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]
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
0.0080(75)y1+ 0.0070(250)y2+0.0085x3+0.0040x4
0.180(75)y1 + 0.032(250)y2 + 1.0 x5
0.180(75)y1 + 0.032(250)y2 + 1.0 x5
0.120(75)y1 + 0.011(250)y2 + 1.0 x6
0.120(75)y1 + 0.011(250)y2 + 1.0 x6
0.001(250)y2 + 1.0 x7
0.001(250)y2 + 1.0 x7
x3…x7  0
Cost = 9967.06
y1, y2 = 0 or 1
(12.1)
 6.5
 7.5
 30.0
 30.5
 10.0
 12.0
 11.0
 13.0
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
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]
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 Tshirts, 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)
(12.2)
x1, x2  0
Net Income = 3450
y1, y2 = 0 or 1
x1* = 200, x2* = 0, y1* = 1, y2* = 0
Constraint Relaxation Scenarios
• Double capacities
1.5x1 + 4x2  600
x1  400y1
x2  150y2
x1, x2  0
y1, y2 = 0 or 1
Net Income = 7450
𝑥 1* = 400, 𝑥 2* = 0, 𝑦1* = 1, 𝑦2 * = 0
• Dropping first constraint
1.5x1 + 4x2  300
x1  200y1
x2  75y2
x1, x2  0
y1, y2 = 0 or 1
Net Income = 4980
𝑥 1* = 200, 𝑥 2* = 75, 𝑦1* = 1, 𝑦2 * = 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, 𝑦1* = 1, 𝑦2 * = 0
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]
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
8
2
9
1
6
4
5
10
3
7
Minimum Cover EMS Model
10
𝑀𝑖𝑛
𝑥𝑗
𝑗=1
s.t.
x2
x1 + x2
x1 + x3
x3
x3
x2
x2 + x4
x3 + x4
x8
1
1
1
1
1
1
1
1
1
(12.3)
x4 + x6
1
x4 + x5
1
x4 + x5 + x6  1
x4 + x5 + x7  1
x8 + x9
1
x2* = x3* = x4* = x6*
x6 + x9
1
= x8* = x10* =1,
x5 + x6
1
x1* = x5* = x7* = x9*
x5 + x7 + x10  1
=0
x8 + x9
1
x9 + x10
1
x10
 1 xi = 0 or 1  j=1,…,10
Minimum Cover EMS Model
with Relaxation
10
𝑀𝑖𝑛
𝑥𝑗
𝑗=1
s.t.
x2
x1 + x2
x1 + x3
x3
x3
x2
x2 + x4
x3 + x4
x8
1
1
1
1
1
1
1
1
1
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
(12.4)
𝑥 2* = 𝑥 3 * = 𝑥 8* = 𝑥 10* =1,
𝑥 4* = 𝑥 5* = 𝑥 6* = 𝑥 9* = 0.5
0xj 1
 j=1,…,10
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]
Rounded Solutions from Relaxation:
EMS Model
Ceiling
𝑥1 = 𝑥1 = 0 = 0
𝑥2 = 𝑥2 = 1 = 1
𝑥3 = 𝑥3 = 1 = 1
𝑥4 = 𝑥4 = 0.5 = 1
𝑥5 = 𝑥5 = 0.5 = 1
𝑥6 = 𝑥6 = 0.5 = 1
𝑥7 = 𝑥7 = 0 = 0
𝑥8 = 𝑥8 = 1 = 1
𝑥9 = 𝑥9 = 0.5 = 1
𝑥10 = 𝑥10 = 1 = 1
10
𝑗=1 𝑥 j= 8
(12.5)
Floor
𝑥1 = 𝑥1 = 0 = 0
𝑥 2= 𝑥 2 = 1 = 1
𝑥 3= 𝑥 3 = 1 = 1
𝑥4 = 𝑥4 = 0.5 = 0
𝑥5 = 𝑥5 = 0.5 = 0
𝑥6 = 𝑥6 = 0.5 = 0
𝑥 7= 𝑥 7 = 0 = 0
𝑥 8= 𝑥 8 = 1 = 1
𝑥9 = 𝑥9 = 0.5 = 0
𝑥10= 𝑥10 = 1 = 1
10
𝑗=1 𝑥𝑗 = 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]
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)
(12.2)
s.t.
1.5x1 + 4x2  300
(Display space)
x1  200y1
(T-shirt if equipment)
x2  75y2
(Sweatshirt if equipment)
x1, x2  0
Net Income = 3450
x1* = 200, x2* = 0, y1* = 1, y2* = 0
y1, y2 = 0 or 1
• Max 20x1 + 30x2 – 550y1 – 720y2
(Net income)
(12.6)
s.t.
1.5x1 + 4x2  300
(Display space)
x1  10000y1
(T-shirt if equipment)
x2  10000y2
(Sweatshirt if equipment)
x1, x2  0
Net Income = 3989
𝑥 1 = 200, 𝑥 2 = 0, 𝑦1 = 0.02, 𝑦2 = 0
y1, y2 = 0 or 1
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
4
6
7
1
2
5
3
8
i
Fixed
Cost
1
2400
2
7000
3
3600
4
1600
5
3000
6
4600
7
9000
8
2000
Tmark Facilities Location Example
with LP Relaxation
8
14
8
𝑚𝑖𝑛
(𝑟𝑖,𝑗 𝑑𝑗 )𝑥𝑖,𝑗 +
𝑖=1 𝑗=1
𝑓𝑖 𝑦𝑖
(12.8)
𝑖=1
8
s.t.
𝑥𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑗 (𝑐𝑎𝑟𝑟𝑦 𝑗 𝑙𝑜𝑎𝑑)
𝑖=1
14
1500𝑦𝑖 ≤
𝑑𝑗 𝑥𝑖,𝑗 ≤ 5000𝑦𝑖 (𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑖)
𝑗=1
𝑥𝑖,𝑗 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗
𝑦𝑖 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖
y4*=y8*= 1
y1*=y2*= y3*=y5*= y6*=y7*= 0
Total Cost = 10153
𝑦1 = 0.230, 𝑦2 = 0.000, 𝑦3 = 0.000, 𝑦4 = 0.301,
𝑦5 = 0.115, 𝑦6 = 0.000, 𝑦7 = 0.000, 𝑦8 = 0.650
Total Cost = 8036.60
LP Relaxation
Tmark Facilities Location Example
with LP Relaxation
8
14
8
𝑚𝑖𝑛
(𝑟𝑖,𝑗 𝑑𝑗 )𝑥𝑖,𝑗 +
𝑖=1 𝑗=1
𝑓𝑖 𝑦𝑖
(12.8)
𝑖=1
8
s.t.
𝑥𝑖,𝑗 = 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑗 (𝑐𝑎𝑟𝑟𝑦 𝑗 𝑙𝑜𝑎𝑑)
𝑖=1
14
1500𝑦𝑖 ≤
𝑑𝑗 𝑥𝑖,𝑗 ≤ 5000𝑦𝑖 (𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑖)
𝑗=1
𝑥𝑖,𝑗 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗
𝒙𝒊,𝒋 ≤ 𝒚𝒊 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒊, 𝒋
𝑦𝑖 = 0 𝑜𝑟 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖
LP Relaxation
𝑦1 = 0.000, 𝑦2 = 0.000, 𝑦3 = 0.000,
𝑦4 = 0.537, 𝑦5 = 0.000, 𝑦6 = 0.000,
𝑦7 = 0.000, 𝑦8 = 1.000
Total Cost = 10033.68