1. 

2. Introduction to Operations Research

3. 1- Definition of Operations Research Operations research is the science of decision making and its applications. 2- Operations Research and Decision Making Operations research or management science, as the name suggests, is the science of managing. As is known, management is the most of the time making decisions. It is thus a decision science which helps management to make better decisions. The essential characteristics of all decision are: (i) Objectives (ii) alternatives,

4. (iii) Influencing factors (constraints). 3- Scope of OR in Modern Management Some of the areas of management where OR techniques have been successfully application are: Allocation and Distribution Production and Facility Planning Procurement Marketing Finance Personnel Research and Development

5. 4- Phases of Operations Research or Operations Research Approach or How Operations Research Works Operations research, like all scientific research, is based on scientific methodology, which proceeds along the following lines: Formulating the problem. Constructing a model to represent the system under study. Deriving a solution from the model. Testing the model and the solution derived from it.

6. Establishing controls over the solution. Putting the solution to work, i.e., implementation. Research and Development. Optimization Problems In an optimization problem one seeks to optimize (i.e., maximize (max) or minimize (min)) a specific quantity, called the objective function, which depends on a finite number of input variables. There variables may be independent of one another, or they may be related through one or more constraints. A mathematical program is an optimization problem in which the objective and constraints are given a

7. mathematical functions and functional relationships. Mathematical programs treated here have the form Objective function Minimize (min) Cost Time Deterioration Maximize (max) Profit Efficiency Effectiveness Constraints

8. Variables Unrestricted in sign rarely most common Optimize subject to (1)

9. Mathematical program (1) is called constrained mathematical program. If each function g i ( i = 1, …, m) is chosen as zero and each constant b i ( i = 1, …, m) is chosen zero, then mathematical program (1) is called unconst- rained mathematical program. Linear Programs A mathematical program (1) is linear if f (x 1, x 2, …, x n ) and each g i (x 1, x 2, …, x n ) ( i = 1, …, m) are linear in each of their arguments-that is, if (2) and (3)

10. where c j and a i j (i = 1, …, m; j = 1, …, n) are known constants. Any other mathematical program is nonlinear. Integer Programs An integer program is a linear program with the additional restriction that the input variables be integers. It is not necessary that the coefficients in (2) and (3), and the constants in (1) (i.e., b i, i = 1, 2, …, m), also be integers, but this will very often be the case. Problem Formulation Optimization problems most often are stated verbally. The solution procedure is to model the problem

11. with mathematical program and then solve the program by the techniques described later. The following approach is recommended for transforming a word problem into a mathematical program. Step 1: Determine the quantity to be optimized and express it as a mathematical function. Doing so serves to define the input variables. Step 2: Identify all stipulated requirements, restrictions, and express them mathematically. These requirem- ents constitute the constraints. Step 2: Express any hidden conditions. Such conditions are not stipulated explicity in the problem but are

12. apparent from the physical situation being modeled. Generally they involve non negativity or integer requirements on the input variables. The Village Butcher Shop traditionally makes its meat loaf from a combination of lean ground beef and ground pork. The ground beef contains 80 percent meat and 20 percent fat, and costs the shop 80 € per pound; the ground pork contains 68 percent meat and 32 percent fat, and costs 60 € per bound. How much of each kind of meat should the shop use in each pound of meat loaf if it wants to minimize its cost and to keep the fat content of the meat loaf to no more than 25 percent? Example 1

13. Let x 1 = poundage of ground beef sued in each pound of meat loaf, and Solution x 2 = poundage of ground pork used in each pound of meat loaf Mathematical modelL. P. P subject to

14. A furniture maker has 6 units of wood and 28 h of free time, in which he will make decorative screens. Two models have sold well in the past, so he will restrict himself to those two. He estimates that model I requires 2 units of wood and 7 h of time, while model II requires 1 unit of wood and 8 h of time. The prices of the models are $ 120 and $ 80, respectively. How many screens of each model should the furniture maker assemble if he wishes to maximize his sales revenue? Example 2 Let x 1 = number of model I screens to be produced, and Solution

15. x 2 = number of model II screen to be produced Mathematical model subject to References [1] Gupta, K. P., and Hira, S. P., (1991), Problems in Operations Research (Principles and Solutions), S. Chand and Company Ltd (Ramanagar, New Delhi, 110055). [2] Richard, B. (1982), Theory and Problems of Operations Research, Schaum ’ s Outline Series (McGraw- Hill Book Company Singapore).

16. (Production Allocation Problem) Example 3 A firm produce three products. These products are processed on three different machines. The time required to manufacture one unit of each of the three products and the daily capacity of the three machines are given in the table below. Machine capacity (minutes/day) Time per unit (minutes) Machine Product 3Product 2Product 1 440232M1M1 4703-4M2M2 430-52M3M3 Table 1

17. it is required to determine the daily number of units to be manufactured for each product. The profit per unit for product 1, 2 and 3 in L. E., 4; L. E., 3 and L. E., 6, respectively. It is assumed that all the amounts produced are consumed in the market. Let x 1 = product 1, x 2 = product 2, and x 3 = product 3 Solution subject to

18. (Diet Problem) A person wants to decide the constituents of a diet which will fulfill his daily requirements of proteins, fats and carbohydrates at the minimum cost. The choose is to made from four different types of foods. The yields per unit of these foods are given in table below Example 4 Cost/unit (L. E.) Yield per unit Food type CarbohydrateFatsProtein 456231 404242 857783 654564 700200800Minimum requirement Table 2

19. subject to Transportation Models Transportation models deal with problems concerning as to what happens to the effectiveness function when we associate each of a number of origins (sources) with each of possibly different number of destination (jobs).

