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