sda 3e chapter 13
TRANSCRIPT
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© 2007 Pearson Education
Chapter 13: OptimizationModeling
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Optimization The process of selecting values of
decision variables that maximize or minimize an objective function.
Optimal solution – the best set of
decision variables
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Constrained Optimization Constraints - limitations or requirements that decision
variables must satisfy.
The amount of material used to produce a set of productscannot exceed the available amount of 850 square feet.
The amount of money spent on research and developmentprojects cannot exceed the assigned budget of $300,000.
Contractual requirements specify that at least 500 units of product must be produced.
A mixture of fertilizer must contain exactly 30 percentnitrogen.
We cannot produce a negative amount of product(nonnegativity).
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Constraint Functions Amount of material used 850 square feet
Amount spent on research and development
$300,000 Number of units of product produced 500
Amount of nitrogen in mixture/total amount
in mixture = 0.30 Amount of product produced 0
The left hand sides are called constraint functions.
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Mathematical Representation Suppose that the material requirements for three
products are 3.0, 3.5, and 2.3 square feet per unit.
Let A, B, and C represent the number of units of each
product to produce. The amount of material used to produce A units of
product A = 3.0 A
The amount of material used to produce B units of
product B = 3.5B The amount of material used to produce C units of
product C = 2.3C
Constraint: 3.0 A + 3.5B + 2.3C 850
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Constraint Categories Simple bounds
Limitations
Requirements
Proportional relationships
Balance constraints
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Solutions Feasible solution: any solution that
satisfies all constraints
A problem that has no feasible solutionsis called infeasible.
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Types of Optimization
Problems Linear
Integer linear
Nonlinear
A linear function is a sum of terms, each of which is
some constant multiplied by a decision variable, forexample:
2x + 3y - 12z
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Transformation to Linear
Functions If two ingredients contain 20 percent and 33 percent
nitrogen, respectively, then the fraction of nitrogen ina mixture of x pounds of the first ingredient and ypounds of the second ingredient is expressed by theconstraint function (0.20x + 0.33y)/(x + y)
If a constraint requires that the fraction to be 0.3,this can be rewritten as a linear function
(0.20x + 0.33y ) = 0.3(x + y)
or
-0.1x + 0.03y = 0
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Common Model Types Product mix
Media selection
Process selection
Blending
Production Planning
Portfolio selection Multiperiod investment planning
Transportation planning
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Example: Product MixComponents Required /Unit
Component A Component B Profit/unit
Gold Player 6 3 $48
Platinum Player 6 5 $70
Maximize Profit = 48G + 70P
6G + 6P
3,000 (Component A limitation) 3G + 5P 1750 (Component B limitation)
G 0, P 0 (nonnegativity )
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Example: Media Selection
Maximize 2000R + 3500T + 2700M
500R + 2000T + 200N 40000
0 R 15
T 12
6 N 30
Medium Cost/ad Exposurevalue/ad
MinimumUnits
Maximum Units
Radio $ 500 2,000 0 15
TV $ 2,000 3,500 12 -
Magazine $ 200 2,700 6 30
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Example: Process Selection A textile mill produces three types of fabrics. The decision facing the
plant manager is on what type of loom to process each fabric duringthe next 13 weeks. The mill has 15 regular looms and 3 dobbie looms.
Dobbie looms can be used to make all fabrics and are the only loomsthat can weave certain fabrics. After weaving, fabrics are sent to thefinishing department and then sold. Any fabrics that cannot be wovenin the mill because of limited capacity will be purchased from anexternal supplier, finished at the mill, and sold at the selling price. Inaddition to determining which looms to process the fabrics, themanager also needs to determine which fabrics to buy externally.
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LP Model D i = number of yards of fabric i to produce on dobbie looms, i = 1, .
. . , 3
R i = number of yards of fabric i to produce on regular looms, i = 1, .
. . , 3 P i = number of yards of fabric i to purchase from an outside
supplier, i = 1, . . . , 3
Min 0.65D 1 + 0.61D 2 + 0.50D 3 ++ 0.61R 2 + 0.50R 3 + 0.85P 1 + 0.75P 2 +0.65P 3
D 1 + P 1 = 45,000 (Demand, fabric 1)D 2 + R 2 + P 2 = 76,500 (Demand, fabric 2)
D 3 + R 3 + P 3 =10,000 (Demand, fabric 3)
0.213D 1 + 0.192D 2 + 0.227D 3 6552 (Dobbie loom production time)
0.192R 2 + 0.227R 3 32,760 (Regular loom production time)
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Example: BlendingIngredient Protein % Fat % Fiber % Cost/lb.
