Transcript decision analysis - University of Redlands

Lecture
3
MGMT 650
Sensitivity Analysis in LP
Chapter 3
1
Example

Consider the following linear program:
Min
s.t.
6x1 + 9x2
($ cost)
x1 + 2x2 < 8
10x1 + 7.5x2 > 30
x2 > 2
x1, x2 > 0
2
The Management Scientist Output
OBJECTIVE FUNCTION VALUE = 27.000
Variable
x1
x2
Value
1.500
2.000
Reduced Cost
0.000
0.000
Constraint
1
2
3
Slack/Surplus
2.500
0.000
0.000
Dual Price
0.000
-0.600
-4.500
3
The Management Scientist Output (continued)
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value
x1
0.000
6.000
x2
4.500
9.000
Upper Limit
12.000
No Limit
RIGHTHAND SIDE RANGES
Constraint Lower Limit Current Value
1
5.500
8.000
2
15.000
30.000
3
0.000
2.000
Upper Limit
No Limit
55.000
4.000
4
Example Optimal Solution
According to the output:
x1 = 1.5
x2 = 2.0
Objective function value = 27.00
5
Range of Optimality

Question
Suppose the unit cost of x1 is decreased to $4. Is
the current solution still optimal? What is the value
of the objective function when this unit cost is
decreased to $4?
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The Management Scientist Output
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value
x1
0.000
6.000
x2
4.500
9.000
Upper Limit
12.000
No Limit
RIGHTHAND SIDE RANGES
Constraint Lower Limit Current Value
1
5.500
8.000
2
15.000
30.000
3
0.000
2.000
Upper Limit
No Limit
55.000
4.000
7
Range of Optimality

Answer
• The output states that the solution remains
optimal as long as the objective function
coefficient of x1 is between 0 and 12.
• Because 4 is within this range, the optimal
solution will not change.
• However, the optimal total cost will be affected
• 6x1 + 9x2 = 4(1.5) + 9(2.0) = $24.00.
8
Range of Optimality

Question
How much can the unit cost of x2 be decreased without
concern for the optimal solution changing?

The Management Scientist Output
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value
x1
0.000
6.000
x2
4.500
9.000
Upper Limit
12.000
No Limit
RIGHTHAND SIDE RANGES
Constraint Lower Limit Current Value
1
5.500
8.000
2
15.000
30.000
3
0.000
2.000
Upper Limit
No Limit
55.000
4.000
9
Range of Optimality

Answer
The output states that the solution remains
optimal as long as the objective function coefficient of
x2 does not fall below 4.5.
10
Range of Optimality and 100% Rule

Question
If simultaneously the cost of x1 was raised to $7.5 and the cost
of x2 was reduced to $6, would the current solution remain
optimal?

Answer
• If c1 = 7.5, the amount c1 changed is 7.5 - 6 = 1.5.
• The maximum allowable increase is 12 - 6 = 6,
• so this is a 1.5/6 = 25% change.
• If c2 = 6, the amount that c2 changed is 9 - 6 = 3.
• The maximum allowable decrease is 9 - 4.5 = 4.5,
• so this is a 3/4.5 = 66.7% change.
• The sum of the change percentages is 25% + 66.7% = 91.7%.
• Since this does not exceed 100% the optimal solution would
not change.
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Range of Feasibility

Question
If the right-hand side of constraint 3 is increased by 1, what will
be the effect on the optimal solution?
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value
x1
0.000
6.000
x2
4.500
9.000
Upper Limit
12.000
No Limit
RIGHTHAND SIDE RANGES
Constraint Lower Limit Current Value
1
5.500
8.000
2
15.000
30.000
3
0.000
2.000
Upper Limit
No Limit
55.000
4.000
12
Range of Feasibility – Contd.
OBJECTIVE FUNCTION VALUE = 27.000
Variable
x1
x2
Value
1.500
2.000
Reduced Cost
0.000
0.000
Constraint
1
2
3
Slack/Surplus
2.500
0.000
0.000
Dual Price
0.000
-0.600
-4.500
13
Range of Feasibility

Answer
• A dual price represents the improvement in the
objective function value per unit increase in the
right-hand side.
• A negative dual price indicates a deterioration
(negative improvement) in the objective, which in
this problem means an increase in total cost
because we're minimizing.
• Since the right-hand side remains within the range
of feasibility, there is no change in the optimal
solution.
• However, the objective function value increases by
$4.50.
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Reduced Cost

