1. A model consisting of linear relationships representing a firm’s objective and resource constraints Linear Programming (LP) LP is a mathematical modeling technique used to determine a level of operational activity in order to achieve an objective, subject to restrictions called constraints

2. Types of LP

3. Types of LP (cont.)

5. LP Model Formulation  Decision variables mathematical symbols representing levels of activity of an operation  Objective function a linear relationship reflecting the objective of an operation most frequent objective of business firms is to maximize profit most frequent objective of individual operational units (such as a production or packaging department) is to minimize cost  Constraint a linear relationship representing a restriction on decision making

6. LP Model Formulation (cont.) Max/min z = c 1 x 1 + c 2 x 2 +... + c n x n subject to: a 11 x 1 + a 12 x 2 +... + a 1n x n (≤, =, ≥) b 1 a 21 x 1 + a 22 x 2 +... + a 2n x n (≤, =, ≥) b 2 : a m1 x1 + a m2 x 2 +... + a mn x n (≤, =, ≥) b m x j = decision variables b i = constraint levels c j = objective function coefficients a ij = constraint coefficients

7. LP Model: Example LaborClayRevenue PRODUCT(hr/unit)(lb/unit)($/unit) Bowl1440 Mug2350 There are 40 hours of labor and 120 pounds of clay available each day Decision variables x 1 = number of bowls to produce x 2 = number of mugs to produce RESOURCE REQUIREMENTS

8. LP Formulation: Example Maximize Z = $40 x 1 + 50 x 2 Subject to x 1 +2x 2  40 hr(labor constraint) 4x 1 +3x 2  120 lb(clay constraint) x 1, x 2  0 Solution is x 1 = 24 bowls x 2 = 8 mugs Revenue = $1,360

9. Solving LP Problems with Excel Click on “Tools” to invoke “Solver.” Objective function Decision variables – bowls (x 1 )=B10; mugs (x 2 )=B11 =C6*B10+D6*B11 =C7*B10+D7*B11 =E6-F6 =E7-F7

10. Solving LP Problems with Excel (cont.) After all parameters and constraints have been input, click on “Solve.” Objective functionDecision variables C6*B10+D6*B11≤40 C7*B10+D7*B11≤120 Click on “Add” to insert constraints

11. Solving LP Problems with Excel (cont.)

12. Sensitivity Analysis

13. Solving a Minimisation Problem Using the Simplex Method Min1X1 + 1X2 Max -1X1 – 1 X2 s.t.1X1> 301X1> 30 1X2>201X2> 20 1X1 + 2 X2>801X1 + 2X2> 80 X1, X2 >0X1, X2 > 0 Max -1X1 – X2 + 0S1 + 0S2 +0S3 - Ma1 – Ma2 – Ma3 s. t. 1X1 -1S1 +1a1 = 30 1X2 -1S2 +1a2 = 20 1X1 + 1 X2 -1S3 +1a3 =80 X1, X2, S1, S2, S3 > 0

14. X1X2S1S2S3a1a2a3 Basis Cj 000-M a1-M100010030 a2-M010001020 a3-M120000180 Zj -2M -3MMMM-M -130M Cj-Zj -1+2M-1+3M -M 000

15. X1X2S1S2S3a1a2a3 Basis Cj 000-M a1-M 100010030 X2010 001020 a3-M10020-2140 Zj -2M M -2M+1 M-M -1+2M -M -70M- 20 Cj-Zj -1+2M 0-M -1+2M -M0- 3M+1 0

16. X1X2S1S2S3a1a2a3 Basis Cj 000-M X110 0010030 X2010 001020 a3-M0012 -2110 Zj -M+1-2M+1 M -1+M -1+2M -M -10M- 50 Cj-Zj 00 -1+M-1+2M -M -2M+1 -3M+1 0

17. X1X2S1S2S3a1a2a3 Basis Cj 000-M X110 0010030 X2010 001 1/2 25 S200012 -2 1/2 5 Zj 1/2 0 0 -1/2 -55 Cj-Zj 00 -1/2 0 - M-1/2 -M -M+1/2

