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Chapter 12

Discrete Optimization Methods

12.1 Solving by

Total Enumeration

• If model has only a few discrete decision variables, the

most effective method of analysis is often the most

direct: enumeration of all the possibilities. [12.1]

• Total enumeration solves a discrete optimization by

trying all possible combinations of discrete variable

values, computing for each the best corresponding

choice of any continuous variables. Among

combinations yielding a feasible solution, those with the

best objective function value are optimal. [12.2]

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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 x3…x7 0 y1, y2 = 0 or 1

(12.1)

Cost = 9967.06

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

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

Swedish Steel Model

with All-or-Nothing Constraints

Discrete

Combination

Corresponding Continuous Solution Objective

Value

y1 y2 x3 x4 x5 x6 x7

0 0 823.11 125.89 30.00 10.00 11.00 10340.89

0 1 646.67 63.33 22.00 7.25 10.75 10304.08

1 0 736.44 160.06 16.50 1.00 11.00 9967.06

1 1 561.56 94.19 8.50 0.00 10.75 10017.94

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Exponential Growth

of Cases to Enumerate

• Exponential growth makes total enumeration impractical

with models having more than a handful of discrete

decision variables. [12.3]

12.2 Relaxation of Discrete

Optimization Models

Constraint Relaxations

• Model (𝑃 ) is a constraint relaxations of model (P) if

every feasible solution to (P) is also feasible in (𝑃 ) and

both models have the same objective function. [12.4]

• Relaxation should be significantly more tractable than

the models they relax, so that deeper analysis is

practical. [12.5]

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Example 12.1

Bison Booster

The Boosters are trying to decide what fundraising projects to

undertake at the next country fair. One option is customized T-

shirts, which will sell for $20 each; the other is sweatshirts selling for

$30. History shows that everything offered for sale will be sold

before the fair is over.

Materials to make the shirts are all donated by local

merchants, but the Boosters must rent the equipment for

customization. Different processes are involved, with the T-shirt

equipment renting at $550 for the period up to the fair, and the

sweatshirt equipment for $720. Display space presents another

consideration. The Boosters have only 300 square feet of display

wall area at the fair, and T-shirts will consume 1.5 square feet each,

sweatshirts 4 square feet each. What plan will net the most income?

Bison Booster Example Model

• Decision variables:

x1 number of T-shirts made and sold

x2 number of sweatshirts made and sold

y1 1 if T-shirt equipment is rented; =0 otherwise

y1 1 if sweatshirt equipment is rented; =0 otherwise

• Max 20x1 + 30x2 – 550y1 – 720y2 (Net income)

s.t. 1.5x1 + 4x2 300 (Display space)

x1 200y1 (T-shirt if equipment)

x2 75y2 (Sweatshirt if equipment)

x1, x2 0

y1, y2 = 0 or 1 Net Income = 3450

x1* = 200, x2* = 0, y1* = 1, y2* = 0

(12.2)

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Constraint Relaxation Scenarios

• Double capacities 1.5x1 + 4x2 600

x1 400y1

x2 150y2

x1, x2 0

y1, y2 = 0 or 1

• Dropping first constraint 1.5x1 + 4x2 300

x1 200y1

x2 75y2

x1, x2 0

y1, y2 = 0 or 1

Net Income = 7450

𝑥 1* = 400, 𝑥 2* = 0, 𝑦 1* = 1, 𝑦 2 * = 0

Net Income = 4980

𝑥 1* = 200, 𝑥 2* = 75, 𝑦 1* = 1, 𝑦 2 * = 1

Constraint Relaxation Scenarios

• Treat discrete variables as continuous 1.5x1 + 4x2 300

x1 200y1

x2 75y2

x1, x2 0

0 y1 1

0 y2 1

Net Income = 3450

𝑥 1* = 200, 𝑥 2* = 0, 𝑦 1* = 1, 𝑦 2 * = 0

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Linear Programming Relaxations

• Continuous relaxations (linear programming relaxations if the given model is an ILP) are formed by treating any discrete variables as continuous while retaining all other constraints. [12.6]

• LP relaxations of ILPs are by far the most used relaxation forms because they bring all the power of LP to bear on analysis of the given discrete models. [12.7]

Proving Infeasibility with Relaxations

• If a constraint relaxation is infeasible, so is the full model it relaxes. [12.8]

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Solution Value Bounds from

Relaxations

• The optimal value of any relaxation of a maximize model yields an upper bound on the optimal value of the full model. The optimal value of any relaxation of a minimize model yields an lower bound. [12.9]

Feasible solutions in

relaxation

Feasible solutions

in true model

True optimum

Example 11.3

EMS Location Planning

1

2

3

4

5

6

7

8

9

10

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Minimum Cover EMS Model

𝑀𝑖𝑛 𝑥𝑗

10

𝑗=1

s.t.

(12.3)

x2 1

x1 + x2 1

x1 + x3 1

x3 1

x3 1

x2 1

x2 + x4 1

x3 + x4 1

x8 1 xi = 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

Minimum Cover EMS Model

with Relaxation

𝑀𝑖𝑛 𝑥𝑗

10

𝑗=1

s.t.

