Ouantitative technics
- 3. LPP – simultaneous equation / in equality condition linear function / object function. i) Optimize linear function. ii) Subjective to some condition or restriction. iii) Non – negativity restrictions max z =x1+x2
- 4. A general LPP includes a set of simultaneous linear equations which represent the conditions of the problem and a linear constrains which express the objective linear problem. The linear functions which is the optimized is called the objective function and the condition of the problem express the simultaneous linear equations (or) the qualities are referred as constrains.
- 5. A general linear programming problem can be started as follows: Max z = 5x1 + 10x2 St = 1x1 +2x2 < 100 2x1 +3x2 < 150 x1 + x3 > 0
- 6. LPP includes: i) Optimize the linear ii) Subject to some constrains / restrictions iii) Non – regativity restrictions / 0,1,2,3…… a) The LPP is applicable when a concern producers 2 (or) more qualities and each quantity shall on 2 (or) more process. b) A large no. of decision problem faced by the business managers involved allocation of resources to varies activities with the objectives of increasing profit or decreasing (or) both.
- 7. c) The decision problem becomes complicated when no. of resources are required to be allocated and there are several activities to perform. d) The decision problems can be formulated solved as a mathematical programming problem. example: Rule of thumb even a experienced manager, in all likely hood may not provide the right answer in such cases.
- 8. e) Mathematical programming involved optimization of a certain functions called the objective function, subject too certain constrains. Example: A manager may faced with the problem of decisioning appropriate product mix of the 4 product with the profitability of product along with their requirements of raw materials, labour extra…. His problem can be formulated as a mathematical programming problem taking the objective funtions as the maximization of product obtainable from the mix the various constrain.
- 9. The availability of raw material, labour, supply market and so on…. The method of mathematical programming can be divided into 3 groups: I) Linear II) Integer III) Non – Linear Programming