18. Infeasibility Infeasibility occurs when there is no solution to the linear program which satisfies all constraints including non-negativity condition ( X1, X2, …….Xn>0) Max 10X1+ 9 X2 st7/10X1 + 1 X2<630 1/2X1 + 5/6 X2<600 1X1 + 2/3 X2<708 10X1 + 1/4 X2<135 1X1>500 New constraint 1X2>360 New constraint

19. X1X2S1S2S3S4S5S6a5a6 Basis Cj109000000-M S107/10110000000630 S20 ½ 5/601000000600 S3012/300100000708 S401/10 ¼ 0001O000135 a5-M100000010500 a6-M010000001360 Zj-M O000MM -860M Cj- Zj 10+M 9+M0000-M 00

20. X1X2S1S2S3S4S5S6a5a6 Basis Cj109000000-M S100110007/100 -7/10 0280 S2005/601001/20-1/20350 S3002/30010100208 S400¼00011/100 -1/10 085 X11011100000010500 a6-M010000001360 Zj10-MO000-10M10-M 5000- 360M Cj-Zj09+M000010-M -M-10 0

21. X1X2S1S2S3S4S5S6a5a6 Basis Cj109000000-M X290110007/101 -7/10 280 S2000-5/6100 -1/12 01/120116( 2/3) S3000-2/3010 116/30 0 -16/30 021(1 /3) S40001/4001 -9/120 0 9/120 015 X110100000010500 a6-M00000 -7/10 7/10180 Zj1099+M000 - 37+7M/ 10 M 37- 7M/10 -M 7520- 80M Cj-Zj00 -9-M 000 37- 7M/10 -M -37- 8M/10 0

22. Unboundedness For maximisation problem we say that a linear program is unbounded if the value of the solution may be made infinitely large without violating any constraints. All the aij<0 in column j and the simplex method indicates the variables Xj is to be introduced into solution. Maxm2X1 + 1X2 St1X1>2 1X2<5 X1, X2 >0

23. X1X2S1a1S2 Basis Cj210-M0 X1 21010 2 S2 001001 5 Zj 20-220 4 Cj-Zj 012-M-20

24. Alternate Optimal Solution 2 or more optimal solutions possible when Cj-Zj=0 for one or more variables not in the solution Let profit of X1 becomes Rs.7/- Maxm. 7X1+10X2 St7/10X1+ 1X2<630 1/2X1+5/6X2<600 1X1+2/3X2<708 1/10X1+1/4X2<135 X1, X2 >0

25. Final X1 Table X2S1S2S3S4 BasisCj7100000 X171010/300-40/3300 S2000-10/1810-20/8100 S3000-22/90164/9128 X21001-4/30028/3420 Zj710 OOO6300 Cj-Zj00-10000

26. X1X2S1S2S3S4 BasisCj7100000 X1710 -5/4 0 120/64 0540 S2000 -30/32 1 10/64 0120 S4000 -22/64 0 9/64 118 X21001 15/18 0 -84/64 0252 Zj710 OOO6300 Cj-Zj00-10000

27. Degenercy A LP is said to be degenerate if one or more of the basis variables has a value zero. Maxm. 10X1+ 9X2 st7/10X1+ 1X2<630 1/2X1+5/6X2<480 sewing capacity reduced to 480 1X1+2/3X2<708 1/10X1+1/4X2<135 X1, X2 >0

28. X1X2S1S2S3S4 BasisCj1090000 S100 16/301 0 -7/10 0134.4 S200½ 0 1 -1/2 0126 X1101 2/30 0 1 0708 S400 22/1200 0 -1/10 164.2 Zj10 20/3 001007080 Cj-Zj0 7/3 00-100

29. Introduce variable : X2 Appropiate ratio to determine the pivotal elements: b1/a12 = 134.4X(30/16) = 252 b2/a22 = 126X2 = 252 b3/a32= 708X(3/2) = 1062 b4/a42= 64.2X ( 120/22) = 350.2

