1. Introduction Operations Research (OR) It is a scientific approach to determine the optimum (best) solution to a decision problem under the restriction of limited resources. Using the mathematical techniques to model, analyze, and solve the problem

2. Basic components of the model 1.Decision Variables 2.Objective Function 3.Constraints

3. Decision Variables It is the unknown to be determined from the solution of a model (what does the model seek to determine). It is one of the specific decisions made by a decision maker (DM).

4. Objective Function It is the end result (goal) desired to be achieved by the system. A common objective is to maximize profit or minimize cost. It is expressed as a mathematical function of the system decision variables.

5. Constraints These are the limitations imposed on the variables to satisfy the restriction of the modeled system. They must be expressed as mathematical functions of the system decision variables.

6. Example 1: two products A&B. 4 & 3 units3 & 2 minutes is 800 hoursat least 10000 units of Amore than6000 units of B A company manufactures two products A&B. with profit 4 & 3 units. A&B take 3 & 2 minutes respectively to be machined. The total time available at machining department is 800 hours. A market research showed that at least 10000 units of A and not more than 6000 units of B are needed. It is required to determine the number of units of A&B to be produced to maximize profit.

7. Decision variables X1= number of units produced of A. X2= number of units produced of B. Objective Function Maximize Z= 4 X 1 + 3 X 2 Constraints 3 X 1 + 2 X 2 ≤ 800x60 X 1 ≥ 10000 X 2 ≤ 6000 X 1, X 2 ≥ 0 Non negativity

8. Example 2: Example 2: Feed mix problem Two feeds are used A&B at least400 grams/day of proteinat least 800 grams/day of carbohydratesnot more than 100 grams/day of fatA contains 10% protein, 80% carbohydrates and 10% fat while B contains 40% protein, 60% carbohydratesno fat A costs 2 L.E/kg B costs 5 L.E/kg. A farmer is interested in feeding his cattle at minimum cost. Two feeds are used A&B. Each cow must get at least 400 grams/day of protein, at least 800 grams/day of carbohydrates, and not more than 100 grams/day of fat. Given that A contains 10% protein, 80% carbohydrates and 10% fat while B contains 40% protein, 60% carbohydrates and no fat. A costs 2 L.E/kg, and B costs 5 L.E/kg. Formulate the problem to determine the optimum amount of each feed to minimize cost.

9. Decision variables X1= weight of feed A kg/day/animal X2= weight of feed B kg/day/animal Objective Function Minimize Z= 2X 1 + 5X 2 Constraints Protein 0.1 X 1 + 0.4 X 2 ≥ 0.4 Carbohydrates 0.8 X 1 + 0.6 X 2 ≥ 0.8 Fats 0.1 X 1 ≤ 0.1 X 1, X 2 ≥ 0

10. Example 3: Example 3: Blending Problem 4 mines 3 basic elements A, B, and C4 mines 3 elements An iron ore from 4 mines will be blended. The analysis has shown that, in order to obtain suitable tensile properties, minimum requirements must be met for 3 basic elements A, B, and C. Each of the 4 mines contains different amounts of the 3 elements (see the table). Formulate to find the least cost (Minimize) blend for one ton of iron ore.see the table

12. Problem Formulation Decision variables X1= Fraction of ton to be selected from mine number 1 X2= Fraction of ton to be selected from mine number 2 X3= Fraction of ton to be selected from mine number 3 X4= Fraction of ton to be selected from mine number 4 Objective Function Minimize Z= 800 X1 + 400 X2 + 600 X3 + 500 X4 Constraints 10 X 1 + 3X 2 + 8X 3 + 2X 4 5 90 X 1 + 150 X 2 + 75 X 3 + 175 X 4 10 45 X 1 + 25 X 2 + 20 X 3 + 37 X 4 30 X 1 + X 2 + X 3 + X 4 1 X 1, X 2, X 3, X 4 0

13. Example 4: Example 4: Inspection Problem 2 grades at least1800 piecesper 8 hour dayGrade 1 rate of 25 per houraccuracy of 98%. Grade 2rate of 15 pieces per houraccuracy of 95%.Grade 1 costs 4 L.E/hourgrade 2 costs 3 L.E/hourEach time an error is made by an inspector costs the company 2 L.E8 grade 110 grade 2 A company has 2 grades of inspectors 1&2. It is required that at least 1800 pieces be inspected per 8 hour day. Grade 1 inspectors can check pieces at the rate of 25 per hour with an accuracy of 98%. Grade 2 inspectors can check at the rate of 15 pieces per hour with an accuracy of 95%. Grade 1 costs 4 L.E/hour, grade 2 costs 3 L.E/hour. Each time an error is made by an inspector costs the company 2 L.E. There are 8 grade 1 and 10 grade 2 inspectors available. The company wants to determine the optimal assignment of inspectors which will minimize the total cost of inspection/day.

