CD S7-1-1
SUPPLEMENT 1 TO CHAPTER 7
ADVANCED FORMULATION TECHNIQUES FOR
BINARY INTEGER PROGRAMMING
Chapter 7 gives various examples of how yes-or-no decisions can arise. Each such decision is
represented by a binary decision variable in a binary integer programming (BIP) model.
In addition to any such binary decision variables, other binary variables sometimes are
introduced simply to help formulate the model. Here is the terminology to distinguish between
the two kinds of binary variables.
A binary decision variable is a binary variable that represents a yes-or-no decision. An
auxiliary binary variable is an additional binary variable that is introduced into the
model, not to represent a yes-or-no decision, but simply to help formulate the model as a
(pure or mixed) BIP problem. Auxiliary binary variables will be denoted by y1, y2, . . . .
This supplement to Chapter 7 illustrates some of the ways in which auxiliary binary
variables can play a crucial role in being able to formulate the model to fit a standard problem so
that the model can be solved. To facilitate focusing on the role of the auxiliary binary variables,
the first two examples are variations of the familiar Wyndor Glass Co. problem introduced in
Section 2.1 and formulated as a linear programming model on a spreadsheet in Section 2.2. To
further refresh your memory, Figure 1 shows the graphical solution originally developed in
Section 2.4 for this problem, where the symbol P represents the weekly profit in dollars.
Production rate
for windows
W
8
Optimal solution
(2, 6)
6
4
Feasible
Region
P = 3,600 = 300 D + 500 W
2
0
Figure 1
4
2
Production rate for doors
6
8
10 D
This graph summarizes the presentation in Section 2.4 of the application of the
graphical method to the original Wyndor problem.
This original Wyndor problem has no yes-or-no decisions and so no binary decision
variables. However, each of the two variations presented below introduces a complication that
can be overcome by using auxiliary binary variables to formulate a model that can be readily
CD S7-1-2
solved. (Because the Wyndor problem has only two decision variables, we will be able to use
graphical analysis to help introduce and analyze each variation before showing how auxiliary
binary variables can be used with any number of decision variables.).
Example 1: The Wyndor Problem with Mutually Exclusive Products
Change for Example 1: The two potential new products (doors and windows) would
compete for the same customers. Therefore, management has decided not to produce both
of them together. At most one can be chosen for production, so
either D = 0 or W = 0 (or both)
Thus, we now are dealing with mutually exclusive products.
Figure 2 shows the feasible region for this problem, namely, the line segment from (0, 0)
to (4, 0) and the line segment from (0, 0) to (0, 6). These are the only solutions from the feasible
region for the original problem for which either x1 = 0 or x2 = 0. For this tiny problem, it can be
seen from the figure that the feasible solution that maximizes P (i.e., the optimal solution) is
(D, W) = (0, 6) with P = 3,000
W
P = 300 D + 500 W
Either D = 0 or W = 0
8
Production
rate for
windows
(0, 6) gives P = 3,000
6
4
2
(0, 0) gives P = 0
0
Figure 2
(4, 0) gives P = 1,200
2
4
Production rate for doors
6
8 D
The dark line segments show the feasible solutions for Example 1.
Linear or integer programming models do not permit an either-or-restriction such as
either D = 0 or W = 0. How can we rewrite this restriction in a standard form to fit such a model
so that the model can be solved by available algorithms (including those in Solver)?
As illustrated by the case study in Section 7.1, if D and W were binary variables, we
would only need to rewrite the restriction that the two products are mutually exclusive
alternatives as D + W = 1. However, D and W represent production rates that can take on various
values besides 0 and 1, so this constraint does not work.
Now watch auxiliary binary variables come to the rescue.