Sunflower seeds 16.90 26 29 $0.22
White millet 12 4.1 8.3 $0.19
Kibble corn 8.5 3.8 2.7 $0.10
Oats 15.4 6.3 2.4 $0.10
Cracked corn 8.50 3.80 2.70 $0.07
Wheat 12 1.7 2.3 $0.05
Safflower 18 17.9 28.8 $0.26
Canary grass seed 11.9 4 10.9 $0.11
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LP ModelMinimize 0.22X1 + 0.19X2 + 0.10X3 + 0.10X4 + 0.07X5 + 0.05X6 +0.26X7 + 0.11X8
X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 = 1 (proportion)
0.169X1 + 0.12X2 + 0.085X3 + 0.154X4 + 0.085X5 +0 .12X6 + 0.18X7 +0.119X8 0.13 (protein)
0.26X1 + 0.041X2 + 0.038X3 + 0.063X4 + 0.038X5 + 0.017X6 +0.179X7 + 0.04X8 0.15 (fat)
0.29X1 + 0.083X2 + 0.027X3 + 0.024X4 + 0.027X5 + 0.023X6 +0.288X7 + 0.109X8 0.14 (fiber)
Xi 0, for i = 1,2,...8
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Production Planning
Autumn SpringWinter
PA PW PS
150 400 50
IA IW IS
Minimize 11P A + 14PW + 12.50PS + 1.20I A + 1.20IW + 1.20IS
P A - I A = 150PW + I A - IW = 400PS + IW - IS = 50Pi 0, for all iIi 0, for all i
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Cash Management A financial manager must ensure that funds are available to pay
company expenditures but would also like to maximize interest income.Three short-term investment options are available over the next six
months: A, a one-month CD that pays 0.5 percent, available eachmonth; B, a three-month CD that pays 1.75 percent, available at thebeginning of the first four months; and C, a six-month CD that pays 2.3percent, available in the first month. The net expenditures for the nextsix months are forecast as $50,000, ($12,000), $23,000, ($20,000),$41,000, ($13,000). Amounts in parentheses indicate a net inflow of cash. The company must maintain a cash balance of at least $100,000at the end of each month. The company currently has $200,000 incash.
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Model Development A i = amount ($) to invest in a one-month CD at the start of month i
B i = amount ($) to invest in a three-month CD at the start of month i
C i = amount ($) to invest in a six-month CD at the start of month i
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LP Model
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Transportation ProblemPlant/D.C. Cleveland Baltimore Chicago Phoenix Capacity
Marietta $12.60 $14.35 $11.52 $17.58 1200
Minneapolis $9.75 $12.63 $8.11 $15.88 800
Demand 150 350 500 1000
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LP ModelMinimize 12.60X11 + 14.35X12 +11.52X13 +17.58X14
+9.75X21 +12.63X22 +8.11X23 +15.88X24
X11 + X12 + X13 + X14 1200
X21 + X22 + X23 + X24 800
X11 + X21 = 150
X12 + X22 = 350X13 + X23 = 500
X14 + X24 = 1000
Xij
0, for all i and j
Supply constraints
Demand constraints
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Spreadsheet Modeling Set up a logical format
Define cells for the decision variables
Define separate cells for the objectivefunction and each constraint function
Avoid Excel functions ABS, MIN, MAX,INT, ROUND, IF, COUNT
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Example: Product Mix Model
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Spreadsheet Formulas
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Example: Transportation
Model
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Integer Optimization Models IP Model: some or all decision variables
are restricted to integer values
Binary variables: 0 or 1
0 x 1 and integer
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Example: Cutting Stock
Problem Suppose that a company makes standard 100-inch-wide
rolls of thin sheet metal, and slits them into smaller rollsto meet customer orders for widths of 12, 15, and 30
inches. The demands for these widths vary from week toweek. Demands this week are 870 12” rolls, 450 15” rolls,and 650 30” rolls.
Cutting patterns:
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IP Model Define Xi to be the number of 100” rolls to cut
using cutting pattern i, for i = 1,…,5.
