Chapter 11 Discrete Optimization Models

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 • 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.

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 = 10000.0080 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 30.00.180 x1 + 0.032 x2 + 1.0 x5

30.5 0.120 x1 + 0.011 x2 + 1.0 x6 10.00.120 x1 + 0.011 x2 + 1.0 x6

12.0 0.001 x2 + 1.0 x7 11.00.001 x2 + 1.0 x7

13.0 x1 75x2 250x1…x7 0

(11.1)

Cost = 9953.67x1* = 75, x2* = 90.91, x3* = 672.28, x4* = 137.31x5* = 13.59, x6* = 0, x7* = 10.91

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

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 = 10000.0080(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 30.00.180(75)y1 + 0.032(250)y2 + 1.0 x5 30.5 0.120(75)y1 + 0.011(250)y2 + 1.0 x6 10.00.120(75)y1 + 0.011(250)y2 + 1.0 x6 12.0 0.001(250)y2 + 1.0 x7 11.00.001(250)y2 + 1.0 x7 13.0 x1 75x2 250x3…x7 0y1, y2 = 0 or 1

(11.2)

Cost = 9967.06y1* = 1, y2* = 0, x3* = 736.44, x4* = 160.06x5* = 16.50, x6* = 1.00, x7* = 11.00

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]

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]

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

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 = 10000.0080 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 30.00.180 x1 + 0.032 x2 + 1.0 x5 30.5 0.120 x1 + 0.011 x2 + 1.0 x6 10.00.120 x1 + 0.011 x2 + 1.0 x6 12.0 0.001 x2 + 1.0 x7 11.00.001 x2 + 1.0 x7 13.0 x1 75 y1 x3 1000 y3

x2 250y2 x4 1000 y4

x1…x7 0 y1,.., y4 = 0 or 1

Cost = 11017.06x1* = 75, x2* = 0, x3* = 736.44, x4* = 160.06x5* = 13.59, x6* = 1, x7* = 11y1* = 1, y2* = 0y3* = 1, y4* = 1

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]

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 year’s car to improve its top speed. The following table lists their estimated costs and speed enhancements.

Proposed Feature, j1 2 3 4 5 6

Cost ($000s) 10.2 6.0 23.0 11.1 9.8 31.6

Speed increase (mph) 8 3 15 7 10 12


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 35x1 ,…, x6 = 0 or 1

Speed increase = 25Cost = 32.8x1* = 0, x2* = 0, x3* = 1, x4* = 0, x5* = 1, x6* = 0


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 30x1 ,…, x6 = 0 or 1

Cost = 43.0Speed increase = 33x1* = 1, x2* = 0, x3* = 1, x4* = 0, x5* = 1, x6* = 0

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]

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

Example 11.2 NASA Capital Budgeting

MissionBudget Requirements ($ billion)

Value Not with

Depends On2000-

20042005-2009

2010-2014

2015-2019

2020-2024

1 Comm. satellite 6 - - - - 200 - -2 Orbital microwave 2 3 - - - 3 - -3 lo lander 3 5 - - - 20 - -4 Uranus orbiter 2020 - - - - 10 50 5 35 Uranus orbiter 2010 - 5 8 - - 70 4 36 Mercury probe - - 1 8 4 20 - 37 Saturn probe 1 8 - - - 5 - 38 Infrared imaging - - - 5 - 10 11 -9 Ground—based SETI 4 5 - - - 200 14 -10 Large orbital struc. - 8 4 - - 150 - -11 Color imaging - - 2 7 - 18 8 212 Medical technology 5 7 - - - 8 - -13 Polar orbital platform - 1 4 1 1 300 - -14 Geosynchronous SETI - 4 5 3 3 185 9 -Budget 10 12 14 14 14


• 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 103x2 + 5x3 + 5x5 + 8x7 + 5x9 + 8x10 + 7x12 + 1x13 + 4x14

128x5 + 1x6 + 4x10 + 2x11 + 4x13 + 5x14 148x6 + 5x8 + 7x11 + 1x13 + 3x14 1410x4 + 4x6 + 1x13 + 3x14 14


• 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 1x8 + x11 1x9 + 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

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 103x2 + 5x3 + 5x5 + 8x7 + 5x9 + 8x10 + 7x12 + 1x13 + 4x14

128x5 + 1x6 + 4x10 + 2x11 + 4x13 + 5x14 148x6 + 5x8 + 7x11 + 1x13 + 3x14 1410x4 + 4x6 + 1x13 + 3x14 14x4 + x5 1x8 + x11 1x9 + 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* =1Value = 765

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]

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]

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 to serve from some combination of the 10 indicated possibilities for EMS stations. Each station can provide service to all adjacent districts. For example, station 2 could service districts 1. 2. 6, and 7. Main decision variables are