- 10. Problem 1: Max z = 6x1 + 11x2 St = 2x1 + x2 < 104 x1 + 2x2 < 76 x1 , x2 > 0 Mathematical Conversion: Max z = 6x1 + 11x2 + x3 + x4 St = 2x1 + x2 + x3 = 104 x1 + 2x2 + + x4 = 76 x1 , x2 = 0
- 11. INITIAL TABLE: B CB XB X1 X2 X4 X5 XB/X2 X3 0 104 2 1 1 0 104/1 =104 X4 0 76 1 2 0 1 76/2= 38 (min) Z 6 11 0 0 CB 0 0 0 0 Z-CB 6 11 (max) 0 0 2
- 12. WORKINGS: 76 1 2 0 1 / 2 38 1/2 1 0 1/2 - 2 (2ND ROW) 38 ½ 1 0 ½ * 1 104 2 1 1 0 38 ½ 1 0 ½ 66 3/2 0 1 -1/2 ( 1ST ROW) I SIMPLEX TABLE: B CB XB X1 X2 X3 X4 XB/X1 X3 0 66 3/2 0 1 -1/2 66*2/3=4 4(MIN) X2 11 38 ½ 1 0 ½ 38*2/1=7 6 Z 6 11 0 0 CB 11/2 11 0 -11/2 CB-Z ½(MAX) 0 0 -11/2 3/ 2
- 13. WORKINGS: 66 3/2 0 1 ½ / 3/2 44 1 0 2/3 -1/3 * ½ (I ST ROW) 22 ½ 0 1/3 -1/3 38 ½ 1 0 ½ (-) 22 ½ 0 1/3 -1/3 16 0 1 -1/3 2/3 (IIND ROW) II SIMPLEX TABLE: B CB XB X1 X2 X3 X4 X1 6 44 1 0 2/3 -1/3 X2 11 16 0 1 -1/3 2/3 Z 6 11 0 0 CB 6 11 -1/3 28/3 Z-CB 0 0 -1/3 -28/3
- 14. THE OPTIMUM SOLUTION IS X1 = 44 X2 = 16 MAXZ = 6X1 + 11X2 = (6*44) + ( 11*16) = 264 + 176 MAXZ = 440 Problem 2: Solve the following LPP problem by applying simplex method. MAX z = 5x1 + 3x2 St = 5x1 + 5x2 < 15 = 5x1 + 2x2 < 10 x1 , x2 > 0
- 15. SOLUTION: MAX z = 5x1 + 3x2 + x3 + x4 St = 3x1 + 5x2 + x3 =15 = 5x1 + 2x2 + x4 = 10 x1 , x2 = 0 INITIAL TABLE: B CB XB X1 X2 X3 X4 XB/X1 X3 0 15 3 5 1 0 15/3=5 X4 0 10 5 2 0 1 15/5=3 (MIN) Z 5 3 0 0 CB 0 0 0 0 CB-Z 5(MAX) 3 0 0 5
- 16. WORKINGS: 10 5 2 0 1 / 5 2 1 2/5 0 1/5 (2ND ROW) * 3 15 3 5 1 0 (-) 6 3 6/5 0 -3/5 9 0 19/5 1 -3/5 I- SIMPLEX TABLE: B CB XB X1 X2 X3 X4 XB/X3 X3 0 9 0 19/5 1 -3/5 9*5/19= 2.3(MIN) X1 5 2 1 2/5 0 1/5 2*5/2=5 Z 5 3 0 0 CB 5 2 0 1 Z-CB 0 1(MAX) 0 0 19/5
- 17. WORKINGS: 9 0 10/5 1 -3/5 / 19/5 45/19 0 1 5/19 -3/19 (1ST ROW) * 2/5 2/1 1 2/5 0 1/5 (-) 18/19 0 2/5 2/19 -6/95 20/19 1 0 -2/19 5/19 II SIMPLEX TABLE: B CB XB X1 X2 X3 X4 X2 3 45/19 0 0 5/19 -3/19 X1 5 20/19 1 1 -2/19 5/19 Z 5 3 0 0 CB 5 3 -5/19 -16/19 Z-CB 0 0 -5/19 -16/19
- 18. THE OPTIMUM SOLUTION IS: X1 = 45/19 X2 = 20/19 MAX Z = 5X1 + 3X2 (5*45/19) + (3*20/19) 225/19 + 60/19 285/19 MAXZ = 15 Problems 3: SOLVE THE PROBLEM BY APPLING THE SIMPLEX METHOD: MAXZ = 30X1 + 23X2 + 29X3 St = 6X1 + 5X2 + 3X3 < 26 4X1 + 2X2 + 5X3 < 7 X1 , X2 , X3 > 0
