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
RECEIVER
0,60
0,20
AUDIO
16,12
ELEC
PLAY
MAXIMUM PROFIT
PLAYERS=X1
0
RECEIVERS
=X2
20
PROFIT=
50X1+120X2
2400=MAX
16
12
2240
20
0
1000
RECEIVER
0,60
MAX
0,20
AUDIO
16,12
ELEC
PLAY
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
0
RECEIVERS
=X2
20
PROFIT=
50X1+80X2
1600
16
12
20
0
1760=NEW
MAX
1000
RECEIVER
0,60
OLD
MAX
0,20
AUDIO
16,12
=NEW MAX
ELEC
PLAY
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
RECEIVER
0,60
0,20
AUDIO
16,12
0,16
NEW
ELEC
OLD
ELEC
17,9.2
PLAY
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
0
RECEIVERS
=X2
16
PROFIT=
50X1+120X2
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
RECEIVER
BACK TO ORIGINAL OPTIMUM
OPTIMUM ON ELEC CONSTR
0,60
OPTIMUM NOT ON AUDIO CONSTR
MAX
0,20
AUDIO
16,12
ELEC
PLAY
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)
RECEIVER
0,75
REDUNDANT CONSTR
0,60
MAX
0,20
NEW ELEC
AUDIO
16,12
OLD
ELEC
PLAY
150,0
RECEIVER
0,75
REDUNDANT CONSTR
0,60
OLD
MAX
0,20
AUDIO
PLAY
150,0
NEW MAXIMUM
X1
X2
PROFIT=
50X1+120X2
0
60
7200=MAX
20
0
1000
RECEIVER
0,75 INFEASIBLE
0,60 =NEW
MAX
OLD
MAX
0,20
REDUNDANT CONSTR
NEW ELEC
AUDIO
PLAY
150,0
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
X2
(24,8) =MAX
(1)
(2)
X1
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
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•
•
•
•
-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