Chapter 8
Integer Linear Programming
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Types of Integer Linear Programming Models
Graphical and Computer Solutions for an AllInteger Linear Program
Applications Involving 0-1 Variables
Modeling Flexibility Provided by 0-1 Variables
© 2005 Thomson/South-Western
Slide 1
Types of Integer Programming Models
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An LP in which all the variables are restricted to be
integers is called an all-integer linear program (ILP).
The LP that results from dropping the integer
requirements is called the LP Relaxation of the ILP.
If only a subset of the variables are restricted to be
integers, the problem is called a mixed-integer linear
program (MILP).
Binary variables are variables whose values are
restricted to be 0 or 1. If all variables are restricted to
be 0 or 1, the problem is called a 0-1 or binary integer
linear program.
© 2005 Thomson/South-Western
Slide 2
Example: All-Integer LP

Consider the following all-integer linear program:
Max 3x1 + 2x2
s.t.
3x1 + x2 < 9
x1 + 3x2 < 7
-x1 + x2 < 1
x1, x2 > 0 and integer
© 2005 Thomson/South-Western
Slide 3
Example: All-Integer LP

LP Relaxation
Solving the problem as a linear program ignoring
the integer constraints, the optimal solution to the
linear program gives fractional values for both x1 and
x2. From the graph on the next slide, we see that the
optimal solution to the linear program is:
x1 = 2.5, x2 = 1.5, z = 10.5
© 2005 Thomson/South-Western
Slide 4
Example: All-Integer LP

LP Relaxation
x2
5
-x1 + x2 < 1
4
3x1 + x2 < 9
3
Max 3x1 + 2x2
LP Optimal (2.5, 1.5)
2
x1 + 3x2 < 7
1
1
2
3
© 2005 Thomson/South-Western
4
5
6
7
x1
Slide 5
Example: All-Integer LP

Rounding Up
If we round up the fractional solution (x1 = 2.5,
x2 = 1.5) to the LP relaxation problem, we get x1 = 3
and x2 = 2. From the graph on the next slide, we
see that this point lies outside the feasible region,
making this solution infeasible.
© 2005 Thomson/South-Western
Slide 6
Example: All-Integer LP

Rounded Up Solution
x2
5
-x1 + x2 < 1
4
3x1 + x2 < 9
Max 3x1 + 2x2
3
ILP Infeasible (3, 2)
2
LP Optimal (2.5, 1.5)
x1 + 3x2 < 7
1
1
2
3
© 2005 Thomson/South-Western
4
5
6
7
x1
Slide 7
Example: All-Integer LP

Rounding Down
By rounding the optimal solution down to x1 = 2,
x2 = 1, we see that this solution indeed is an integer
solution within the feasible region, and substituting
in the objective function, it gives z = 8.
We have found a feasible all-integer solution, but
have we found the OPTIMAL all-integer solution?
--------------------The answer is NO! The optimal solution is x1 = 3
and x2 = 0 giving z = 9, as evidenced in the next two
slides.
© 2005 Thomson/South-Western
Slide 8
Example: All-Integer LP

Complete Enumeration of Feasible ILP Solutions
There are eight feasible integer solutions to this
problem:
x1
x2 z
1. 0
0
0
2. 1
0
3
3. 2
0
6
4. 3
0
9
optimal solution
5. 0
1
2
6. 1
1
5
7. 2
1
8
8. 1
2
7
© 2005 Thomson/South-Western
Slide 9
Example: All-Integer LP
x2
-x1 + x2 < 1
5
3x1 + x2 < 9
4
Max 3x1 + 2x2
3
ILP Optimal (3, 0)
2
x1 + 3x2 < 7
1
1
2
© 2005 Thomson/South-Western
3
4
5
6
7
x1
Slide 10
Example: All-Integer LP
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Partial Spreadsheet Showing Problem Data
A
B
C
1
LHS Coefficients
2 Constraint
X1
X2
3
#1
3
1
4
#2
1
3
5
#3
-1
1
6 Obj.Coefficients
3
2
© 2005 Thomson/South-Western
D
RHS
9
7
1
Slide 11
Example: All-Integer LP
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Partial Spreadsheet Showing Formulas
A
8
9
10
11
12
13
14
15
B
Optimal Decision Variable Values
Maximized Objective Function
Constraints
#1
#2
#3
LHS
=B3*$C$9+C3*$D$9
=B4*$C$9+C4*$D$9
=B5*$C$9+C5*$D$9
© 2005 Thomson/South-Western
C
X1
D
X2
=B6*C9+C6*D9
<=
<=
<=
RHS
9
7
1
Slide 12
Example: All-Integer LP
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Partial Spreadsheet Showing Optimal Solution
A
8
9
10
11
12
13
14
15
B
Optimal Decision Variable Values
Maximized Objective Function
Constraints
#1
#2
#3
LHS
9
3
-3
© 2005 Thomson/South-Western
C
X1
3.000
9.000
<=
<=
<=
D
X2
2.9419E-12
RHS
9
7
1
Slide 13
Special 0-1 Constraints

