Linear Programming

1. Linear Programming B PowerPoint presentation to accompany Heizer and Render Operations Management, 10e Principles of Operations Management, 8e PowerPoint slides by Jeff Heyl

2. Outline • Requirements of a Linear Programming Problem • Formulating Linear Programming Problems • Shader Electronics Example • Graphical Solution to a Linear Programming Problem • Graphical Representation of Constraints • Iso-Profit Line Solution Method • Corner-Point Solution Method • Solving Minimization Problems • Linear Programming Applications • Production-Mix Example • Diet Problem Example

3. Learning Objectives Formulate linear programming models, including an objective function and constraints Graphically solve an LP problem with the iso-profit line method Graphically solve an LP problem with the corner-point method Construct and solve a minimization problem

4. Why Use Linear Programming? • A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources • Will find the minimum or maximum value of the objective • Guarantees the optimal solution to the model formulated

5. LP Applications Scheduling school buses to minimize total distance traveled Allocating police patrol units to high crime areas in order to minimize response time to 911 calls Scheduling tellers at banks so that needs are met during each hour of the day while minimizing the total cost of labor

6. LP Applications Selecting the product mix in a factory to make best use of machine- and labor-hours available while maximizing the firm’s profit Picking blends of raw materials in feed mills to produce finished feed combinations at minimum costs Determining the distribution system that will minimize total shipping cost

7. Requirements of an LP Problem LP problems seek to maximize or minimize some quantity The presence of restrictions, or constraints There must be alternative courses of action to choose from The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities

8. Formulating LP Problems The product-mix problem at Shader Electronics • Two products • Shader x-pod, a portable music player • Shader BlueBerry, an internet-connected color telephone • Determine the mix of products that will produce the maximum profit

9. Hours Required to Produce 1 Unit x-pods BlueBerrys Available Hours Department (X1) (X2) This Week Electronic 4 3 240 Assembly 2 1 100 Profit per unit $7 $5 Formulating LP Problems Table B.1 Decision Variables: X1 = number of x-pods to be produced X2 = number of BlueBerrys to be produced

10. Formulating LP Problems Objective Function: Maximize Profit = $7X1 + $5X2 • There are three types of constraints • Upper limits where the amount used is ≤ the amount of a resource • Lower limits where the amount used is ≥ the amount of the resource • Equalities where the amount used is = the amount of the resource

11. Electronic time used Assembly time used Assembly time available Electronic time available is ≤ is ≤ Formulating LP Problems First Constraint: 4X1 + 3X2 ≤ 240 (hours of electronic time) Second Constraint: 2X1 + 1X2 ≤ 100 (hours of assembly time)

12. Graphical Solution • Can be used when there are two decision variables • Plot the constraint equations at their limits by converting each equation to an equality • Identify the feasible solution space • Create an iso-profit line based on the objective function • Move this line outwards until the optimal point is identified

13. X2 – – 80 – – 60 – – 40 – – 20 – – – Assembly (Constraint B) Number of BlueBerrys Electronics (Constraint A) | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of x-pods Graphical Solution Feasible region Figure B.3

14. X2 – – 80 – – 60 – – 40 – – 20 – – – Choose a possible value for the objective function Assembly (Constraint B) $210 = 7X1 + 5X2 Solve for the axis intercepts of the function and plot the line Number of BlueBerrys Electronics (Constraint A) X2 = 42 X1 = 30 | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of x-pods Graphical Solution Iso-Profit Line Solution Method Feasible region Figure B.3

15. X2 – – 80 – – 60 – – 40 – – 20 – – – Number of BlueBerrys | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of x-pods Graphical Solution $210 = $7X1 + $5X2 (0, 42) (30, 0) Figure B.4

16. X2 – – 80 – – 60 – – 40 – – 20 – – – Number of BlueBerrys | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of x-pods Graphical Solution $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 $210 = $7X1 + $5X2 $420 = $7X1 + $5X2 Figure B.5

17. X2 – – 80 – – 60 – – 40 – – 20 – – – Maximum profit line Optimal solution point (X1 = 30, X2 = 40) Number of BlueBerrys $410 = $7X1 + $5X2 | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of x-pods Graphical Solution Figure B.6

18. X2 – – 80 – – 60 – – 40 – – 20 – – – Number of BlueBerrys 2 3 1 | | | | | | | | | | | 0 20 40 60 80 100 X1 Number of x-pods 4 Corner-Point Method Figure B.7

19. Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350 Corner-Point Method • The optimal value will always be at a corner point • Find the objective function value at each corner point and choose the one with the highest profit

