1. Exam Feb 28: sets 1,2 Set 2 due Thurs

2. LP SENSITIVITY Ch 3

3. Sensitivity Analysis I.GRAPHICAL A.Objective Function B.Left-hand side of constraint C.Right-hand side of constraint II. ALGEBRA

4. SENSITIVITY ANALYSIS Does optimal solution change if input data changes?

5. INSENSITIVE

6. SENSITIVE

7. SENSITIVITY ANALYSIS Estimation error Change over time? Input might be random variable Should we remove constraint? “What if?” questions

8. I. GRAPHICAL NEW EXAMPLE: RENDER AND STAIR QUANTITATIVE ANALYSIS X1 = NUMBER OF CD PLAYERS X2 = NUMBER OF RECEIVERS

9. ORIGINAL PROBLEM MAX PROFIT = 50X1 + 120X2 SUBJECT TO CONSTRAINTS (1) ELECTRICIAN CONSTRAINT: 2X1 + 4X2 < 80 (2) AUDIO TECHNICIAN CONSTR: 3X1+ X2 < 60 NEXT SLIDE: REVIEW OF LAST WEEK

10. PLAYER RECEIVER AUDIO 0,60 20,0 ELEC 0,20 40,0 16,12

11. MAXIMUM PROFIT PLAYERS=X1RECEIVERS =X2 PROFIT= 50X1+120X2 0202400=MAX 16122240 2001000

12. PLAYER RECEIVER AUDIO 0,60 20,0 ELEC 0,20 40,0 16,12 MAX

13. ORIGINAL PROBLEM MAKE RECEIVERS ONLY NOW WE WILL BEGIN TO CONSIDER “WHAT IF” QUESTIONS IF NEW OPTIMUM IS “MAKE RECEIVERS ONLY”, WE CALL IT “OUTPUT INSENSITIVE” IF NEW OPTIMUM IS A DIFFERENT CORNER POINT, “OUTPUT SENSITIVE”

14. SENSITIVITY ANALYSIS I.A. OBJECTIVE FUNCTION

15. NEW OBJECTIVE FUNCTION 50X1 + 80X2 NEW PROFIT PER RECEIVER = $80 OLD PROFIT “ “ WAS $120 IS OPTIMUM SENSITIVE TO REDUCED PROFIT, PERHAPS DUE TO LOW PRICE (INCREASED COMPETITION) OR HIGHER COST?

16. NEW MAXIMUM PROFIT PLAYERS=X1RECEIVERS =X2 PROFIT= 50X1+80X2 0201600 16121760=NEW MAX 2001000

17. PLAYER RECEIVER AUDIO 0,60 20,0 ELEC 0,20 40,0 16,12 MAX OLD =NEW MAX

18. OUTPUT SENSITIVE WE NOW MAKE BOTH RECEIVERS AND PLAYERS (MIX) ORIGINAL: RECEIVERS ONLY REASON: LOWER PROFIT OF EACH RECEIVER IMPLIES PLAYERS MORE DESIRABLE THAN BEFORE

19. SENSITIVITY ANALYSIS I.B LEFT –HAND SIDE OF CONSTRAINT

20. ELECTRICIAN CONSTRAINT OLD: 2X1 + 4X2 < 80 NEW: 2X1 + 5X2<80 REASON: NOW TAKES MORE TIME FORELECTRICIAN TO MAKE ONE RECEIVER(5 NOW VS 4 BEFORE), LOWER PRODUCTIVITY PERHAPS DUE TO WEAR AND TEAR

21. BACK TO ORIGINAL OBJECTIVE FUNCTION WE COMPARE WITH ORIGINAL PROBLEM, NOT FIRST SENSITIVITY BUT FEASIBLE REGION WILL CHANGE

22. PLAYER RECEIVER AUDIO 0,60 20,0 ELEC 0,20 40,0 16,12 OLD NEW ELEC 0,16 17,9.2

23. SMALLER FEASIBLE REGION CAN MAKE FEWER RECEIVERS THAN BEFORE NEW INTERCEPT (0,16) REPLACES (0,20) (0,20) NOW INFEASIBLE ALSO, NEW MIX CORNER POINT: (17,9.2)

24. NEW MAXIMUM PROFIT PLAYERS=X1RECEIVERS =X2 PROFIT= 50X1+120X2 0161920 179.21954=NEW MAX 2001000

25. OUTPUT SENSITIVE COMPARED TO ORIGINAL PROBLEM, A NEW OPTIMUM INCREASED LABOR TO MAKE RECEIVER MAKES PLAYER MORE DESIRABLE

26. I.C. RIGHT-HAND SIDE OF CONSTRAINT

27. SLACK VARIABLES S1 = NUMBER OF HOURS OF ELECTRICIAN TIME NOT USED S2 =NUMBER OF HOURS OF AUDIO TIME NOT USED (1) ELEC CONSTR: 2X1+4X2+S1=80 (2) AUDIO CONSTR: 3X1+X2+S2=60

28. PLAYER RECEIVER AUDIO 0,60 20,0 ELEC 0,20 40,0 16,12 MAX BACK TO ORIGINAL OPTIMUM OPTIMUM ON ELEC CONSTR OPTIMUM NOT ON AUDIO CONSTR

