Dr. Ayham Jaaron First semester 2013/2014 August 2013 Operations Research I Chapter 02 (continued) Modeling with Linear Programming Dr. Ayham Jaaron First semester 2013/2014 August 2013

Model formulation Example 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 I Chapter 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 X2 Let’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 problem Graphical Solution of Maximization Model (1 of 12) X2 is mugs Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 X1 is bowls Figure 2.2 Coordinates for Graphical Analysis

Graphical Solution of Maximization Model (2 of 12) Labor Constraint Graphical Solution of Maximization Model (2 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.3 Graph of Labor Constraint

Graphical Solution of Maximization Model (3 of 12) Labor Constraint Area Graphical Solution of Maximization Model (3 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.4 Labor Constraint Area

Graphical Solution of Maximization Model (4 of 12) Clay Constraint Area Graphical Solution of Maximization Model (4 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.5 Clay Constraint Area

Graphical Solution of Maximization Model (5 of 12) Both Constraints Graphical Solution of Maximization Model (5 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.6 Graph of Both Model Constraints

Feasible Solution Area Graphical Solution of Maximization Model (6 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.7 Feasible Solution Area

Objective Function Solution = $800 Graphical Solution of Maximization Model (7 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.8 Objection Function Line for Z = $800

Alternative Objective Function Solution Lines Graphical Solution of Maximization Model (8 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.9 Alternative Objective Function Lines

Graphical Solution of Maximization Model (9 of 12) Optimal Solution Graphical Solution of Maximization Model (9 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.10 Identification of Optimal Solution Point

Optimal Solution Coordinates Graphical Solution of Maximization Model (10 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.11 Optimal Solution Coordinates

Extreme (Corner) Point Solutions Graphical Solution of Maximization Model (11 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.12 Solutions at All Corner Points

Optimal Solution for New Objective Function Graphical 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