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Chapter 11 Discrete Optimization Models. 11.1 Lumpy Linear Programs and Fixed Charges. Lumpy linear problems add either/or constraints or objective functions to what is otherwise a linear programs. ILP Modeling of All-or-Nothing Requirements - PowerPoint PPT Presentation

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Chapter 1 Making Economic Decisions

Chapter 11Discrete Optimization Models

1

11.1 Lumpy Linear Programs and Fixed Charges

Lumpy linear problems add either/or constraints or objective functions to what is otherwise a linear programs.

ILP Modeling of All-or-Nothing Requirements

All-or-nothing variable requirements of the form

xj = 0 or uj

Can be modeled by substituting xj = uj yj , with new discrete variable yj = 0 or 1. [11.1]

The new yj can be interpreted as the fraction of limit uj chosen.

2

Swedish Steel Blending Example

min 16 x1+10 x2 +8 x3+9 x4 +48 x5+60 x6 +53 x7

s.t.x1+ x2 + x3+ x4 + x5+ x6 + x7 = 1000

0.0080 x1 + 0.0070 x2 + 0.0085 x3 + 0.0040 x4 6.5 0.0080 x1 + 0.0070 x2 + 0.0085 x3 + 0.0040 x4 7.5

0.180 x1 + 0.032 x2 + 1.0 x5 30.0

0.180 x1 + 0.032 x2 + 1.0 x5 30.5

0.120 x1 + 0.011 x2 + 1.0 x6 10.0

0.120 x1 + 0.011 x2 + 1.0 x6 12.0

0.001 x2 + 1.0 x7 11.0

0.001 x2 + 1.0 x7 13.0

x1 75

x2 250

x1x7 0

(11.1)

Cost = 9953.67

x1* = 75, x2* = 90.91, x3* = 672.28, x4* = 137.31

x5* = 13.59, x6* = 0, x7* = 10.91

3

Swedish Steel Model with All-or-Nothing Constraints

Suppose that the first two ingredients (x1, x2) had this lumpy character. We use either none or all 75 kg of ingredient 1 and none or all 250 kg of ingredient 2.

Let yj represents the discrete alternatives

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Swedish Steel Model with All-or-Nothing Constraints

min 16(75)y1+10(250)y2 +8 x3+9 x4 +48 x5+60 x6 +53 x7

s.t.75y1+ 250y2 + x3+ x4 + x5+ x6 + x7 = 1000

0.0080(75)y1+ 0.0070(250)y2+0.0085x3+0.0040x4 6.5 0.0080(75)y1+ 0.0070(250)y2+0.0085x3+0.0040x4 7.5

0.180(75)y1 + 0.032(250)y2 + 1.0 x5 30.0

0.180(75)y1 + 0.032(250)y2 + 1.0 x5 30.5

0.120(75)y1 + 0.011(250)y2 + 1.0 x6 10.0

0.120(75)y1 + 0.011(250)y2 + 1.0 x6 12.0

0.001(250)y2 + 1.0 x7 11.0

0.001(250)y2 + 1.0 x7 13.0

x1 75

x2 250

x3x7 0

y1, y2 = 0 or 1

(11.2)

Cost = 9967.06

y1* = 1, y2* = 0, x3* = 736.44, x4* = 160.06

x5* = 16.50, x6* = 1.00, x7* = 11.00

5

ILP Modeling of Fixed Charges

Another common source of lumpy phenomena arises when the objective function involves fixed charges.

Minimize objective functions with non-negative fixed charges for making variable xj>0 can be modeled by introducing new fixed charge variables

The objective coefficient of yj is the fixed cost of xj, and the coefficient of xj is its variable cost. [11.2]

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ILP Modeling of Fixed Charges

Switching constraints model the requirement that continuous variable xj0 can be used only if a corresponding binary variable yj =1 by

Where is a given or derived upper bound on the value of xj in any feasible solution. and the coefficient of xj is its variable cost. [11.3]

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Swedish Steel Model with Fixed Charges

Assume that ingredients 1 to 4 (x1, x2, x3, x4) can be used in the furnace only after injection mechanisms are setup at a cost of 350 kroner each. Let yj represents the discrete alternatives

8

Swedish Steel Examplewith Fixed Charges

min 16 x1+10 x2 +8 x3+9 x4 +48 x5+60 x6 +53 x7 + 350 y1+ 350 y2 + 350 y3 + 350 y4

s.t.x1+ x2 + x3+ x4 + x5+ x6 + x7 = 1000

0.0080 x1 + 0.0070 x2 + 0.0085 x3 + 0.0040 x4 6.5 0.0080 x1 + 0.0070 x2 + 0.0085 x3 + 0.0040 x4 7.5

0.180 x1 + 0.032 x2 + 1.0 x5 30.0

0.180 x1 + 0.032 x2 + 1.0 x5 30.5

0.120 x1 + 0.011 x2 + 1.0 x6 10.0

0.120 x1 + 0.011 x2 + 1.0 x6 12.0

0.001 x2 + 1.0 x7 11.0

0.001 x2 + 1.0 x7 13.0

x1 75 y1x3 1000 y3

x2 250y2x4 1000 y4

x1x7 0y1,.., y4 = 0 or 1

Cost = 11017.06

x1* = 75,

x2* = 0,

x3* = 736.44,

x4* = 160.06

x5* = 13.59,

x6* = 1, x7* = 11

y1* = 1, y2* = 0

y3* = 1, y4* = 1

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11.2 Knapsack and Capital Budgeting Models

Knapsack and capital budgeting problems are completely discrete.

