Operations Research
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  • 1. OPERATIONS RESEARCH CDR NK NATARAJAN
  • 2. ORIGIN OF OR
    • DATES BACK TO 1800s
    • EARLY CONTRIBUTORS INCLUDE FW TAYLOR, HENRY L GANTT, AK ERLANG ETC.
    • DURING WW II, IT WAS WIDELY USED TO FIND OUT EFFECTIVE ALLOCATION OF LIMITED MILITARY RESOURCES TO VARIOUS OPERATIONS.
  • 3. ORIGIN OF OR
    • POST WW II, INDUSTRIES STARTED ADOPTING THE TECHNIQUE FOR OPTIMIZATION PROBLEMS
    • IN INDIA OR CAME IN TO EXISTANCE IN 1949 AT REGIONAL RESEARCH LABORATORY IN HYERABAD.
  • 4. CHARECTERISTICS OF OR
    • SYSTEM ORIENTATION
    • USES INTER DISCIPLINARY TEAMS
    • APPLICATION OF SCIENTIFIC METHODS
    • IMPROVEMENT IN THE QUALUTY OF DECISION
    • USE OF COMPUTER
    • QUANTITATIVE SOLUTIONS
  • 5. SCOPE OF OR
    • Industry – product mix, blending, inventory control, demand forecasting, transportation, repair and maintenance, scheduling and sequencing, control of projects etc.
    • Defence – Allocation of resources to various operations, logistics, movement of personnel etc.
    • Planning – growth of per capita income, greatest impact of various projects, employment etc.
  • 6. SCOPE OF OR
    • Agriculture – Optimum allocation of land for various crops, water, etc to maximize yield
    • Public Utilities – Hospitals to reduce waiting time, telephone services, public transportation, insurance sector etc.
    • Management –
      • Allocation and Distribution
      • Production and facility Planning
      • Procurement
      • Marketing
      • Finance
      • Personnel
      • Research and Development
  • 7. OBJECTIVES OF OR
    • To provide a scientific basis to the managers of an organization for solving problems involving interaction of the components of the system, by employing a systems approach by a team of scientists drawn from different disciplines for finding a solution which is in the best interest of the organization as a whole.
  • 8. PHASES OF OR
    • Formulate the problem
    • Construct a model to represent the system under study
    • Deriving a solution from the model
    • Testing the model and the solution derived from it
    • Establishing controls over the solution
    • Implementing the solution
  • 9. CONSIDER THIS
    • You have five week business commitment between Mumbai and Kochi. You fly out of Mumbai on Mondays and return on Wednesdays. A regular round trip ticket costs Rs 4000, but a 20% discount is granted if the dates of the ticket span a weekend. A one way ticket in either direction costs 75% of the regular price. How should you buy the tickets for the five week period.
      • What are the decision alternatives?
      • Under what restrictions is the decision made?
      • What is an appropriate objective criteria for evaluating the alternatives?
  • 10. CONSIDER THIS
    • Decision Alternatives
      • 5 regular Mumbai- Kochi- Mumbai tickets for departure on Monday and return on Wednesday of the same week
      • One Mumbai – Kochi, Four Kochi – Mumbai _ Kochi that span weekends and one Kochi – Mumbai ticket
      • One Mumbai – Kochi – Mumbai to cover Monday of the first Week and Wednesday of the last week and four Kochi - Mumbai - Kochi to cover the remaining legs.
  • 11. CONSIDER THIS
    • Cost Alternatives
      • Alternative 1 - 5X4000 = 20000
      • Alternative 2 - .75 X 4000 + 4 (.80 X 4000) + .75 X 4000 = 18800
      • Alternative 3 – 5 X (.8 X 4000) = 16000
    • Alternative 3 is the best option as it minimizes the travel cost.
  • 12. CONSIDER THIS
    • Consider making a maximum area rectangle out of a piece of wire of length ‘L’ inches. What should be the width and height of the rectangle.
      • Let ‘W’ be the width of the rectangle in inches and
      • ‘ H” be the height of the rectangle in inches
    • Based on these
      • Width + Height = Half the length of the wire
      • Width and Height can not be negative
    • Algebraically
      • 2(W+H)=L
      • W ≥ 0; H ≥ 0
  • 13. CONSIDER THIS
    • What is the objective?
      • Maximization of the area of the rectangle. Let ‘Z’ be the area of the rectangle. Then the model becomes
      • Maximize Z =WH
      • Subject to
        • 2(W+H) = L
        • W,H ≥ 0
  • 14. APPROACH OF OR
    • In OR we either maximize or minimize an objective function subject to the various constraints.
    • A solution of the model is feasible only if it satisfied all the constraints.
    • It is optimal if in addition to being feasible, it yields the best (Max or Min) value of the objective function
  • 15. TYPES OF OR MODELS
    • Mathematical models
    • Statistical models
    • Inventory Models
    • Allocation Models
    • Sequencing Models
    • Project Scheduling Models
    • Routing Models
    • Competitive Models
    • Queuing Models
    • Simulation Techniques
    • Decision Theory
    • Replacement Models
    • Reliability Theory
    • Markov Analysis
    • Advanced OR Models
    • Combined Methods
  • 16. LINEAR PROGRAMMING
    • Can be used only if
      • There is a well defined objective function (Profit, Cost or quantities produced) which is to be maximized or minimized and which can be expressed as a linear function of decision variables.
      • There must be constraints on the amount or extent of attainment of the objective and these constraints must be capable of being expressed as linear equations or inequalities in terms of variables.
  • 17. LINEAR PROGRAMMING
    • Can be used only if
      • There must be alternative courses of action. For example, a given product may be processed by two different machines and problem may be as to how much of the product to allocate to which machine.
      • Decision variables should be interrelated and non-negative. The non-negativity condition shows that the linear programming deals with real life situations
      • Resources must be in limited supply
  • 18. FORMULATION OF LP PROBLEM
    • Production allocation problem
    • A firm produces 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 next slide.