14. Problem Formulation Decision variables X1= Number of grade 1 inspectors/day. X2= Number of grade 2 inspectors/day. Objective Function Cost of inspection = Inspector salary/day + Cost of error Cost of grade 1/hour = 4 + (2 X 25 X 0.02) = 5 L.E Cost of grade 2/hour = 3 + (2 X 15 X 0.05) = 4.5 L.E Minimize Z= 8 (5 X1 + 4.5 X2) = 40 X1 + 36 X2 Constraints X 1 ≤ 8 X 2 ≤ 10 8(25) X 1 + 8(15)X 2 ≥ 1800 200 X 1 + 120 X 2 ≥ 1800 X 1, X 2 ≥ 0

15. Example 5: Example 5: Trim-loss Problem. A company produces paper rolls with a standard width of 20 feet. Each special customer orders with different widths are produced by slitting the standard rolls. Typical orders are summarized in the following tables.

16. Possible knife settings Formulate to minimize the trim loss and the number of rolls needed to satisfy the order. Figure 2.9

18. Problem Formulation Decision variables X j = Number of standard rolls to be cut according to setting j j = 1, 2, 3, 4, 5, 6 Number of 5 feet rolls produced Number of 5 feet rolls produced = 2 X 2 + 2 X 3 + 4 X 4 + X 5 Number of 7 feet rolls produced Number of 7 feet rolls produced = X 1 + X 2 + 2 X 5 Number of 9 feet rolls produced Number of 9 feet rolls produced = X 1 + X 3 + 2 X 6 Let Y 1, Y 2, Y 3 be the number of surplus rolls of the 5, 7, 9 feet rolls thus Y 1 = 2 X 2 + 2 X 3 + 4 X 4 + X 5 - 150 Y 2 = X 1 + X 2 + 2 X 5 - 200 Y 3 = X 1 + X 3 + 2 X 6 - 300 The total trim losses = L (4X 1 +3 X 2 + X 3 + X 5 + 2 X 6 + 5Y 1 + 7Y 2 + 9Y 3 )* Where L is the length of the standard roll.

19. Objective Function Minimize Z= L(4X 1 +3 X 2 + X 3 + X 5 + 2 X 6 + 5Y 1 + 7Y 2 + 9Y 3 ) Constraints 2 X 2 + 2 X 3 + 4 X 4 + X 5 - Y 1 = 150 X 1 + X 2 + 2 X 5 - Y 2 = 200 X 1 + X 3 + 2 X 6 - Y 3 = 300 X1, X2, X3, X4, X5, X6 ≥ 0 Y1, Y2, Y3 ≥ 0

20. Terminology of solutions for a LP model: A Solution Any specifications of values of X 1, X 2, …, X n is called a solution. A Feasible Solution Is a solution for which all the constraints are satisfied. A n Optimal Solution Is a feasible solution that has the most favorable value of the objective function (largest of maximize or smallest for minimize)

21. Graphical Solution Construction of the LP model Example 1: The Reddy Mikks Company Reddy Mikks produces both interior and exterior paints from two raw materials, M1&M2. The following table provides the basic data of the problem.

22. A market survey indicates that the daily demand for interior paint cannot exceed that of exterior paint by more than 1 ton. Also, the maximum daily demand of interior paint is 2 ton. Reddy Mikks wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit

23. Problem Formulation Decision variables X 1 = Tons produced daily of exterior paint. X 2 = Tons produced daily of interior paint. Objective Function Maximize Z= 5 X 1 + 4 X 2 Constraints 6 X 1 +4 X 2 24 X 1 +2 X 2 6 - X 1 + X 2 1 X 2 2 X 1, X 2 0

24. Any solution that satisfies all the constraints of the model is a feasible solution. For example, X1=3 tons and X2=1 ton is a feasible solution. We have an infinite number of feasible solutions, but we are interested in the optimum feasible solution that yields the maximum total profit.

25. Graphical Solution The graphical solution is valid only for two- variable problem which is rarely occurred. The graphical solution includes two basic steps: 1.The determination of the solution space that defines the feasible solutions that satisfy all the constraints. 2.The determination of the optimum solution from among all the points in the feasible solution space.

28. ABCDEF consists of an infinite number of points; we need a systematic procedure that identifies the optimum solutions. The optimum solution is associated with a corner point of the solution space. To determine the direction in which the profit function increases we assign arbitrary increasing values of 10 and 15 5 X 1 + 4 X 2 =10 And5 X 1 + 4 X 2 =15 The optimum solution is mixture of 3 tons of exterior and 1.5 tons of interior paints will yield a daily profit of 21000$.