1. 1 Teaching Innovation - Entrepreneurial - Global The Centre for Technology enabled Teaching & Learning, N Y S S, India DTEL DTEL (Department for Technology Enhanced Learning)

2. DEPARTMENT OF MECHANICAL ENGINEERING VI-SEMESTER OPERATION RESEARCH 2 CHAPTER NO.1 INTRODUCTION TO OPERATION RESEARCH

3. CHAPTER 1:- SYLLABUSDTEL. Introduction to OR & Basic OR Models, Definition Characteristics and limitations of OR 1 Introduction of linear programming 2 Numericals on LPP by graphical method 3 4 3 Numericals on LPP by simplex method 5

4. CHAPTER 1:- SYLLABUSDTEL. Numericals on LPP by simplex method 6 Introduction to Dual of LPP. 7 Numericals on Sensitivity analysis & formulation of Dual of LPP. 8 4

5. CHAPTER-1 SPECIFIC OBJECTIVE / COURSE OUTCOMEDTEL Understand the Basics of operation research. 1 Understand various OR programming solutions. 2 5 The student will be able to: How to use Graphical and Simplex method in OR. 3

6. LECTURE 1:- INTRODUCTION TO ORDTEL 6 6  Operations Research is an Art and Science  It had its early roots in World War II and is flourishing in business and industry with the aid of computer  Primary applications areas of Operations Research include forecasting, production scheduling, inventory control, capital budgeting, and transportation. INTRODUCTION

7. LECTURE 1:- INTRODUCTION TO ORDTEL 7 7 Operations The activities carried out in an organization. Research The process of observation and testing characterized by the scientific method. Situation, problem statement, model construction, validation, experimentation, candidate solutions. Operations Research is a quantitative approach to decision making based on the scientific method of problem solving. INTRODUCTION

8. LECTURE 1:- INTRODUCTION TO ORDTEL 8 8 What is Operations Research?  Operations Research is the scientific approach to execute decision making, which consists of: The art of mathematical modeling of complex situations The science of the development of solution techniques used to solve these models The ability to effectively communicate the results to the decision maker What is OR?

9. LECTURE 1:- INTRODUCTION TO ORDTEL 9 9 Operations Research Models Deterministic ModelsStochastic Models Linear Programming Discrete-Time Markov Chains Network Optimization Continuous-Time Markov Chains Integer Programming Queuing Theory (waiting lines) Nonlinear Programming Decision Analysis Inventory Models Game Theory Inventory models Simulation Models

10. LECTURE 1:- LINEAR PROGRAMINGDTEL 10 Objective of OR Operational Research always try to find the best and optimal solution to the problem. For this purpose objectives of the organization are defined and analyzed. These objectives are then used as the basis to compare the alternative courses of action. Limitations of Operational Research  Magnitude of Computation  Non-Quantifiable Factors  Distance between User and Analyst  Time and Money Costs  Implementation Objectives and Limitations

11. DTEL 11 THANK YOU LECTURE 1:- LINEAR SYSTEM

12. LECTURE 2:- LINEAR PROGRAMINGDTEL 12 Linear Programming Real-World Problem Recognition and Definition of the Problem Formulation and Construction of the Mathematical Model Solution of the Model Interpretation Validation and Sensitivity Analysis of the Model Implementation Modeling Process Introduction

13. LECTURE 2:- LINEAR PROGRAMINGDTEL 13 What Is a Linear Programming Problem? A linear programming problem (LP) is an optimization problem for which we do the following: 1. Attempt to maximize (or minimize) a linear function (called the objective function) of the decision variables. 2. The values of the decision variables must satisfy a set of constraints. 3. Each constraint must be a linear equation or inequality. 4. A sign restriction is associated with each variable. For each variable x i, the sign restriction specifies either that x i must be nonnegative (x i ≥ 0) or that x i may be unrestricted in sign. Introduction

14. LECTURE 2:- LINEAR PROGRAMINGDTEL 14 Contd…. For a maximization problem, an optimal solution to an LP is a point in the feasible region with the largest objective function value. Similarly, for a minimization problem, an optimal solution is a point in the feasible region with the smallest objective function value. Most LPs have only one optimal solution. However, some LPs have no optimal solution, and some LPs have an infinite number of solutions. Introduction

