2. An-Najah N. University Faculty of Engineering and Information Technology Department of Management Information systems Operations Research and Applications MIS:10676213 Prepared by Mr.Maher Abubaker Fall 2015/2016 Resources Daniel J. Epstein Department of Industrial and Systems Engineering University of Southern California Andrew and Erna Viterbi School of Engineering http://www.eecs.qmul.ac.uk/~eniale/teaching/ise330/index.html Operations Research: An Introduction, 9/EHamdy A. Taha, University of ArkansasISBN-10: 013255593X ISBN-13: 9780132555937©2011 Prentice Hall Cloth, 832 ppPublished 08/29/2010 http://www.pearsonhighered.com/educator/product/Operations-Research-An- Introduction/9780132555937.page#downlaoddiv ™ INFORMS – www.informs.org ™ ORMS - www.lionhrtpub.com/ORMS.shtml ™ Science of Better - www.scienceofbetter.org

3. Chapter 2 Modeling with Linear Programming (LP)

4. What is Linear Programming (LP)?  Linear Programming provides methods for allocating limited resources among competing activities in an optimal way.  „ Linear → All mathematical functions are linear  „ Programming → Involves the planning of activities  ‡ Any problem whose model fits the format for the linear programming model is a linear programming problem.

5. Example 1 The Opti Mize Company manufactures two products A and B that compete for the same (limited) resources labor hours, machine hours and daily budget. Relevant information is: Available Resources BAProduct Resources 20 hrs/day21Labor-hrs / unit 30 hrs/day22Machine-hrs / unit $180/day$20$6Cost / unit $20$5Profit / unit Example Solution : See the video, An-Najah N. University e-learning tool on Zajel ( Moodle, Operations Research and Applications, Mr. Maher AbuBaker )

6. Example 1 cont. The LP model, as in any OR model, has three basic components: 1. Decision variables that we seek to determine. 2. Objective (goal) that we need to optimize (maximize or minimize). 3. Constraints that the solution must satisfy.

7. The Model What is the decision?  Decision Variables Let X = number of units of product A to manufacture Y = number of units of product B to manufacture What is the Objective?  Objective Function Max Profit = z = 5 X + 15 Y$ What are the restrictions ?  Constraints subject to: X + 2 Y <= 20 (labor-hours) 2 X + 2 Y <= 30 (machine hours) 6 X + 20Y <= 180 ($ - budget) X >= 0, Y >= 0

8. Example 1 cont. Graphical LP Solution : it includes two steps 1. Determination of the feasible solution space. 2. Determination of the optimum solution from among all the points in the solution space Solution of the Model: Max Z = 5 X + 15 Y$ subject to: X + 2 Y <= 20 (labor-hours) 2 X + 2 Y <= 30 (machine hours) 6 X + 20Y <= 180 ($ - budget) X >= 0, Y >= 0

9. The Graphical Solution X Y 5 1015252030 5 10 20 15 X + 2 Y = 20 9  Because of x>=0 and y>= 0 nonnegativity constraints the drawing restricted to the first quadrant.  To gragh the constraints do the following: x + 2 y <=20  Replace each inequality with an equation. x+2 y =20  Graph the straight line by locating two distinct points.  Setting x=0 to obtain y = 10  ( 0,10)  Setting y=0 to obtain x= 20  (20,0)  The line passes through (0,10) and ( 20,0)

10. The Graphical Solution X Y 5 1015252030 5 10 20 15 X + 2 Y = 20 2X + 2Y = 30 9 6X + 20Y = 180 The feasible region

11. The Graphical Solution (continued) X Y 5 1015252030 10 20 15 9 Z = 5X + 15Y = 30 2 6 4 Z = 5X + 15Y = 60 12 (5, 7.5) Z = 5 (5) + 15 (7.5) = 137.5

12. The Graphical Solution Alternate Approach X Y 5 1015252030 5 10 20 15 9 (x = 5, y = 7.5; z = 137.5) (x = 10, y = 5; z = 125) Z = 5X + 15Y (x = 15, y = 0; z = 75) (x = 0, y = 9; z = 135) (x = 0, y = 0; z = 0)

13. Example 2 The Wyndor Glass Co.  The company produces glass products and owns 3 plants.  Management decides to produce two new products.  Product 1  „ 1 hour production time in Plant 1  „ 3 hours production time in Plant 3 „  $3,000 profit per batch ‡  Product 2  „ 2 hours production time in Plant 2  „ 2 hours production time in Plant 3  „ $5,000 profit per batch  ‡ Production time available each week  „ Plant 1: 4 hours  „ Plant 2: 12 hours „  Plant 3: 18 hours

14. Example 2 The Wyndor Glass Co. Production Time Available per Week, Hours Production Time per Batch, Hours Plant Products 21 4011 12202 18233 $5000$3000Profit Per Batch

15. The Model Let X1 = number of batches of product 1 to manufacture X2 = number of batches of product 2 to manufacture Max Profit = Z = 3 X1 + 5 X2$ subject to: X1 <= 4 (Plant 1-hours) 2 X2 <= 12 (Plant 2-hours) 3 X1 + 2 X2 <= 18 (Plant 3-hours) X1 >= 0, X2 >= 0 Example 2 The Wyndor Glass Co.