Example 11.3 EMS Location Planning

1

2

3

4

5

6

7

8

9

10

Minimum Cover EMS Model

s.t. (11.8)

x2 1

x1 + x2 1x1 + x3 1x3 1x3 1x2 1x2 + x4 1x3 + x4 1x8 1

xj = 0 or 1 j=1,…,10

x4 + x6 1x4 + x5 1

x4 + x5 + x6 1x4 + x5 + x7 1x8 + x9 1x6 + x9 1x5 + x6 1x5 + x7 + x10 1x8 + x9 1x9 + x10 1x10 1

x2* = x3* = x4* = x6* = x8* = x10* =1, x1* = x5* = x7* = x9* = 0

Example 11.3 EMS Location Planning with Maximum Coverage

Suppose that we have funds for only 4 EMS locations. How can we find the collection of 4 that minimize coverage insufficiency?

• Set Packing, covering, and partitioning models can be modified to penalize uncovered items i by introducing new variables into each constraint i. [11.12]

Dist. Value Dist. Value Dist. Value Dist. Value1 5.2 6 5.7 11 30.4 16 25.62 4.4 7 10.0 12 30.9 17 11.03 7.1 8 12.2 13 12.0 18 5.34 9.0 9 7.6 14 9.3 19 7.95 6.1 10 20.3 15 15.5 20 9.9

Maximum Coverage EMS Model

Min 5.2y1+4.4y2 +7.1y3+9.0y4 +6.1y5+5.7 y6 +10.0y7+12.2y8+7.6y9 +20.3y10

+30.4y11+30.9y12+12y13+9.3y14 +15.5y15 +25.6y16+11y17+5.3y18 +7.9y19 +9.9y20 s.t. (11.9)x2 + y1 1

x1 + x2 + y2 1x1 + x3 + y3 1x3 + y4 1x3 + y5 1x2 + y6 1x2 + x4 + y7 1x3 + x4 + y8 1x8 + y9 1x4 + x6 + y10 1x4 + x5 + y11 1

xi = 0 or 1 j=1,…,10

x4 + x5 + x6 + y12 1x4 + x5 + x7 + y13 1x8 + x9 + y14 1x6 + x9 + y15 1x5 + x6 + y16 1x5 + x7 + x10 + y17 1x8 + x9 + y18 1x9 + x10 + y19 1x10 + y20 1

x3*=x4*=x5*=x9*= 1, y1*=y2*=y6*=y9*=y20*= 1Value = 32.8

Column Generation Models

• Column generation approaches deal with complex combinatorial problems by first enumerating a sequence of columns representing viable solutions to parts of the problem, and then solving a set of partitioning (or covering or packing) model to select an optimal collection of these alternatives fulfilling all problem requirements. [11.13]

Example 11.4 AA Crew Scheduling

A classic application of column generation arises in the enormously complex problem of scheduling crews for airlines. For example, American Airlines reports spending over $1.3 billion per year on salaries, benefits, and travel expenses of air crews. Careful scheduling or crew pairing can produce enormous savings.

Figure 11.2 illustrates a (tiny) sequence of flights to be crewed in our fictitious case. Each pairing is a sequence of flights to be covered by a single crew over a 2- to 3-day period. It must begin and end in the base city where the crew resides.

In real applications, intricate government and union rules regulate exactly which sequences of flights constitute a reasonable pairing. Complex software is employed to generate a list such as that of Table 11.3. Here we simply allow all closed sequences of 3 or 4 flights in Figure 11.2.Pairing


MIA

CHI

CHR

DFW

101

402

203

204

305406

407308

310

109

211

212

j Flight Sequence Cost1 101-203-406-308 29002 101-203-407 27003 101-204-305-407 26004 101-204-308 30005 203-406-310 26006 203-407-109 31507 204-305-407-109 25508 204-308-109 25009 305-407-109-212 260010 308-109-212 205011 402-204-305 240012 402-204-310-211 360013 406-308-109-211 255014 406-310-211 265015 407-109-211 2350


Pairing costs are equally complex. On duty crews are guaranteed pay for minimum periods, regardless of the part of that time they are actually flying. If the pairing requires overnight stays away from home, hotel and other expenses are added. Values for our illustration are shown in Table 11.3.

The decision variables are defined as

Column Generation Modelfor AA Example

max 2900x1+2700x2 +2600x3+3000x4 +2600x5+3150x6+2550x7 +2500x8

+2600x9+2050x10+2400x11+3600x12+2550x13+2650x14 +2350x15

s.t.