When xi and xj represent binary variables designating
whether projects i and j have been completed, the
following special constraints may be formulated:
•
•
•
•
At most k out of n projects will be completed:
xj < k
j
Project j is conditional on project i:
xj - xi < 0
Project i is a corequisite for project j:
xj - xi = 0
Projects i and j are mutually exclusive:
xi + xj < 1
© 2005 Thomson/South-Western
Slide 14
Example: Metropolitan Microwaves
Metropolitan Microwaves, Inc. is planning to
expand its operations into other
electronic appliances. The company
has identified seven new product lines
it can carry. Relevant information
about each line follows on the next slide.
© 2005 Thomson/South-Western
Slide 15
Example: Metropolitan Microwaves
Product Line
1.
2.
3.
4.
5.
6.
7.
TV/VCRs
Color TVs
Projection TVs
VCRs
DVD Players
Video Games
Home Computers
© 2005 Thomson/South-Western
Initial Floor Space Exp. Rate
Invest.
(Sq.Ft.)
of Return
$ 6,000
12,000
20,000
14,000
15,000
2,000
32,000
125
150
200
40
40
20
100
8.1%
9.0
11.0
10.2
10.5
14.1
13.2
Slide 16
Example: Metropolitan Microwaves
Metropolitan has decided that they should not
stock projection TVs unless they stock either
TV/VCRs or color TVs. Also, they will not stock
both VCRs and DVD players, and they will stock
video games if they stock color TVs. Finally, the
company wishes to introduce at least three new
product lines.
If the company has $45,000 to invest and 420 sq.
ft. of floor space available, formulate an integer linear
program for Metropolitan to maximize its overall
expected return.
© 2005 Thomson/South-Western
Slide 17
Example: Metropolitan Microwaves

Define the Decision Variables
xj = 1 if product line j is introduced;
= 0 otherwise.
where:
Product line 1 = TV/VCRs
Product line 2 = Color TVs
Product line 3 = Projection TVs
Product line 4 = VCRs
Product line 5 = DVD Players
Product line 6 = Video Games
Product line 7 = Home Computers
© 2005 Thomson/South-Western
Slide 18
Example: Metropolitan Microwaves
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Define the Decision Variables
xj = 1 if product line j is introduced;
= 0 otherwise.