    • It is required to determine the daily number of units to be manufactured for each product. The profit per unit of product 1,2 and 3 is Rs 4,3&6 respectively. It is assumed that all the amounts produced are consumed in the market. Formulate the LP model that will maximize the daily profit.
  • 19. FORMULATION OF LP PROBLEM Machine Time per unit (minutes) Machine Capacity (Minutes/Day) Product 1 Product 2 Product 3 M1 M2 M3 2 4 2 3 - 5 2 3 - 440 470 430
  • 20. FORMULATION OF LP PROBLEM Step 1 Find the key decision to be made. The key decision is to decide the extent of product 1,2&3 to be produced as this can vary. Step 2 Assume symbols for the extent of production. Let the extent of Product 1,2&3 be X1, X2 & X3. Step 3 Express the feasible alternatives mathematically in terms of variables. Feasible alternatives are those which are physically, economically and financially possible. In this example, feasible alternatives are sets of values of x1, x2 & x3, where x1,x2 &x3 ≥ 0 since negative production has no meaning and is not feasible.
  • 21. FORMULATION OF LP PROBLEM Step 4 Mention the object quantitatively and express it as a linear function of variables. IN the present example, objective is to maximize the profit. i.e. Maximize Z = 4x1+3x2+6x3 Step 5 Express the constraints as linear equations/inequalities in terms of variables. Here, constraints are o the machine capacities and can be mathematically expressed as 2x1 + 3x2 + 2x3 ≤ 440, 4x1 + 0x2 + 3x3 ≤ 470, 2x1 + 5x2 + 0x3 ≤ 430.
  • 22. FORMULATION OF LP PROBLEM
    • Diet Problem
    • A person wants to decide the constituents of a diet which will fulfill daily requirements of proteins, fats and carbohydrates at the minimum cost. The choice is to be made from four different types of foods. The yields per unit of these foods are given in a table in the next slide.
    • Formulate a linear programming model for the problem.
  • 23. FORMULATION OF LP PROBLEM Food Type Yield per unit Cost per unit (Rs) Protein Fats Carbohydrates 1 2 3 4 3 4 8 6 2 2 7 5 6 4 7 4 45 40 85 65 Min Requirement 800 200 700
  • 24. FORMULATION OF LP PROBLEM Let x1, x2, x3 and x4 denote the number of units of food of type 1,2,3 & 4 respectively. Objective is to minimize the cost i.e. Minimize Z = 45x1+40x2+85x3+65x4 Constraints are on the fulfillment of the daily requirements of various constituents i.e. Proteins - 3x1 + 4x2 + 8x3 + 6x4 ≥ 800 Fats - 2x1 + 2x2 + 7x3 + 5x4 ≥ 200, Carbohydrates - 6x1 + 4x2 + 7x3 + 4x4 ≥ 700. Where x1,x2,x3,x4 each ≥ 0
  • 25. FORMULATION OF LP PROBLEM
    • Blending Problem
    • A firm produces an alloy having the following specifications:
      • Specific gravity ≤ 0.98
      • Chromium ≥ 8%
      • Melting Point ≥ 450ºC
    • Raw materials A,B and C having the properties shown in the table can be used to make the alloy. Cost of various raw materials per ton are : Rs. 90 for A, Rs. 280 for B and Rs. 40 for C. Formulate the LP model to find the proportions in which A,B and C be used to obtain an alloy of desired properties while the costs of raw materials is minimum.
  • 26. FORMULATION OF LP PROBLEM Property Properties of raw materials A B C Specific Gravity Chromium Melting Point 0.92 7% 440 ºC 0.97 13% 490 º C 1.04 16% 480 ºC
  • 27. FORMULATION OF LP PROBLEM Let the percentage contents of raw materials A,B and C to be used for making the alloy be x1, x2 and x3 respectively. Objective is to minimize the cost i.e. Minimize Z = 90x1+280x2+40x3 Constraints are imposed by the specifications required for the alloy. i.e. 0.92x1 + 0.97x2 + 1.04x3 ≤ 0.98 7x1 + 13x2 + 16x3 ≥ 8, 440x1 + 490x2 + 480x3 ≥ 450. x1+x2+x3 = 100 as x1,x2 and x3 are the percentage contents of materials A,B and C in making the alloy. Also x1,x2,x3 each ≥ 0
  • 28. FORMULATION OF LP PROBLEM
    • Product Mix Problem
    • A truck company has Rs. 50 Lakhs to invest and is to choose among three types of trucks A,B & C. Truck A has 12 tonne payload and is expected to average 50 Km per hour. It costs Rs 80,000. Truck B has a 20 tonne payload, is expected to average 45 Km per hour and costs Rs. 1,00,000. Truck C is a modified form of B. It has sleeping space for the driver, which reduces its payload capacity to 17 tonnes, while raising the cost to Rs 1,20,000.
    • Truck A requires a crew of one man and if driven on three shifts per day, could run for an average of 20 hours a day. Truck B & C require a crew of two men each and if driven on three shifts a day could be run for an average of 19 hours and 22 hours respectively. The company has a fleet of 120 crewmen available to it. If the total number of trucks are not to exceed 40, how many trucks of each type should be purchased if the company wants to maximize its capacity in Tonne - KM per day? Formulate the problem as LP problem.