(12.4)

x2 1

x1 + x2 1

x1 + x3 1

x3 1

x3 1

x2 1

x2 + x4 1

x3 + x4 1

x8 1

0xj 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

𝑥 2* = 𝑥 3 * = 𝑥 8* = 𝑥 10* =1,

𝑥 4* = 𝑥 5* = 𝑥 6* = 𝑥 9* = 0.5

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Optimal Solutions from Relaxations

• If an optimal solution to a constraint relaxation is also feasible in the model it relaxes, the solution is optimal in that original model. [12.10]

Rounded Solutions from Relaxations

• Many relaxations produce optimal solutions that are easily “rounded” to good feasible solutions for the full model. [12.11]

• The objective function value of any (integer) feasible solution to a maximizing discrete optimization problem provides a lower bound on the integer optimal value, and any (integer) feasible solution to a minimizing discrete optimization problem provides an upper bound. [12.12]

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Rounded Solutions from Relaxation:

EMS Model

Ceiling

𝑥 1 = 𝑥 1 = 0 = 0

𝑥 2 = 𝑥 2 = 1 = 1

𝑥 3 = 𝑥 3 = 1 = 1

𝑥 4 = 𝑥 4 = 0.5 = 1

𝑥 5 = 𝑥 5 = 0.5 = 1

𝑥 6 = 𝑥 6 = 0.5 = 1

𝑥 7 = 𝑥 7 = 0 = 0

𝑥 8 = 𝑥 8 = 1 = 1

𝑥 9 = 𝑥 9 = 0.5 = 1

𝑥 10 = 𝑥 10 = 1 = 1

𝑥 j10𝑗=1 = 8

(12.5)

Floor

𝑥 1 = 𝑥 1 = 0 = 0

𝑥 2= 𝑥 2 = 1 = 1

𝑥 3= 𝑥 3 = 1 = 1

𝑥 4 = 𝑥 4 = 0.5 = 0

𝑥 5 = 𝑥 5 = 0.5 = 0

𝑥 6 = 𝑥 6 = 0.5 = 0

𝑥 7= 𝑥 7 = 0 = 0

𝑥 8= 𝑥 8 = 1 = 1

𝑥 9 = 𝑥 9 = 0.5 = 0

𝑥 10= 𝑥 10 = 1 = 1 𝑥 𝑗

10𝑗=1 = 4

12.3 Stronger LP Relaxations, Valid Inequalities,

and Lagrangian Relaxation

• A relaxation is strong or sharp if its optimal value

closely bounds that of the true model, and its optimal

solution closely approximates an optimum in the full

model. [12.13]

• Equally correct ILP formulations of a discrete problem

may have dramatically different LP relaxation optima.

[12.14]

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Choosing Big-M Constants

• Whenever a discrete model requires sufficiently large

big-M’s, the strongest relaxation will result from models

employing the smallest valid choice of those

constraints. [12.15]

Bison Booster Example Model

with Relaxation in Big-M Constants

• Max 20x1 + 30x2 – 550y1 – 720y2 (Net income)

s.t. 1.5x1 + 4x2 300 (Display space)

x1 200y1 (T-shirt if equipment)

x2 75y2 (Sweatshirt if equipment)

x1, x2 0

y1, y2 = 0 or 1

• Max 20x1 + 30x2 – 550y1 – 720y2 (Net income)

s.t. 1.5x1 + 4x2 300 (Display space)

x1 10000y1 (T-shirt if equipment)

x2 10000y2 (Sweatshirt if equipment)

x1, x2 0

y1, y2 = 0 or 1

Net Income = 3450

x1* = 200, x2* = 0, y1* = 1, y2* = 0

(12.2)

(12.6)

Net Income = 3989

𝑥 1 = 200, 𝑥 2 = 0, 𝑦 1 = 0.02, 𝑦 2 = 0

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Valid Inequalities

• A linear inequality is a valid inequality for a given

discrete optimization model if it holds for all (integer)

feasible solutions to the model. [12.16]

• To strengthen a relaxation, a valid inequality must cut

off (render infeasible) some feasible solutions to the

current LP relaxation that are not feasible in the full ILP

model. [12.17]

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

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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

with LP Relaxation

(12.8)

s.t.

y4*=y8*= 1

y1*=y2*= y3*=y5*= y6*=y7*= 0

Total Cost = 10153

LP Relaxation 𝑦 1 = 0.230, 𝑦 2

= 0.000, 𝑦 3 = 0.000, 𝑦 4

= 0.301,

𝑦 5 = 0.115, 𝑦 6

= 0.000, 𝑦 7 = 0.000, 𝑦 8

= 0.650

Total Cost = 8036.60

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 𝒙𝒊,𝒋 ≤ 𝒚𝒊 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒊, 𝒋

𝑦𝑖 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

Tmark Facilities Location Example

with LP Relaxation

(12.8)

s.t.

LP Relaxation

𝑦 1 = 0.000, 𝑦 2

= 0.000, 𝑦 3 = 0.000,

𝑦 4 = 0.537, 𝑦 5

= 0.000, 𝑦 6 = 0.000,

𝑦 7 = 0.000, 𝑦 8

= 1.000

Total Cost = 10033.68