30. X1X2S1S2S3S4 BasisCj1090000 X290 130/16 0 -210/160 0252 S2000 -15/16 1 25/160 00 X1101 0-20/16 0 300/160 0540 S400 0-11/32 0 45/320 118 Zj10 970/16 0 111/16 07668 Cj-Zj0 0-70/16 0 -111/16 0

31. Sensitivity Analysis Is the study of how optimal solution and the value of the optimal solution behaves given changes in the various coefficients of the problems: 1. What will be the effect on the optimal solution if we change coefficient of the objective function (Cj) 2. What will be the effect on the optimal solution if we change right hand side constraint value(bj)? 3. What will be the effect on the optimal solution if we change coefficient of a constraint equation (aij) Examples :- Changes: - Price of raw material changes - Demand of various product shifts. - Companies purchase new machinery to replace old - Stock prices fluctuate - Employee turnover/ absenteeism

32. From graphical solution: Rotating the objective function line clockwise decreases the slope: 7/10X1 + 1X2 = 6301X1 + 2/3 X2 = 708 X2 = -7/10 X1 + 630X2 = -3/2 X1 + 1062 -3/2 < slope of objective function < -7/10 Z = C1X1 + C2 X2 X2 = -C1/C2 X1 + Z/C2 -3/2 C1/9 ; C1< 27/2 <13.5 -C1/9 7/10 ; C1 > 6.3 6.3 <C1 < 13.5 Standard bag profit is within 6.3 to 13,5 ; Production quantity : Std Bag- 540 and Deluxe bag - 252 remain same.

33. Right Hand side :  Cutting & Dyeing Constraint 630  640  (Feasible area will enlarge) 7/10 X1 + X2 < 640 X1 = 527.5 X2 = 270.75 Value of objective function : 527.5 X 10 + 9 X 270.75 = 7711.75 Profit Increased by 43.75 Profit increased by 43.75/10 = 4.375 per hour The change in the value of the objective function per unit increase in the RHS is called shadow price. As more and more resources are obtained and RHS value continues to increase, other constraints may become binding and limit the change in the value of the objective function

34. Sensitivity Analysis with the Simplex tableau: objective function coefficients X1X2S1S2S3S4 BasisCj1090000 X290130/160-21/160252 S2000-15/1615/320120 X1(C1)1010-20/16030/160540 S4000-11/3209/64118 ZJ10970/160111/1607668 Optimal solution Cj-Zj00-70/160 -111/16 0

35. Sensitivity Analysis with the Simplex tableau X1X2S1S2S3S4 BasisCj1090000 X290130/160-21/160252 S2000-15/1615/320120 X1C110-20/16030/160540 S4000-11/3209/64118 ZJC19 270- 20C1/16 030C1- 189/16 0 2268+54 0C1 Cj-Zj00 20C1- 270/16 0 189- 30C1/16 0

36. (20C1-270)/16 < 0 and (189 -30c1)/16 < 0 20C1 189 C1 6.3 6.3 < C1 < 13.5 Profit of the standard bag reduced to Rs 7 per unit

37. Sensitivity Analysis with the Simplex tableau: objective function coefficients X1X2S1S2S3S4 BasisCj790000 X290130/160-21/160252 S2000-15/1615/320120 X1710-20/16030/160540 S4000-11/3209/64118 Zj79130/16021/1606048 Cj-Zj00-130/160 -21/16 0

38. Sensitivity Analysis with the Simplex tableau (profit:Rs5/-) X1X2S1S2S3S4 BasisCj590000 X290130/160-21/160252 S2000-15/1615/320120 X1510-20/16030/160540 S4000-11/3209/64118 Zj79 170/16 0-39/160 4968 Cj-Zj00 -170/16 0 39/16 0

