Integer Linear Programming
(ILP)
Prof KG Satheesh Kumar
Asian School of Business
Types of ILP Models
ILP:
A linear program in which some or all
variables are restricted to integer values.
Types:
1. All-integer LP or a pure ILP
2. Mixed-Integer LP
3. 0-1 integer LP
An All-Integer IP or Pure ILP
Maximise 2 x1 + 3 x2
Subject to
3 x1 + 2 x2 ≤ 12
¼ x1 + 1 x2 ≤ 4
1 x1 + 2 x2 ≤ 6
x1, x2 ≥ 0 and integer
Illustration of All-Integer LP
Sweeny: Eastborne Reality is considering investing in
townhouses (T) and apartments(A). Determine the
number of T’s and A’s to be purchased. (Integers)
Funds available: $2 million.
Cost: $282k per T and $400k per A
Numbers available: 5 T’s and any number of A’s.
Management time available: 140 h/mo
Time needed: 4 hrs/mo for T and 40 h/mo for A.
Contribution: $10k for T and $15k for A.
Sweeny Eastborne: ILP Model
Maximise 10T + 15A
Subject to: 282T + 400A ≤ 2000
4T + 40A ≤ 140
T
≤
5
T, A ≥ 0 and integer
Sweeny Eastborne Relaxed LP:
Rounded Solution
If we relax the integer restriction, the optimum
solution is
T=2.48, A = 3.25, Z = 73.57
This is not acceptable because townhouses and
apartments cannot be purchased in fractions!
Rounding down gives
integer solution with T = 2, A
= 3 and Z = 65
Such rounding down may
sometimes yield an
optimum solution, but one
cannot be sure!
Sweeny Eastborne ILP
Optimal Integer Solution
The optimal solution
is T = 4, A = 2, Z = 70
and not the roundeddown solution which
gives Z = 65.
ILP Algorithms
The ILP algorithms are based on exploiting the
tremendous computational success of LP. The strategy
involves three steps:
1. Relax the ILP: Remove integer restrictions; replace any
binary variable y with continuous range 0  y  1.
2. Solve the relaxed LP as a regular LP.
3. Starting with the relaxed optimum, add constraints that
iteratively modify the solution space to satisfy the
integer requirements.
B&B and Cutting Plane Methods
The two commonly used methods are:
1. Branch and bound method
2. Cutting Plane Method
Neither method is consistently effective;
but B&B is far more successful.
Branch-and-Bound (B&B)
• Developed in 1960 by A Land and G Doig
• Relax the integer restrictions in the
problem and solve it as a regular LP. Let’s
call this LP0 (to imply node-zero LP)
• Test if integer requirements are met. Else
branch to get sub-problems LP1 and LP2.
Branching
• If LP0 (in general LPi) fails to yield integer
solution, branch on any variable that fails to
meet this requirement. The process of branching
is illustrated below.
If LPi yields x1 = 3.5 and x1 is taken as the
branching variable, we get two sub-problems,
LPi+1 = LPi & (x1 3) and LPi+2 = LPi & (x1 4).
Note: In mixed integer problems, a continuous variable is
never selected for branching.
Bounding / Fathoming
• Select LP1 (in general LPi) and solve. Three
conditions arise.
– Infeasible solution, declare fathomed (no further
investigation of LPi).
– Integer solution. If it is superior to the current best
solution update the current best. Declare fathomed.
– Non-integer solution. If it is inferior to the current best,
declare fathomed. Else branch again.
Best Bound
• In maximisation, the solution to a sub-problem is
superior if it raises the current lower bound.
• In minimisation, the solution to a sub-problem is
superior if it lowers the current upper bound.
• When all sub-problems have been fathomed,
stop. The current bound is the best bound.
