Operations Research 1 Dr. El-Sayed Badr

1. Operations Research 1 Dr. El-Sayed Badr Associate Professor of Computers & Informatics - Benha University Alsayed.badr@fsc.bu.edu.eg Dr. El-Sayed Badr 2014

2. Example 2.1-1 (The Reddy Mikks Company) Reddy Mikks produces both interior and exterior paints from two raw materials M1 and M2 Tons of raw material per ton of Exterior paint Interior paint Maximum daily availability (tons) Raw material M1 6 4 24 Raw material M2 1 2 6________ Profit per ton ($1000) 5 4 ___________________________________________________________________________ 1-Daily demand for interior paint cannot exceed that of exterior paint by more than 1 ton 2-Maximum daily demand of interior paint is 2 tons 3-Reddy Mikks wants to determine the optimum product mix of interior and exterior paints that maximizes the total daily profit Dr. El-SayedBadr 2012

3. Solution: Let x1 = tons produced daily of exterior paint x2 = tons produced daily of interior paint Let z represent the total daily profit (in thousands of dollars) Objective: Maximize z = 5 x1 + 4 x2 (Usage of a raw material by both paints) < (Maximum raw material availability) Usage of raw material M1 per day = 6x1 + 4x2 tons Usage of raw material M2 per day = 1x1 + 2x2 tons - daily availability of raw material M1 is 24 tons - daily availability of raw material M2 is 6 tons Dr. El-Sayed Badr 2014

4. Restrictions: 6x1 + 4x2 < 24 (raw material M1) x1 + 2x2 < 6 (raw material M2) Difference between daily demand of interior (x2) and exterior (x1) paints does not exceed 1 ton, so x2 - x1 < 1 Maximum daily demand of interior paint is 2 tons, so x2 < 2 - Variables x1 and x2 cannot assume negative values, so x1 > 0 , x2 > 0 Dr. El-Sayed Badr 2014

5. Complete Reddy Mikks model: Maximize z = 5 x1 + 4 x2 (total daily profit) subject to 6x1 + 4x2 < 24 (raw material M1) x1 + 2x2 < 6 (raw material M2) x2 - x1 < 1 x2 < 2 x1 > 0 x2 > 0 - Objective and the constraints are all linear functions in this example. Dr. El-Sayed Badr 2014

6. 6 2.2 GRAPHICAL LP SOLUTION 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.

7. 1 2 3 4 5 6 7 2.2.1 Solution of a Maximization model Example 2.2-1 (Reddy Mikks model) Step 1: 1) Determination of the feasible solution space: - Find the coordinates for all the 6 equations of the restrictions (only take the equality sign) 6x1 + 4x2 <24 x1 + 2x2 < 6 x2 -x1 < 1 x2 <2 x1> 0 x2 > 0

8. 1 2 3 4 5 6 8 - Change all equations to equality signs 6x1 + 4x2 =24 x1 + 2x2 = 6 x2 -x1 =1 x2 = 2 x1 = 0 x2 = 0

9. 9 - Plot graphs of x1 = 0 and x2 = 0 - Plot graph of 6x1 + 4x2 = 24 by using the coordinates of the equation -Plot graph of x1 + 2x2 = 6 by using the coordinates of the equation -Plot graph of x2 -x1 =1 by using the coordinates of the equation -Plot graph of x2 = 2 by using the coordinates of the equation

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11. 11 - Now include the inequality of all the 6 equations - Inequality divides the (x1, x2) plane into two half spaces , one on each side of the graphed line - Only one of these two halves satisfies the inequality - To determine the correct side , choose (0,0) as a reference point - If (0,0) coordinate satisfies the inequality, then the side in which (0,0) coordinate lies is the feasible half-space , otherwise the other side is - If the graph line happens to pass through the origin (0,0) , then any other point can be used to find the feasible half-space

12. 12 Step 2: 2) Determination of the optimum solution from among all the feasible points in the solution space: - After finding out all the feasible half-spaces of all the 6 equations, feasible space is obtained by the line segments joining all the corner points A, B, C, D ,E and F - Any point within or on the boundary of the solution space ABCDEF is feasible as it satisfies all the constraints - Feasible space ABCDEF consists of infinite number of feasible points

13. 2 1 1 2 13 - To find optimum solution identify the direction in which the maximum profit increases , that is z = 5x1 + 4x2 - Assign random increasing values to z , z = 10 and z = 15 5x1 + 4x2 = 10 5x1 + 4x2 = 15 - Plot graphs of above two equations - Thus in this way the optimum solution occurs at corner point C which is the point in the solution space - Any further increase in z that is beyond corner point C will put points outside the boundaries of ABCDEF feasible space - Values of x1 and x2 associated with optimum corner point C are determined by solving the equations and 6x1 + 4x2 = 24 x1 + 2x2 = 6 - x1 = 3 and x2 = 1.5 with z = 5 X 3 + 4 X 1.5 = 21 - So daily product mix of 3 tons of exterior paint and 1.5 tons of interior paint produces the daily profit of $21,000 .

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15. 1 15 - Important characteristic of the optimum LP solution is that it is always associated with a corner point of the solution space (where two lines intersect) - This is even true if the objective function happens to be parallel to a constraint - For example if the objective function is, z = 6x1 + 4x2 - The above equation is parallel to constraint of equation - So optimum occurs at either corner point B or corner point C when parallel - Actually any point on the line segment BC will be an alternative optimum - Line segment BC is totally defined by the corner points B and C

16. 16 - Since optimum LP solution is always associated with a corner point of the solution space, so optimum solution can be found by enumerating all the corner points as below:- ______________Corner point (x1,x2) z_________________ A (0,0) 0 B (4,0) 20 C (3,1.5) 21 (optimum solution) D (2,2) 18 E (1,2) 13 F (0,1) 4 - As number of constraints and variables increases , the number of corner points also increases

17. Solution of a Minimization Model Dr. El-Sayed Badr 2012

18. Geometric Solution 18

19. Software : 1- TORA (Prof. Hamdy Taha ) 2- Matlab ( Cleve Barry Moler ) 3- Excell ( Microsoft ) 19

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