SDA 3E Chapter 13

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© 2007 Pearson Education

Chapter 13: OptimizationModeling



Optimization The process of selecting values of

decision variables that maximize or minimize an objective function.

Optimal solution – the best set of

decision variables



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).



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.



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



Constraint Categories Simple bounds

Limitations

Requirements

Proportional relationships

Balance constraints



Solutions Feasible solution: any solution that

satisfies all constraints

A problem that has no feasible solutionsis called infeasible.



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



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



Common Model Types Product mix

Media selection

Process selection

Blending

Production Planning

Portfolio selection Multiperiod investment planning

Transportation planning



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 )



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



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.



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)



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



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



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



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.



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



LP Model



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



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



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



Example: Product Mix Model



Spreadsheet Formulas



Example: Transportation

Model



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



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:



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



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



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)



Project Selection Model -

Spreadsheet



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



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)



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



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



Nonlinear Optimization Either objective function or constraint

functions are not linear

Models are unique in structure

Solution techniques are different fromlinear and integer optimization



Example: Hotel Pricing With

Elastic Demand A 450-room hotel has the following history:



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



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)



Spreadsheet Model



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



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.

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SDA 3E Chapter 13