CD S7-1-3
Formulation with Auxiliary Binary Variables
For each product, there are just two possibilities regarding the decision of whether it can
be produced. Either it can or it cannot. Therefore, we can associate each of the two values of an
auxiliary binary variable with one of these possibilities. Specifically, let the auxiliary binary
variables be
⎧1, if D > 0 can hold (can produce doors)
y1 = ⎨
⎩0, if D = 0 must hold (cannot produce doors)
⎧1, if W > 0 can hold (can produce windowss)
y2 = ⎨
⎩0, if W = 0 must hold (cannot produce windows)
The relationships between y1 and D, as well as between y2 and W, are identical to the ones
shown in Section 7.5 for another variation of the Wyndor problem.. Therefore, proceeding as in
Figure 7.8 (including using the safely large coefficient of 99). Figure 3 demonstrates that the
analogous constraints involving these variables, UnitsProduced (C13:D13) ≤ OnlyIfProduce
(C15:D15) and Produce? (C16:D16) = binary, can be used to ensure that these relationships hold.
To make the products mutually exclusive, we now add the usual kind of constraint for mutually
exclusive alternatives with regard to these two binary variables,
y1 + y2 ≤ 1
which gives the constraint, TotalProduced (E16) ≤ MaximumToProduce (G16), in the
spreadsheet model. This forces either C16 = 0 or D16 = 0 (or both).
CD S7-1-4
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
B
C
D
E
F
G
<=
<=
<=
Hours
Available
4
12
18
<=
Maximum
To Produce
1
Wyndor Glass Co. with Mutually Exclusive Products
Doors
$300
Unit Profit
Plant 1
Plant 2
Plant 3
Hours Used Per Unit Produced
1
0
0
2
3
2
Units Produced
Only If Produce
Produce?
Doors
0
<=
0
0
Windows
6
<=
99
1
Hours
Used
0
12
12
Total
Produced
1
Total Profit
$3,000
Solver Parameters
Set Objective Cell: TotalProfit
To: Max
By Changing (Variable) Cells:
UnitsProduced, Produce?
Subject to the Constraints:
Produce? = binary
HoursUsed <= HoursAvailable
TotalProduced <= MaximumToProduce
UnitsProduced <= OnlyIfProduce
B
C
Only If Produce =99*C16
Range Name
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
MaximumToProduce
OnlyIfProduce
Produce?
TotalProduced
TotalProfit
UnitProfit
UnitsProduced
Cells
G8:G10
E8:E10
C8:D10
G16
C15:D15
C16:D16
E16
G19
C4:D4
C13:D13
E
6
Hours
7
Used
8 =SUMPRODUCT(C8:D8,UnitsProduced)
9 =SUMPRODUCT(C9:D9,UnitsProduced)
10 =SUMPRODUCT(C10:D10,UnitsProduced)
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
15
Windows
$500
D
=99*D16
E
14
Total
15
Produced
16 =SUM(Produce?)
G
18
Total Profit
19 =SUMPRODUCT(UnitProfit,UnitsProduced)
Figure 3
A spreadsheet model for Example 1, where Solver provides the optimal solution
shown in the changing cells, UnitsProduced (C13:D13) and Produce? (C16:D16).
CD S7-1-5
These are the only new constraints needed along with the constraints of the original
model. Since D and W are production rates, these variables do not need to have integer values, so
the model in Figure 3 is a mixed BIP model.
There are no extra costs associated with any values of y1 and y2. The original objective
function, without y1 and y2, still applies, as indicated by the equation entered into the objective
cell TotalProfit (G19).
Solver gives the optimal solution shown in the changing cells, namely, the windows are
the product chosen to be produced, and then they are produced at the maximum rate (W = 6)
allowed by the original constraints.
For such a small problem, we were able to find this optimal solution from Figure 3
without introducing auxiliary binary variables. However, auxiliary binary variables become
necessary when dealing with larger problems. For example, if this pair of mutually exclusive
products is just part of a larger group of products under consideration, then the larger model
would need to add the constraints involving y1 and y2.
You will see this same approach included again later in Example 3 when there are three
potential new products and at most two can be chosen to be produced.
