solution of Maximization Model (9 of 12)
Figure 2.10 Identification of Optimal Solution Point
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Optimal Solution CoordinatesGraphical Solution of Maximization Model (10 of 12)
Figure 2.11 Optimal Solution Coordinates
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Extreme (Corner) Point SolutionsGraphical Solution of Maximization Model (11 of 12)
Figure 2.12 Solutions at All Corner Points
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Optimal Solution for New Objective FunctionGraphical Solution of Maximization Model (12 of 12)
Maximize Z = $70x1 + $20x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.13 Optimal Solution with Z = 70x1 + 20x2
Graphical Solution for Minimization Problems
To apply the Graphical method to solve a minimization LP Model, consider the following problem
Minimize: Z= 0.3X1 + 0.9X2
Subject to: x1+x2 ≥ 800 (0, 800), (800,0)
0.21x1- 0.30x2 ≤ 0 (0,0) , (100,70)
0.03x1-0.01x2 ≥0 (0,0), (25,75)
x1,x2 ≥ 0
Example (2) minimization problem using Graphical solution
• Solve the following LP model using graphical solution:
• Minimize: z = 8X1 + 6X2
• Subject to:4X1 + 2X2 >=20
-6X1+4X2 <=12
X1 + X2 >=6
X1, X2>=0
Home work No.1
Home work Number 1 is due to be submitted on Wednesday 11th
September 2013. No late submissions will be accepted
under any circumstances.Page 15: Problems 1, 2
Page 20: Problems 4, 5, 6