Define the Objective Function
Maximize total expected return:
Max .081(6000)x1 + .09(12000)x2 + .11(20000)x3
+ .102(14000)x4 + .105(15000)x5 + .141(2000)x6
+ .132(32000)x7
© 2005 Thomson/South-Western
Slide 19
Example: Metropolitan Microwaves
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Define the Constraints
1) Money:
6x1 + 12x2 + 20x3 + 14x4 + 15x5 + 2x6 + 32x7 < 45
2) Space:
125x1 +150x2 +200x3 +40x4 +40x5 +20x6 +100x7 < 420
3) Stock projection TVs only if
stock TV/VCRs or color TVs:
x1 + x2 > x3 or x1 + x2 - x3 > 0
© 2005 Thomson/South-Western
Slide 20
Example: Metropolitan Microwaves
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Define the Constraints (continued)
4) Do not stock both VCRs and DVD players:
x4 + x5 < 1
5) Stock video games if they stock color TV's:
x2 - x6 > 0
6) Introduce at least 3 new lines:
x1 + x2 + x3 + x4 + x5 + x6 + x7 > 3
7) Variables are 0 or 1:
xj = 0 or 1 for j = 1, , , 7
© 2005 Thomson/South-Western
Slide 21
Example: Metropolitan Microwaves
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Partial Spreadsheet Showing Problem Data
A
B
C
1
2 Constraints
X1
3
#1
6
4
#2
125
5
#3
1
6
#4
0
7
#5
0
8
#6
1
9 Obj.Func.Coeff. 486
D
E
F
LHS Coefficients
X2
X3
X4
X5
12
20
14
15
150 200
40
40
1
-1
0
0
0
0
1
1
1
0
0
0
1
1
1
1
1080 2200 1428 1575
© 2005 Thomson/South-Western
G
X6
2
20
0
0
-1
1
282
H
I
X7 RHS
32 45
100 420
0
0
0
1
0
0
1
3
4224
Slide 22
Example: Metropolitan Microwaves
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Partial Spreadsheet Showing Example Formulas
A
12
13
14
15
16
17
18
19
20
21
22
B
C
D
X1
X2
X3
Dec.Values
0
0
0
Maximized Total Expected Return
Constraints
Money
Space
TVs
VCRs
Video
Lines
© 2005 Thomson/South-Western
LHS
0
0
0
0
0
0
E
X4
0
F
X5
0
0
<=
<=
>=
<=
>=
>=
G
X6
0
H
X7
0
RHS
45
420
0
1
0
3
Slide 23
Example: Metropolitan Microwaves
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Solver Parameters Dialog Box
© 2005 Thomson/South-Western
Slide 24
Example: Metropolitan Microwaves
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Solver Options Dialog Box
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Slide 25
Example: Metropolitan Microwaves
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Integer Options Dialog Box
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Slide 26
Example: Metropolitan Microwaves
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Optimal Solution
A
12
13
14
15
16
17
18
19
20
21
22
B
C
D
X1
X2
X3
Dec.Values
1
0
1
Maximized Total Expected Return
Constraints
Money
Space
TVs
VCRs
Video
Lines
© 2005 Thomson/South-Western
LHS
41
365
0
1
0
3
E
X4
0
F
X5
1
4261
<=
<=
>=
<=
>=
>=
G
X6
0
H
X7
0
RHS
45
420
0
1
0
3
Slide 27
Example: Metropolitan Microwaves
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Optimal Solution
Introduce:
TV/VCRs, Projection TVs, and DVD Players
Do Not Introduce:
Color TVs, VCRs, Video Games, and Home Computers
Total Expected Return:
$4,261
© 2005 Thomson/South-Western
Slide 28
Example: Tina’s Tailoring
Tina's Tailoring has five idle tailors and four
custom garments to make. The estimated time (in
hours) it would take each tailor to make each garment
is shown in the next slide. (An 'X' in the table indicates
an unacceptable tailor-garment assignment.)
Garment
Wedding gown
Clown costume
Admiral's uniform
Bullfighter's outfit
1
19
11
12
X
© 2005 Thomson/South-Western
Tailor
2 3 4 5
23 20 21 18
14 X 12 10
8 11 X 9
20 20 18 21
Slide 29
Example: Tina’s Tailoring
Formulate an integer program for determining
the tailor-garment assignments that minimize
the total estimated time spent making the four
garments. No tailor is to be assigned more than one
garment and each garment is to be worked on by only
one tailor.
-------------------This problem can be formulated as a 0-1 integer
program. The LP solution to this problem will
automatically be integer (0-1).
© 2005 Thomson/South-Western
Slide 30
Example: Tina’s Tailoring
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Define the decision variables
xij = 1 if garment i is assigned to tailor j
= 0 otherwise.
Number of decision variables =
[(number of garments)(number of tailors)]
- (number of unacceptable assignments)
= [4(5)] - 3 = 17
© 2005 Thomson/South-Western
Slide 31
Example: Tina’s Tailoring
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Define the objective function
Minimize total time spent making garments:
Min 19x11 + 23x12 + 20x13 + 21x14 + 18x15 + 11x21
+ 14x22 + 12x24 + 10x25 + 12x31 + 8x32 + 11x33
+ 9x35 + 20x42 + 20x43 + 18x44 + 21x45
© 2005 Thomson/South-Western
Slide 32
Example: Tina’s Tailoring

Define the Constraints
Exactly one tailor per garment:
1) x11 + x12 + x13 + x14 + x15 = 1
2) x21 + x22 + x24 + x25 = 1
3) x31 + x32 + x33 + x35 = 1
4) x42 + x43 + x44 + x45 = 1
© 2005 Thomson/South-Western
Slide 33
Example: Tina’s Tailoring
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Define the Constraints (continued)
No more than one garment per tailor:
5) x11 + x21 + x31 < 1
6) x12 + x22 + x32 + x42 < 1
7) x13 + x33 + x43 < 1
8) x14 + x24 + x44 < 1
9) x15 + x25 + x35 + x45 < 1
Nonnegativity: xij > 0 for i = 1, . . ,4 and j = 1, . . ,5
© 2005 Thomson/South-Western
Slide 34
Cautionary Note About Sensitivity Analysis
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Sensitivity analysis often is more crucial for ILP
problems than for LP problems.
A small change in a constraint coefficient can cause a
relatively large change in the optimal solution.
Recommendation: Resolve the ILP problem several
times with slight variations in the coefficients before
choosing the “best” solution for implementation.
© 2005 Thomson/South-Western
Slide 35
End of Chapter 8
© 2005 Thomson/South-Western
Slide 36