15. LECTURE 2:- LINEAR PROGRAMINGDTEL 15 Assumptions of Linear Programming Proportionality and Additive Assumptions The objective function for an LP must be a linear function of the decision variables has two implications: 1. The contribution of the objective function from each decision variable is proportional to the value of the decision variable. 2. The contribution to the objective function for any variable is independent of the other decision variables. Assumptions

16. LECTURE 2:- LINEAR PROGRAMINGDTEL 16 Contd…. Each LP constraint must be a linear inequality or linear equation has two implications: 1. The contribution of each variable to the left-hand side of each constraint is proportional to the value of the variable. 2. The contribution of a variable to the left-hand side of each constraint is independent of the values of the variable. Assumptions

17. LECTURE 2:- LINEAR PROGRAMINGDTEL 17 Contd…. Divisibility Assumption The divisibility assumption requires that each decision variable be permitted to assume fractional values. The Certainty Assumption The certainty assumption is that each parameter (objective function coefficients, right-hand side, and technological coefficients) are known with certainty. Assumptions

18. DTEL 18 THANK YOU LECTURE 2:- LINEAR PROGRAMING

19. LECTURE 3:- NUMERICALS ON LPP BY GRAPHICAL METHODDTEL 19 The linear programming model for this example can be summarized as: Numericals

20. LECTURE 3:- NUMERICALS ON LPP BY GRAPHICAL METHODDTEL 20 Graphical Solution to LP Problems Solution

21. LECTURE 3:- NUMERICALS ON LPP BY GRAPHICAL METHODDTEL 21 Contd…… An equation of the form 4x1 + 5x2 = 1500 defines a straight line in the x1-x2 plane. An inequality defines an area bounded by a straight line. Therefore, the region below and including the line 4x1 + 5x2 = 1500 in the Figure represents the region defined by 4x1 + 5x2  1500. Same thing applies to other equations as well. The shaded area of the figure comprises the area common to all the regions defined by the constraints and contains all pairs of xI and x2 that are feasible solutions to the problem. This area is known as the feasible region or feasible solution space. The optimal solution must lie within this region. There are various pairs of x1 and x2 that satisfy the constraints such as: Solution

22. DTEL 22 THANK YOU LECTURE 3:- NUMERICALS ON LPP BY GRAPHICAL METHOD

23. LECTURE 4:- NUMERICALS ON LPP BY GRAPHICAL METHODDTEL 23 Contd…… Trying different solutions, the optimal solution will be: X1 = 270 X2 = 75 This indicates that maximum income of $4335 is obtained by producing 270 units of product I and 75 units of product II. In this solution, all the raw material and available time are used, because the optimal point lies on the two constraint lines for these resources. However, 1500- [4(270) + 5(75)], or 45 ft2 of storage space, is not used. Thus the storage space is not a constraint on the optimal solution; that is, more products could be produced before the company ran out of storage space. Thus this constraint is said to be slack. Solution

24. LECTURE 4:- NUMERICALS ON LPP BY GRAPHICAL METHODDTEL 24 Contd…… If the objective function happens to be parallel to one of the edges of the feasible region, any point along this edge between the two extreme points may be an optimal solution that maximizes the objective function. When this occurs, there is no unique solution, but there is an infinite number of optimal solutions. The graphical method of solution may be extended to a case in which there are three variables. In this case, each constraint is represented by a plane in three dimensions, and the feasible region bounded by these planes is a polyhedron. Solution

25. DTEL 25 THANK YOU LECTURE 4:- NUMERICALS ON LPP BY GRAPHICAL METHOD

26. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 26 Problem-The N. Dustrious Company produces two products: I and II. The raw material requirements, space needed for storage, production rates, and selling prices for these products are given below: The total amount of raw material available per day for both products is 15751b. The total storage space for all products is 1500 ft2, and a maximum of 7 hours per day can be used for production. The company wants to determine how many units of each product to produce per day to maximize its total income. Problem

27. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 27 Solution- Step 1: Convert all the inequality constraints into equalities by the use of slack variables. Let: As already developed, the LP model is: Solution

28. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 28 Contd.. Introducing these slack variables into the inequality constraints and rewriting the objective function such that all variables are on the left-hand side of the equation. Equation 4 can be expressed as: Solution

29. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 29 Contd.. Since the coefficients of x1 and x2 are both negative, the value of Z can be increased by giving either x1 or x2 some positive value in the solution. if x2 = S1 = 0, then x1 = 1500/4 = 375. That is, there is only sufficient storage space to produce 375 units at product I. there is only sufficient raw materials to produce 1575/5 = 315 units of product I. there is only sufficient time to produce 420/1 = 420 units of product I. Therefore, considering all three constraints, there is sufficient resource to produce only 315 units of x1. Solution

30. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 30 Contd.. Step 2: From Equation CI, which limits the maximum value of x1. Substituting this equation into Eq. (5) yields the following new formulation of the model. Solution

31. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 31 Contd.. It is now obvious from these equations that the new feasible solution is: x1 = 315, x2 = 0, S1 = 240, S2 = 0, S3 = 105, and Z = 4095 It is also obvious from Eq.(A2) that it is also not the optimum solution. The coefficient of x1 in the objective function represented by A2 is negative ( -16/5), which means that the value of Z can be further increased by giving x2 some positive value. Solution

32. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 32 Contd..  Following the same analysis procedure used in step 1, it is clear that:  In Eq. (B2), if S1 = S1 = 0, then x2 = (5/13)(240) = 92.3.  From Eq. (C2), x2 can take on the value (5/3 )(315) = 525 if x1 = S2 = 0  From Eq. (D2), x2 can take on the value (5/7)(105) = 75 if S2 = S3 = 0  Therefore, constraint D2 limits the maximum value of x2 to 75. Thus a new feasible solution includes x2 = 75, S2 = S3 = 0. Solution

33. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 33 Contd.. Step 3: From Equation D2: Substituting this equation into Eq. (7) yield: From these equations, the new feasible solution is readily found to be: x1 = 270, x2 = 75, S1 = 45, S2 = 0, S3 = 0, Z = 4335. Solution

34. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 34 Contd.. Because the coefficients in the objective function represented by Eq. (A3) are all positive, this new solution is also the optimum solution. Solution

35. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 35 Simplex Tableau for Maximization Contd.. Step I: Set up the initial tableau using Eq. (5). In any iteration, a variable that has a nonzero value in the solution is called a basic variable. Solution

36. LECTURE 5:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 36 Contd.. Step II: Identify the variable that will be assigned a nonzero value in the next iteration so as to increase the value of the objective function. This variable is called the entering variable. It is that nonbasic variable which is associated with the smallest negative coefficient in the objective function. If two or more nonbasic variables are tied with the smallest coefficients, select one of these arbitrarily and continue. Step III: Identify the variable, called the leaving variable, which will be changed from a nonzero to a zero value in the next solution. Solution

37. DTEL 37 THANK YOU LECTURE 5 :- NUMERICALS ON LPP BY SIMPLEX METHOD

38. DTEL 38 Contd.. Step IV: Enter the basic variables for the second tableau. The row sequence of the previous tableau should be maintained, with the leaving variable being replaced by the entering variable Solution

39. LECTURE 6:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 39 Contd.. Step V: Compute the coefficients for the second tableau. A sequence of operations will be performed so that at the end the x1 column in the second tableau will have the following coefficients: The second tableau yields the following feasible solution: x1 = 315, x2 = 0, SI = 240, S2 = 0, S3 = 105, and Z = 4095 Solution

40. LECTURE 6:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 40 Contd.. Step VI: Check for optimality. The second feasible solution is also not optimal, because the objective function (row A2) contains a negative coefficient. Another iteration beginning with step 2 is necessary. In the third tableau (next slide), all the coefficients in the objective function (row A3) are positive. Thus an optimal solution has been reached and it is as follows: x1 = 270, x2 = 75, SI = 45, S2 = 0, S3 = 0, and Z = 4335 Solution

41. LECTURE 6:- NUMERICALS ON LPP BY SIMPLEX METHODDTEL 41 Contd.. Solution

42. DTEL 42 THANK YOU LECTURE 6 :- NUMERICALS ON LPP BY SIMPLEX METHOD

43. LECTURE 7:- INTRODUCTION TO DUEL OF LPPDTEL 43  For every LP formulation there exists another unique linear programming formulation called the 'Dual' (the original formulation is called the 'Primal').  The - Dual formulation can be derived from the same data from which the primal was formulated.  The Dual formulated can be solved in the same manner in which the Primal is solved since the Dual is also a LP formulation. Introduction