39. Sensitivity Analysis with the Simplex tableau X1X2S1S2S3S4 BasisCj1090000 X2C20130/160-21/160252 S2000-15/1615/320120 X11010-20/16030/160540 S4000-11/3209/64118 ZJC19 (30C2- 200)/16 0 (300- 21C2)/16 0 25C2 +5400 Cj-Zj00 (200- 30C2)/16 0 (21C2- 3000/16 0

40. (200 -30C2)/16 < 0 and (21C2-300)/16 < 0 30C2 >200 21C2 < 300 C2 > 6.67 C2 < 14.29 6.67 < C2 < 14.29 We see as long as the profit per deluxe bag is between Rs 6.67 and Rs 14.29 the production quantities of 540 standard bags and 252 deluxe bags will be optimal. Range of optimality for non-basic variables can be found out.

41. X1X2S1S2S3S4 BasisCj109Cs1000 X290 130/16 0 -210/160 0252 S2000 -15/16 1 25/160 00 X1101 0-20/16 0 300/160 0540 S400 0-11/31 0 45/320 118 Zj10 970/16 0 111/16 07668 Cj-Zj0 0Cs1- 70/16 0 -111/16 0

42. Cs1< 70/16 < 0; Cs1,< 70/16 The range of optimaility for Cs1 is : - Infinity < Cs1 < 70/16 As long as the objective function coefficient for S1 < 70/16 the current solution will be optimal. In addition, there is no lower bound on how much the coefficient may be decreased; The same analysis holds for all non basic variables. There is no lower limit on how much the coefficient of a nonbasic variable can be decreased. In addition the upper limit on how much it can be increased before it is profitable to bring the nonbasic variable into solution is given by Zj. Thus the range of optimality for the objective function coefficient of each nonbasic variable is : -infinity < Cj < Zj

43. Steps to compute : Range of optimality for the objective function 1. Replace the numerical value for the objective function coefficient of Xk with Ck everywhere it appears in the final simplex tableau. 2. Recompute the Cj – Zj entries for each non basic variable ( if variable Xk is non basic, it is only necessary to recompute (Ck-Zk) 3. Apply inequality to each recomputed Cj-Zj 4. Solve the resulting inequalities for the upper and lower bounds on the range of optimality

44. Sensitivity Analysis with the Simplex tableau: (optimal solution) X1X2S1S2S3S4 BasisCj1090000 X290130/160-21/160252 S2000-15/1615/320120 X11010-20/16030/160540 S4000-11/3209/64118 Zj10970/160111/1607668 Optimal solution Cj-Zj00-70/160 -111/16 0

45. Zj values of the four slack variables are 70/16, 0, 111/16 and 0. Thus the shadow price of cutting & Dyeing time is Rs 70/16 = Rs 4.375 Change the value of the objective function resulting from one unit increase in the value of the right hand side. Shadow price = 70/16 = 4.375 : Cutting & Dyeing = 111/16 = 6.9375 : Fininshing If a slack variable is a basic variable in the optimal solution; the shadow price of the corresponding resource is zero. Range of Feasibility for RHS Values: Cutting & dyeing time available ( bi) from 630-650 hrs We expect a increase in the value of the objective function 20. (70/16 ) = 87.50 New solution = 252 30/16= 289.5 120-15/16= 101.25 540 + 20-20/16= 515 18-11/32= 11.125 7668 70/16= 7755.5

46. X1X2S1S2S3S4 BasisCj590000 X290130/160-21/160289.5 S2000-15/1615/320101.25 X1510-20/16030/160515 S4000-11/3209/64111.25 Zj5970/160111/1607755.5 Cj-Zj00-70/160 -111/16 0