20. The total movement from each origin and the total movement to each destination is given and its desired to find two the associations be made subject to the limitions on totals. In such problems, sources can be divided among the jobs or jobs may be done with a combination of sources. The distinct features of transportation problems is that sources and jobs must be expressed in terms of only one kind of unit. Suppose that three are m sources and n destinations. Let a i ( i = 1, 2, …, m) be the number of supply units available to source i and let b j ( j = 1, 2, …, n) be the number of demand units required at destination j. Let c i j represent the per unit transportation cost for transporting the unite

21. from source i to destination j. The objective is to determine the number of units to be transported from source i to destination j so that the total transportation cost is minimum. If x i j (x i j ≥ 0) is the number of units shipped from source i to destination j, then the equivalent L. P. model will be subject to

22. s1s1 s2s2 smsm D1D1 D2D2 DnDn Stores/Supplies/Origins Customers/Demands/Destinations

23. Example 5 A factory has three stores of a certain product P: S 1, S 2, and S 3 which contain 800, 700 and 1000 units of P, respectively. It received an order to supply two customers C 1 and C 2 which an amount of 1100 and 1400 units, respectively. The distance between stores and customers in k m are given in the following table: C2C2 C1C1 C S 800400300S1S1 700300200S2S2 1000500800S3S3 14001100 Table 3

24. If it is assumed that the cost of transportation is a fixed amount per unit per k m. Construct the mathematical model for minimizing the total transportation cost. Solution 800 700 1000 1100 1400 s1s1 s2s2 smsm C1C1 C2C2 2500 Supply = demand (balanced)

25. Let x i j (x i j ≥ 0) be the amount of the product transported from store i (i = 1, 2 and 3) to customer (j = 1, and 2) subject to Since the amount of units available in the stores is equal to he amount heeded by the customers, therefore we can commete one of these equations. Remark

26. Consider a L. P. P in a compact form subject to (4) Objective function (5) (6) feasible domain (region or area) (Solution): Any n-real numbers satisfy the constraints (5) is called a solution. Definition 1 (Feasible solution): Any solution which satisfies the constraints (5) and (6) is called a feasible solution. Or, Definition 2

27. any solution of system (5) for which all the variables are non-negative is called a feasible solution. (Basic solution): Any solution of system (5) for which at least n – m of its variables are zero is called a basic solution. Definition 3 (Basic feasible solution): It is a solution of system (5) for which at least n – m of its variable are zero and the non-zero variables are positive. Definition 4 (Degenerate basic feasible solution): It is a solution of system (5) for which the numbers of its zero variables are greater than n – m and the non-zero variables are positive. Definition 5

28. (Non degenerate basic feasible solution): The basic feasible solution which has got exactly m positive x i is called non degenrate basic feasible solution Definition 6 (Optimal solution): A feasible solution which optimize the objective function is called an optimal solution. Definition 7 (Optimal basic feasible solution): A basic feasible solution for which the objective function is optimal is called optimal basic feasible solution. Definition 8 Any basic feasible solution is an extreme point (vertix) of the closed convex set of the feasible solutions. Theorem 1

29. Find the extreme points of the following set analytically: Example 6 Solution where x 3 and x 4 are called slack variables.

30. Extreme pointx4x4 x3x3 x2x2 x1x1 (0, 0)43001 -20302 (0, 2)01203 (3, 0)10034 0045 (2, 1 )00126 Table 4

31. Find the extreme points of the following set analytically: Assignment 1

32. Geometry of L. P. P. (Graphical solution) Solve the following L. P. P. graphically: Example 7 subject to

33. 2 4 6 8 10 12 14 16 18 20 22 24 30 28 26 (1) 2 04 6 8 10 12 14161820 2224 26 2830 x1x1 x2x2 Solution (0, 30) (0, 12) (0, 3) (4) (5) (3) (2) Feasible region

34. There are five extreme points (vertices): (3, 3), (12, 12), (18, 12), (20, 10), and (20, 3) subject to The optimal solution is

35. (2) There are five extreme points: (0, 0), (0, 5), (2, 4), (4, 2), and (4, 0) Solution 1 2 3 4 5 6 (0, 6) 02 3 4 5 6 78910 1 x1x1 x2x2 Feasible region (1) (3) (10, 0) (0, 5) (4, 2) (4, 0) (6, 0)

36. subject to The optimal solution is (non-unique optimal solution or multiple optimal solution.

