1. 1 Chapter 8 Sensitivity Analysis  Bottom line:   How does the optimal solution change as some of the elements of the model change?  For obvious reasons we shall focus on Models.  For obvious reasons we shall focus on Linear Programming Models.

2. 2 Ingredients of LP Models  Linear objective function  A system of linear constraints – RHS values – Coefficient matrix (LHS) – Signs (=, =)   How does the optimal solution change as these elements change?

3. 3 8.1 Introduction  What kind of changes are to be considered?  How do we handle such changes?  To motivate the discussion it is instructive to classify the changes into two categories: – changes – Structural changes – changes – Parametric changes

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5. 5 Structural Changes  eg.  New decision variable  New constraint  Loss of a decision variable  Relaxation of a constraint

6. 6 Parametric Changes  Changes in one or more of the coefficients of the objective function (c j )  Changes in the RHS values (b i )  Changes in the coefficients of the LHS matrix {a ij }.

7. 7 our emphasis will be on  rather than investigation of many specific cases (as done in many textbooks....)  Basic principles and ideas rather than investigation of many specific cases (as done in many textbooks....)  Thus, in the exam you may have to demonstrate that you know how to apply the principles to solve “new” problems.

8. 8 it is time for a concrete example!

9. 9 Question?Question?  Suppose that the RHS value of the second constraint is changing.  What kind of changes should be expect in the optimal solution?

10. 10 Geometric Analysis

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12. 12 Back to algebra....

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20. 20   How do we conduct such an analysis in higher dimensions?

21. 21 8.2 The Basic Principles  Impact of changes: – – Feasibility – – Optimality  Terminology:  = before the change  Old = before the change  = after the change  New = after the change

22. 22 Two Cases  The old optimal solution remains optimal  The old optimal solution remains feasible but is no longer optimal  The second case has two sub-cases depending on the impact of the changes on the :  The second case has two sub-cases depending on the impact of the changes on the basis: – All the old basic variables remain in the basis – There is a change in the basis

23. 23  The latter means that a number of pivot operations will be required to construct the new optimal solution from the old optimal solution.  In the former case, the new optimal solution can be easily computed from the new RHS values (why?).

24. 24 General Observation  Changes in the objective function coefficients (c) are manifested in changes in the reduced costs of the non-basic variables of the (old) optimal solution. r = c B B -1 D - c (optimality criterion!)

25. 25 ObservationObservation  Changes in vector b are manifested in changes in the RHS values of the final simplex tableau. b’ = B -1 b (feasibility)

26. 26   Structural changes, as well as changes in the coefficient matrix (A), may require restoration of the canonical form of the simplex tableau.

27. 27 In short,... the tasks are  Checking whether the new RHS values are non-negative  Checking the signs of the new reduced costs  Pivot operations to restore the canonical form of the simplex tableau

28. 28 RecipeRecipe  Given the final tableau of the old solution and the changes to the model, compute the new tableau.  If the new tableau is not in a canonical form, restore the canonical form by appropriate pivot operations.  Check, if necessary, that the RHS values are non-negative.  Check, if necessary, that the reduced costs are of the appropriate sign (non-negative for opt=max, non-positive for opt=min).

29. 29 Implications....   to accomplish these tasks we need to be able to quickly compute new RHS and/or reduced costs resulting from the changes in the problem.

30. 30 8.3 Overview  The formulas used by the Revised Simplex Method suggest what factors should be under consideration when parameters of the LP model change  These formulas are constructive in developing recipes for sensitivity analysis of a number of important cases.  Thus,   read again Chapter 6....... ???

31. 31 Case 1: All the components of the new RHS are non- negative.  In this case the old optimal solution is still feasible, namely the changes in the problem do not have any impact on the feasibility of the "old" optimal solution. We do not have to take any "corrective action" as far as feasibility is concerned.  The of the basic variables are simply equal to the of the RHS.  The new values of the basic variables are simply equal to the new values of the RHS.

32. 32 Case 2. At least one of the components of the new RHS is negative.  Clearly in this case the "old" optimal solution is no longer feasible, as the non- negativity constraint is violated by at least one decision variable.  Thus, in this case the changes in the parameters of the model caused the "old" solution to become infeasible and therefore corrective actions involving must be taken.  Thus, in this case the changes in the parameters of the model caused the "old" solution to become infeasible and therefore corrective actions involving changes in the basis, must be taken.

33. 33 Case 3. All the components of r satisfy the optimality condition.  Since the optimality conditions are satisfied, the basic variables remain basic, thus no corrective actions are required.  Since the optimality conditions are satisfied, (by) the basic variables remain basic, thus no corrective actions are required.  Observe, however, that the value of x may have changed due to changes in the vector b.

34. 34 Case 4. At least one of the components of r violates the optimality conditions  Obviously, if we exclude degeneracy (why?), there is a need to changes the basis itself.  Thus, the changes in the problem results in a new basis.

35. 35 8.4 Common Cases   8.4.1 Changes in the RHS values, b. _ _ Suppose that we change one of the elements of b, say b k, by . so that the new b is equal to the old one except that the new value of b k is equal to b k + . _ _ In short, _ _ b (new) = b +  e k _ _ where e k is the kth column of the identity matrix (i.e. e k =(0,0,...0,1,0,...,0) where the 1 is in the kth position).