Definition
• How much the objective function coefficient of
each variable would have to improve before it
would be possible for that variable to assume a
positive value in the optimal solution
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(Alternatively) LINDO output Report (of Diet Problem)
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Integer Programming Applications
Chapter 8
17
Example: Metropolitan Microwaves
 Metropolitan Microwaves, Inc. is planning to expand its
operations into other electronic appliances. The company
has identified seven new product lines it can carry.
Relevant information about each line follows on the next
slide.
Initial Floor Space Exp. Rate
Product Line
Invest.
(Sq.Ft.)
of Return
1.
2.
3.
4.
5.
6.
7.
TV/DVDs
Color TVs
Projection TVs
VCRs
DVD Players
Video Games
Home Computers
$ 6,000
12,000
20,000
14,000
15,000
2,000
32,000
125
150
200
40
40
20
100
8.1%
9.0
11.0
10.2
10.5
14.1
13.2
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Example: Metropolitan Microwaves
 Metropolitan has decided that they should not stock
projection TVs unless they stock either TV/DVDs or color
TVs.
 Also, they will not stock both VCRs and DVD players, and
they will stock video games if they stock color TVs.
 Finally, the company wishes to introduce at least three new
product lines.
 The company has $45,000 to invest and
 420 sq. ft. of floor space available
 Determine Metropolitan’s optimal expansion policy to
maximize its overall expected return.
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Problem Formulation
 Define Decision Variables
xj = 1 if product line j is introduced;
= 0 otherwise.
where:
Product line 1 = TV/DVDs
Product line 2 = Color TVs
Product line 3 = Projection TVs
Product line 4 = VCRs
Product line 5 = DVD Players
Product line 6 = Video Games
Product line 7 = Home Computers
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Problem Formulation

Define the Decision Variables
xj = 1 if product line j is introduced;
= 0 otherwise.

Define the Objective Function
Maximize total expected return:
Max .081(6000)x1 + .09(12000)x2 + .11(20000)x3
+ .102(14000)x4 + .105(15000)x5 + .141(2000)x6
+ .132(32000)x7
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Problem Formulation
 Define the Constraints
1) Budget:
6x1 + 12x2 + 20x3 + 14x4 + 15x5 + 2x6 + 32x7 < 45
2) Space:
125x1 +150x2 +200x3 +40x4 +40x5 +20x6 +100x7 < 420
3) Stock projection TVs only if stock TV/VCRs or color TVs:
x1 + x2 > x3 or x1 + x2 - x3 > 0
4) Do not stock both VCRs and DVD players:
x4 + x5 < 1
5) Stock video games if they stock color TV's:
x2 - x6 > 0
6) Introduce at least 3 new lines:
x1 + x2 + x3 + x4 + x5 + x6 + x 7 > 3
7) Variables are 0 or 1:
xj = 0 or 1 for j = 1, , , 7
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Optimal Solution in LINDO
Expected return
Introduce
TV/DVDs
Projection TVs
DVD players
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Optimal Solution in Management Scientist
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Integer Programming Application in
Distribution System Design




Dell Computers operates a plant in St. Louis; annual capacity = 30000 units
Computers are shipped to regional distribution centers located in Boston, Atlanta and Houston;
anticipated demands = 30,000, 20,000, 20,000 respectively
Because of anticipated increase in demand, plan to increase capacity by constructing a new plant in
one or more locations in Detroit, Columbus, Denver or Kansas City.
Proposed Location
Annual Fixed Cost ($)
Annual Capacity
Detroit
175,000
10,000
Columbus
300,000
20,000
Denver
375,000
30,000
Kansas City
500,000
40,000
Develop a model for choosing the best plant locations
 Determine the optimal amounts to transport from each plant to each distribution center such that
all demand is satisfied.
Plant Site
Boston
Atlanta
Houston
Detroit
5
2
3
Columbus
4
3
4
Denver
9
7
5
Kansas City
10
4
2
St. Louis
8
4
3
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Formulation