A knapsack model is a pure integer linear program with a single main constraint. [11.4]

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Example 11.1 Indy Car Knapsack

The mechanics in the Indy Car racing team face a dilemma. Six different features might still be added to this years car to improve its top speed. The following table lists their estimated costs and speed enhancements.

Proposed Feature, j123456Cost ($000s)10.26.023.011.19.831.6Speed increase (mph)83157101211

Example 11.1 Indy Car Knapsack

Suppose first that Indy Car wants to maximize the performance gain without exceeding a budget of $35,000. Using decision variables

Max 8 x1 + 3 x2 + 15 x3 + 7 x4 + 10 x5 + 12 x6

s.t.10.2x1 + 6.0x2 + 23.0x3+ 11.1x4 + 9.8x5 + 31.6x6 35

x1 ,, x6 = 0 or 1

Speed increase = 25

Cost = 32.8

x1* = 0, x2* = 0, x3* = 1,

x4* = 0, x5* = 1, x6* = 0

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Example 11.1 Indy Car Knapsack

Suppose now that the Indy Car team decides they simply must increase speed by 30 miles per hour to have any chance of winning the next race. Ignoring the budget, they wish to find the minimum cost way to achieve at least that much performance.

Min 10.2x1 + 6.0x2 + 23.0x3+ 11.1x4 + 9.8x5 + 31.6x6

s.t.8 x1 + 3 x2 + 15 x3 + 7 x4 + 10 x5 + 12 x6 30

x1 ,, x6 = 0 or 1

Cost = 43.0

Speed increase = 33

x1* = 1, x2* = 0, x3* = 1,

x4* = 0, x5* = 1, x6* = 0

13

Capital Budgeting Models

The typical maximize form of a knapsack problem has a single main constraint enforcing a budget.

Capital budgeting models (or multi-dimensional knapsack) select a maximum value collection of project, investments, and so on, subject to limitations on budgets or other resources consumed. [11.5]

14

Example 11.2 NASA Capital Budgeting

The U.S. space agency, NASA, must deal constantly with such decision problems in choosing how to divide its limited budgets among many competing missions proposed. The following Table shows a fictitious list of alternatives.

We must decide which of the 14 indicated missions to include in program plans for the 2000-2024 era. Thus it should be clear that the needed decision variables are

15

Example 11.2 NASA Capital Budgeting

MissionBudget Requirements ($ billion)ValueNot withDepends On2000-20042005-20092010-20142015-20192020-20241 Comm. satellite6----200--2 Orbital microwave 23---3--3 lo lander 35---20--4 Uranus orbiter 2020 ----1050535 Uranus orbiter 2010 -58--70436 Mercury probe --18420-37 Saturn probe 18---5-38 Infrared imaging ---5-1011-9 Groundbased SETI 45---20014-10 Large orbital struc.-84--150--11 Color imaging --27-188212 Medical technology 57---8--13 Polar orbital platform -1411300--14 Geosynchronous SETI -45331859-Budget 101214141416

Capital Budgeting Models

Budget constraints limit the total funds or other resources consumed by selected projects, investments, and so on, in each time period not to exceed the amount available. [11.6]

6x1 + 2x2 + 3x3 + 1x7 + 4x9 + 5x12 10

3x2 + 5x3 + 5x5 + 8x7 + 5x9 + 8x10 + 7x12 + 1x13 + 4x14 12

8x5 + 1x6 + 4x10 + 2x11 + 4x13 + 5x14 14

8x6 + 5x8 + 7x11 + 1x13 + 3x14 14

10x4 + 4x6 + 1x13 + 3x14 14

17

Capital Budgeting Models

Mutually exclusiveness conditions allowing at most one of a set of choices are modeled by constraints summing over each choice set. [11.7]

x4 + x5 1

x8 + x11 1

x9 + x14 1

Dependence of choice j on choice i can be enforced on corresponding binary variables by constraint [11.8]

x11 x2

x4 x3

x5 x3

x6 x3

x7 x3

18

NASA Example Model

max 200x1+3x2 +20x3+50x4 +70x5+20 x6 +5x7+10x8+200x9 +150x10+18x11 +8x12 +300x13+185x14

s.t. 6x1 + 2x2 + 3x3 + 1x7 + 4x9 + 5x12 10

3x2 + 5x3 + 5x5 + 8x7 + 5x9 + 8x10 + 7x12 + 1x13 + 4x14 12

8x5 + 1x6 + 4x10 + 2x11 + 4x13 + 5x14 14

8x6 + 5x8 + 7x11 + 1x13 + 3x14 14

10x4 + 4x6 + 1x13 + 3x14 14

x4 + x5 1

x8 + x11 1

x9 + x14 1

xi = 0 or 1 j=1,,14

(11.7)

x11 x2

x4 x3

x5 x3

x6 x3

x7 x3

x1* = x3* = x4* = x8* = x13* = x14* =1

Value = 765

19

11.3 Set Packing, Covering, and Partitioning Models

Set covering constraints requiring that at least one member of sub-collection J belongs to a solution are expressed [11.9]

Set packing constraints requiring that at most one member of sub-collection J belongs to a solution are expressed [11.10]

20

11.3 Set Packing, Covering, and Partitioning Models

Set partitioning constraints requiring that exactly one member of sub-collection J belongs to a solution are expressed [11.11]

21

Example 11.3 EMS Location Planning

Austin, Texas undertook a study of the positioning of its emergency medical service (EMS) vehicles. That city was divided into service districts needing EMS services, and vehicle stations selected from a list of alternatives so that as much of the population as possible would experience a quick response to calls for help.

Our city is divided into 20 service districts that we wish t