Example 2: The Wyndor Problem with Either-Or Constraints
Now suppose that the only change from the original Wyndor problem is the one spelled out
below.
Change for Example 2: The company has just opened a new plant (plant 4) that is similar
to plant 3, so the new plant can perform the same operations as plant 3 to help produce the
two new products (doors and windows). However, for administrative reasons, management
wants just one of the plants to be chosen to work on these products. The plant chosen should
be the one that provides the most profitable product mix.
Table 1 gives the data for this problem. This table is identical to Table 2.1 for the original
problem except for the addition of the data for plant 4. Although the hours of production time are
different for plants 3 and 4 (because of differences in the types of production facilities being
used), the costs of the operations for each product are essentially the same for the two plants.
Therefore, the unit profits in the last row of the table are unaffected by the choice of which plant
to use for these products.
CD S7-1-6
Table 1
Data for Example 2
Production Time Used for
Each Unit Produced (Hours)
Plant
Doors
Windows
Production Time
Available
per Week (Hours)
1
1
0
4
2
0
2
12
3
3
2
18
4
2
4
28
Unit Profit
$300
$500
The data for plant 4 indicate that if this plant is chosen, then we must satisfy the
constraint
2D + 4W ≤ 28
when solving for the most profitable product mix. However, if plant 3 is chosen instead, then this
constraint is irrelevant and we must instead satisfy the original constraint for plant 3,
3D + 2W ≤ 18
In other words, the relevant restriction is the following pair of either-or constraints:
Either 3D + 2W ≤ 18
Or 2D + 4W ≤ 28
Choosing one of these two constraints as the one that must be satisfied corresponds to choosing
one of the two plants to help produce the doors and windows. The choice of which plant depends
on which one allows the largest total profit when considering all the constraints of the model.
Figure 4 shows the effect of these two choices. If plant 3 were chosen to help produce
these two products (so 3D + 2W ≤ 18 is relevant but 2D + 4W ≤ 28 is not), then we would have
the linear programming problem on the left side of the figure. Since this problem is identical to
the original Wyndor problem shown in Figure 1, the best available solution would be
(D, W) = (2, 6)
with
P = 3,600
However, if plant 4 were chosen instead (so 2D + 4W ≤ 28 is relevant but 3D + 2W ≤ 18 is not),
then we would have the linear programming problem on the right side of the figure. The best
available solution for this problem would be
(D, W) = (4, 5)
with
P = 3,700
Since P = 3,700 is larger than P = 3,600, the largest possible weekly profit is $3,700, which is
only obtainable by choosing plant 4 instead of plant 3 to help produce the two new products.
CD S7-1-7
(a) Choose Plant 3
(b) Choose Plant 4
W
W
8
8
(2, 6) gives
P = 3,600
6
6
4
4
(4, 5) gives
P = 3,700
Feasible
region
Feasible
region
2
2
0
2
Figure 4
4
D
0
2
4
D
These two graphs for Example 2 show the linear programming problem and its
optimal solution that would result if the plant chosen to help produce the two new
products were (a) plant 3 or (b) plant 4.
Despite its either-or constraints, we have just managed to solve the complete model for
variation 3 by solving and comparing two linear programming problems. However, solving larger
models with several pairs of either-or constraints in this way would require solving and
comparing numerous linear programming problems. We would much prefer to be able to apply a
standard algorithm (such as those used by Solver) just once to solve the model.
Unfortunately, the model for Example 2 is not a linear programming model, since eitheror constraints are not allowed in linear or integer programming. In fact, this model does not fit the
format for any kind of standard model. Therefore, we cannot use a standard algorithm once to
find an optimal solution for this model in its current form.
How can we reformulate this model into a standard format where a standard algorithm
can be used one time to find an optimal solution? Once again, auxiliary binary variables come to
the rescue.