  • 29. FORMULATION OF LP PROBLEM Let the number of trucks of type A,B and C to be purchased be x1, x2 and x3 respectively. Constraints are : On the number of crew: x1+2x2+2x3 ≤ 120 On the number of trucks: x1 + x2 + x3 ≤ 40 On the money to be invested: 80,000x1 + 1,00,000x2 + 1,20,000x3 ≤ 50,00000 Objective is to maximize the tonne Km/per day. Tonne Km per Day = Pay load in tonne X Km/Hour X hours / Day A = 20 x 50 x 20 = 12,000 B = 20 x 45 x 19 = 17,100 C = 17 x 45 x 22 = 16,830 Objective function is Maximize Z = 12,000 x1 + 17,100 x2 + 16830 x3 Where x1,x2,x3 ≥ 0
  • 30. GRAPHICAL METHOD A firm manufactures two products A & B on which the profits earned per unit are Rs. 3 and Rs. 4 respectively. Each product is processed on two machines M1 and M2. Product A requires one minute of processing time on M1 and two minutes on M2. b requires 2 minute on M1 and M2. Machine M1 is available for not more than 7 hrs. 30 minutes. While machine M2 is available for 10 hrs. during any working day. Find the number of units of product A and B to be manufactured to get maximum profit.
  • 31. FORMULATION OF LP PROBLEM Let x1 and x2 denote the number of units of products A and B to be produced per day. Objective Function Maximize Z = 3 x1+ 4 x2 Subject to X1+x2 ≤ 450 2X1 + x2 ≤ 600 Where x1,x2 ≥ 0
  • 32. GRAPHICAL SOLUTION
  • 33. GRAPHICAL METHOD Find the maximum value of Z = 2x1 + 3x2 Subject to x1 + x2 ≤ 30 x2 ≥ 3 x2 ≤ 12 X1 – x2 ≥ 0 0 ≤ x1 ≤ 20
  • 34. GRAPHICAL METHOD Find the maximum value of Z = 2x1 + x2 Subject to x1 + 2x2 ≤ 10 x1 + x2 ≤ 6 x1 - x2 ≤ 2 X1 – 2x2 ≤ 1 x1,x2 ≥ 0
  • 35. GRAPHICAL SOLUTION
  • 36. GRAPHICAL METHOD Find the minimum value of Z = 5x1 - 2x2 Subject to 2x1 + 3x2 ≥ 1 x1,x2 ≥ 0
  • 37. GRAPHICAL METHOD Find the minimum value of Z = -x1 + 2x2 Subject to -x1 + 3x2 ≤ 10 x1+x2 ≤ 6 x1-x2 ≤ 2 x1,x2 ≥ 0
  • 38. GRAPHICAL SOLUTION
  • 39. SIMPLEX METHOD A firm manufactures two products A & B on which the profits earned per unit are Rs. 3 and Rs. 4 respectively. Each product is processed on two machines M1 and M2. Product A requires one minute of processing time on M1 and two minutes on M2. b requires minute on M1 and M2. Machine M1 is available for not more than 7 hrs. 30 minutes. While machine M2 is available for 10 hrs. during any working day. Find the number of units of product A and B to be manufactured to get maximum profit.
  • 40. SIMPLEX METHOD Maximize Z = 3x1 + 4x2 Subject to x1 + x2 ≤ 450 2x1 + x2 ≤ 600 x1,x2 ≥ 0 Step 1- Express the problem in standard form Maximize Z = 3x1 + 4x2 + 0s1 + 0s2 Subject to x1 + x2 +s1= 450 2x1 + x2 +s2 = 600 x1,x2, s1, s2 ≥ 0
  • 41. SIMPLEX METHOD
    • Step 2- Find initial basic feasible solution
      • We start with a initial basic feasible solution which we get by assuming that the profit earned is zero.
      • Assume decision variable x1 andx2 as zero.
      • Substituting the values of x1 and x2 in the objective function we get s1= 450 and s2= 600.
      • this is called the initial basic feasible solution and s1 and s2 will form the basis.
      • This means that neither of the products have been produced and the time available fo both machines are unused.
  • 42. SIMPLEX TABLE Contribution/unit cj 3 4 0 0 Basis Body Matrix identity Matrix F.R CB Basic variables x1 x2 s1 s2 b θ 0 s1 1 (1) 1 0 450 450 0 s2 2 1 0 1 600 600 Zj 0 0 0 0 0 Cj-Zj 3 4 0 0 K
  • 43. SIMPLEX TABLE Contribution/unit cj 3 4 0 0 Basis Body Matrix identity Matrix F.R CB Basic variables x1 x2 s1 s2 b 4 X2 1 1 1 0 450 0 s2 1 0 -1 1 150 Zj 4 4 4 0 1800 Cj-Zj -1 0 -4 0 Optimum Solution
