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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

• 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.

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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 x1…x7 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

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

𝑦𝑗 ≡ 1 if scrap j is part of the blend0 otherwise

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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 x3…x7 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

ILP Modeling of Fixed Charges

• Another common source of lumpy phenomena arises

when the objective function involves fixed charges.

𝜃(𝑥) ≡ 𝑓 + 𝑐𝑥 if x >0 0 otherwise

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

𝑦𝑗 1 if xj >0 0 otherwise

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

𝑥𝑗 ≤ 𝑢𝑗𝑦𝑗 Where 𝑢𝑗 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

𝑦𝑗 ≡ 1 if setup for j is performed0 otherwise

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Swedish Steel Example

with 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 y1 x3 1000 y3

x2 250y2 x4 1000 y4

x1…x7 0 y1,.., 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

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 year’s car to improve its

top speed. The following table lists their estimated costs and speed

enhancements.

Proposed Feature, j

1 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

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

𝑥𝑗 ≡ 1 if feature j is added0 otherwise

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

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]

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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

𝑥𝑗 ≡ 1 if mission j is selected0 otherwise

Example 11.2

NASA Capital Budgeting

Mission

Budget Requirements ($ billion)

Value Not

with

Depends

On 2000-

2004

2005-

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 3

5 Uranus orbiter 2010 - 5 8 - - 70 4 3

6 Mercury probe - - 1 8 4 20 - 3

7 Saturn probe 1 8 - - - 5 - 3

8 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 2

12 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

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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

Capital Budgeting Models

• Mutually exclusiveness

conditions allowing at

most one of a set of

choices are modeled by 𝑥𝑗 ≤ 1 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

𝑥𝑗 ≤ 𝑥𝑖 [11.8]

x11 x2

x4 x3

x5 x3

x6 x3

x7 x3

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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

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]

𝑥𝑗 ≥ 1

𝑗∈𝐽

• Set packing constraints requiring that at most one

member of sub-collection J belongs to a solution are

expressed [11.10]

𝑥𝑗 ≤ 1

𝑗∈𝐽

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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]

𝑥𝑗 = 1

𝑗∈𝐽

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

𝑥𝑗 ≡ 1 if location j is selected0 otherwise

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Example 11.3

EMS Location Planning

1

2

3

4

5

6

7

8

9

10

Minimum Cover EMS Model

𝑀𝑖𝑛 𝑥𝑗

10

𝑗=1

s.t.

(11.8)

x2 1

x1 + x2 1

x1 + x3 1

x3 1

x3 1

x2 1

x2 + x4 1

x3 + x4 1

x8 1 xj = 0 or 1 j=1,…,10

x4 + x6 1

x4 + x5 1

x4 + x5 + x6 1

x4 + x5 + x7 1

x8 + x9 1

x6 + x9 1

x5 + x6 1

x5 + x7 + x10 1

x8 + x9 1

x9 + x10 1

x10 1

x2* = x3* = x4* = x6*

= x8* = x10* =1,

x1* = x5* = x7* = x9*

= 0

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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]

𝑦𝑖 ≡ 1 if item i is uncovered in the solution0 otherwise

Dist. Value Dist. Value Dist. Value Dist. Value

1 5.2 6 5.7 11 30.4 16 25.6

2 4.4 7 10.0 12 30.9 17 11.0

3 7.1 8 12.2 13 12.0 18 5.3

4 9.0 9 7.6 14 9.3 19 7.9

5 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 1

x1 + x3 + y3 1

x3 + y4 1

x3 + y5 1

x2 + y6 1

x2 + x4 + y7 1

x3 + x4 + y8 1

x8 + y9 1

x4 + x6 + y10 1

x4 + x5 + y11 1 xi = 0 or 1 j=1,…,10

x4 + x5 + x6 + y12 1

x4 + x5 + x7 + y13 1

x8 + x9 + y14 1

x6 + x9 + y15 1

x5 + x6 + y16 1

x5 + x7 + x10 + y17 1

x8 + x9 + y18 1

x9 + x10 + y19 1

x10 + y20 1

𝑥𝑗 ≤ 4

10

𝑗=1

x3*=x4*=x5*=x9*= 1,

y1*=y2*=y6*=y9*=y20*= 1

Value = 32.8

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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