Minimize
5x11+2x12+3x13+4x21+3x22+4x23+9x31+7x32+5x33+10x41+4x42+2x43
+8x51+4x52+3x53+175000y1+300000y2+375000y3+500000y4
Subject to
 x11+x12+x13-10000y1<=0
Detroit capacity
 x21+x22+x23-20000y2<=0
Columbus capacity
 x31+x32+x33-30000y3<=0
Denver capacity
 x41+x42+x43-40000y4<=0
Kansas City capacity
 x51+x52+x53 <=30000
St. Louis capacity
 x11+x21+x31+x41+x51=30000
Boston demand
 x12+x22+x32+x42+x52=20000
Atlanta demand
 x13+x23+x33+x43+x53=20000
Houston demand
 xij>=0
Non-negativity constraints
 y1, y2, y3, y4 = {0,1}
Integrality constraints
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Optimal Solution




Locate 1 new plant in Kansas City (capacity = 40000) {y4=1}
Supply 30000 from existing St. Louis plant to Boston DC {x51=30000}
Supply 20000 from existing St. Louis plant to Atlanta DC {x42=20000}
Supply 20000 from existing St. Louis plant to Houston DC {x43=20000}
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Example: Tina’s Tailoring





Tina's Tailoring has five idle tailors and four custom
garments to make.
The estimated time (in hours) it would take each tailor to
make each garment is shown below.
An 'X' indicates an unacceptable tailor-garment assignment.
Formulate an integer program for determining the tailorgarment assignments that minimize the total estimated time
spent making the four garments.
No tailor is to be assigned more than one garment and each
garment is to be worked on by only one tailor.
Tailor
Garment
Wedding gown
Clown costume
Admiral's uniform
Bullfighter's outfit
1
19
11
12
X
2
23
14
8
20
3
20
X
11
20
4
21
12
X
18
5
18
10
9
21
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Formulation: Tina’s Tailoring


Define the decision variables
xij = 1 if garment i is assigned to tailor j
= 0 otherwise.
Number of decision variables = [(# of garments)(# of
tailors)] - (# of unacceptable assignments) = [4(5)] - 3 = 17
Define the objective function
Minimize total time spent making garments:
Min 19x11 + 23x12 + 20x13 + 21x14 + 18x15 + 11x21
+ 14x22 + 12x24 + 10x25 + 12x31 + 8x32 + 11x33
+ 9x35 + 20x42 + 20x43 + 18x44 + 21x45
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Constraints: Tina’s Tailoring


Exactly one tailor per garment:
1) x11 + x12 + x13 + x14 + x15 = 1
2) x21 + x22 + x24 + x25 = 1
3) x31 + x32 + x33 + x35 = 1
4) x42 + x43 + x44 + x45 = 1
No more than one garment per tailor:
5) x11 + x21 + x31 <= 1
6) x12 + x22 + x32 + x42 <= 1
7) x13 + x33 + x43 <= 1
8) x14 + x24 + x44 <= 1
9) x15 + x25 + x35 + x45 <= 1
Integrality: xij > {0,1} for i = 1, . . ,4 and j = 1, . . ,5
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Optimal Solution
Estimated total time
Wedding Gown to Tailor #5
Clown costume to Tailor #1
Admiral’s uniform to Tailor #2
Bullfighter’s outfit to Tailor #4
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Airline Crew Scheduling


Southeast Airlines needs to assign crews to cover all of its upcoming flights.
We will focus on assigning 3 crews based in San Francisco to the flights listed below.
Flight
Flight Leg
1 2 3 4 5 6 7 8 9 10
1
SF to LA
2
SF to Denver
3
SF to Seattle
4
LA to Chicago
5
LA to SF
6
Chicago to Denver
7
Chicago to Seattle
8
Denver to SF
9
Denver to Chicago
10
Seattle to SF
11
Seattle to LA
Cost (000’s)



1
1
1
1
1
1
1
3
3
3
3
2
4
3
4
4
7
5
3
3
4
5
7
2
4
2
6
3
5
2
2
4
2
5
2
3
1
4
3
2
1
1
2
2
12
1
1
2
11
8
5
2
4
4
2
9
9
8
9
Exactly 3 of the feasible sequences need to be chosen such that every flight is covered.
It is permissible to have more than 1 crew per flight, where the extra crews would fly as passengers, but union
contracts require that the extra crews would still need to be paid for their time as if they were working.
Schedule one crew for every feasible sequence to minimize the total cost of the 3 crew assignments that cover
all the flights.
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