37. Solution -5 -4 -3 -2 0 1234 -6 x1x1 1 2 3 4 x2x2 (1) (2) (-6, 0) (0, 2) (0, 3) (-2, 0) unbounded Unbounded solution

38. subject to

39. x2x2 (1) (3) (2) 0 1 2 3 4567 x1x1 (0, 4) (2, 0) 1 2 3 4 5 6 7 (0, 7) Infeasible solution or non feasible solution

40. Solve the following L. P. Ps. Graphically: Assignment 2 subject to

41. The simplex Method Phase I Phase II Objective function min constraints ≤ tue variables ≥ 0. Solve the following L. P. Ps using the simplex: Example 8 subject to

42. The problem can be stated as: Solution subject to By introducing the stock variables x 4, x 5 and x 6 ≥ 0, then the system of equations can be stated in the canonical forms

43. non basic variables: (x 1, x 2, x 3 ) = (0, 0, 0), basic variables: (x 4, x 5, x 6 ) = (2, 6, 6), initial basic feasible solution. 1- Unique optimal solution: When all the coefficients of the non-basic variables in the z-equation are positive (i.e., c j > 0 ).

44. x6x6 x5x5 x4x4 x3x3 x2x2 x1x1 B. Vs 2001(1)2x4x4 601052x5x5 100114x6x6 0000 -2(-z) Table 5 Round (0) 2- Non-unique optimal solution: When at least one of the coefficients of the non-basic variables in the z-equation say c k = 0, while all the other coefficients are positive.

45. x6x6 x5x5 x4x4 x3x3 x2x2 x1x1 B. Vs 200112x2x2 8011(4)04x5x5 4101202x6x6 002-303(-z) Table 6 Round (1) 3- Improving optimal solution: When at least one of the coefficients of the non-basic variables in the z-equation is negative.

46. x6x6 x5x5 X4X4 x3x3 x2x2 x1x1 B. Vs 40¼5/4013x2x2 20¼1/4101x5x5 01-1/2-3/2000x6x6 1003/4 * 11/4 * 006*6* (-z) Table 7 Round (2) 4- Unbounded solution (No solution): when one of the coefficients of the non-basic variables in the z-equations say c k < 0, while all the elements above this column less than or equal to zero, in this case we have unbounded solution or we can say that the optimal solution obes not exist (D. N. E). The optimal solution is z * = 10 at:

47. subject to Solution subject to non basic variables: (x 1, x 2 ) = (0, 0), basic variables: (x 3, x 4 ) = (1, 6).

48. x4x4 x3x3 x2x2 x1x1 B. Vs 101(1)x3x3 610-23x4x4 000 -3z Table 8 Round (0) x4x4 x3x3 x2x2 x1x1 B. Vs 1011x1x1 31-3(1)0x4x4 303-50z Table 9 Round (1)

49. x4x4 x3x3 x2x2 x1x1 B. Vs 41-301x1x1 31-210x2x2 185-12 * 00z Table 10 Round (2) Unbounded solution. subject to

50. Solution Non basic variables: (x 1, x 2 ) = (0, 0), basic variables: (x 3, x 4, x 5 ) = (2500, 2000, 900). x5x5 x4x4 x3x3 x2x2 x1x1 B. Vs 25000015[10]x3x3 2000010(10)4x4x4 90010032x5x5 0000-100-40(-z) Table 11 Round (0)

51. x5x5 x4x4 x3x3 x2x2 x1x1 B. Vs 15000-1/2108x3x3 20001/10014/10x2x2 3001-3/10008/10x5x5 -20000010000*0* (-z) Table 12 Round (1) x5x5 x4x4 x3x3 x2x2 x1x1 B. Vs 1500/801/161/801x1x1 12501/8-1/2010x2x2 1501-1/4-1/1000x3x3 2000010000(-z) Table 13 Round (2)

52. An infinite number of optimal solutions can be obtained by taking any weighted average of the two solutions as: It can be verified that solution x * will always give the same value of -20000 for z for all 0 ≤ w ≤ 1.

53. Solve the following L. P. Ps. Using simplex method: Assignment 2 subject to

54. Duality Theory Duality is an extremely important and interesting feature and L. P. The various useful aspects of this property are: 1- If the primal problem contains a large number of rows (constraints) and a smaller number of columns (variab- les), the computational procedure can be considerably reduced by converting it into dual and then solving it. Hence it offers an advantage in many applications, 2- It gives additional information as to how the optimal solution changes as a result of the changes in the coefficients and the formulation of the problem. This termed as post optimality or sensitivity analysis,

55. 3- duality in LP has certain for reaching consequences of economic nature, 4- calculation of the dual checks the accuracy of the primal solution. subject to The general L. P. p in canonical form is written as: If the above problem is referred to as primal, then its associated dual will be

56. subject to The above pair of programs can be written as

57. Dual (D)Primal (P) subject to and y j ≥ 0, i = 1, 2, …, m. subject to and x j ≥ 0, j = 1, 2, …, n.

58. Write the dual of the primal problems given below or consider the dual to the primal problems Example 9 subject to Dual (D)Primal (P) (1)

59. subject to Dual (D)Primal (P) (2)

60. Construct the dual of the problem Assignment 3 subject to