(11.10)

x1 + x2 + x3 + x4 = 1

x6 + x7 + x8 + x9 + x10 + x13 + x15 = 1x1 + x2 + x5 + x6 = 1

x3 + x4 + x7 + x8 + x11 + x12 = 1x12 + x13 + x14 + x15 = 1

x9 + y10 = 1x3 + x7 + x9 + x11 = 1

x1 + x4 + x8 + x10 + x13 = 1x5 + x12 + x14 = 1

x11 + x12 = 1x1 + x5 + x13 + x14 = 1x2 + x3 + x6 + x7 + x9 + x15 = 1

xi = 0 or 1 j=1,…,10

x1*=x9*=x12*= 1Cost = 9100

• It is standard to model all assignment forms with the decision variables

• Assignment Constraints take the form

Where all sums are limited to () combinations allowed in the problem. [11.14]

11.4 Assignment and Matching Models

CAM Linear Assignment ExampleRevisited

CAM model in Section 10.5 illustrates one assignment case. The following table repeats total transportation, queuing, and processing time for 8 pending jobs on 10 workstations to which they might next be routed. Each workstation can accommodate only one job at a time. We want to find a minimum total time routing.

Job, iNext Workstation, j

j=1 j=2 j=3 j=4 j=5 j=6 j=7 j=8 j=9 j=10i=1 8 23 5i=2 4 12 15i=3 20 13 6 8i=4 19 10i=5 8 12 16i=6 14 8 3i=7 6 27 12i=8 5 15 32

CAM Linear Assignment Model

min 8x1,1+23x1,3 +5x1,9+4x2,2 +12x2,4+15x2,5+20x3,3 +13x3,5 +6x3,6+8x3,8

+19x4,5 +10x4,6 +8x5,4+12x5,7+16x5,10+14x6,1+8x6,7 +3x6,9

+6x7,2+27x7,8+12x7,10 +5x8,2+15x8,3 +32x8,8

s.t.

(11.11)

x1,1 + x1,3 + x1,9 = 1

x2,2 + x2,4 + x2,5 = 1

x10,1 + x10,2 + x10,3 + x10,4 + x10,5 + x10,6

+ x10,7 + x10,8 + x10,9 + x10,10 = 1

Xi,j = 0 or 1 i=1,…,10; j=1,…,10

x1,1*=x2,2*=x3,8*=x4,6*=x5,4*=x6,9*=x7,10*=x8,3*= 1Total Time = 68 x1,1 + x6,1 + x9,1 + x10,1 = 1

x2,2 + x7,2 + x8,2 + x9,2 + x10,2 = 1

x5,10 + x7,10 + x9,10 + x10,10 = 1

• Linear assignment models minimize or maximize a linear objective function of the form

Subject to assignment constraints in [11.14], where is the cost (or benefit) of assigning . [11.15]

Linear Assignment Models

• Quadratic assignment models minimize or maximize a quadratic objective function of the form

Subject to assignment constraints in [11.14], where is the cost (or benefit) of assigning and . [11.16]

Quadratic Assignment Models

Example 11.5 Mall Layout Quadratic Assignment

Some of the most common cases producing quadratic assignment models arise in facility layout. We are given a collection of machines, offices, departments, stores, and so on, to arrange within a facility, and a set of locations within which they must fit. The problem is to decide which unit to assign to each location.

Figure 11.3 illustrates with 4 possible locations for stores in a shopping mall. Walking distances (in feet) between the shop locations are displayed in the adjacent table. The 4 prospective tenants for the shop locations are listed in Table 11.5. The table also shows the number of customers each week (in thousands) who might wish to visit various pairs of shops.


1

2

3

4

L=1 L=2 L=3 L=4J=1 -- 80 150 170

J=2 80 -- 130 100

J=3 150 130 -- 120

J=4 170 100 120 --

Distance (feet)

Common Customers (1000)i=1 i=2 i=3 i=4

i=1: Clothes Are -- 5 2 7

i=2: Computers Aye 5 -- 3 8

i=3: Toy Parade 2 3 -- 3

i=4: Book Bazaar 7 8 3 --


Mall managers want to arrange the stores in the 4 locations to minimize customer inconvenience. One very common measure is flow-distance, the product of flow volumes between facilities and the distances between their assigned locations. • Using the decision variables

Mall Layout Quadratic Model

min 5(80x1,1x2,2+150x1,1x2,3+170x1,1x2,4+80x1,2x2,1+130x1,2x2,3+100x1,2x2,4

+150x1,3x2,1+130x1,3x2,2+120x1,3x2,4+170x1,4x2,1+100x1,4x2,2+120x1,4x2,3) + 2(80x1,1x3,2+150x1,1x3,3+170x1,1x3,4+80x1,2x3,1+130x1,2x3,3+100x1,2x3,4





+150x3,3x4,1+130x3,3x4,2+120x3,3x4,4+170x3,4x4,1+100x3,4x4,2+120x3,4x4,3) (11.12)

Mall Layout Quadratic Model

s.t.