44. LECTURE 7:- INTRODUCTION TO DUEL OF LPPDTEL 44  The Dual can be considered as the 'inverse' of the Primal in every respect.  The column coefficients in the Primal constrains become the row coefficients in the Dual constraints.  The coefficients in the Primal objective function become the right hand side constraints in the Dual constraints.  The column of constants on the right hand side of the Primal constraints becomes the row of coefficients of the dual objective function.  The 'direction of the inequalities are reversed. If the primal objective function is a 'Maximization' function then the dual objective function is a 'Minimization' function and vice-versa. Introduction

45. LECTURE 7:- INTRODUCTION TO DUEL OF LPPDTEL 45 Steps Dual Formation  For each constraint in primal problem there is an associated variable in dual problem.  The elements of right hand side of the constraints will be taken as the co-efficient of the objective function in the dual problem.  If the primal problem is maximization, then its dual problem will be minimization and vice versa.  The inequalities of constraints should be interchanged from >= to <= and vice versa and the variables in both the problems and non-negative.  The row of primal problem are changed to columns in the dual problem. In other words the matrix A of the primal problem will be changed ti its transpose (A) for the dual problem.  The co-efficient of the objective function will be taken the right hand side of the constraints of the dual problem. Steps

46. LECTURE 7:- INTRODUCTION TO DUEL OF LPPDTEL 46  With every linear programming problem, there is associated another linear programming problem which is called the dual of the original (or the primal) problem. Duality Formulating the Dual problem  Consider again the production mix problem of N. Dustrious Company.  Suppose that the company is considering leasing out the entire production facility to another company, and it must decide on the minimum daily rental price that will be acceptable.  This decision problem can also be formulated as a linear programming problem. Duality

47. LECTURE 7:- INTRODUCTION TO DUEL OF LPPDTEL 47 Duality Let y1, y2 and y3 represent the unit price of each unit of storage space, raw materials, and production time, respectively. The unit prices are in fact the income values of each unit of resource to the N. Dustrious Company. There are available 500 ft2 of storage space, 1575 lb of raw materials, and 420 minutes of production time per day. Thus the total income value (P) of all the available resources may be expressed as follows : P = 1500y1 + 1575y2 + 420y3 The objective of the problem is to minimize P subject to the condition that the N. Dustrious Company will earn at least as much income as when it operates the production facility itself Duality

48. LECTURE 7:- INTRODUCTION TO DUEL OF LPPDTEL 48 Contd…  Since the market value (or selling price) of 1 unit of product I is $13 and it requires 4 ft2 of storage space, 5 lbs of raw materials, and 1 minute of production time, the following constraint must be satisfied: 4y1 + 5y2 + 5y3  13  Similarly, for Product II: 5y1 + 3y2 + 2y3  11  In addition, the unit prices y1, y2 and y3 must all be greater than or equal to zero. Duality

49. DTEL 49 THANK YOU LECTURE 7 :- INTRODUCTION TO DUEL OF LPP

50. LECTURE 8:- NUMERICAL ON DUEL OF LPPDTEL 50 Consider the following primal problem : Solution-The first inequality requires no modification. Problem

51. LECTURE 8:- NUMERICAL ON DUEL OF LPPDTEL 51 : The second inequality can be changed to the less-than- or-equal-to type by multiplying both sides of the inequality by -1 and reversing the direction of the inequality; that is, The equality constraint can be replaced by the following two inequality constraints: If both of these inequality constraints are satisfied, the original equality constraint is also satisfied Solution

52. LECTURE 8:- NUMERICAL ON DUEL OF LPPDTEL 52 : Multiplying both sides of the inequality by –1 and reversing the direction of the inequality yields The primal problem can now take the following standard form: Solution

53. LECTURE 8:- NUMERICAL ON DUEL OF LPPDTEL 53 : The dual of this problem can now be obtained as follows: Solution

54. DTEL 54 THANK YOU LECTURE 8 :- NUMERICAL ON DUEL OF LPP