1. 1© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 The Wyndor Glass Company Problem (Hillier and Liberman) The Wyndor Glass Company is planning to launch two new products. Product 1 is an 8-foot glass door with aluminum framing and Product 2 a 4x6 foot double-hung wood-framed window Aluminum frames are made in Plant 1, wood frames are made in Plant 2, and Plant 3 produces the glass and assembles the products. Product 1 requires some of the production capacity in Plants 1 and 3, but none in Plant 2. Product 2 needs only Plants 2 and 3. The marketing division has concluded that the company could sell as much of either product as could be processed by these plants. The management of the company wants to determine what mixture of both products would be the most profitable. The following table provides the information available.

2. 2© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 The Wyndor Glass Company Problem (Hillier and Liberman)

3. 3© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 The Wyndor Glass Company Problem Formulation

4. 4© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Montana Wood Products Problem Montana Wood Products manufacturers two-high quality products, tables and chairs. Its profit is $15 per chair and $21 per table. Weekly production is constrained by available labor and wood. Each chair requires 4 labor hours and 8 board feet of wood while each table requires 3 labor hours and 12 board feet of wood. Available wood is 2400 board feet and available labor is 920 hours. Management also requires at least 40 tables and at least 4 chairs be produced for every table produced. To maximize profits, how many chairs and tables should be produced?

5. 5© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Montana Wood Products Problem Formulation Objective Function Max Z = 15x 1 + 21x 2 Subject To 4x 1 + 3x 2 < 920 ( labor constraint) 8x 1 + 12x 2 < 2400 ( wood constraint) x 2 - 40 > 0 (make at least 40 tables) x 1 - 4 x 2 > 0 (at least 4 chairs for every table) x 1, x 2 > 0 (non-negativity constraints)

6. 6© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 The Sureset Concrete Company Problem The Sureset Concrete Company produces concrete. Two ingredients in concrete are sand (costs $6 per ton) and gravel (costs $8 per ton). Sand and gravel together must make up exactly 75% of the weight of the concrete. Also, no more than 40% of the concrete can be sand and at least 30% of the concrete be gravel. Each day 2000 tons of concrete are produced. To minimize costs, how many tons of gravel and sand should be purchased each day?

7. 7© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 The Sureset Concrete Company Problem Formulation Objective Function Min Z = 6x 1 + 8x 2 Subject To x 1 + x 2 = 1500 ( mix constraint) x 1 < 800 ( mix constraint) x 2 > 600 ( mix constraint ) x 1, x 2 > 0 (non-negativity constraints)

8. 8© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Widgets Problem Suppose a company produces two types of widgets, manual and electric. Each requires in its manufacture the use of three machines; A, B, and C. A manual widget requires the use of the machine A for 2 hours, machine B for 1 hour, and machine C for 1 hour. An electric widget requires 1 hour on A, 2 hours on B, and 1 hour on C. Furthermore, suppose the maximum numbers of hours available per month for the use of machines A, B, and C are 180, 160, and 100, respectively. The profit on a manual widget is $4 and on electric widget it is $6. See the table below for a summary of data. If the company can sell all the widgets it can produce, how many of each type should it make in order to maximize the monthly profit?

9. 9© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Widgets Problem Formulation X 1 = number of manual widgets X 2 = number of electric widgets Objective Function max Z = 4X 1 + 6X 2 ٍٍ Subject To 2X 1 + X 2  180 X 1 + 2X 2  160 X 1 + X 2  100 X 1  0, X 2  0

10. 10© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Product-Mix Problem The Handy-Dandy Company wishes to schedule the production of a kitchen appliance which requires two resources – labor and material. The company is considering three different models of this appliance and its engineering department has furnished the following data: The supply of raw materials is restricted to 200 pounds per day. The daily availability of manpower is 150 hours. Formulate a linear programming model to determine the daily production rate of the various models of appliances in order to maximize the total profit.