Formulation with an Auxiliary Binary Variable
There are just two possibilities: either 3D + 2W ≤ 18 must hold (due to choosing plant 3) or 2D +
4W ≤ 28 must hold (due to choosing plant 4). Therefore, we can introduce an auxiliary binary
variable y to indicate which possibility is chosen by defining y as
⎧1, if 2D + 4W ≤ 28 must hold (choose plant 4)
y=⎨
⎩0, if 3D + 2W ≤ 18 must hold (choose plant 3)
CD S7-1-8
This definition is enforced by introducing an extremely large positive number (we will use 99
again) and then making the following changes in the model.
Replace
by
Either
3D + 2W ≤ 18
3D + 2W ≤ 18 + 99y
Or
2D + 4W ≤ 28
2D + 4W ≤ 28 + 99(1 - y)
y is binary
To see why these new constraints work, check what happens when y = 0.
y=0
gives
3D + 2W ≤ 18
and
2D + 4W ≤ 28 + 99 (a relatively huge number)
so
3D + 2W ≤ 18 must hold
but
2D + 4W ≤ 28 does not need to hold
Since the other constraints in the model prevent 2D + 4W from being much larger than 28, having
y = 0 give 2D + 4W ≤ 28 + 99 has the same effect as eliminating this constraint from the model.
Similarly,
y=1
gives
and
3D + 2W ≤ 18 + 99 (a relatively huge number)
2D + 4W ≤ 28
so
3D + 2W ≤ 18 does not need to hold
but
2D + 4W ≤ 28 must hold
Again, adding 99 to the right-hand side of 3D + 2W ≤ 18 is equivalent to eliminating the
constraint.
Figure 5 shows how this approach can be incorporated into a spreadsheet model. An
additional changing cell (E16) displays the value of y. As indicated in the figure by the equations
entered into cells G10 and G11, these cells give the modified right-hand sides of the plants 3 and
4 constraints that result from the value of y. With the constraints included in the Solver dialog
box, we now have a mixed BIP model that can be solved by Solver.
Clicking on the Solve button causes Solver to simultaneously choose the value of y and
the production rates (D and W) that maximize the total profit given in the objective cell
TotalProfit (H14). Since y = 1 in this optimal solution, plant 4 should be chosen to help produce
the new products, with D = 4 and W = 5. This choice of plant provides a weekly profit of $3,700
rather than the $3,600 obtainable by choosing plant 3 instead.
CD S7-1-9
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
B
C
D
E
F
G
H
Modified
Hours
Available
4
12
117
28
Hours
Available
4
12
18
28
Wyndor Glass Co. Problem with Either-Or Constraints
Doors
$300
Unit Profit
Plant 1
Plant 2
Plant 3
Plant 4
Windows
$500
Hours
Used
4
10
22
28
Hours Used Per Unit Produced
1
0
0
2
3
2
2
4
Doors
4
Units Produced
Windows
5
Total Profit
Which Plant to Use? (0=Plant 3, 1=Plant 4)
Solver Parameters
Set Objective Cell: TotalProfit
To: Max
By Changing Variable Cells:
UnitsProduced, WhichPlantToUse?
Subject to the Constraints:
WhichPlantToUse? = binary
HoursUsed <= ModifiedHoursAvailable
<=
<=
<=
<=
$3,700
1
Range Name
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
ModifiedHoursAvailable
TotalProfit
UnitProfit
UnitsProduced
WhichPlantToUse?
Cells
H8:H11
E8:E11
C8:D11
G8:G11
H14
C4:D4
C14:D14
E16
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
5
6
7
8
9
10
11
E
F
Hours
Used
=SUMPRODUCT(C8:D8,UnitsProduced)
=SUMPRODUCT(C9:D9,UnitsProduced)
=SUMPRODUCT(C10:D10,UnitsProduced)
=SUMPRODUCT(C11:D11,UnitsProduced)
14
<=
<=
<=
<=
G
Modified
Hours
Available
=H8
=H9
=H10+99*WhichPlantToUse?
=H11+99*(1-WhichPlantToUse?)