B&B Tree for Eastborne
LP0
T
= 2.48, A = 3.25, Z = 73.57
Noninteger, non-inferior to current best, branch on T
LP1 = LP0 & T ≤ 2
T = 2, A = 3.3, Z = 69.5
Non-integer, can’t give better
solution than LP5, fathomed
LP2 = LP0 & T ≥ 3
T = 3, A = 2.89 Z = 73.28
Non-integer, non-inferior to
current best, branch on A
LP3 = LP0 & T ≥ 3 & A ≤ 2
T
= 4.26, A = 2, Z = 72.55 Noninteger, non-inferior to current
best, branch on T
LP5 = LP0 & T  [3,4] & A ≤ 2
T = 4, A = 2, Z = 70
Integer, Lower (best) bound
LP4 = LP0 & T ≥ 3 & A ≥ 3
Infeasible, fathomed
LP6 = LP0 & T ≥ 5 & A ≤ 2
T = 5, A = 1.48, Z = 72.13
Can’t give better solution than LP5, fathomed
Note: Z is a multiple of 5 and hence only Z ≥ 75 can be better than z = 70
LP0
T = 2.48, A = 3.25, Z = 73.57
Non-integer, non-inferior to current best, branch on T
LP1 = LP0 & T ≤ 2
T = 2, A = 3.3, Z = 69.5
Non-integer, can’t give better solution than LP5, fathomed
LP2 = LP0 & T ≥ 3
T = 3, A = 2.89 Z = 73.28
LP3 = LP0 & T ≥ 3 & A ≤ 2
Non-integer, non-inferior to current best, branch on A
T = 4.26, A = 2, Z = 72.55
Non-integer, non-inferior to current best, branch on T
LP4 = LP0 & T ≥ 3 & A ≥ 3
Infeasible, fathomed
LP5 = LP0 & T  [3,4] & A ≤ 2
LP6 = LP0 & T ≥ 5 & A ≤ 2
T = 4, A = 2, Z = 70
T = 5, A = 1.48, Z = 72.13
Integer, Lower (best) bound
Can’t give better solution than LP5, fathomed
See more illustrations of
Branch-and-Bound
algorithm in the Excel sheet
0-1 Integer LP
“0 -1” decision variables are used in
problems where an Yes-No decision
is to be taken regarding multiple
choices.
If the variable is 1, the corresponding
choice is selected; if the variable is 0,
it is not selected.
0-1 ILP: Capital Budgeting
Five projects are being evaluated over a three year planning horizon.
The table gives the expected returns for each project and associated
yearly expenditures. Which project should be selected over the 3-year
horizon?
Expenditure in million $/year
Project
Available
funds,
million $
Year 1
Year 2
Returns,
million $
Year 3
1
5
1
8
20
2
4
7
10
40
3
3
9
2
20
4
7
4
1
15
5
8
6
10
30
25
25
25
0-1 ILP Model
Maximise: 20 P1 + 40 P2 + 20 P3 + 15 P4 + 30 P5
subject to:
5 P1 + 4 P2 + 3 P3 + 7 P4 + 8 P5 <= 25
1 P1 + 7 P2 + 9 P3 + 4 P4 + 6 P5 <= 25
8 P1 + 10 P2 + 2 P3 + 1 P4 + 10 P5 <= 25
P1, P2, P3, P4, P5 = [0,1]
The Ice –Cold refrigerator Company is considering
investing in several projects that have varying
capital requirements over the next 4 years. Faced
with limited capital each year, management would
like to select the most profitable projects. The
estimated net present value for each project, the
capital requirements, and the available capacity
over the four year period are shown:
Projects
Plant
expansion
Warehouse
expansion
New
Machinery
New product
research
Present
value
90000
40000
10000
37000
Yr1
15000
10000
10000
15000
40000
Yr2
20000
15000
10000
50000
Yr3
20000
20000
10000
40000
yr4
15000
5000
10000
35000
4000
Total capital
available
Fixed Cost
Three raw materials are used for a company to produce
three products: a fuel additive, a solvent base, a carpet
cleaning. The profit contribution are $40 per ton for the
fuel additive, $30 per ton for the solvent base, and $50
per ton for the cleaning fluid. Each ton of fuel additive is
a blend of 0.4 tons of material 1 and 0.6 tons of material
3. Each ton of solvent base is a blend of 0.5 of material
1, 0.2 of material 2, 0.3 of material 3, Each ton of carpet
cleaning fluid is a blend of 0.6 of material 1, 0.1 of
material 2, 0.3 of material 3.
F = tons of fuel additive used
S = tons of solvent base used
C = tons of carpet cleaning used.
Max 40F+30S+50C
s.t.
0.4F+0.5S+0.6C<=20
0.2S+0.1C<=5
0.6F+0.3S+0.3C<=21
F,S,C>=0
product
Set up cost
Maximum
Production
Fuel additive
$200
50 tons
Solvent base
$50
25 tons
Carpet cleaning $400
fluid
40 tons
SF = 1 if fuel additive is produced ,0 else
SS = 1 if fuel additive is produced ,0 else
SC = 1 if fuel additive is produced ,0 else
Max 40F+30S+50C-200SF-50SS-400SC (net profit)
s.t .
