1. OPERATIONS RESEARCH CDR NK NATARAJAN

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.

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

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

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

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

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

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

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

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

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

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)

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.

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

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