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

LP SENSITIVITY Ch 3

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

SENSITIVITY ANALYSIS Does optimal solution change if input data changes?

INSENSITIVE

SENSITIVE

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

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

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

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

MAXIMUM PROFIT PLAYERS=X1RECEIVERS =X2 PROFIT= 50X1+120X =MAX

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

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”

SENSITIVITY ANALYSIS I.A. OBJECTIVE FUNCTION

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?

NEW MAXIMUM PROFIT PLAYERS=X1RECEIVERS =X2 PROFIT= 50X1+80X =NEW MAX

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

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

SENSITIVITY ANALYSIS I.B LEFT –HAND SIDE OF CONSTRAINT

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

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

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

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)

NEW MAXIMUM PROFIT PLAYERS=X1RECEIVERS =X2 PROFIT= 50X1+120X =NEW MAX

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

I.C. RIGHT-HAND SIDE OF CONSTRAINT

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

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

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

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

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

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

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

NEW MAXIMUM X1X2PROFIT= 50X1+120X =MAX

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

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

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

INTERPRET INCREASE IN ELECTRICIAN AVAILABILITY TOO MANY ELECTRICIANS THEREFORE ELEC SLACK

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

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

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

SHADOW PRICE = =30 Electrician “worth” up to $30/hr “Dual” value

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

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

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

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

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

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

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

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

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

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

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