  • 44. SIMPLEX METHOD Maximize Z = 2x1 + 5x2 Subject to x1 + 4x2 ≤ 24 3x1 + x2 ≤ 21 x1 + x2 ≤ 9 x1,x2 ≥ 0 Step 1- Express the problem in standard form Maximize Z = 2x1 + 5x2 + 0s1 + 0s2+0s3, Subject to x1 + 4x2 +s1 = 2 4 3x1 + x2 +s2 = 21 x1+x2+s3 = 9 x1,x2, s1, s2, s3 ≥ 0
  • 45. SIMPLEX TABLE cj 2 5 0 0 0 F.R CB Basis x1 x2 s1 s2 s3 b θ 0 s1 1 (4) 1 0 0 24 6 1/4 0 s2 3 1 0 1 0 21 21 1/4 0 s3 1 1 0 0 1 9 9 Zj 0 0 0 0 0 0 Cj-Zj 2 5 0 0 0 K Initial Basic Feasible Solution
  • 46. SIMPLEX METHOD Values of s2 Row Values of s3 Row 3 – ¼ x 1= 11/4 1 – ¼ x 1= ¾ 1 - ¼ x 4 = 0 1 - ¼ x 4 = 0 0 – ¼ x 1 = -1/4 0 – ¼ x 1 = -1/4 1 – ¼ x 0 = 1 0 – ¼ x 0 = 0 0 – ¼ x 0 = 0 1 – ¼ x 0 = 1 21 – ¼ x 24 = 15 9 – ¼ x 24 = 3
  • 47. SIMPLEX TABLE cj 2 5 0 0 0 F.R CB Basis x1 x2 s1 s2 s3 b θ 1/3 5 x2 1/4 1 1/4 0 0 6 24 11/3 0 s2 11/4 0 -1/4 1 0 15 60/11 0 s3 3/4 0 -1/4 0 1 3 4 Zj 5/4 5 5/4 0 0 30 Cj-Zj 3/4 0 -5/4 0 0 K Second Feasible Solution
  • 48. SIMPLEX TABLE cj 2 5 0 0 0 CB Basis x1 x2 s1 s2 s3 b 5 x2 0 1 1/3 0 -1/3 5 0 s2 0 0 2/3 1 -11/3 4 2 x1 1 0 -1/3 0 4/3 4 Zj 2 5 1 0 1 33 Cj-Zj 0 0 -1 0 -1 K Third Feasible Solution Optimal Solution
  • 49. SIMPLEX METHOD Maximize Z = 4x1 + 3x2 + 6x3 Subject to 2x1 + 3x2 +2x3 ≤ 440 4x1 + 3x3 ≤ 470 2x1 + 5x2 ≤ 430 x1,x2, x3 ≥ 0 Step 1- Express the problem in standard form Maximize Z = 4x1 + 3x2 + 6x3 + 0s1 + 0s2 + 0s3, Subject to 2x1 + 3x2 + 2x3 + s1 = 440 4x1 + 3x3 + s2 = 470 2x1 + 5x2 + s3 = 430 x1,x2, x3, s1, s2, s3 ≥ 0
  • 50. SIMPLEX TABLE cj 4 3 6 0 0 0 FR CB Basis x1 x2 x3 s1 s2 s3 b θ 2/3 0 s1 2 3 2 1 0 0 440 220 0 s2 4 0 (3) 0 1 0 470 470/3 0 s3 2 5 0 0 0 1 430 ∞ Zj 0 0 0 0 0 0 0 Cj-Zj 4 3 6 0 0 0 K Third Feasible Solution Optimal Solution
  • 51. SIMPLEX TABLE cj 4 3 6 0 0 0 FR CB Basis x1 x2 x3 s1 s2 s3 b θ 0 s1 -2/3 (3) 0 1 -2/3 0 380/3 380/9 6 x3 4/3 0 1 0 1/3 0 470/3 ∞ 5/3 0 s3 2 5 0 0 0 1 430 86 Zj 8 0 6 0 2 0 940 Cj-Zj -4 3 0 0 -2 0 K Second Feasible Solution
  • 52. SIMPLEX TABLE cj 4 3 6 0 0 0 CB Basis x1 x2 x3 s1 s2 s3 b 3 x2 -2/9 1 0 1/3 -2/9 0 380/9 6 x3 4/3 0 1 0 1/3 0 470/3 0 s3 28/9 0 0 -5/3 10/9 1 1970/9 Zj 22/3 3 6 1 4/3 0 3200/3 Cj-Zj -10/3 0 0 -1 -4/3 0 Optimal solution
  • 53. SIMPLEX METHOD Minimize Z = x1 - 3x2 + 3x3 Subject to 3x1 - x2 + 2x3 ≤ 7, 2x1 + 4x2 ≥ -12, -4x1 + 3x2 + 8 x3 ≤ 10 x1,x2, x3 ≥ 0 Step 1- Express the problem in standard form Maximize Z = x1 - 3x2 + 3x3 + 0s1 + 0s2 + 0s3, Subject to 3x1 - x2 + 2x3 + s1 = 7 -2x1 - 4x2 + s2 = 12 -4x1 + 3x2 + 8x3 + s3 = 10 x1,x2, x3, s1, s2, s3 ≥ 0
  • 54. SIMPLEX TABLE cj 1 -3 3 0 0 0 FR CB Basis x1 x2 x3 s1 s2 s3 b θ 1/3 0 s1 3 -1 2 1 0 0 7 -7 4/3 0 s2 -2 -4 0 0 1 0 12 -3 0 s3 -4 (3) 8 0 0 1 10 10/3 Zj 0 0 0 0 0 0 Cj-Zj 1 -3 3 0 0 0 K Initial Basic Feasible Solution
  • 55. SIMPLEX TABLE cj 1 -3 3 0 0 0 FR CB Basis x1 x2 x3 s1 s2 s3 b θ 0 s1 (5/3) 0 14/3 1 0 1/3 31/3 31/3 22/5 0 s2 -22/3 0 32/3 0 1 4/3 76/3 -38/11 4/5 -3 x2 -4/3 1 8/3 0 0 1/3 10/3 -5/2 Zj 4 -3 -8 0 0 -1 -10 Cj-Zj -3 0 11 0 0 1 K Second Feasible Solution
  • 56. SIMPLEX TABLE C j 1 -3 3 0 0 0 CB Basis x1 x2 x3 s1 s2 s3 b 1 x1 1 0 14/5 3/5 0 1/5 31/5 0 s2 0 0 156/5 22/5 1 14/5 354/5 -3 x2 0 1 32/5 4/5 0 3/5 58/5 Zj 1 -3 -82/5 -9/5 0 -8/5 -143/5 Cj-Zj 0 0 97/5 9/5 0 8/5 Optimal solution
  • 57. BIG M METHOD
    • Minimize Z = 12x1 + 20x2
    • Subject to
    • 6x1 + 8x2 ≥ 100,
    • 7x1 + 12x2 ≥ 120,
    • X1,x2, ≥ 0
    • Step 1- Express the problem in standard form
    • Maximize Z = 12x1 + 20x2 + 0s1 + 0s2,
    • Subject to
    • 6x1 + 8x2 - s1 = 100
    • 7x1 + 12x2 - s2 = 120
    • x1,x2, s1, s2 ≥ 0
  • 58. BIG M METHOD
    • Initial Basic Feasible Solution
    • Putting x1=x2 = 0, we get s1 = -100 and s2 = - 120 which is not feasible, as it does not fulfill the non negativity restriction.