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Example 11.4

AA Crew Scheduling

MIA

CHI

CHR

DFW

101

402

203

204

305 406

407

308

310

109

211

212

j Flight Sequence Cost

1 101-203-406-308 2900

2 101-203-407 2700

3 101-204-305-407 2600

4 101-204-308 3000

5 203-406-310 2600

6 203-407-109 3150

7 204-305-407-109 2550

8 204-308-109 2500

9 305-407-109-212 2600

10 308-109-212 2050

11 402-204-305 2400

12 402-204-310-211 3600

13 406-308-109-211 2550

14 406-310-211 2650

15 407-109-211 2350

Example 11.4

AA Crew Scheduling

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

𝑥𝑗 ≡ 1 if pairing j is selected0 otherwise

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Column Generation Model

for 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 = 1

x1 + x2 + x5 + x6 = 1

x3 + x4 + x7 + x8 + x11 + x12 = 1

x12 + x13 + x14 + x15 = 1

x9 + y10 = 1

x3 + x7 + x9 + x11 = 1

x1 + x4 + x8 + x10 + x13 = 1

x5 + x12 + x14 = 1

x11 + x12 = 1

x1 + x5 + x13 + x14 = 1

x2 + x3 + x6 + x7 + x9 + x15 = 1

xi = 0 or 1 j=1,…,10

x1*=x9*=x12*= 1

Cost = 9100

• It is standard to model all assignment forms with the

decision variables

𝑥𝑖,𝑗 ≡ 1 if i of the 1st set is matched with j of the 2nd0 otherwise

• Assignment Constraints take the form

𝑥𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

𝑗

𝑥𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑗; 𝑥𝑖,𝑗 = 0 𝑜𝑟 1 ∀ 𝑖, 𝑗

𝑖

Where all sums are limited to (𝑖, 𝑗) combinations allowed in

the problem. [11.14]

11.4 Assignment

and Matching Models

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CAM Linear Assignment Example

Revisited

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, i Next Workstation, j

j=1 j=2 j=3 j=4 j=5 j=6 j=7 j=8 j=9 j=10

i=1 8 23 5

i=2 4 12 15

i=3 20 13 6 8

i=4 19 10

i=5 8 12 16

i=6 14 8 3

i=7 6 27 12

i=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*= 1

Total 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

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• Linear assignment models minimize or maximize a linear

objective function of the form

𝑐𝑖,𝑗𝑗

𝑥𝑖,𝑗

𝑖

Subject to assignment constraints in [11.14], where 𝑐𝑖,𝑗 is the

cost (or benefit) of assigning 𝑖 to 𝑗. [11.15]

Linear Assignment Models

• Quadratic assignment models minimize or maximize a

quadratic objective function of the form

𝑐𝑖,𝑗,𝑘,𝑙𝑙≠𝑗𝑘>𝑖

𝑥𝑖,𝑗

𝑗

𝑥𝑘,𝑙

𝑖

Subject to assignment constraints in [11.14], where 𝑐𝑖,𝑗,𝑘,𝑙 is

the cost (or benefit) of assigning 𝑖 to 𝑗 and 𝑘 to 𝑙. [11.16]

Quadratic Assignment Models

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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.

Example 11.5

Mall Layout Quadratic Assignment

1

2

3

4

L=1 L=2 L=3 L=4

J=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 --

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Example 11.5

Mall Layout Quadratic Assignment

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

𝑥𝑖,𝑗 ≡ 1 if shop i is assigned to location j0 otherwise

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

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

+ 7(80x1,1x4,2+150x1,1x4,3+170x1,1x4,4+80x1,2x4,1+130x1,2x4,3+100x1,2x4,4

+150x1,3x4,1+130x1,3x4,2+120x1,3x4,4+170x1,4x4,1+100x1,4x4,2+120x1,4x4,3)

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

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

+ 8(80x2,1x4,2+150x2,1x4,3+170x2,1x4,4+80x2,2x4,1+130x2,2x4,3+100x2,2x4,4

+150x2,3x4,1+130x2,3x4,2+120x2,3x4,4+170x2,4x4,1+100x2,4x4,2+120x2,4x4,3)

+ 3(80x3,1x4,2+150x3,1x4,3+170x3,1x4,4+80x3,2x4,1+130x3,2x4,3+100x3,2x4,4

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

(11.12)

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21

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 = 1

x3,1 + x3,2 + x3,3 + x3,4 = 1

x4,1 + x4,2 + x4,3 + x4,4 = 1

x1,1 + x2,1 + x3,1 + x4,1 = 1

x1,2 + x2,2 + x3,2 + x4,2 = 1

x1,3 + x2,3 + x3,3 + x4,3 = 1

x1,4 + x2,4 + x3,4 + x4,4 = 1

Xi,j = 0 or 1 i=1,…,4; j=1,…,4

x1,1*=x2,4*=x3,3*=x4,2*=1

Total Distance = 3260

• Suppose each object 𝑖 must be assigned to some 𝑗, but that

𝑗’s may received several 𝑖.