11. 11© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Product-Mix Problem Formulation X A = daily production of model A X B = daily production of model B X C = daily production of model C Objective Function Maximize Z = 4X A + 2X B + 3X C subject to the constrains 7X A + 3X B + 6X C  150 4X A + 4X B + 5X C  200 X A  0, X B  0, X C  0

12. 12© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 An Electric Company Problem An electric company manufacturers two radio models, each on a separate rate production line. The daily capacity of the first line is 60 radios and that of the second is 75 radios. Each unit of the first model uses 10 pieces of a certain electronic component, whereas each unit of the second model requires 8 pieces of the same component. The maximum daily availability of the special component is 800 pieces. The profit per unit of models 1 and 2 is $30 and $20, respectively. Determine the optimum daily production of each model.

13. 13© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 An Electric Company Problem Formulation X 1 = number of radios of model 1 X 2 = number of radios of model 2 Objective Function max Z =30X 1 + 20X 2 ٍٍ Subject To X 1  60 X 2  75 10 X 1 +8X 2  800 X 1  0, X 2  0

14. 14© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Furniture Factory Problem A small furniture factory manufacturers tables and chairs. It takes 2 hours to assemble a table and 30 minutes to assemble a chair. Assembly is carried out by four workers on the basis of a single 8-hour shift per day. Customers usually buy at least four chairs with each table, meaning that the factory must produce at least four times as many chairs as tables. The sale price is $150 per table and $50 per chair. Determine the daily production mix of chairs and tables that would maximize the total daily revenue to the factory.

15. 15© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Furniture Factory Problem Formulation X 1 = number of tables X 2 = number of chairs Objective Function max Z =150X 1 + 50X 2 ٍٍ Subject To X 2 - 4X 1  0 120 X 1 +30X 2  4x8x60 X 1  0, X 2  0

16. 16© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Giapetto’s Woodcarving Problem Giapetto’s, Inc., manufactures wooden soldiers and trains. Each soldier built: Sell for $27 and uses $10 worth of raw materials. Giapetto’s variable labor/overhead costs by $14. Requires 2 hours of finishing labor. Requires 1 hour of carpentry labor Each train built Sell for $21 and used $9 worth of raw materials. Giapetto’s variable labor/overhead costs by $10. Requires 1 hour of finishing labor. Requires 1 hour of carpentry labor. Each week Giapetto can obtain: all needed raw material, only 100 finishing hours & only 80 carpentry hours. Demand for the trains is unlimited, at most 40 soldiers are bought each week. Giapetto wants to maximize weekly profit. Formulate a mathematical model of Giapetto’s situation that can be used to maximize weekly profit.

17. 17© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Giapetto’s Woodcarving Problem Formulation x1 = number of soldiers produced each week x2 = number of trains produced each week Weekly profit = weekly revenue – weekly raw material costs – the weekly variable costs = 3x 1 + 2x 2 objective function Maximize z = 3x 1 + 2x 2 Subject to 2 x1 + x2 ≤ 100(finishing constraint) x1 + x2 ≤ 80(carpentry constraint) x1 ≤ 40(constraint on demand for soldiers) x1 ≥ 0, x2 ≥ 0

18. 18© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Graphical Solution of an LP Problem Used to solve LP problems with two (and sometimes three) decision variables Consists of two phases Finding the values of the decision variables for which all the constraints are met (feasible region of the solution space) Determining the optimal solution from all the points in the feasible region (from our knowledge of the nature of the optimal solution)

19. 19© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Finding the Feasible Region (2D) Steps Use the axis in a 2-dimensional graph to represent the values that the decision variables can take For each constraint, replace the inequalities with equations and graph the resulting straight line on the 2- dimensional graph For the inequality constraints, find the side (half-space) of the graph meeting the original conditions (evaluate whether the inequality is satisfied at the origin) Find the intersection of all feasible regions defined by all the constraints. The resulting region is the (overall) feasible region.