47. Upper and lower bounds for the maximisation amount that b1 can be changed before the current optimal basis becomes infeasible. If we change bi to b1 : The new value for the basic variables : X225230/16252 +30/16 b1 S2 =120-15/16 =120-15/16 b1 X1540+ b1-20/16540-20/16 b1 S4 18-11/32 18-11/32 b1 252+30/16 b1 > 0 b1 > -134.4 120-15/16 b1 > 0 b1 < 128 540 – 20/16 bi >0 b1 < 432 18-11/32 bi >0 b1 < (52)4/11 -134.4 < b1 < (52)4/11 495.6 < b1 < (682)4/11 Net profit increase = 7668X 70/16 X (52)4/11 = (7897) 1/11

48. Duality Every LP Problem has an associated LP problem called duality. Primal --  Dual Maxm. 10X1+ 9X2 st7/10X1+ 1X2<630 1/2X1+5/6X2<480 1X1+2/3X2<708 1/10X1+1/4X2<135 X1, X2 >0 Dual : Minm. 630u1+600u2+708u3+135u4 s t 7/10u1 + 1/2u2 + 1u3 + 1/10u4>10 1u1 + 5/6u2 + 2/3 u3 + ¼ u4 >9 u1, u2, u3, u4 > 0

49. General Points about the dual of a maximisation problem o Maximisation --  minimisation o Dual has all > constraints o Where primal has n decision variables the dual will have n constraints o When primal has m constraints dual has m decision variables o The RHS values of primal becomes objective function coefficients in the dual o The objective function coefficients of the primal become the RHS values in the dual o The constraint coefficient of the ith primal variable becomes the coefficient of the ith constraints o Both have non negativity constraints

50. Simplex tableau U1U2U3U4S1S2a1a2 Basis Cj-630-600-708-13500-M a1-M7/10½11/1001010 a2-M15/62/3¼0019 Zj -17/M-4M/3-5M/3- 42M/12 0 MM-M -19M Cj-Zj - 630+17/ M - 600+4M /3 - 708+5M /3 - 135+42 M/120 -M 00

51. Simplex tableau U1U2U3U4S1S2a1a2 Basis Cj-630-600-708-13500-M a1-M0-5/128/15- 9/12 0 7/101-7/10 37/10 U1-63015/62/3¼0019 Zj -630 - 525+M/ 12 -420- 8M/15 - 1571//2 +9M/12 0 M 630- 7M/10 -M 7M/10- 630 - 37M/1 0-5670 Cj-Zj 0 -75- M/12 - 288+8M /15 22 ½- 9M/120 -M 7M/10- 630 0 630- 17M/1 0

52. Simplex tableau U1U2U3U4S1S2a1a2 Basis Cj-630-600-708-13500-M S200 -10/84 16/21-9/84 -10/7110/737/7 U1-6301 60/84 30/2112/84 -10/7010/70100/ 7 Zj -630-450-900-90 9000-9000 -9000 Cj-Zj0-150192-45-9000900- M -M

53. Simplex tableau U1U2U3U4S1S2a1a2 Basis Cj-630-600-708-13500-M U3-7080 -5/32 1-9/64 -15/8 21/16 15/8 - 21/16 111/1 6 U1-6301 15/16 011/32 5/4-15/8-5/4 15/870/16 Zj -630-480-708-117 540252-540-252 -7668 Cj-Zj0-1200-18-540-252540- M 252-M

54. Since all the Cj-Zj values in the net evaluation row are < 0 this iteration provides the optimal solution. U1 = 70/16 U2=0 U3 = 111/16 U4=0 S1=0 S2=0 a1=0 a2=0 Since we have been maximising the negative of the dual objective function, the value of the objective function for the dual solution must be –(-7668) or 7668 The optimal solution for the primal problem is X1=540; X2 =252; S1=0 S2=120; S3=0 and S4=18. The optimal value of the objective function is 7668

55. Simplex tableau: primal X1= 540; X2=252, S1=0; S2=120, S3=0 and S4= 18; optimal value of objective function=7668 X1X2S1S2S3S4 BasisCj1090000 X290130/160-21/160252 S2000-15/1615/320120 X1(C1)1010-20/16030/160540 S4000-11/3209/64118 ZJ10970/160111/1607668 Cj-Zj00-70/160 -111/16 0