F<=50SF or F-50SF <=0 Maximum F
S<=25SS or S-25SS <=0 Maximum S
C<=40SC or C-40SC <=0 Maximum C
Distribution Systems Design
The Martin-Beck Company operates a plant in
St. Louis with an annual capacity of 30,000 units.
Product is shipped to regional centers located in
Boston, Atlanta, Houston. Bcos of an anticipated
increase in demand, Martin-Beck plans to
increase the capacity by constructing a new
plant in one or more of the following cities :
Detroit, Toledo, Denver or Kansas City. The
estimated annual fixed cost and the annual
capacity for the 4 proposed plants is as follows:
Proposed plant Annual fixed
cost
Detroit
175000
Annual
Capacity
10000
Toledo
300000
20000
Denver
375000
30000
Kansas city
500000
40000
Annual Demand of the Distribution
Centers:
Distribution Centers
Annual Demand
Boston
30000
Atlanta
20000
Houston
20000
Plant size
Boston
Atlanta
Houston
Detroit
5
2
3
Toledo
4
3
2
Denver
9
7
5
Kansas city 10
4
2
St.louis
4
3
8
Max:
5x11+2x12+3x13+4x21+3x22+4x23+9x31+7x32+5x33+10x
41+4x42+2x43+8x51+4x52+3x53+175y1+300y2+375y3
+500y4
y1=1 if a plant is constructed in Detroit, 0 else
y2=1 if a plant is constructed in Toledo, 0 else
y3=1 if a plant is constructed in Denver, 0 else
y4=1 if a plant is constructed in Kansas city, 0 else
s.t.
x11+x12+x13<=10y1 or x11+x12+x13-10y1<=0
x21+x22+x23<=20y1 or x21+x22+x23-20y1
x31+x32+x33<=30y1 or x31+x32+x33-30y1
x41+x42+x43<=40y1 or x41+x42+x43-40y1
x51+x52+x53<=30
x11+x21+x31+x41+x51=30
x12+x22+x32+x42+x52=20
x13+x23+x33+x43+x53=20
Branch Location
• The long-range planning department for the Ohio Trust
Company is considering expanding its operation into a
20-county region in NE Ohio. Currently, Ohio Trust does
not have a principal place of business in any of the 20
counties. According to the banking laws in Ohio, if a
bank establishes a principal place of business (PPB) in
any county, branch banks can be estd in that county and
in any adjacent county. However, to establish a new
PPB, Ohio Trust must either obtain approval for a new
bank from the state’s superintendent of banks or
purchase an existing bank.
COUNTIES IN THE OHIO TRUST EXPANSION REGION
COUNTIES UNDER CONSIDERATION
ADJACENT COUNTIES
1
Ashtabula
2,12,16
2
Lake
1,3,12
3
Cuyahoga
2,4,9,10,12,13
4
Lorain
3,5,7,9
5
Huron
4,6,7
6
Richland
5,7,17
7
Ashland
4,5,6,8,9,17,18
8
Wayne
7,9,10,11,18
9
Medina
3,4,7,8,10
10
Summit
3,8,9,11,12,13
11
Stark
8,10,13,14,15,18,19,20
12
Geauga
1,2,3,10,13,16
13
portage
3,10,11,12,15,16
14
Columbians
11,15,20
15
Mahoning
11,13,14,16
16
Trumbull
1,12,13,15
17
Knox
6,7,18
18
Holmes
7,8,11,17,19
19
Tuscarawas
11,18,20
20
Carroll
11,14,19
• Decision variables
Let xi= 1 if a PBB is estd in county i; 0 otherwise
• Objective Function
Minimise: Z= x1+x2+x3.,.,.x20
• Subject to:
x1 + x2 + x12 + x16 >= 1 (Ashtabula)
x1 + x2 + x3 + x12 >= 1 (Lake)
.,.,,.,.,.,(20 constraints)
• Non-negativity
Xi = 0,1
i = 1,2,.,., 20
See more illustrations of
0-1 ILP
and solution using
Excel Solver
Reference
• Hamdy A Taha, “Operations Research: An
Introduction”; Pearson
• Anderson, Sweeny and Williams, “An
Introduction to Management Science”;
Thomson
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