20. 20© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Equation Form Basic data Tons of raw material per ton of Maximum daily Exterior paint Interior paint availability (tons) Raw material, M1 6 4 24 Raw material, M2 1 2 6 Profit per ton ($1000) 5 4 *Decision variables: Need to determine the amounts to be produced of exterior and interior paints. x 1 = tons produced daily of exterior paint x 2 = tons produced daily of interior paint *Objective (goal) aims to optimize: The company wants to increase its profit as much as possible. z represents the total daily profit (in thousands of dollars) Maximize z = 5x 1 + 4x 2 *Constraints: Restrict raw materials usage and demand (Usage of a raw material) ≤ (Maximum raw material) by both paints availability 6x 1 + 4x 2 ≤ 24 (Raw material M1) x 1 + 2x 2 ≤ 6 (Raw material M2) -x 1 + x 2 ≤ 1 (Demand Limit) x 2 ≤ 2 (Demand Limit) Nonnegativity restrictions: x 1, x 2 ≥ 0

21. 21© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Graphical LP Solution Graphical procedure includes 2 steps Determination: 1) The solution space that defines all feasible solutions of the model 2) The optimum solution from among all the feasible points in the solution space Step 1: Determination of the Feasible Solution Space x2 x2 0 x1x1 123465 1 2 4 3 5 6 Nonnegativity constraints x 1 ≥ 0 x 2 ≥ 0 Replace each inequality (≤) with equations (=) 6x 1 + 4x 2 ≤ 24 6x 1 + 4x 2 = 24 x 1 + 2x 2 ≤ 6 x 1 + 2x 2 = 6 -x 1 + x 2 ≤ 1 -x 1 + x 2 = 1 x 2 ≤ 2 x 2 = 2 Determine 2 points 6x 1 + 4x 2 = 24 let x 1 = 0, x 2 = 6 let x 2 = 0, x 1 = 4 2 points: (0,6) and (4,0) 1 2 3 4 first quadrant Graph of equations

22. 22© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Feasible Space 1 2 3 4 5 6 5 6 4 3 2 1 01 3 4562 Reference point Convenient computationally: (0,0) Note: select another point if the line passes through the origin, can’t use (0,0) Ex: 6x 1 + 4x 2 ≤ 24 Feasible half-space ( 0,0) 6 x 0 + 4 x 0 = 0 (0 ≤ 24) Not a feasible side (6,0) 6 x 6 + 4 x 0 = 36 (36 ≥ 24) Constraints: 6x 1 + 4x 2 ≤ 24 x 1 + 2x 2 ≤ 6 -x 1 + x 2 ≤ 1 x 2 ≤ 2 x 1 ≥ 0 x 2 ≥ 0 Feasible space of the Reddy Mikks model x2 x2 x1 x1 Solution space

23. 23© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Optimum Solution Step 2: Determination of the Optimum Solution The feasible space consists of an infinite number of points Optimum solution: identify the piont in which the profit function z = 5x 1 + 4x 2 increases (maximize z).. 01234 1 2 3 x1x1 x2x2 Values of x 1 and x 2 associated with the optimum point 6x 1 + 4x 2 = 24 solve x 1 + 2x 2 = 6 simultaneously x 1 = 3 and x 2 = 1.5 z = 21 Optimum: x 1 + 2x 2 ≤ 6 6x 1 + 4x 2 ≤ 24 x 1 = 3 tons x 2 = 1.5 tons z = $21,000

24. 24© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Reddy Mikks Problem Formulation

25. 25© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Reddy Mikks Problem Solution  I 1 2 3 4 123  E 4567 0 + 2(0)  6 Constraint 4: X I  2 Constraint 3: -X E + X I  1 Constraint 2: 2X E + X I  8 FeasibleRegion Constraint 1: X E + 2X I  6

26. 26© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Example(2) Objective Function Minimize z = 3x – y Subject to x – y  1 x + y  5 x  0, and y  0.

27. 27© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458 Example(2)\solution y x (1, 0) (3, 2) (0, 5) vertexvalue of z at vertex (0, 0)z = 3(0) – (0) = 0 (1, 0)z = 3(1) – (0) = 3 (3, 2)z = 3(3) – (2) = 7 (0, 5)z = 3(0) – (5) = –5 The minimum value of z is –5 and this occurs at (0, 5).