G
H
Total Profit =SUMPRODUCT(UnitProfit,UnitsProduced)
Figure 5
A spreadsheet model for Example 2, where Solver gives the optimal solution
shown in the changing cells, UnitsProduced (C14:D14) and WhichPlantToUse?
(E16).
CD S7-1-10
Example 3: Imposing Managerial Restrictions
The Research and Development Division of the Good Products Company has developed three
possible new products. However, to avoid undue diversification of the company’s product line,
management has imposed the following restriction:
Restriction 1: From the three possible new products, at most two should be chosen to be
produced.
Each of these products can be produced in either of two plants. For administrative reasons,
management has imposed a second restriction in this regard:
Restriction 2: Just one of the two plants should be chosen to be the sole producer of the
new products.
The production cost per unit of each product would be essentially the same in the two plants.
However, because of differences in their production facilities, the number of hours of production
time needed per unit of each product might differ between the two plants. These data are given in
Table 2 along with other relevant information, including marketing estimates of the number of
units of each product that could be sold per week if it is produced. According to management, the
objective is to choose the products, the plant, and the production rates of the chosen products so
as to maximize the total profit.
Table 2 Data for Example 3 (The Good Products Co. Problem)
Production Time Used for Each
Unit Produced (Hours)
Plant
Product 1
Product 2
Product 3
Production Time
Available per
Week (Hours)
1
3
4
2
30
2
4
6
2
40
Unit Profit
5
7
3
($thousands)
Sales potential
7
5
9
(units per week)
In some ways, this problem resembles a standard product-mix problem such as the
Wyndor Glass Co. case study described in Section 2.1. In fact, if we changed the problem by
dropping the two restrictions and by requiring each unit of a product to use the production hours
given in Table 2 in both plants (so the two plants now perform different operations needed by the
products), it would become just such a problem. In particular, let x1, x2, and x3 be the production
rates of the respective products. Displaying the values of these decision variables in changing
cells UnitsProduced (C12:E12), the spreadsheet model then would become the one shown in rows
1–13 and 15 of Figure 6 if column H and the other rows were omitted except for TotalProfit
(H21). MaximumSales (C15:E15) provides upper bounds on the production rates for the three
products, so
x1 ≤ 7 ,
x2 ≤ 5 ,
are needed as constraints in the model.
x3 ≤ 9
CD S7-1-11
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
B
C
D
E
F
G
H
I
<=
<=
Modified
Hours
Available
129
40
Hours
Available
30
40
<=
Maximum
To Produce
2
Good Products Co. with Managerial Restrictions
Unit Profit ($thousands)
Product 1
5
Only If Produce
Maximum Sales
Product 3
3
Product 1
5.5
<=
7
7
Product 2
0
<=
0
5
Product 3
9
<=
9
9
1
0
1
Produce?
Which Plant to Use? (0=Plant 1, 1=Plant 2)
Solver Parameters
Set Objective Cell: TotalProfit
To: Max
By Changing Variable Cells:
UnitsProduced, Produce?, WhichPlantToUse?
Subject to the Constraints:
Produce? = binary
WhichPlantToUse? = binary
HoursUsed <= ModifiedHoursAvailable
TotalProduced <= MaximumToProduce
UnitsProduced <= OnlyIfProduce
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
5
6
7
8
9
F
F
15
Total
16
Produced
17 =SUM(Produce?)
Figure 6
Total
Produced
2
Total Profit
($thousands)
54.5
1
Range Name
HoursAvailable
HoursUsed
HoursUsedPerUnitProduced
MaximumSales
MaximumToProduce
ModifiedHoursAvailable
OnlyIfProduce
Produce?
TotalProduced
TotalProfit
UnitProfit
UnitsProduced
WhichPlantToUse?