- 19. SOLUTION: MAXZ = 30X1 + 23X2 + 29X3 + X4 + X5 St = 6X1 + 5X2 + 3X3 + X4 4X1 + 2X2 + 5X3 + + X5 X1 , X2 , X3 = 0 INITIAL TABLE: B CB XB X1 X2 X3 X4 XB/X5 X4 0 26 6 5 3 1 26/6= 4.3 X5 0 7 2 5 0 7/4= 1.75(MIN) Z 30 23 29 0 CB 0 0 0 0 Z-CB 30(MAX) 23 29 0 4
- 20. WORKINGS: 7 4 2 5 0 1 / 4 7/4 1 2/4 5/4 0 1/4 (2ND ROW) * 6 26 6 5 3 1 0 (-) 21/2 6 6/2 15/2 0 -3/2 31/2 0 4/2 -9/2 1 -3/2 (1ST ROW) I SIMPLEX TABLE: B CB XB X1 X2 X3 X4 X5 XB/X2 X4 0 31/2 0 2 -9/2 1 -3/2 3/2*2 =62 X1 30 7/4 1 2/4 5/4 0 1/4 7/4*2/4 =7/2(MIN) Z 30 23 29 0 0 CB 30 15 75/2 0 -15/2 Z-CB 0 8(MAX) -17/2 0 -15/2 2/4
- 21. WORKINGS: 7/4 1 2/4 5/4 0 1/4 / 2/4 7/2 2 1 5/2 0 ½(2ND ROW) * 2 31/2 0 2 -9/2 1 -3/2 (-) 7 4 2 5 0 1 17/2 -4 0 19/2 1 -5/2 II SIMPLEX TABLE: B CB XB X1 X2 X3 X4 X5 X4 0 17/2 -4 0 19/2 1 -5/2 X2 23 7/2 2 1 5/2 0 1/2 Z 30 23 29 0 0 CB 46 23 115/2 0 -23/2 Z-CB -16 0 -57/2 0 -23/2
- 22. THE OPTIMUM SOLUTION IS: MAXZ = 30X1 + 23X2 + 29X3 (30*0) + (23*7/2) + (29*0) 0 + 161/2 + 0 MAXZ = 161/2 Problem 4: Use the simplex to solve the following LPP: MAXZ = 4X1 + 10X2 St = 2X1 + X2 < 50 = 2X1 + 5X2 < 100 = 2X1 + 3X2 < 90 X1 , X2 > 0
- 23. WORKINGS: MAXZ = 4X1 + 10X2 + X3 + X4 + X5 St = 2X1 + X2 + X3 = 2X1 + 5X2 + + X4 2X1 + 3X2 + X5 X1 , X2 = 0 INITIAL TABLE: B CB XB X1 X2 X3 X4 X5 XB/X2 X3 0 50 2 1 1 0 0 50 X4 0 100 2 5 0 1 0 100/5= 20(MIN) X5 0 90 2 3 0 0 1 90/3= 30 Z 4 10 0 0 0 CJ 0 0 0 0 0 Z-CJ 4 10(MAX) 0 0 0 5
- 24. WORKINGS: 100 2 3 0 1 0 / 5 20 2/5 1 0 1/5 0(2nd ROW) * 1 50 2 1 1 0 0 (-) 20 2/5 1 0 1/5 0 30 8/5 0 1 3/5 0 (1ST ROW) 20 2/5 1 0 1/5 0 * 3 90 2 3 0 0 1 (-) 60 6/5 3 0 3/5 0 30 4/5 0 0 -3/5 1 (3RD ROW) I SIMPLEX TABLE: B CB XB X1 X2 X3 X4 X5 X3 0 30 8/5 0 1 1/5 0 X2 10 20 2/5 1 0 1/5 0 X5 0 30 4/5 0 0 -3/5 1 Z 4 10 0 0 0 Cj 4 10 0 -2 0 Z-Cj 0 0 0 -2 0
- 25. THE OPTIMUM SOLUTION IS: MAXZ = 4X1 + 10X2 X1 = 30 X2 = 20 X3 = 30 = (0*30) + (10*20) + (0*30) = 0 + 200 + 0 MAXZ = 200 Problems 5: SOLVE THE FOLLOWING PROBLEM BY APPLYING SIMPLEX METHOD: MAXZ = 5X1 + 7X2 St = X1 + X2 < 4 = 3X1 + 8X2 < 24 = 10X1 + 7X2 < 35 X1 , X2 > 0
- 26. SOLUTION: MAXZ = 5X1 + 7X2 + X3 + X4 + X5 St = X1 + X2 + X3 = 4 3X1 + 8X2 + X4 = 24 10X1 + 7X2 + X5 = 35 X1 , X2 = 0 INITIAL TABLE: B CB XB X1 X2 X3 X4 X4 XB/X 1 X3 0 4 1 1 1 0 0 4 X4 0 24 3 0 1 0 24/8=3 X5 0 35 10 7 0 0 1 35/7=5 Z 5 7 0 0 0 Cj 0 0 0 0 0 Z-Cj 5 7 0 0 0