16. The Graphical Solution Example 2 The Wyndor Glass Co.

17. General Linear Programming Problems  ‡ Allocating resources to activities GeneralExample Resources m resources Production capacity of plants 3 plants Activities n activities Production of products 2 products Level of activity j,xjProduction rate of product j,xj Overall measure of performance ZProfit Z

18. General Linear Programming Problems

20. Other Forms of Linear Programming Problems

21. Linear Programming Solutions

22. Max Z = x1 + x2 Subject to: x1 <= 4 2 x2 >= 12 3x1 + 2x2 <= 18 x1, x2 >=0

23. Linear Programming Solutions Max Z = 6 x1 + 4 x2 Subject to: x1 <= 4 2 x2 <= 12 3x1 + 2x2 = 18 x1, x2 >=0

24. Linear Programming Solutions Max Z = x1 + x2 Subject to: x1 <= 4 2 x2 >= 12 3x1 + 2x2 >= 18 x1, x2 >=0 Unbounded Solution

25. Linear Programming Solutions

26. Linear Programming Assumptions

27. In-Class Example

28. Radiation Therapy Example

32. Regional Planning Example

39. Real Example: Personal Scheduling at United Airline

40. More Linear Programming Applications  The Classical Diet Problem.  The Blending Problem.  Marketing Problem  Portfolio Analysis Problem  Trim-Loss Problem  Job Training Problem  Bus Scheduling Problem

41. The Classical Diet Problem Mr. U. R. Fatte has been placed on a diet by his Doctor (Dr. Ima Quack) consisting of two fruit: Banana and Apple. The doctor warned him to insure proper consumption of nutrients to sustain life. Relevant information is: NutrientsBananaAppleWeekly Requirement I2 mg/oz3 mg/oz3500 mg II6 mg/oz2 mg/oz7000 mg cost/oz10 cents4.5 cents

42. The Mathematical Model LetX = ounces of banana consumed per week Y = ounces of apple consumed per week Min cost = z = 10 X + 4.5 Y subject to: 2X + 3Y >= 3500 6X + 2Y >= 7000 X, Y >= 0

43. Graphical Solution to the Diet Problem Solution of Minimization Problem X Y 1000 3000 2000 30002000 1000 4000 6x + 2y = 7000 2x + 3y = 3500 Z = 10x + 4.5y = 18000 cents (x = 1000, y = 500; z = 122.50)

44. A Blending Problem The B. A. Nutt Company sells mixed nuts of two quality levels. The expensive mix should not contain more than 25% peanuts nor less than 40% cashews. The cheap mix should not have more than 60% peanuts and no less than 20% cashews. Cashews cost 50 cents a pound and peanuts cost 20 cents a pound. The expensive mix sells for 80 cents a pound and the cheap mix for 40 cents a pound. What should the blend of each mix be in order to maximize profit. The company has $100 a day with which to purchase nuts.

45. The Model Letx 1 = pounds of cashews in expensive mix x 2 = pounds of peanuts in expensive mix y 1 = pounds of cashews in cheap mix y 2 = pounds of peanuts in cheap mix Max z =.80 (x 1 + x 2 ) +.40 (y 1 + y 2 ) -.5(x 1 +y 1 ) -.2(x 2 + y 2 ) subject to:.5 (x 1 + y 1 ) +.2(x 2 + y 2 ) <= 100 x 2 / (x 1 +x 2 ) <=.25y 2 /(y 1 + y 2 ) <=.6 x 1 / (x 1 +x 2 ) >=.40y 1 /(y 1 + y 2 ) >=.20

46. Re-formulate the Model Max z = =.3x 1 +.6x 2 -.1y 1 +.2y 2 subject to:.5x 1 +.2x 2 +.5y 1 +.2 y 2 <= 100 -.25x 1 +.75x 2 <= 0 -.60y 1 +.40y 2 <= 0.60x 1 -.40x 2 >= 0.80y 1 -.20y 2 >= 0

47. A Marketing Example The I. B. Adman Advertising Company is planning a large media blitz covering television, radio, and magazines to sell management science to the public. The company’s objective is to reach as many people as possible. Results of a market survey show: Television Day timePrime TimeRadioMagazines cost per unit$40,00075,00030,00015,000 # people400,000900,000500,000200,000 # business300,000400,000200,000100,000 The company has a budget of $800,000 to spend on the campaign. It requires at least two million exposures among the business community. Television must be limited to $500,000, at 3 units of day time and 2 units of prime time must be purchased. Advertising units on both radio and magazines should be between 5 and 10.