    • Introduce artificial variables in the constraints
    • 6x1 + 8x2 - s1 + A1 = 100
    • 7x1 + 12 x2 - s2 + A2 = 120
    • x1,x2, s1, s2,A1, A2 ≥ 0
  • 59. BIG M METHOD
    • Since they are artificial variables A1 and A2 should not arrear in the final solution. Hence they are assigned a large unit penalty ( a large + value M) in he objective function which can be written as
    • Minimize Z = 12x1+20x2+0s1+0s2+MA1+MA2
    • Now there are 6 variables and two constraints and hence we have to zeroise four variables to get the initial basic feasible solution. Assume x1=x2=s1=s2=0
    • A1 = 100, A2 = 120, Z = 220 M
  • 60. SIMPLEX TABLE C j 12 20 0 0 M M FR CB Basis x1 x2 s1 s2 A1 A2 b θ 2/3 M A1 6 8 -1 0 1 0 100 25/2 M A2 7 (12) 0 -1 0 1 120 10 Zj 13M 20M -M -M M M 220M Cj-Zj 12-13M 20-20M M M 0 0 K Initial Basic Feasible Solution
  • 61. SIMPLEX TABLE C j 12 20 0 0 M FR CB Basis x1 x2 s1 s2 A1 b θ M A1 4/3 0 -1 2/3 1 20 15 7/16 20 X2 7/12 1 0 -1/12 0 10 120/7 Zj 35/4+4/3M 20 -M -5/3+2/3M M 200+20M Cj-Zj 1/3-4/3M 0 M 5/3 – 2/3M 0 K Initial Basic Feasible Solution
  • 62. SIMPLEX TABLE C j 12 20 0 0 CB Basis x1 x2 s1 s2 b 12 X1 1 0 -3/4 1/2 15 20 X2 0 1 7/16 -3/4 5/4 Zj 12 20 -1/4 -9 205 Cj-Zj 0 0 ¼ 9 Optimum Solution
  • 63. BIG M METHOD
    • Maximize Z = 3x1 - x2
    • Subject to
    • 2x1 + x2 ≤ 2,
    • x1 + 3x2 ≥ 3,
    • x2 ≤ 4
    • X1,x2, ≥ 0
    • Step 1- Express the problem in standard form
    • Maximize Z = 3x1 - x2 + 0s1 + 0s2 + 0s3 – MA1,
    • Subject to
    • 2x1 + x2 + s1 = 2
    • x1 + 3x2 - s2 + A1 = 3
    • x2 + s3 = 4
    • x1,x2, s1, s2, s3, A1 ≥ 0
  • 64. SIMPLEX TABLE C j 3 -1 0 0 0 -M FR CB Basis x1 x2 s1 s2 s3 A1 b θ 0 s1 2 1 1 0 0 0 2 2 -M A1 1 (3) 0 -1 0 1 3 1 0 s3 0 1 0 0 1 0 4 4 Zj -M -3M 0 M 0 -M -3M Cj-Zj 3+M -1+3M 0 -M 0 0 K Initial Basic Feasible Solution
  • 65. SIMPLEX TABLE C j 3 -1 0 0 0 FR CB Basis x1 x2 s1 s2 s3 b θ 0 s1 (5/3) 0 1 1/3 0 1 3/5 -1 x2 1/3 1 0 -1/3 0 1 3 0 s3 -1/3 0 0 1/3 1 3 -9 Zj -1/3 -1 0 1/3 0 -1 Cj-Zj 10/3 0 0 -1/3 0 K second Feasible Solution
  • 66. SIMPLEX TABLE C j 3 -1 0 0 0 FR CB Basis x1 x2 s1 s2 s3 b 3 x1 1 0 3/5 1/5 0 3/5 -1 x2 0 1 -1/5 -2/5 0 4/5 0 s3 0 0 1/5 2/5 1 16/5 Zj 3 -1 2 1 0 1 Cj-Zj 0 0 -2 -1 0 Optimal Solution
  • 67. BIG M METHOD
    • Maximize Z = x1 + 2x2 + 3x3 – x4
    • Subject to
    • x1 + 2x2 + 3x3 = 15,
    • 2x1 + x2 + 5x3 = 20,
    • x1 + 2x2 + x3 + x4 = 10
    • x1,x2, x3, x4 ≥ 0
    • Step 1- Express the problem in standard form
    • Maximize Z = x1 + 2x2 + 3x3 – x4 - MA1 – MA2 – MA3,
    • Subject to
    • x1 + 2x2 + 3x3 +0x4 +A1 + 0A2 + 0A3 = 15
    • 2x1 + x2 + 5 X3 + 0A1 +A2 + 0A3 = 20
    • X1 + 2x2 + X3 + X4 + 0A1 + 0A2 + A3 = 10
    • x1,x2, X3, X4, A1, A2, A3 ≥ 0
  • 68. SIMPLEX TABLE C j 1 2 3 -1 -M -M -M CB Basis x1 x2 X3 X4 A1 A2 A3 b θ -M A1 1 2 3 0 1 0 0 15 5 -M A2 2 1 (5) 0 0 1 0 20 4 -M A3 1 2 1 1 0 0 1 10 10 Zj -4M -5M -9M -M -M -M -M -45M Cj-Zj 1+4M 2+5M 3+9M -1+M 0 0 0 K Initial Basic Feasible Solution
  • 69. SIMPLEX TABLE C j 1 2 3 -1 -M -M FR CB Basis x1 x2 X3 X4 A1 A3 b θ -M A1 -1/5 (7/5) 0 0 1 0 3 15/7 3 X3 2/5 1/5 1 0 0 0 4 20 -M A3 3/5 9/5 0 1 0 1 6 10/3 Zj 6-2M/5 3-16M/5 3 -M -M -M 12-9M Cj-Zj -1+2M/5 7+16M/5 0 -1+M 0 0 K Second Feasible Solution
  • 70. SIMPLEX TABLE C j 1 2 3 -1 -M FR CB Basis x1 x2 X3 X4 A3 b θ 2 x2 -1/7 1 0 0 0 15/7 ∞ 3 X3 3/7 0 1 0 0 25/7 ∞ -M A3 6/7 0 0 (1) 1 15/7 15/7 Zj 7-6M/7 2 3 -M -M 105-15M/7 Cj-Zj 6M/7 0 0 -1+M 0 K Third Feasible Solution
  • 71. SIMPLEX TABLE C j 1 2 3 -1 FR CB Basis x1 x2 X3 X4 b θ 2 x2 -1/7 1 0 0 15/7 -15 3 X3 3/7 0 1 0 25/7 25/3 -1 x4 (6/7) 0 0 1 15/7 5/2 Zj 1/7 2 3 -1 90/7 Cj-Zj 6/7 0 0 0 K Fourth Feasible Solution
  • 72. TANSPORTATION MODEL
    • A dairy has three plants located throughout the state. Daily milk production at each plant is as follows:
      • Plant 1 - 6 million litres
      • Plant 2 – 1 million litres
      • Plant 3 – 10 million litres
    • Each day the firm must fulfill the needs of its four distribution centres. Milk requirement at each centre is as ollows:
      • Distribution centre 1 – 7 million litres
      • Distribution centre 2 – 5 million litres
      • Distribution centre 3 – 3 million litres
      • Distribution centre 4 – 2 million litres
    • Cost of shipping one million litres of milk from each plant to each distribution centre is given in the following table in hundreds of rupees.