20. 4X1 + 3X2 ≤ 240 (electronics time) 2X1 + 1X2 ≤ 100 (assembly time) 4X1 + 3X2 = 240 - 4X1 - 2X2 = -200 + 1X2 = 40 Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350 Corner-Point Method • The optimal value will always be at a corner point • Find the objective function value at each corner point and choose the one with the highest profit Solve for the intersection of two constraints 4X1 + 3(40) = 240 4X1 + 120 = 240 X1 = 30

21. Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350 Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410 Corner-Point Method • The optimal value will always be at a corner point • Find the objective function value at each corner point and choose the one with the highest profit

22. Solving Minimization Problems • Formulated and solved in much the same way as maximization problems • In the graphical approach an iso-cost line is used • The objective is to move the iso-cost line inwards until it reaches the lowest cost corner point

23. Minimization Example X1 = number of tons of black-and-white picture chemical produced X2 = number of tons of color picture chemical produced Minimize total cost = 2,500X1 + 3,000X2 Subject to: X1 ≥ 30 tons of black-and-white chemical X2 ≥ 20 tons of color chemical X1 + X2 ≥ 60 tons total X1, X2 ≥ $0 nonnegativity requirements

24. X2 60 – 50 – 40 – 30 – 20 – 10 – – X1 + X2 = 60 X1 = 30 X2 = 20 | | | | | | | 0 10 20 30 40 50 60 X1 Minimization Example Table B.9 Feasible region b a

25. Minimization Example Total cost at a = 2,500X1 + 3,000X2 = 2,500 (40) + 3,000(20) = $160,000 Total cost at b = 2,500X1 + 3,000X2 = 2,500 (30) + 3,000(30) = $165,000 Lowest total cost is at point a

26. Department Product Wiring Drilling Assembly Inspection Unit Profit XJ201 .5 3 2 .5 $ 9 XM897 1.5 1 4 1.0 $12 TR29 1.5 2 1 .5 $15 BR788 1.0 3 2 .5 $11 Capacity Minimum Department (in hours) Product Production Level Wiring 1,500 XJ201 150 Drilling 2,350 XM897 100 Assembly 2,600 TR29 300 Inspection 1,200 BR788 400 LP Applications Production-Mix Example

27. LP Applications X1 = number of units of XJ201 produced X2 = number of units of XM897 produced X3 = number of units of TR29 produced X4 = number of units of BR788 produced Maximize profit = 9X1 + 12X2 + 15X3 + 11X4 subject to .5X1 + 1.5X2 + 1.5X3 + 1X4 ≤ 1,500 hours of wiring 3X1 + 1X2 + 2X3 + 3X4 ≤ 2,350 hours of drilling 2X1 + 4X2 + 1X3 + 2X4 ≤ 2,600 hours of assembly .5X1 + 1X2 + .5X3 + .5X4 ≤ 1,200 hours of inspection X1 ≥ 150 units of XJ201 X2 ≥ 100 units of XM897 X3 ≥ 300 units of TR29 X4 ≥ 400 units of BR788

28. Feed Product Stock X Stock Y Stock Z A 3 oz 2 oz 4 oz B 2 oz 3 oz 1 oz C 1 oz 0 oz 2 oz D 6 oz 8 oz 4 oz LP Applications Diet Problem Example

29. LP Applications X1 = number of pounds of stock X purchased per cow each month X2 = number of pounds of stock Y purchased per cow each month X3 = number of pounds of stock Z purchased per cow each month Minimize cost = .02X1 + .04X2 + .025X3 Ingredient A requirement: 3X1 + 2X2 + 4X3 ≥ 64 Ingredient B requirement: 2X1 + 3X2 + 1X3 ≥ 80 Ingredient C requirement: 1X1 + 0X2 + 2X3 ≥ 16 Ingredient D requirement: 6X1 + 8X2 + 4X3 ≥ 128 Stock Z limitation: X3 ≤ 80 X1, X2, X3 ≥ 0 Cheapest solution is to purchase 40 pounds of grain X at a cost of $0.80 per cow

30. In-Class Problems from the Lecture Guide Practice Problems Problem 1: Chad’s Pottery Barn has enough clay to make 24 small vases or 6 large vases. He has only enough of a special glazing compound to glaze 16 of the small vases or 8 of the large vases. Let X1 = the number of small vases and X2 = the number of large vases. The smaller vases sell for $3 each, and the larger vases would bring $9 each. (a) Formulate the problem (b) Solve the problem graphically

31. In-Class Problems from the Lecture Guide Practice Problems Problem 2: A fabric firm has received an order for cloth specified to contain at least 45 pounds of cotton and 25 pounds of silk. The cloth can be woven out of any suitable mix of two yarns A and B. They contain the proportions of cotton and silk (by weight) as shown in the following table: Material A costs $3 per pound, and B costs $2 per pound. What quantities (pounds) of A and B yarns should be used to minimize the cost of this order?