Cells
I8:I9
F8:F9
C8:E9
C15:E15
H17
H8:H9
C14:E14
C17:E17
F17
H21
C4:E4
C12:E12
E21
G
Hours
Used
=SUMPRODUCT(C8:E8,UnitsProduced)
=SUMPRODUCT(C9:E9,UnitsProduced)
14
Hours
Used
34.5
40
Hours Used Per Unit Produced
3
4
2
4
6
2
Plant 1
Plant 2
Units Produced
Product 2
7
H
Modified
Hours
Available
<= =I8+99*WhichPlantToUse?
<= =I9+99*(1-WhichPlantToUse?)
B
C
Only If Produce =C15*C17
D
=D15*D17
E
=E15*E17
H
19
Total Profit
20
($thousands)
21 =SUMPRODUCT(UnitProfit,UnitsProduced)
A spreadsheet formulation of the BIP model for the Good Products Co. problem,
where Solver provides the optimal solution given in the changing cells,
UnitsProduced (C12:E12), Produce? (C17:E17), and WhichPlanToUse? (E21)
CD S7-1-12
For the real problem, however, restriction 1 necessitates adding to the model the
constraint:
No more than two of the decision variables (x1, x2, x3) can have a value greater than zero.
This constraint does not fit into a linear or integer programming format, so the key question is
how to convert it to such a format so that a corresponding algorithm can be used to solve the
overall model. If the decision variables were binary variables, then the constraint would be
expressed in this format as x1 + x2 + x3 ≤ 2. However, with continuous decision variables, a more
complicated approach involving the introduction of auxiliary binary variables is needed.
Restriction 2 necessitates replacing the first two functional constraints (3x1 + 4x2 + 2x3 ≤
30 and 4x1 + 6x2 + 2x3 ≤ 40) by the restriction
Either
3x1 + 4x2 + 2x3 ≤ 30
Or
4x1 + 6x2 + 2x3 ≤ 40
must hold, where the choice of which constraint must hold corresponds to the choice of which
plant will be used to produce the new products. Example 2 illustrated how such either-or
constraints can be converted to a linear or integer programming format, again with the help of an
auxiliary binary variable.
Formulation with Auxiliary Binary Variables Except for involving more products and choices, restriction 1 is similar to the restriction imposed
in Example 1. For Example 1, there were just two new products, and the restriction was that at
most one could be chosen to be produced. Following the formulation approach used there, we can
deal with restriction 1 by introducing three auxiliary binary variables (y1, y2, y3) with the
interpretation that
⎧1, if x j > 0 can hold (can produce product j)
yj = ⎨
⎩0, if x j = 0 must hold (cannot produce product j)
for j = 1, 2, 3. To enforce this interpretation in the model, we replace the constraints on the
maximum production rates of the three products—x1 ≤ 7, x2 ≤ 5, and x3 ≤ 9—by the new
constraints,
x1 ≤ 7y1 ,
x2 ≤ 5y2 ,
x3 ≤ 9y3
Therefore, y1 = 1 allows any feasible value of x1, whereas y1 = 0 forces x1 = 0, and both y2 and y3
have the same effect on x2 and x3, respectively. We also add the constraints,
y1 + y2 + y3 ≤ 2
yj is binary,
for j = 1, 2, 3
Consequently, when these constraints force choosing at most two of the yj to equal 1, this
amounts to choosing at most two of the new products as the ones that can be produced.