(11.12)

x1,1 + x1,2 + x1,3 + x1,4 = 1

x2,1 + x2,2 + x2,3 + x2,4 = 1x3,1 + x3,2 + x3,3 + x3,4 = 1x4,1 + x4,2 + x4,3 + x4,4 = 1x1,1 + x2,1 + x3,1 + x4,1 = 1x1,2 + x2,2 + x3,2 + x4,2 = 1x1,3 + x2,3 + x3,3 + x4,3 = 1x1,4 + x2,4 + x3,4 + x4,4 = 1Xi,j = 0 or 1 i=1,…,4; j=1,…,4

x1,1*=x2,4*=x3,3*=x4,2*=1Total Distance = 3260

• Suppose each object must be assigned to some , but that ’s may received several .

• b capacity of • size of ’s capacity consumed if is assigned to • cost (or benefit) of assigning to

Generalized Assignment Models

• Generalized assignment models, which encompass cases where allocation of to requires fixed size within capacity b have the form

s.t.

Here is the cost (or benefit) of assigning to and all sums are limited to (,) combinations allowed in the problem. [11.17]

Generalized Assignment Models

Example 11.6 CDOT Generalized Assignment

The Canadian Department of Transportation encountered a problem of the generalized assignment form when reviewing their allocation of coast guard ships on Canada’s Pacific coast. The ships maintain such navigational aids as lighthouses and buoys. Each of the districts along the coast is assigned to one of a smaller number of coast guard ships. Since the ships have different home bases and different equipment and operating costs, the time and cost for assigning any district varies considerably among the ships. The task is to find a minimum cost assignment.

Table 11.6 shows data for our (fictitious) version of the problem. Three ships-the Estevan, the Mackenzie, and the Skidegate--are available to serve 6 districts. Entries in the table show the number of weeks each ship would require to maintain aides in each district, together with the annual cost (in thousands of Canadian dollars). Each ship is available SG weeks per year.

Example 11.6 CDOT Generalized Assignment

District, iShip, j 1 2 3 4 5 61.Estevan Cost 130 30 510 30 340 20

Time 30 50 10 11 13 92.Mackenzie Cost 460 150 20 40 30 450

Time 10 20 60 10 10 173.Skidegate Cost 40 370 120 390 40 30

Time 70 10 10 15 8 12

𝑥𝑖 , 𝑗≡ {1 if district i is assigned to ship j0 otherwise

CDOT Assignment Model

min 130x1,1+460x1,2 +40x1,3 +30x2,1+150x2,2 +370x2,3 +510x3,1+20x3,2

+120x3,3 +30x4,1+40x4,2 +390x4,3 +340x5,1+30x5,2 +40x5,3 +20x6,1+450x6,2

+30x6,3

s.t.

(11.13)

x1,1 + x1,2 + x1,3 = 1

x2,1 + x2,2 + x2,3 = 1

x3,1 + x3,2 + x3,3 = 1

x4,1 + x4,2 + x4,3 = 1

x5,1 + x5,2 + x5,3 = 1

x6,1 + x6,2 + x6,3 = 1

Xi,j = 0 or 1 i=1,…,6; j=1,…,3

x1,1*=x4,1*=x6,1*=x2,2*=x5,2*=x3,3*=1Total Cost = 480

30x1,1 + 50x2,1 + 10x3,1 + 11x4,1 + 13x5,1 + 9x6,1 50

10x1,2 + 20x2,2 + 60x3,2 + 10x4,2 + 10x5,2 + 17x6,2 50

70x1,3 + 10x2,3 + 10x3,3 + 15x4,3 + 8x5,3 + 12x6,3 50

• Matching models, which seek an optimal pairing of like objects have the form

s.t.

Here is the cost (or benefit) of pairing with and all sums are limited to (,) combinations allowed in the problem. [11.18]

Matching Models

(11.14)

Example 11.7 Superfi Speaker Matching

We may illustrate matching with the task faced by fictitious high-fidelity speaker manufacturer Superfi. Superfi sells its speakers in pairs. Even though the manufacturing process maintains the most rigid quality standards, any two speakers produced will still interfere slightly with each other when connected to the same stereo system.

To improve its product quality even more, Superfi has measured the distortion dw for each pair of speakers in the current lot. They wish to determine how to pair the speakers so that total distortion is minimized.

Notice that any two speakers may be paired. There is no distinction between large and small, or left and right.