    • Formulate the mathematical model for the problem.
    • Determine the optimal transportation policy.
  • 73. SIMPLEX TABLE C j 1 2 3 -1 FR CB Basis x1 x2 X3 X4 b 2 x2 0 1 0 1/6 5/2 3 X3 0 0 1 -1/2 5/2 1 x1 1 0 0 7/6 5/2 Zj 1 2 3 0 15 Cj-Zj 0 0 0 -1 Optimal Solution
  • 74. TANSPORTATION MODEL
    • Deals with transportation of a product available at several sources a number of different destinations in such a way that the total transportation cost is minimum.
    • The origin of this model dates back to 1941 when FL Hitchcock presented a study on the subject.
    • Assumptions in the transportation model:-
      • Total quantities of an item item available at different sources is equal to total requirement at different destinations.
      • Items can be transported conveniently from all sources to destinations.
      • The unit transportation cost of the item from all sources to destinations is certainly and precisely known.
      • The transportation cost on a given route is directly proportional to the number of units shipped on that route.
      • The objective is to minimize the total transportation cost for the organization as a whole and not for individual supply and distribution centres.
  • 75. TANSPORTATION MODEL Distribution Center 1 2 3 Plant 1 2 3 4 2 3 11 7 1 0 6 1 5 8 15 9
  • 76. TANSPORTATION MODEL Distribution Center 1 2 3 Plant 1 2 3 4 X11 X12 X13 X14 X21 X22 X23 X24 X31 X32 X33 X34
  • 77. TANSPORTATION MODEL
    • Objective is to minimize the cost of transportation
    • i.e. Minimize Z = 2X 11 + 3X12 + 11X13 + 7 X14 + X21 + 0X22 + 6X23 + X24 + 5X31 + 8X32 + 15X33 + 9X34
    • Subject to
    • X11 + X12 + X13 + X14 = 6 (For plant 1)
    • X21 + X22 + X23 + X24 = 1 (For plant 2)
    • X31 + X32 + X33 + X34 = 10 (For plant 3)
    • X11 + X21 + X31 = 7, (For DC 1)
    • X12 + X22 + X32 = 5, (For DC 2)
    • X14 + X24 + X34 = 2 (For DC 3)
  • 78. TANSPORTATION MODEL Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 6 5 3 2 1 10 17 2 3 11 7 1 0 6 1 5 8 15 9
  • 79. NORTH WEST CORNER METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 6 5 3 2 1 10 17 2 (6) 3 11 7 1 (1) 0 (0) 6 1 5 8 (5) 15 (3) 9 (2)
  • 80. ROW MINIMA METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 6 5 3 2 1 10 17 2 (6) 3 11 7 1 0 (1) 6 1 5 (1) 8 (4) 15 (3) 9 (2)
  • 81. COLUMN MINIMA METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 6 5 3 2 1 10 17 2 (6) 3 11 7 1 (1) 0 6 1 5 8 (5) 15 (3) 9 (2)
  • 82. VAM METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 (1) 6 (1) 5 (3) 3 (5) 2 (6) 1 (1) 10 (3) 17 2 3 11 7 1 0 6 1 (1) 5 8 15 9
  • 83. VAM METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 (3) 6 (1) 5 (5) 3 (4) 2 (2) 1 (1) 10 (3) 17 2 3 (5) 11 7 1 0 6 1 (1) 5 8 15 9
  • 84. VAM METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 (3) 6 (5) 5 (5) 3 (4) 2 (2) 1 (1) 10 (3) 17 2 (1) 3 (5) 11 7 1 0 6 1 (1) 5 8 15 9
  • 85. VAM METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 (3) 6 (5) 5 (5) 3 (4) 2 (2) 1 (1) 10 (3) 17 2 (1) 3 (5) 11 7 1 0 6 1 (1) 5 (6) 8 15 (3) 9 (1)
  • 86. VAM METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 (3) 6 (5) 5 (5) 3 (4) 2 (2) 1 (1) 10 (3) 17 2 (1) 3 (5) 11 7 1 0 6 1 (1) 5 (6) 8 15 (3) 9 (1)
  • 87. TRANSPORTATION MODEL
    • Important Note:
    • In case of tie among the highest penalties, select the row or column having minimum cost. In case of tie in the minimum cost also, select the cell which can have maximum allocation. If there is tie among maximum allocation cells also, select the cell arbitrarily or allocation. This will give the best initial basic feasible solution requiring less number of iterations subsequently.