- 27. WORKINGS: 24 3 8 0 1 0 / 8 3 3/8 1 0 1/8 0 (2 ROW) * 1 4 1 1 1 0 0 (-) 3 3/8 1 0 1/8 0 1 5/8 0 1 -1/8 0 (1 ROW) 3 3/8 1 0 1/8 0 35 10 7 0 0 1 (-) 21 21/8 7 0 -7/8 0 14 59/8 0 0 -7/8 1 (3 ROW) I SIMPLEX TABLE: B CB XB X1 X2 X3 X4 X5 XB/X1 X3 0 1 5/8 0 1 -1/8 0 5/8(MIN) X2 7 3 3/8 1 0 ½ 0 8 X5 0 14 59/8 0 0 -7/8 1 812/8 Z 5 7 0 0 0 Cj 21/8 7 0 7/8 0 Z-Cj 19/8(MAX) 0 0 -7/8 0 5/ 8
- 28. WORKINGS: 1 5/8 0 1 -1/8 0 / 5/8 8/5 1 0 8/5 -1/5 0 (1 row) * 3/8 3 3/8 1 0 1 /8 0 (-) 3/5 3/8 0 3/5 -3/5 0 12/5 0 1 -3/5 29/40 0 (2 row) 8/5 1 0 8/5 7/5 0 * 59/8 14 59/8 0 0 -7/8 1 (-) 59/5 59/8 0 -59/5 59 0 11/5 0 0 -59/5 -479/5 1 (3 row)
- 29. Problem 6: SOLVE THE SIMPLEX METHOD BY LPP PROBLEM: MINZ = X1 - 3X2 + 2X3 St = 3X1 - X2 + 2X3 < 7 -2X1 + 4X2 < 12 -4X1 + 3X2 + 8X3 < 10 X1 , X2 < 0 METHAMETICAL CONVERTION: MINZ = -X1 + 3X2 - 2X3 + X4 + X5 + X6 St = 3X1 + X2 + 2X3 + X4 = 7 -2X1 + 4X2 + + X5 = 12 -4X1 + 3X2 + 8X3 + X6 = 10 X1 , X2 = 0
- 30. INITIAL TABLE: WORKINGS: 12 -2 4 0 0 1 0 / 4 3 -1/2 1 0 0 ¼ 0 (2 ROW) * -1 7 3 -1 2 1 0 0 (-) -3 -1/2 1 0 0 -1/4 0 10 5/2 0 2 1 -1/4 0 ( 1 ROW) B CB XB X1 X2 X3 X4 X5 X6 XB/X2 X3 0 7 3 -1 2 1 0 0 7/-1=-7 X4 0 12 -2 0 0 1 0 12/4=3 X5 0 10 -4 3 8 0 0 1 10/3=3.33 Z -1 3 -2 0 0 0 Cj 0 0 0 0 0 0 Z-Cj -1 3 -2 0 0 0
- 31. 3 -1/2 1 0 0 ¼ 0 * 3 10 -4 3 8 0 0 1 (-) 9 -1/2 13 0 0 -3/4 0 1 -5/2 0 8 0 -3/4 1 (3 ROW) I SIMPLEX TABLE: B CB XB X1 X2 X3 X4 X5 X6 XB/X1 X4 0 10 5/2 0 2 1 -1/4 0 10/(5/2) =4(MIN) X2 3 3 -1/2 1 0 0 ¼ 0 3/(1/2) =6 X6 0 1 -5/2 0 8 0 -3/4 1 1/(-5/2) =2/-5 Z -1 3 -2 0 0 0 Cj 3/2 3 0 0 -3/4 0 Z-Cj 5/2*(MA X) 0 2 0 -3/4 0 5/ 2
- 32. WORKINGS: 10 5/2 0 2 1 -1/4 0 / 5/2 4 1 0 4/5 2/5 -1/10 0(1 ROW) * -1/2 3 -1/2 1 0 0 ¼ 0 (-) -2 -1/2 0 -2/5 1/5 -1/20 0 5 0 1 -2/5 1/5 6/20 0 (2 ROW) 4 1 0 4/5 2/5 1/10 0 * -5/2 1 -5/2 0 8 0 -3/4 1 -10 -5/2 0 -2 -1 -1/4 0 -11 0 0 10 1 -2/4 1
- 33. II SIMPLEX TABLE: THE OPTIMUM SOLUTION IS: X1 = -1 , X2 = 3 , X3 = 2 (-1*4) + (3*5) * (0*-1) -4 + 15 = 11 B CB XB X1 X2 X3 X4 X5 X6 X1 -1 4 1 0 4/5 2/5 1/10 0 X2 3 5 0 1 2/5 1/5 6/20 0 X6 0 -11 0 0 10 1 2/4 1 Z -1 3 -2 0 0 0 Cj -1 3 2 1 1 0 Z-Cj -2 0 -4 -1 -1 0
- 50. Thank you