• b𝑗 capacity of 𝑗

• 𝑠𝑖,𝑗 size of 𝑗’s capacity consumed if 𝑖 is assigned to 𝑗

• 𝑐𝑖,𝑗 cost (or benefit) of assigning 𝑖 to 𝑗

Generalized Assignment Models

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22

• Generalized assignment models, which encompass cases

where allocation of 𝑖 to 𝑗 requires fixed size 𝑠𝑖,𝑗 within

capacity b𝑗 have the form

min𝑜𝑟 𝑚𝑎𝑥 𝑐𝑖,𝑗𝑗

𝑥𝑖,𝑗

𝑖

s.t. 𝑥𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖𝑗

𝑥𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑗; 𝑥𝑖,𝑗 = 0 𝑜𝑟 1 ∀ 𝑖, 𝑗𝑖

Here 𝑐𝑖,𝑗 is the cost (or benefit) of assigning 𝑖 to 𝑗 and all

sums are limited to (𝑖,𝑗) 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.

11/15/2011

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Example 11.6

CDOT Generalized Assignment

District, i

Ship, j 1 2 3 4 5 6

1.Estevan Cost 130 30 510 30 340 20

Time 30 50 10 11 13 9

2.Mackenzie Cost 460 150 20 40 30 450

Time 10 20 60 10 10 17

3.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*=1

Total 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

11/15/2011

24

• 𝑥𝑖,𝑖` ≡ 1 if 𝑖 is paired with 𝑖`0 otherwise

• Matching models, which seek an optimal pairing of like

objects 𝑖, have the form

min𝑜𝑟 𝑚𝑎𝑥 𝑐𝑖,𝑖`𝑖`>𝑖

𝑥𝑖,𝑖`

𝑖

s.t. 𝑥𝑖`,𝑖 + 𝑥𝑖,𝑖`𝑖`>𝑖𝑖`<𝑖 = 1

𝑥𝑖,𝑖` = 0 𝑜𝑟 1 ∀ 𝑖, 𝑖` > 𝑖

Here 𝑐𝑖,𝑖` is the cost (or benefit) of pairing 𝑖 with 𝑖` and all

sums are limited to (𝑖,𝑖`) 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.

11/15/2011

25

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]

11/15/2011

26

• 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 𝑖 to any other point 𝑗 is the

same as the distance from 𝑗 to 𝑖. 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 26.4

2 3.6 3.6 6.4 12.1 18.1 13.2 10.6 19.7 23.0

3 5.1 3.6 7.1 10.6 15.0 15.8 10.8 18.4 21.9

4 10.0 6.4 7.1 7.0 15.7 10.0 4.2 13.9 17.0

5 15.3 12.1 10.6 7.0 9.9 15.3 5.0 7.8 11.3

6 20.0 18.1 15.0 15.7 9.9 25.0 14.9 12.0 15.0

7 16.0 13.2 15.8 10.0 15.3 25.0 10.3 19.2 21.0

8 14.2 10.6 10.8 4.2 5.0 14.9 10.3 10.2 13.0

9 23.0 19.7 18.4 13.9 7.8 12.0 19.2 10.2 3.6

10 26.4 23.0 21.9 17.0 11.3 15.0 21.0 13.0 3.6

1

2

3

4 5

6

7 8 9

10

11/15/2011

27

• Most ILP models of the symmetric case employ decision

variables, 𝑖 < 𝑗

𝑥𝑖,𝑗 ≡ 1 if the route includes a leg between i and j0 otherwise

• Total length of a route can be calculated by

𝑑𝑖,𝑗

𝑗>𝑖

𝑥𝑖,𝑗

𝑖

• Constraints for symmetric TSP

𝑥𝑗,𝑖 +

𝑗<𝑖

𝑥𝑖,𝑗 = 2 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

𝑗>𝑖

Formulating the Symmetric TSP

(11.15)

(11.16)

• Definition of subtours

S proper subset of points/cities to

be routed

• Subtour elimination constraints

𝑥𝑖,𝑗 +

𝑖∉𝑆𝑖∈𝑆

𝑥𝑖,𝑗 ≥ 2

𝑖∈𝑆𝑖∉𝑆

• Number of legs between points in S

and points not in S must be at least 2.