    • Optimality Test
    • Optimality test can be performed only on that feasible solution in which
    • Number of allocations is m + n – 1, where m is the number of rows and n is the number of columns.
    • These (m + n - 1) allocations should be in independent positions.
  • 88. TRANSPORTATION MODEL Optimality Test The test procedure for optimality involves examination of each vacant cell to find whether or not making an allocation in it reduces the total transportation cost. Two Methods: Stepping Stone method: Modified Distribution Method (MODI)
  • 89. STEPPING STONE METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 (3) 6 (5) 5 (5) 3 (4) 2 (2) 1 (1) 10 (3) 17 2 (1) 3 (5) 11 7 1 0 6 1 (1) 5 (6) 8 15 (3) 9 (1)
  • 90. TRANSPORTATION MODEL Optimality Test Cell Evaluations Cell (1,3) = 11-15+5-2 = -1 Cell (1,4) = 7-9+5-2 = +1 Cell (2,1) = 1-1+9-5 = 4 Cell (2,2) = 0-1+9-5+2-3 = 2 Cell (2,3) = 6-1+9-15 = -1 Cell (3,2) = 8-5+2-3 =2
  • 91. MODI METHOD Distribution Center 1 2 3 Plant 1 2 3 4 Supply Requirement 7 (3) 6 (5) 5 (5) 3 (4) 2 (2) 1 (1) 10 (3) 17 ui vj 2 (1) 3 (5) 11 7 1 0 6 1 (1) 5 (6) 8 15 (3) 9 (1)
  • 92. MODI METHOD U1+V1=2 U1+V2=3 U2+V4=1 U3+V1=5 U3+V3=15 U3+V4=9 Let v1 = 0 (Arbitrary) – Then U1=2, v2=1, u3=5, v3=10, v4=4, u2=-3
  • 93. MODI METHOD Distribution Center 2 -3 5 Plant 0 1 10 4 ui vj 2 3 1 5 15 9
  • 94. MODI METHOD Distribution Center 2 -3 5 Plant 0 1 10 4 ui vj Compute ui+vj for each empty cell 12 6 -3 -2 7 6
  • 95. MODI METHOD Distribution Center 2 -3 5 Plant 0 1 10 4 ui vj Compute cij-(ui+vj) for each empty cell 11-12=-1 7-6=1 1+3=4 0+2=2 6-7=-1 8-6=2
  • 96. MODI METHOD Optimality Test A –Ve value in an unoccupied cell indicates that a better solution can be obtained by allocating units to this cell. A +ve value in an unoccupied cell indicates that a poorer solution will result by allocating units to the cell. A zero value in an unoccupied cell indicates that another solution of the total value can be obtained by allocating units to this cell. Iterate for optimality by allocating to most –ve cell. Recheck for optimality. If still notoptimal allocate to most –ve cell and go on.
  • 97. EXAMPLE Is An optimal solution for the transportation problem: 70 55 90 85 35 50 45 If not iterate it for optimal solution 50 20 55 30 35 25 6 1 9 3 11 5 2 8 10 12 4 7
  • 98. EXAMPLE Solve the following transportation problem D1 D2 D3 D4 D5 O1 O2 03 O4 O5 Available 18 17 19 13 15 Required 16 18 20 14 14 82 (P-264) 68 35 4 74 15 57 88 91 3 8 91 60 75 45 60 52 53 24 7 82 51 18 82 13 7
  • 99. EXAMPLE A product is produced by four factories A,B,C,D. The unit production costs in them are Rs. 2, 3, 1 and 5 respectively. Their production capacities are 50,70,30,50 units respectively. These factories supply the product to four stores, demands of which are 25,35,105,20 units respectively. Unit transportation cost in rupees from each factory to each store is given in the table below: Determine the extent of deliveries from each of the factories to each of the stores so that the total production and transportation cost is minimum. (p-267) 2 4 6 11 10 8 7 5 13 3 9 12 4 6 8 3
  • 100. ASSIGNMENT PROBLEM ASSIGNMENT MODEL CAN BE REGARDED AS A SPECIAL CASE OF TRANSPORTATION PROBLEM. Determine the assignment that will optimize the total processing cost. (325) X Y Z A B C JOBS 25 15 22 31 20 19 35 24 17
  • 101. ASSIGNMENT PROBLEM Make sure the matrix is balanced. If not add a dummy row or column to balance it. Subtract each row element from the lowest value in that row to arrive at the machine opportunity cost. X Y Z A B C JOBS 25 – 15 = 10 0 7 12 1 0 18 7 0
  • 102. ASSIGNMENT PROBLEM Subtracting each column element from the lowest value in that column will give job opportunity cost. The total opportunity cost (combined job and machine opportunity cost) is given by subtraction the lowest column value in table 2 from each of the column values. X Y Z A B C JOBS 10 – 10 = 0 0 7 2 1 0 8 7 0
  • 103. ASSIGNMENT PROBLEM Checking for optimality. Hungarian Method Draw vertical and horizontal lines to cover all zeros in the matrix. If the number of lines are equal to n then the assignment is optimal. If no subtract the smallest value of the uncovered cells from each uncovered cell and add it to all intersecting values. Again draw vertical and horizontal lines to cover all zeros. Repeat till the number of lines are equal to n.
  • 104. EXAMPLE A machine tool company decides to make four subassemblies through four contractors. Each contractor is to receive only one sub assembly. The cost of each sub assembly is determined by the bids submitted by each contractor and is shown in table below in hundreds of rupees.
    • Formulate the mathematical model for the problem
    • Show that the assignment model is special case of the transportation model
    • Assign the different subassemblies to contractors so as to minimize the total cost (329)
    15 13 14 17 11 12 15 13 13 12 10 11 15 17 14 16
  • 105. EXAMPLE A machine tool company decides to make four subassemblies through four contractors. Each contractor is to receive only one sub assembly. The cost of each sub assembly is determined by the bids submitted by each contractor and is shown in table below in hundreds of rupees.