Min 10X1 + 10X2 + 4X3 + 1X4 + 1X5
0X1 + 0X2 + 8X3 + 2X4 + 7X5 870 (12” rolls)
6X1 + 0X2 + 0X3 + 1X4 + 1X5 450 (15” rolls)
0X1 + 3X2 + 0X3 + 2X4 + 0X5 650 (30” rolls) Xi 0 and integer
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IP Models With Binary
Variables A binary variable x is simply a general
integer variable that is restricted to
being between 0 and 1:0 x 1 and integer
We usually just write this as
x = 0 or 1
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Example: Project Selection
Maximize $180,000x 1 + $220,000x 2 + $150,000x 3 + $140,000x 4 +$200,000x 5
$55,000x 1
+ $83,000x 2
+ $24,000x 3
+ $49,000x 4+ $61,000x
5 $150,000
(cash limitation)
5x 1 + 3x 2 + 2x 3 + 5x 4 + 3x 5 12 (personnel limitation)
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Project Selection Model -
Spreadsheet
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Modeling Logical ConditionsLogical Condition Constraint Model Form
If A then B B A or B – A 0
If not A then B B 1 – A or A + B 1
If A then not B B 1 – A or B + A 1
At most one of A and B A + B 1
If A then B and C (B A and B A) or B + C 2A
If A and B then C C A + B – 1 or A + B – C 1
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Example: Supply Chain Facility
Location Xij = 1 if customer zone j is assigned to DC i, and 0 if not,
and Y i = 1 if CD i is chosen from among a set of k potential locations.
Cij = the total cost of satisfying the demand in customerzone j from DC i.
Min CijXij
Xij =1, for every j (each customer assigned to exactly one DC) Y i = k, for every i (choose k DC’s)
Xij Y i, for every i and j (only assign zone j to DC i if DC i isselected)
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Example: Distribution Center
LocationPlant/D.C. Cleveland Baltimore Chicago Phoenix Capacity
Marietta $12.60 $14.35 $11.52 $17.58 1200
Minneapolis $9.75 $12.63 $8.11 $15.88 800Fayetteville $10.41 $11.54 $9.87 $8.32 1500
Chico $13.88 $16.95 $12.51 $11.64 1500
Demand 300 500 700 1800
Select a new plant from among Fayetteville and Chico
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IP ModelMinimize 12.60X11 + 14.35X12 +11.52X13 +17.58X14 +9.75X21 +12.63X22
+8.11X23 +15.88X24 + 10.41X31 + 11.54X32 + 9.87X33 + 8.32X34 + 13.88X41
+ 16.95X42 + 12.51X43 + 11.64X44
X11 + X12 + X13 + X14 1200
X21 + X22 + X23 + X24 800
X31 + X32 + X33 + X34 1500Y1
X41 + X42 + X43 + X44 1500Y2
X11 + X21 + X31 + X41 = 300X12 + X22 + X32 + X42 = 500
X13 + X23 + X33 + X43 = 700
X14 + X24 + X34 + X44 = 1800
Y1 + Y2 = 1
Xij 0, for all i and j
Y1, Y2 = 0,1
Ensures thatexactly one DC is
selected. Y 1corresponds toFayetteville; Y 2 corresponds to
Chico
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Nonlinear Optimization Either objective function or constraint
functions are not linear
Models are unique in structure
Solution techniques are different fromlinear and integer optimization
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Example: Hotel Pricing With
Elastic Demand A 450-room hotel has the following history:
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Model Development Projected number of rooms of a given type sold =
(Historical Average Number of Rooms Sold) +(Elasticity)(New Price - Current Price)(Historical Average
Number of Rooms Sold)/(Current Price)
Define S = price of a standard room, G = price of a goldroom, and P = price of a platinum room.
Total Revenue = S (625 - 4.41176S ) + G (300 - 2.04082G ) + P (100 -0.35971P )
= 625S + 300G + 100P - 4.41176S 2 - 2.04082G 2 -0.35971P 2
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ModelMaximize 625S + 300G + 100P - 4.41176S 2 - 2.04082G 2 -0.35971P 2
70 S 90 (price range restrictions)
90 G 110
120 P 149
(625 - 4.41176S ) + (300 = 2.04082G ) + (100 = 0.35971P ) 450
or 1025 - 4.41176S - 2.04082G - 0.35971P 450 (room limitation)
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Spreadsheet Model
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Example: Markowitz Portfolio
Model Select stocks to minimize portfolio
variance
and ensure a specified expected return
k
1i i j jiij
2
i
k
1i
2
i xx2sxs
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Example Variance-Covariance MatrixStock 1 Stock 2 Stock 3
Stock 1 .025 .015 -.002
Stock 2 .030 .005
Stock 3 .004
Exp. return 10% 12% 7%
Minimize Variance =
x1 + x2 + x3 = 1
10x1 + 12x2 + 7x3 10 (required return)
323121
2
3
2
2
2
1 xx010.0xx0.004-xx0.03x004.x030.x025.