29. X1=0,X2=20 (1)ELEC: 2(0)+4(20)+S1=80 S1 = 0 NO IDLE ELECTRICIAN (2) AUDIO: 3(0)+20+S2=60 S2 = 60 –20= 40 AUDIO SLACK

30. INPUT SENSITIVE INPUT SENSITIVE IF NEW OPTIMUM CHANGES SLACK ORIGINAL ELEC SLACK = 0 IF NEW ELEC SLACK > 0, INPUT SENSITIVE IF NEW ELEC SLACK STILL ZERO, INPUT INSENSITIVE

31. CHANGE IN RIGHT SIDE OLD ELEC CONSTR: 2X1+4X2<80 NEW ELEC CONSTR 2X1+4X2<300 NEW INTERCEPTS: (0,75) & (150,0)

32. PLAYER RECEIVER AUDIO 0,60 20,0 ELEC 0,20 40,0 16,12 MAX OLD 150,0 0,75 REDUNDANT CONSTR NEW ELEC

33. PLAYER RECEIVER AUDIO 0,60 20,0 0,20 40,0 MAX 150,0 0,75 REDUNDANT CONSTR OLD

34. NEW MAXIMUM X1X2PROFIT= 50X1+120X2 0607200=MAX 2001000

35. PLAYER RECEIVER AUDIO 0,60 20,0 0,20 40,0 MAX 150,0 0,75 REDUNDANT CONSTR OLD INFEASIBLE =NEW MAX NEW ELEC

36. NEW OPTIMUM OUTPUT INSENSITIVE SINCE WE STILL MAKE RECEIVERS ONLY (SAME AS ORIGINAL) BUT INPUT SENSITIVE ORIGINAL: OPTIMUM ON ELEC CONSTR NEW: OPTIMUM ON AUDIO CONSTR

37. SLACK SLACK VARIABLE ORIGNEW ELECS1=0S1>0 AUDIOS2>0S2=0

38. INTERPRET INCREASE IN ELECTRICIAN AVAILABILITY TOO MANY ELECTRICIANS THEREFORE ELEC SLACK

39. SHADOW PRICE VALUE OF 1 ADDITIONAL UNIT OF RESOURCE INCREASE IN PROFIT IF WE COULD INCREASE RIGHT-HAND SIDE BY 1 UNIT

40. THIS EXAMPLE SHADOW PRICE = MAXIMUM YOU WOULD PAY FOR 1 ADDITIONAL HOUR OF ELECTRICIAN OLD ELEC CONSTR: 2X1+4X2<80 NEW ELEC CONSTR: 2X1+4X2<81

41. OPTIMUM X1X2PROFIT= 50X1+120X2 080/4=202400=OLD MAX 081/4=20.252430=NEW MAX

42. SHADOW PRICE = 2430-2400=30 Electrician “worth” up to $30/hr “Dual” value

43. II. ALGEBRA OBJECTIVE FUNCTION: Z = C1X1 + C2X2, Where C1 and C2 are unit profits

44. II. Algebra For what range of values of the objective function coefficient C1 does the optimum stay at the current corner point?

45. Example Source: Taylor, Bernard, INTO TO MANAGEMENT SCIENCE, p 73 X1 = NUMBER OF BOWLS TO MAKE X2 = NUMBER OF MUGS TO MAKE MAX PROFIT = C1X1+C2X2=40X1+50X2 CONSTRAINTS (1) LABOR: X1 + 2X2 < 40 (2) MATERIAL: 4X1+ 3X2 < 120

46. X1 X2 (1) (2) (24,8)=MAX

47. Old Optimum Make both bowls and mugs Output insensitive if new solution is also bowls and mugs

48. STEP 1: SOLVE FOR X2 IN OBJECTIVE FUNCTION WE WANT A RANGE FOR C1, SO WE SOLVE FOR X2 C1 VARIABLE, C2 CONSTANT PROFIT= Z=C1X1 + 50X2 50X2= Z – C1X1 X2 = (Z/50) –(C1/50)X1 COEFFICIENT OF X1 IS –C1/50

49. STEP2: SOLVE FOR X2 IN CONSTRAINT (1) (1) X1 + 2X2=40 2X2=40-X1 X2=20-0.5X1 COEFFICIENT OF X1 IS –0.5

50. STEP 3: STEP 1 = STEP 2 -C1/50= -0.5 C1 = 25 OLD C1 = 40 SENSITIVITY RANGE: SAME CORNER POINT OPTIMUM SENSITIVITY RANGE SHOULD INCLUDE OLD C1 C1 > 25

51. STEP 4: SOLVE FOR X2 IN CONSTRAINT (2) (2) 4X1+3X2=120 3X2= 120-4X1 X2=40-(4/3)X1=40-1.33X1 COEFICIENT OF X1 IS –1.33

52. STEP 5: STEP 1 = STEP 4 -C1/50 = -1.33 C1=67 OLD C1 = 40 RANGE INCLUDES 40 C1 < 67

53. Step 3 and step 5 25 < C1 < 67 If you strongly believe that C1 is between 25 and 67, optimal solution is same corner point as C1 =40. Make both bowls and mugs if profit per bowl is between $ 25 and $ 67