Subtours

(11.17)

1

2

3

4 5

6

7

8 9

10

11/15/2011

28

min 𝑑𝑖,𝑗

𝑗>𝑖

𝑥𝑖,𝑗

𝑖

s.t.

𝑥𝑗,𝑖 +

𝑗<𝑖

𝑥𝑖,𝑗 = 2 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

𝑗>𝑖

𝑥𝑖,𝑗 +

𝑖∉𝑆𝑖∈𝑆

𝑥𝑖,𝑗 ≥ 2 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑆, 𝑆 ≥ 3

𝑖∈𝑆𝑖∉𝑆

𝑥𝑖,𝑗 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗 > 𝑖

ILP Model of the Symmetric TSP

[11.25]

min 𝑑𝑖,𝑗

𝑗≠𝑖

𝑥𝑖,𝑗

𝑖

s.t.

𝑥𝑗,𝑖 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

𝑗

𝑥𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

𝑗

𝑥𝑖,𝑗 ≥ 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑆, 𝑆 ≥ 2

𝑖∉𝑆𝑖∈𝑆

𝑥𝑖,𝑗 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗

ILP Model of the Asymmetric TSP

[11.26]

11/15/2011

29

𝑦𝑘,𝑖 ≡ 1 if kth point visited is i0 otherwise

min 𝑑𝑖,𝑗

𝑗

𝑦𝑘,𝑖

𝑘

𝑥𝑘+1,𝑗

𝑖

(Total length)

𝑦𝑘,𝑖 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 (each position occupied)

𝑖

𝑥𝑘,𝑖 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

𝑘

(each point visited)

𝑦𝑘,𝑖 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘, 𝑖

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 fi

1 0.25 11 0.21

2 0.33 12 0.68

3 0.39 13 0.16

4 0.40 14 0.19

5 0.27 15 0.22

6 0.70 16 0.38

7 0.28 17 0.26

8 0.43 18 0.29

9 0.50 19 0.17

10 0.22 20 0.31

1

2

3

4

5

6

7 8 9

12 19 18

16 13

11

10

15 14 17

20

11/15/2011

30

Example 11.9

KI Truck Routing

1

2

3

4

5

6

7

8 9

12

19 18

16 13

11

10

15

14 17

20

𝑧𝑖,𝑗 ≡ 1 if stop i is assigned to route j0 otherwise

min 𝜃𝑗

7

𝑗=1

(𝒛)

𝑧𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 = 1,… , 20 (each i to some j)

7

𝑗=1

𝑓𝑖𝑧𝑖,𝑗 ≤ 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 = 1,… , 7 (truck capacities)

20

𝑖=1

𝑧𝑖,𝑗 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗

KI Truck Routing Example Model

s.t.

Let

(11.18)

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31

𝜃𝑗(𝒛) ≡ 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 and

Network Design Models

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32

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.

Example 11.10

Tmark Facilities Location

1

2 3

4

5

6

7

8

i

Fixed

Cost

1 2400

2 7000

3 3600

4 1600

5 3000

6 4600

7 9000

8 2000

11/15/2011

33

Example 11.10

Tmark Facilities Location

Zone

j

Possible Center Location, i Call

Demand 1 2 3 4 5 6 7 8

1 1.25 1.40 1.10 0.90 1.50 1.90 2.00 2.10 250

2 0.80 0.90 0.90 1.30 1.40 2.20 2.10 1.80 150

3 0.70 0.40 0.80 1.70 1.60 2.50 2.05 1.60 1000

4 0.90 1.20 1.40 0.50 1.55 1.70 1.80 1.40 80

5 0.80 0.70 0.60 0.70 1.45 1.80 1.70 1.30 50

6 1.10 1.70 1.10 0.60 0.90 1.30 1.30 1.40 800

7 1.40 1.40 1.25 0.80 0.80 1.00 1.00 1.10 325

8 1.30 1.50 1.00 1.10 0.70 1.50 1.50 1.00 100

9 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 0.70 220

11 2.10 2.90 2.40 1.90 1.10 2.00 0.80 1.20 900

12 1.80 2.60 2.20 1.80 0.95 0.50 2.00 1.00 1500

13 1.60 2.00 1.90 1.90 1.40 1.00 0.90 0.80 430

14 2.00 2.40 2.00 2.20 1.50 1.20 1.10 0.80 200

𝑦𝑖 ≡ 1 if facility i is open0 otherwise

𝑥𝑖,𝑗 ≡ fraction of demand j fulfilled from facility i

min 𝑐𝑖,𝑗𝑑𝑗𝑥𝑖,𝑗

𝑗

+ 𝑓𝑖𝑦𝑖

𝑖𝑖

𝑥𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑗 (𝑓𝑢𝑙𝑓𝑖𝑙𝑙 𝑗 𝑑𝑒𝑚𝑎𝑛𝑑)

