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Sensitivity Analysis and Optimization in Management Science

This exam covers topics related to sensitivity analysis, graphical representation, objective function, constraints, and algebraic analysis in management science. It includes examples and discussions on whether the optimal solution changes with input data changes, estimation error, change over time, and the impact of removing constraints.

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Sensitivity Analysis and Optimization in Management Science

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

  2. LP SENSITIVITY Ch 3

  3. Sensitivity Analysis GRAPHICAL Objective Function Left-hand side of constraint 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. RECEIVER 0,60 AUDIO 0,20 16,12 ELEC PLAYER 20,0 40,0

  11. MAXIMUM PROFIT

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

  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

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

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

  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

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

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

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

  34. NEW MAXIMUM

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

  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

  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

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

  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

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