Tractability of Assignment and Matching Models

• Linear assignment models are highly tractable because they can be viewed as network flow problems, and thus as special cases of linear programming. Even more efficient special-purpose algorithms exist. [11.19]

• It is very difficult to compute global optima in quadratic assignment models because the nonlinear objective function precludes even the ILP method of Sections 12.2 to 12.5. Improving search heuristics like those of Sections 12.6 and 12.7 are usually applied. [11.20]

Tractability of Assignment and Matching Models

• Generalized assignment models, which are ILPs, can be solved in moderate size by the methods of Sections 12.2 to 12.5. Still, they are far less tractable than the linear assignment case. [11.21]

• Matching models are more difficult to solve than linear assignment case but efficient special-purpose algorithms are known that can solve quite large instances. [11.22]

• The Traveling Salesman Problem (TSP) seeks a minimum-total-length route visiting every points in a given set exactly once. [11.23]

• A traveling salesman problem is symmetric if the distance or cost of passing from any point to any other point is the same as the distance from to . Otherwise, the problem is asymmetric. [11.24]

11.5 Traveling Salesman and Routing Models

Example 11.8 NCB Circuit Board TSP

Figure 11.4 shows the tiny example that we will investigate for fictional board manufacturer NCB. We seek an optimal route through the 10 hole locations indicated. Table 11.7 reports straight-line distances di,j between hole locations i and j. Lines in Figure 11.4 show a fair quality solution with total length 92.8 inches. The best route is 11 inches shorter (see Section 12.6). 1 2 3 4 5 6 7 8 9 10

1 3.6 5.1 10.0 15.3 20.0 16.0 14.2 23.0 26.42 3.6 3.6 6.4 12.1 18.1 13.2 10.6 19.7 23.03 5.1 3.6 7.1 10.6 15.0 15.8 10.8 18.4 21.94 10.0 6.4 7.1 7.0 15.7 10.0 4.2 13.9 17.05 15.3 12.1 10.6 7.0 9.9 15.3 5.0 7.8 11.36 20.0 18.1 15.0 15.7 9.9 25.0 14.9 12.0 15.07 16.0 13.2 15.8 10.0 15.3 25.0 10.3 19.2 21.08 14.2 10.6 10.8 4.2 5.0 14.9 10.3 10.2 13.09 23.0 19.7 18.4 13.9 7.8 12.0 19.2 10.2 3.610 26.4 23.0 21.9 17.0 11.3 15.0 21.0 13.0 3.6

1

23

4 5

6

78 9

10

• Most ILP models of the symmetric case employ decision variables,

• Total length of a route can be calculated by

• Constraints for symmetric TSP

Formulating the Symmetric TSP

(11.15)

(11.16)

• Definition of subtoursS proper subset of points/cities to be routed

• Subtour elimination constraints

• Number of legs between points in S and points not in S must be at least 2.

Subtours

(11.17)

1

23

4 5

6

7

8 9

10

s.t.

ILP Model of the Symmetric TSP

[11.25]

s.t.

ILP Model of the Asymmetric TSP

[11.26]

Quadratic Assignment Formulation of the TSP

[11.27]s.t.

Let

Example 11.9 KI Truck Routing

Kraft Incorporated confronts such multiple-route design problems in planning truck delivery of its food products to over 100,000 commercial, industrial, and military customers in North America. Known customer requirements must be grouped into truckloads and then routes planned. Our tiny fictitious version of the KI case has 20 stops to be serviced from a single depot.

Stop, i fi Stop, i fi1 0.25 11 0.212 0.33 12 0.683 0.39 13 0.164 0.40 14 0.195 0.27 15 0.226 0.70 16 0.387 0.28 17 0.268 0.43 18 0.299 0.50 19 0.1710 0.22 20 0.31

12

3

45

6

78 9

121918

16 13

1110

1514 17

20

Example 11.9 KI Truck Routing

1

2

3

4

5

6

78 9

12

1918

16 13

11

10

1514 17

20

KI Truck Routing Example Model

s.t.

Let

(11.18)

length of the best route through stops assigned to truck j by decision vector z• Routing problems are characteristically difficult to represent

concisely in optimization models. [11.28]

KI Truck Routing Example Model

• Fixed charges can be modeled with new binary variables and switching constraints. Facility Location and Network Design are common cases of fixed charges.

Facility Location Models• Facility/plant/warehouse location models choose which of a

proposed list of facilities to open in order to service specified customer demands at minimum total cost. [11.29]

11.6 Facility Location andNetwork Design Models

Example 11.10 Tmark Facilities Location

AT&T has confronted many facility location problems in recommending sites for the toll-free call-in centers of its telemarketing customers. Such centers handle telephone reservations and orders arising in many geographic zones. Since telephone rates vary dramatically depending on the zone of call origin and the location of the receiving center, site selection is extremely important. A well-designed system should minimize the total of call charges and center setup costs.

Our version of this scenario will involve fictional firm Tmark. Figure 11.7 shows the 8 sites under consideration for Tmark’s catalog order centers embedded in a map of the 14 calling zones. Table 11.9 shows corresponding unit calling charges, ri,j, from each zone j to various centers i, and the zone’s anticipated call load, dj.