𝑖

𝑑𝑗𝑥𝑖,𝑗 ≤ 𝑢𝑖𝑦𝑖 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

𝑗

(𝑖 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑠𝑤𝑖𝑡𝑐𝑕𝑖𝑛𝑔)

𝑥𝑖,𝑗 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗

𝑦𝑖 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

ILP Model of Facilities Location

[11.30] s.t.

Let

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34

𝑚𝑖𝑛 (𝑟𝑖,𝑗𝑑𝑗)𝑥𝑖,𝑗

14

𝑗=1

+ 𝑓𝑖𝑦𝑖

8

𝑖=1

8

𝑖=1

𝑥𝑖,𝑗

8

𝑖=1

= 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑗 (𝑐𝑎𝑟𝑟𝑦 𝑗 𝑙𝑜𝑎𝑑)

1500𝑦𝑖 ≤ 𝑑𝑗𝑥𝑖,𝑗 ≤ 5000𝑦𝑖 (𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑖)

14

𝑗=1

𝑥𝑖,𝑗 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗

𝑦𝑖 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

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]

𝑚𝑖𝑛 𝑐𝑖,𝑗𝑥𝑖,𝑗

(𝑖,𝑗)∈𝐴

+ 𝑓𝑖,𝑗𝑦𝑖,𝑗

(𝑖,𝑗)∈𝐴

𝑥𝑖,𝑘 − 𝑥𝑘,𝑗

𝑘,𝑗 ∈𝐴

= 𝑏𝑘 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘 ∈ 𝑉

(𝑖,𝑘)∈𝐴

0 ≤ 𝑥𝑖,𝑗 ≤ 𝑢𝑖,𝑗𝑦𝑖,𝑗 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑖, 𝑗) ∈ 𝐴

𝑦𝑖,𝑗 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑖, 𝑗) ∈ 𝐴

Network Design Models

s.t.

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35

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).

Example 11.11

Wastewater Network Design

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.

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36

Example 11.11

Wastewater Network Design 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

1 2

3 4

5

6 9

7

8

27 3

14

36

13

21

8

gravity

pumped

plant 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 = -27

x1,2 – x2,3 – 5x2,4 = -3

x1,3 + x2,3 + x4,3 –x3,4 –x3,9 = -14

x2,4 + x3,4 + x7,4 –x4,3 –x4,8 = -36

–x5,6–x5,7 = -21

x5,6–x6,7 –x6,8 = -8

x5,7 + x6,7–x7,4 –x7,9 = -13

x4,8 + x6,8–x8,9 = 0 x3,9 + x7,9 + x8,9 = 122

yi,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

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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 and

Sequencing Models

Example 11.12

Nifty Notes Single-Machine Scheduling

We 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.

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Example 11.12

Nifty Notes Single-Machine Scheduling

pj estimated process time (in hours) job j will require to bind rj 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, j

1 2 3 4 5 6

Process time, pj 12 8 3 10 4 18

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

Due date, dj 10 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)

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• 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

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• 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

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• 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

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• 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 k For 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.

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Example 11.13

Custom Metalworking Job Shop

WS1

3

WS2

10

WS3

8

WS4

45

WS6

1 1. Die

WS7

50

WS1

6

WS2

11

WS3

6 2. Camshaft

WS2

5

WS3

9

WS5

2

WS6

1

WS4

25 3. Fuel

Injector

k Workstations

1 Forging

2 Machining

center

3 Grinding

4 Polishing

5 Drilling

6 Electric

discharge

7 Heat

treatment

Custom Metalworking Example

Decision Variables and Objective

• Decision variables 𝑥𝑗,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]

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Custom Metalworking Example

Decision Variables and Objective

𝑦𝑗,𝑗′ ,𝑘 ≡ 1 if j is scheduled before job j′ on processor k0 otherwise

• 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 0

yj,j’,k = 0 or 1