Exam Feb 28: sets 1,2

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 lt 80
• (2) AUDIO TECHNICIAN CONSTR 3X1 X2 lt 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
PLAYERSX1 RECEIVERS X2 PROFIT 50X1120X2
0 20 2400MAX
16 12 2240
20 0 1000
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
PLAYERSX1 RECEIVERS X2 PROFIT 50X180X2
0 20 1600
16 12 1760NEW MAX
20 0 1000
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 lt 80
• NEW 2X1 5X2lt80
• 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
PLAYERSX1 RECEIVERS X2 PROFIT 50X1120X2
0 16 1920
17 9.2 1954NEW MAX
20 0 1000
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 2X14X2S180
• (2) AUDIO CONSTR 3X1X2S260

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
X10,X220

• (1)ELEC 2(0)4(20)S180
• S1 0
• NO IDLE ELECTRICIAN
• (2) AUDIO 3(0)20S260
• S2 60 20 40
• AUDIO SLACK

30
INPUT SENSITIVE

• INPUT SENSITIVE IF NEW OPTIMUM CHANGES SLACK
• ORIGINAL ELEC SLACK 0
• IF NEW ELEC SLACK gt 0, INPUT SENSITIVE
• IF NEW ELEC SLACK STILL ZERO, INPUT INSENSITIVE

31
CHANGE IN RIGHT SIDE

• OLD ELEC CONSTR 2X14X2lt80
• NEW ELEC CONSTR 2X14X2lt300
• 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
X1 X2 PROFIT 50X1120X2
0 60 7200MAX
20 0 1000
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
SLACK VARIABLE ORIG NEW
ELEC S10 S1gt0
AUDIO S2gt0 S20
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 2X14X2lt80
• NEW ELEC CONSTR 2X14X2lt81

41
OPTIMUM
X1 X2 PROFIT 50X1120X2
0 80/420 2400OLD MAX
0 81/420.25 2430NEW MAX
42
SHADOW PRICE 2430-240030

• 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 C1X1C2X240X150X2
• CONSTRAINTS
• (1) LABOR X1 2X2 lt 40
• (2) MATERIAL 4X1 3X2 lt 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 ZC1X1 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 2X240
• 2X240-X1
• X220-0.5X1
• 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 gt 25

51
STEP 4 SOLVE FOR X2 IN CONSTRAINT (2)

• (2) 4X13X2120
• 3X2 120-4X1
• X240-(4/3)X140-1.33X1
• COEFICIENT OF X1 IS 1.33

52
STEP 5 STEP 1 STEP 4

• -C1/50 -1.33
• C167
• OLD C1 40
• RANGE INCLUDES 40
• C1 lt 67

53
Step 3 and step 5

• 25 lt C1 lt 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