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 = 50X1 + 120X2

• 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=X1 RECEIVERS

=X2

PROFIT=

50X1+120X2

0 20 2400=MAX

16 12 2240

20 0 1000

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=X1 RECEIVERS

=X2

PROFIT=

50X1+80X2

0 20 1600

16 12 1760=NEW

MAX

20 0 1000

PLAYER

RECEIVER

AUDIO

0,60

20,0

ELEC

0,20

40,0

16,12

MAXOLD

=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

16,12OLD

NEWELEC

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=X1 RECEIVERS

=X2

PROFIT=

50X1+120X2

0 16 1920

17 9.2 1954=NEW MAX

20 0 1000

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

X1 X2 PROFIT=

50X1+120X2

0 60 7200=MAX

20 0 1000

PLAYER

RECEIVER

AUDIO

0,60

20,0

0,20

40,0

MAX

150,0

0,75

REDUNDANT CONSTR

OLD

INFEASIBLE

=NEWMAX

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

ORIG NEW

ELEC S1=0 S1>0

AUDIO S2>0 S2=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

X1 X2 PROFIT=

50X1+120X2

0 80/4=20 2400=OLD MAX

0 81/4=20.25 2430=NEW MAX

SHADOW PRICE = 2430-2400=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=40-1.33X1

• COEFICIENT OF X1 IS –1.33

STEP 5: STEP 1 = STEP 4

• -C1/50 = -1.33

• 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