Any Tmark center selected can handle between 1500 and 5000 call units per day. However, their fixed costs of operation vary significantly because of differences in labor and real estate prices. Estimated daily fixed cost, fi, for the 8 centers are displayed in Figure 11.7.


1

2 3

4

5

6

7

8

iFixed Cost

1 24002 70003 36004 16005 30006 46007 90008 2000


Zonej

Possible Center Location, i Call Demand1 2 3 4 5 6 7 8

1 1.25 1.40 1.10 0.90 1.50 1.90 2.00 2.10 2502 0.80 0.90 0.90 1.30 1.40 2.20 2.10 1.80 1503 0.70 0.40 0.80 1.70 1.60 2.50 2.05 1.60 10004 0.90 1.20 1.40 0.50 1.55 1.70 1.80 1.40 805 0.80 0.70 0.60 0.70 1.45 1.80 1.70 1.30 506 1.10 1.70 1.10 0.60 0.90 1.30 1.30 1.40 8007 1.40 1.40 1.25 0.80 0.80 1.00 1.00 1.10 3258 1.30 1.50 1.00 1.10 0.70 1.50 1.50 1.00 1009 1.50 1.90 1.70 1.30 0.40 0.80 0.70 0.80 475

10 1.35 1.60 1.30 1.50 1.00 1.20 1.10 0.70 22011 2.10 2.90 2.40 1.90 1.10 2.00 0.80 1.20 90012 1.80 2.60 2.20 1.80 0.95 0.50 2.00 1.00 150013 1.60 2.00 1.90 1.90 1.40 1.00 0.90 0.80 43014 2.00 2.40 2.00 2.20 1.50 1.20 1.10 0.80 200

j fulfilled from facility iILP Model of Facilities Location

[11.30]s.t.

Let

Tmark Facilities Location Example Model

(11.20)

s.t.

• The fixed-charge network flow or network design model on a digraph on nodes kV with net demand bk, and arcs (i,j)A with capacity ui,j, unit cost ci,j, and non-negative fixed cost fi,j is [11.31]

Network Design Models

s.t.

Example 11.11 Wastewater Network Design

Network design applications may involve telecommunications, electricity, water, gas, coal slurry, or any other substance that flows in a network. We illustrate with an application involving regional wastewater (sewer) networks.

As new areas develop around major cities, entire networks of collector sewers and treatment plants must be constructed to service growing population. Figure 11.8 displays our particular (fictional) instance.Nodes 1 to 8 of the network represent population centers where smaller sewers feed into the main regional network, and locations where treatment plants might be built. Wastewater loads are roughly proportional to population, so the inflows indicated at nodes represent population units (in thousands).


Arcs joining nodes 1 to 8 show possible routes for main collector sewers. Most follow the topology in gravity flow, but one pumped line (4. 3) is included. A large part of the construction cost for either type of line is fixed: right-of-way acquisition, trenching, and so on. Still, the cost of a line also grows with the number of population units carried, because greater flows imply larger-diameter pipes. The table in Figure 11.8 shows the fixed and variable cost for each arc in thousand of dollars.

Treatment plant costs actually occur at nodes—here nodes 3, 7, and 8. Figure 11.8 illustrates, however, that such costs can be modeled on arcs by introducing an artificial “supersink" node 9. Costs shown for arcs (3, 9), (7. 9), and (8, 9) capture the fixed and variable expense of plant construction as flows depart the network.


Arc Fixed Cost Var. Cost(1,2) 240 21

(1,3) 350 30

(2,3) 200 22

(2,4) 750 58

(3,4) 610 43

(3,9) 3800 1

(4,3) 1840 49

(4,8) 780 63

(5,6) 620 44

(5,7) 800 51

(6,7) 500 56

(6,8) 630 94

(7,4) 1120 82

(7,9) 3800 1

(8,9) 2500 2

12

34

56

9

7

8

273

1436

13

21

8

gravitypumpedplant arcs

Wastewater Network Design Example Model

min 21x1,2+30x1,3 +22x2,3+58x2,4 +43x3,4+ x3,9 +49x4,3+63x4,8 +44x5,6 +51x5,7+56x6,7 +94x6,8 +82x7,4+x7,9 +2x8,9+240y1,2 +350y1,3

+200y2,3+750y2,4 +610y3,4+ 3800y3,9 +1840y4,3 +780y4,8 +620y5,6 +800y5,7+500y6,7 +630y6,8 +1120y7,4 +3800y7,9 +2500y8,9

s.t. –x1,2 –x1,3 = -27x1,2 – x2,3 – 5x2,4 = -3x1,3 + x2,3 + x4,3 –x3,4 –x3,9 = -14x2,4 + x3,4 + x7,4 –x4,3 –x4,8 = -36–x5,6–x5,7 = -21x5,6–x6,7 –x6,8 = -8x5,7 + x6,7–x7,4 –x7,9 = -13x4,8 + x6,8–x8,9 = 0x3,9 + x7,9 + x8,9 = 122yi,j = 0 or 1 all arcs (i,j)