48. The Model Let x 1, x 2, x 3, and x 4 be the number of advertising units bought in daytime TV, primetime TV, radio and magazines. Max z = 400x 1 + 900x 2 + 500x 3 + 200x 4 (in thousands) subj to: 40,000x 1 + 75,000x 2 + 30,000x 3 + 15,000x 4 <_ 800,000 300,000x 1 + 400,000x 2 + 200,000x 3 + 100,000x 4 >= 2,000,000 40,000x 1 + 75,000x 2 <= 500,000 x 1 >= 3; x 2 >= 2;5 <= x 3 <= 10; 5 <= x 4 <= 10

49. Portfolio Analysis E. Z. Credit, the president of the 33rd National Bank and Trust Company desires to maximize the annual return the bank receives on its loans. The following interest rates are charged to customers: Type of LoanPercent Interest commercial 15 home mortgage (first)10 home improvement13.6 home mortgage (second)14 short-term revolving load18 The bank has $53,000,000 available from passbook savings (earning 5%) to invest in the above loans.

50. Dollars inv. In home improvement should be less than or equal 20% of first mortgages inv. Dollars inv. In commercial loans should be less than or equal to dollars inv. In second mortgages. First and second mortgages should be greater than or equal 60% of all investments Dollars inv. In first mortgages should be greater than or equal twice of the second mortgages.

51. The Model Letx 1 = dollars invested in commercial loans x 2 = dollars invested in first mortgages x 3 = dollars invested in home improvement loans x 4 = dollars invested in second mortgages x 5 = dollars invested in short-term signature loans Max z =.15x 1 +.10x 2 +.136x 3 +.14x 4 +.18x 5 subject to: x 1 + x 2 + x 3 + x 4 + x 5 <= 53,000,000 x 3 <=.2x 2 x 1 <= x 4 x 2 + x 4 >=.6 (x 1 + x 2 + x 3 + x 4 + x 5 ) x 2 >= 2x 4 x 5 <= 5,000,000

52. Trim-Loss Problem The I. M. Torn Paper Company produces rolls of paper 12 inches wide by 1000 feet in length. These standard rolls are purchased by many of their customers. However, some customers prefer to receive special sizes, namely 2-inch, 3.5 inch, and 5-inch rolls, all 1000 feet long. The special sized rolls are cut from the standard 12-inch roll. Alternative size of rolls cuts 2-in3.5-in 5-inwaste 16000 21020 32201 40210 53011 6410.5 required/mo50020001500

53. The Model Let x i = the number of rolls to cut according to alternative I Min z = 0x 1 + 0x 2 + 1x 3 + 0x 4 + 1x 5 +.5x 6 subject to: 6x 1 + x 2 + 2x 3 + 3x 5 + 4x 6 >= 500 2x 3 + 2x 4 + x 6 >= 2000 2x 2 + x 4 + x 5 >= 1500 x i >= 0; i = 1,2,3,4,5,6

54. Job-Training The Never-Say-Die Life Insurance Company hires and trains a large number of salespersons each month to replace those who have departed. Trained salespersons must be used to train new salespersons. Training takes one month and there is a 20 percent attrition rate by the end of the month. While a salesperson is training a new employee, that person cannot be used in the field selling insurance. The monthly demand for experienced salespeople is: MonthDemand (in the field) January100 February150 March200 April225 May175 Trainees receive $400 per month while it costs the company $850 per month for an experienced salesperson. Ten percent of all experienced salespeople will leave the company by the end of each month.

55. The Model Let x i = the number of trainees during month I y i = the number of experienced salespeople in month I Min z = 1250 (x 1 + x 2 + x 3 + x 4 + x 5 ) subj to:y 1 - x 1 >= 100 y 2 - x 2 >= 150 y 3 - x 3 >= 200 y 4 - x 4 >= 225 y 5 - x 5 >= 175 y i =.9y i-1 +.8x i-1 for i = 1, 2, 3, 4, 5 example: y feb =.9 y jan +.8x jan

56. Bus Scheduling Problem Progress City is studying the feasibility of introducing a mass transit bus system that will alleviate the smog problem by reducing in-city driving. The study seeks the determination of the minimum number of buses that can handle the transportation needs. The following graph shows some important info. to handle. Each bus can operate 8 hrs. a day 4 8 12 12:00 AM 481248 4 8 10 7 12 4 x1 x2 x3 x4 x5 x6

57. Let x1 = # of buses starting at 12:01 am x2 = # == 4:01 am x3 = # == 8:01 am x4 = # == 12:01 pm x5 = #= 4:01 pm x6 = # == 8:01 pm Min Z = x1 + x2 + x3 + x4 + x5 + x6 S.To: x1 + x6 >= 4 ( 12:01 am – 4:00 am) x1 + x2 >= 8 ( 4:01 am – 8:00 am ) x2 + x3 >= 10( 8:01 am – 12:00 noon ) x3 + x4 >= 7( 12:01 pm – 4:00 pm ) x4 + x5 >= 12( 4:01 pm - 8:00 pm ) x5 + x6 >= 4( 8:01 pm – 12:00 pm ) The Model