    • Formulate the mathematical model for the problem
    • Show that the assignment model is special case of the transportation model
    • Assign the different subassemblies to contractors so as to minimize the total cost (329)
    15 13 14 17 11 12 15 13 13 12 10 11 15 17 14 16
  • 106. EXAMPLE Four different jobs can be done on four different machines. The set up and take down time costs are assumed to prohibitively high for change over. The matrix below gives the cost in rupees of producing job i on machine j.
    • How should the jobs be assigned to the various machines so that the total cost is minimum. Also formulate the mathematical model for the problem. (332)
    machines 5 7 11 6 8 5 9 6 4 7 10 7 10 4 8 3
  • 107. EXAMPLE Four different jobs can be done on four different machines. The set up and take down time costs are assumed to prohibitively high for change over. The matrix below gives the cost in rupees of producing job i on machine j.
    • How should the jobs be assigned to the various machines so that the total cost is minimum. Also formulate the mathematical model for the problem. (332)
    machines 5 7 11 6 8 5 9 6 4 7 10 7 10 4 8 3
  • 108. EXAMPLE Solve the following assignment problem. machines 11 17 8 16 20 9 7 12 6 15 13 16 15 12 16 21 24 17 28 26 14 10 12 11 13
  • 109. EXAMPLE Five wagons are available at stations 1,2,3,4&5. These are required at five stations I, II , III , IV & V. The mileages between various between various stations are given below. How should the wagons be transferred so that the total mileage covered is minimum. Stations 10 5 9 18 11 13 9 6 12 14 3 2 4 4 5 18 9 12 17 15 11 6 14 19 10
  • 110. EXAMPLE Unbalanced Matrix A company has one surplus truck in each of the cities A,B,C,D and E and one deficit truck in each of the cities 1,2,3,4,5&6. The distance between the cities in kilometers is shown in the matrix below. Find the assignment of trucks from cities in surplus to cities in deficit so that the total distance covered by the vehicles is minimum. (341) 1,2,3,4,5&6 12 10 15 22 18 8 10 18 25 15 16 12 11 10 3 8 5 9 6 14 10 13 13 12 8 12 11 7 13 10
  • 111. EXAMPLE Maximization problem A company has a team of four salesmen and there are four districts where the company wants to start its business. After taking into account the capabilities of salesmen and the nature of districts, the company estimates that the profit per day in rupees for each salesman in each district is as below. Find the assignment of salesmen to various districts which will yield maximum profit. District 16 10 14 11 14 11 15 15 15 15 13 12 136 12 14 15
  • 112. DECISION THEORY
    • Decision taking environments
    • Certainty – Various out comes are known with certainty (CVP, LP, Assignment Tpt Model etc)
    • Uncertainty – where information required to assign probability is not available, though various possible outcomes are known (
    • Risk – Information for assigning probability is available
    • Conflict – Where two opponents are involved.
  • 113. DECISION THEORY Decision making under conditions of uncertainty MaxiMax Criterion Alternatives States of Nature (Product Demand) Maximum of Row High Moderate Low Nil Expand Construct Subcontract 50000 70000 30000 25000 30000 15000 -25000 -40000 -1000 -45000 -80000 -10000 50000 70000 30000
  • 114. DECISION THEORY Decision making under conditions of uncertainty MaxiMin Criterion Alternatives States of Nature (Product Demand) Minimum of Row High Moderate Low Nil Expand Construct Subcontract 50000 70000 30000 25000 30000 15000 -25000 -40000 -1000 -45000 -80000 -10000 -45000 -80000 -10000
  • 115. DECISION THEORY Decision making under conditions of uncertainty MiniMax Regret Criterion Alternatives States of Nature (Product Demand) Maximum of Row High Moderate Low Nil Expand Construct Subcontract 20000 0 40000 5000 0 15000 24000 39000 0 35000 70000 0 35000 70000 40000
  • 116. DECISION THEORY Decision making under conditions of uncertainty Hurwicz Criterion (Criterion of realism or Weighted Average Criterion) α = Appropriate degree of optimism Alternatives States of Nature (Product Demand) Maximum of Row Minimum of Row P= α . Max + (1 – α ) min α = .8 High Moderate Low Nil Expand Construct Subcontract 50000 70000 30000 25000 30000 15000 -25000 -40000 -1000 -45000 -80000 -10000 50000 70000 30000 -45000 -80000 -10000 31000 40000 22000
  • 117. DECISION THEORY Decision making under conditions of uncertainty Laplace Criterion (Criterion of rationality or Bayes’ Criterion OR Equal Probability) Alternatives States of Nature (Product Demand) Expected Pay Offs Rs = 1/n (P1+P2+P3+…….Pn) High Moderate Low Nil Expand Construct Subcontract 50000 70000 30000 25000 30000 15000 -25000 -40000 -1000 -45000 -80000 -10000 1000/4 (50 + 25 - 25 – 45) = -1250 1000/4 (70+30-40-80) = -5000 1000/4 (30 + 15 - 1 -10) = 8500
  • 118. EXAMPLE The following matrix gives the payoff of different strategies (alternatives) S1,S2,S3 against conditions (events) N1, N2, N3 & N4. Indicate the decision taken under the following approaches (a) Pessimistic (b) Optimistic (c) Regret (d) Equal Probability N1 N2 N3 N4 S1 S2 S3 4000 20000 20000 -100 5000 15000 6000 400 -2000 18000 0 1000
  • 119. EXAMPLE The following matrix gives the payoff of different strategies (alternatives) S1,S2,S3 against conditions (events) N1, N2, N3 & N4. Indicate the decision taken under the following approaches (a) Pessimistic (b) Optimistic (c) Regret (d) Equal Probability N1 N2 N3 N4 S1 S2 S3 4000 20000 20000 -100 5000 15000 6000 400 -2000 18000 0 1000