0 x1,2 27y1,2

0 x1,3 27y1,3

0 x2,3 30y2,3

0 x2,4 30y2,4

0 x3,4 44y3,4

0 x3,9 122y3,9

0 x4,3 108y4,3

0 x4,8 122y4,8

0 x5,6 21y5,6

0 x5,7 21y5,7

0 x6,7 29y6,7

0 x6,8 29y6,8

0 x7,4 42y7,4

0 x7,9 42y7,9

0 x8,9 122y8,9

Single-Processor Scheduling Problems• Single-processor (or single-machine) scheduling problems

seek an optimal sequence in which to complete a given collection of jobs on a single processor that can accommodate only one job at a time. [11.32]

11.7 Processor Scheduling andSequencing Models

Example 11.12 Nifty Notes Single-Machine SchedulingWe begin with the binder scheduling problem confronting a

fictitious Nifty Notes copy shop. Just before the start of each semester, professors at the nearby university supply Nifty Notes with a single original of their class handouts, a projection of the class enrollment, and a due date by which copies should be available. Then the Nifty Notes staff must rush to print and bind the required number of copies before each class begins.

During the busy period each semester, Nifty Notes operates its single binding station 24 hours per day. Table 11.10 shows process times, release times, and due dates for the jobs j = 1,..., 6 now pending at the binder.:

We wish to choose an optimal sequence in which to accomplish these jobs. No more than one can be in process at a time; and once started. a job must be completed before another can begin.

Example 11.12 Nifty Notes Single-Machine Schedulingpj estimated process time (in hours) job j will require to

bindrj release time (hour) at which job j has/will become available for processing (relative to time 0 = now)dj = due date (hour) by which job j should be completed (relative to time 0 = now)Notice that two jobs are already late (dj < 0).

Binder Job, j1 2 3 4 5 6

Process time, pj12 8 3 10 4 18

Release time, rj-20 -15 -12 -10 -3 2

Due date, dj10 2 72 -8 -6 60

• A set of (continuous) decision variables in processor scheduling models usually determines either the start or the completion time(s) of each job on the processor(s) it requires. [11.33]

• xj time binding starts for job j (relative to time 0 = now)• Constraints on the starting time:

xj max {0, rj}• Completion time:

(Start time) + (Process time) = (Completion time)

Time Decision Variables

(11.22)

• The central issue in processor scheduling is that only one job should be in progress on any processor at any time.

• For a pair of jobs j and j’ that might conflict on a processor, the appropriate conflict constraint is either (11.23)

(start time of j) + (process time of j) (start time of j’), or(start time of j’) + (process time of j’) (start time of j)

• Conflict avoidance requirement can be modeled explicitly with the aid of additional disjunctive variables.

• A set of (discrete) disjunctive variables in processor scheduling models usually determines the sequence in which jobs are started on processors by specifying whether each job j is scheduled before or after each other j’ with which it might conflict. [11.34]

Conflict Constraints and Disjunctive Variables

• A processor scheduling model with job start time xj and process time pj can prevent conflicts between jobs j and j’ with disjunctive constraint pairs

xj + pj xj’ + M(1 – yj,j’)xj’ + pj’ xj + Myj,j’

Where M is a large positive constant, and binary disjunctive variable yj,j’=1when j is schedule before j’ on the processor and =0 if j’ is first. [11.35]

Conflict Constraints and Disjunctive Variables

• Due dates in processor scheduling models are usually handled as goals to be reflected in the objective function rather than as explicit constraints. Dates that must be met are termed deadlines to distinguish. [11.36]

Handling of Due Dates

• Denoting the start time of job j=1,…, n by xj, the process time by pj, the release time by rj and the due date by dj, processor scheduling objective functions often minimize one of the following: [11.37]

• Maximum completion time (makespan) maxj{xj + pj}Mean completion time (makespan) (1/n)j(xj

+ pj)• Maximum flow time maxj{xj + pj - rj}

Mean flow time (1/n)j(xj + pj - rj)• Maximum lateness maxj{xj + pj - dj}

Mean lateness (1/n)j(xj + pj - dj)• Maximum tardiness maxj{max[0, xj + pj – dj]}

Mean tardiness (1/n)j(max{0, xj + pj – dj})

Processor Scheduling Objective Functions

• When any of the minmax objective are being optimized, the problem is an integer non-linear program (INLP).