36. 36  In this case the new RHS value is given by b’ = B -1 b (new) = B -1 (b+  e k ) b’ = B -1 b (new) = B -1 (b+  e k ) This yields b’ = B -1 b +  B -1 e k = B -1 b +(  B -1 ). k b’ = B -1 b +  B -1 e k = B -1 b +(  B -1 ). kHence, b’ = old RHS +(  B -1 ). k b’ = old RHS +(  B -1 ). k

37. 37 RecipeRecipe  If the old basis is to remain optimal after the change occurs, the new RHS value must be non-negative, i.e b’>=0.  This will be the case if we require  (B -1 ).k >= - old RHS  (B -1 ).k >= - old RHS or equivalently  (B -1 ) i,k >= - old RHS i  (B -1 ) i,k >= - old RHS i for i=1,2,...,m.

38. 38 This produces.... .....  Don’t take it seriously..... min correction

39. 39 8.4.2 Example k=2

40. 40 final

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42. 42  We therefore conclude that the old basis (whatever it is), will remain optimal if the value of b 2 is in the interval [20- 4,20+4]=[16,24].   Comment: You do not have to use the recipe: Compute the RHS after the change and determine the critical values of .

43. 43 Direct Approach (NILN) k=2

44. 44 New final RHS = B -1 b’

45. 45  Thus, the non-negativity constraint requires 48+4  >= 0, hence  >= -12 16+4  >= 0, hence  >= - 4 4 -  >= 0, hence  = 0, hence  <= 4 In short, -4 <=  <= 4 -4 <=  <= 4 (16 <= b 2 <= 24) (16 <= b 2 <= 24)

46. 46 8.4.3 Changes in the elements of the cost vector, c.  Suppose that the value of c k changes for some k. How will this affect the optimal solution to the LP problem?  We can distinguish between two cases: (1) x k is not in the old basis (2) x k is in the old basis r j  c B B  1 D.j  c j,j  1,2,...,n

47. 47 Case 1: x k is not in the old basis  Thus : Recipe: r k >= , if opt=max r k <= , if opt = min r' k  c B B  1 D.k  (c k  )  (c B B  1 D.k  c)  r k 

48. 48 Case 2: x k is in the old basis

49. 49 ObservationsObservations  r’ j = 0 for basic variables x j.  (e k ) j = 0 for all nonbasic variables x j.  if opt=max all the old reduced costs are non-negative  if opt=min all the old reduced costs are non- positive.

50. 50 RecipeRecipe if opt=max if opt=min

51. 51 RemarkRemark   It is very unfortunate that sometime (often?) mathematical notation tends to obscure the essential elements of the situation under investigation. This is a typical example!   As an exercise (on your own) try to translate this to the language of the simplex tableau

52. 52 8.4.3 Example Suppose that the reduced costs in the final simplex tableau are as follows: r = (0,0,0,2 3 4) r = (0,0,0,2 3 4) with I B =(2,3,1), namely with x,x and x comprising the basis. with I B =(2,3,1), namely with x 2,x 3 and x 1 comprising the basis. What would happen if we change e value of c 4 ? What would happen if we change the value of c 4 ? First we observe that x 4 is not in the basis and that the opt=max (why?)

53. 53  The recipe for this case, namely (8.20) is that the old optimal solution remains optimal as long as r 4 ≥ , or in our case, 2 ≥ .  Note that we do not need to know the current (old) value of c 4 to reach this conclusion.  Next, suppose that consider changes in c 1, recalling that x 1 is in the basis.

54. 54 recipe for this case:  However, it will be instructive to apply the basic ideas directly! Let's do it.

55. 55 Preliminary Analysis  We see that in order to analyze this case we have to know the entries in the row of the final tableau (t p. ) that is represented by x 1 in the basis.  What is the value of p?  Since I B =(2,3,1), this is row p=3.  Suppose that this row is as follows:  t 3. = (0,0,1,3,-4,0)

56. 56 We can display this in a "tableau" form as follows: If we add  to the old c 1, we would have instead So we now have to restore the canonical form of the x 1 column. correction

57. 57 end result....  To ensure that the current basis remains optimal we have to make sure that all the reduced costs are non-negative (opt=max). Hence,  2+3  ≥ 0 and 3-4  ≥ 0  Thus, 3/4 ≥  ≥ -2/3 3/4 ≥  ≥ -2/3

58. 58 in words,....  the old optimal solution will remain optimal if we keep the increase in c 1 in the interval [-2/3, 3/4]. If  is too small it will be better to enter x 4 into the basis, if  is too large it will better to put x 5 into the basis.

59. 59   It is strongly recommended that you consider the following structural change on your own:   A new decision variable is introduced. Its cost coefficient is given and its constraint coefficients are given.   How does this affect the given final tableau of the old problem?

60. 60   Warning If you are asked about the range of admissible changes in say b 1 then it is not sufficient to report on the admissible changes in  1. That is, you have to translate the range of admissible changes in  1 into the range of admissible changes in b 1. Example: Suppose that b 1 = 12, and -2 <=  1 <= 3. Then the admissible range of b 1 is [12-2, 12+3 ] = [10,15].