• Any of the min max objectives can be linearized by introducing a new decision variable f to represent the objective function value, then minimizing f subject to new constraints of the form f each element in the maximize set. A similar construction can model tardiness by introducing new non-negative tardiness variables for each job and adding constraints keeping each tardiness variable the corresponding lateness. [11.38]

ILP Formulation of Minmax Scheduling Objective

• The mean completion time, mean flow time, and mean lateness scheduling objective functions are equivalent in the sense that an optimal schedule for one is also optimal for the others. [11.39]

• An optimal schedule for the maximum lateness objective function is also optimal for maximum tardiness. [11.40]

Equivalences among Scheduling Objective Functions

• Job shop scheduling problems seek an optimal schedule for a given collection of jobs, each of which requires a known sequence of processors that can accommodate only one job at a time. [11.41]

Job Shop Scheduling

Example 11.13 Custom Metalworking Job Shop

We illustrate job shop scheduling with a fictitious Custom Metalworking company which fabricates prototype metal parts for a nearby engine manufacturer. Figure 11.9 provides details on the 3 jobs waiting to be scheduled. First is a die requiring work on a sequence of 5 workstations: 1 (forging), then 2 (machining), then 3 (grinding), then 4 (polishing), and finally, 6 (electric discharge cutting). Job 2 is a cam shaft requiring 4 stations, and job 3 a fuel injector requiring 5 steps. Numbers in boxes indicate process times

Pj,k process time (in minutes) of job j on processor kFor example, job 1 requires 45 minutes at polishing workstation 4.

Any of the objective function forms in [11.37] could be appropriate for Custom Metalworking’s scheduling. We will assume that the company wants to complete all 3 jobs as soon as possible (minimize maximum completion), so that workers can leave for a holiday.

Example 11.13 Custom Metalworking Job Shop

WS13

WS210

WS38

WS445

WS61

1. Die

WS750

WS16

WS211

WS36

2. Camshaft

WS25

WS39

WS52

WS61

WS425

3. FuelInjector

k Workstations1 Forging2 Machining

center3 Grinding4 Polishing5 Drilling6 Electric

discharge7 Heat

treatment

Custom Metalworking ExampleDecision Variables and Objective

• Decision variables𝑥𝑗,k ≡ start time of job j on processor k• Objective function

min max {x1,6+1, x2,3+6, x3,4+25}• Precedence constraints

The precedence requirement that job j must complete on processor k before activity on k’ begins can be expressed as

xj,k + pj,k xj,k’

where xj,k denotes the start time of job j on processor k, pj,k

is the process time of j on k, and xj,k’ is the start time of job j on processor k’. [11.42]

Custom Metalworking ExampleDecision Variables and Objective

• Job shop models can prevent conflicts between jobs by introducing new disjunctive variables yj,j’,k and constraint pair

xj,k + pj,k xj,k’ + M (1 - yj,j’,k )xj’,k + pj’,k xj,k + M yj,j’,k )

For each j, j’ that both require any processor k. Here xj,k

denotes the start time of job j on processor k, pj,k is the process time, M is a large positive constant, and binary yj,j’,k =1 when j is scheduled before j’ on k, and =0 if j’ is first. [11.43]

Custom Metalworking Example Model

min max {x1,6+1, x2,3+6, x3,4+25}s.t. x1,1 +3 x1,2

x1,2 +10 x1,3

x1,3 +8 x1,4

x1,4 +45 x1,6

x2,7 +50 x2,1

x2,1 +6 x2,2

x2,2 +11 x2,3

x3,2 +5 x3,3

x3,3 +9 x3,5

x3,5 +2 x3,6

x3,6 +1 x3,4

x1,1 +3 x2,1+M(1-y1,2,1)

x2,1 +6 x1,1+My1,2,1

x1,2 +10 x2,2 +M(1-y1,2,2)

x2,2 +11 x1,2 +My1,2,2

x1,2 +10 x3,2 +M(1-y1,3,2)

x3,2 +5 x1,2 +My1,3,2

x2,2 +11 x3,2 +M(1-y2,3,2)

x3,2 +5 x1,2 +My2,3,2)x1,3 +8 x2,3 +M(1-y1,2,3)x2,3 +6 x1,3 +My1,2,3)x1,3 +8 x3,3 +M(1-y1,3,3)x3,3 +9 x1,3 +My1,3,3)

x2,3 +6 x3,3+M(1-y2,3,3)

x3,3 +9 x2,3+My2,3,3

x1,4 +45 x3,4 +M(1-y1,3,4)

x3,4 +25 x1,4 +My1,3,4

x1,6 +1 x3,6 +M(1-y1,3,6)x3,6 +1 x1,6 +My1,3,6

xj,k 0yj,j’,k = 0 or 1

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Chapter 11 Discrete Optimization Models