To deal with restriction 2, we use the same approach as for Example 2. Therefore, we
introduce another auxiliary binary variable y4 with the interpretation that
⎧1, if 4x1 + 6x2 + 2x3 ≤ 40 must hold (choose plant 2)
y4 = ⎨
⎩0, if 3x1 + 4x2 + 2x3 ≤ 30 must hold (choose plant 1)
CD S7-1-13
This interpretation is enforced by adding the constraints
3x1 + 4x2 + 2x3 ≤ 30 + 99y4
4x1 + 6x2 + 2x3 ≤ 40 + 99(1 - y4)
y4 is binary
Both column H and rows 13–21 of Figure 6 show how all of this can be incorporated into
the spreadsheet model in an intuitive way. The additional changing cells, Produce? (C17:E17)
and WhichPlantToUse? (E21), give the values of the four auxiliary binary variables, so all four
cells are constrained to be binary. The constraint that TotalProduced (F17) ≤
MaximumToProduce (H17) forces choosing at most two of the new products to be produced. The
equations entered into OnlyIfProduce (C14:E14)—as shown at the bottom of the figure—give the
values of 7y1, 5y2, and 9y3, so the constraints that UnitsProduced (C12:E12) ≤ OnlyIfProduce
(C14:E14) force the production rate of a product to be 0 in row 12 if the decision has been made
in row 17 not to produce that product. With the equations that are shown for ModifiedHours
Available (H8:H9), the constraints that HoursUsed (F18:F19) ≤ ModifiedHoursAvailable
(H8:H9) correspond to the algebraic constraints given at the end of the preceding paragraph.
The spreadsheet model now is a mixed BIP model, with three continuous decision
variables [UnitsProduced (C12:E12)] and four auxiliary binary variables [Produce? (C17:E17)
and WhichPlantToUse? (E21)], so now the problem is formulated in a form that can be solved.
Using Solver gives the optimal solution shown in the changing cells in Figure 6, namely, choose
products 1 and 3 to produce, choose plant 2 for the production, and choose the production rates of
51⁄2 units per week for product 1 and 9 units per week for product 3. The resulting total profit
given in the objective cell TotalProfit (H21) is $54,500 per week.
Example 4: Violating Proportionality
The Supersuds Corporation is developing its marketing plans for next year’s new products. For
three of these products, the decision has been made to purchase a total of five TV spots for
commercials on national television networks. Each spot will feature a single product. Therefore,
the problem on which we will focus is how to allocate the five spots to these three products, with
a maximum of three spots (and a minimum of zero) for each product.
Table 3 shows the estimated impact of allocating zero, one, two, or three spots to each
product. This impact is measured in terms of the profit from the additional sales that would result
from the spots, considering also the cost of producing the commercial and purchasing the spots.
The objective is to allocate five spots to the products so as to maximize the total profit.
This problem is small enough that it can be solved easily by trial and error. (The optimal
solution is to allocate two spots to product 1, no spots to product 2, and three spots to product 3.)
However, we will show one formulation with auxiliary binary variables for illustrative purposes.
Such a formulation would become necessary if this small problem needed to be incorporated into
a larger model involving the allocation of resources to marketing activities for all the
corporation’s new products.
CD S7-1-14
Table 3
Data for Example 4 (the Supersuds Corp. Problem)
Profit (Millions)
Number of TV Spots
Product 1
Product 2
Product 3
0
$0
$0
$0
1
1
0
–1
2
3
2
2
3
3
3
4
A Formulation with Auxiliary Binary Variables
A natural formulation would be to let
x1 = Number of TV spots allocated to product 1
x2 = Number of TV spots allocated to product 2
x3 = Number of TV spots allocated to product 3
P = Total profit (in millions of dollars)
The contribution of each of these integer decision variables (x1, x2, x3) to P then would be given
by the corresponding column in Table 3. However, each column indicates that profit is not
proportional to the number of TV spots allocated to that product. Therefore, we cannot write a
legitimate objective function in terms of these decision variables to fit integer programming.
Using the algebraic form, the best that we can do with these decision variables is to formulate an
incomplete integer programming model (not a binary integer programming model) that includes
all the needed constraints but not an objective function.
Maximize P = ? subject to and x1 ≤ 3 x2 ≤ 3 x3 ≤ 3 x1 + x2 + x3 = 5 x1 ≥ 0 x2 ≥ 0 x3 ≥ 0 x1, x2, x3 are integers CD S7-1-15
Now see what happens when we introduce nine auxiliary binary variables with the following
interpretations:
⎧⎪
y11 = ⎨
⎪⎩
⎧⎪
y21 = ⎨
⎪⎩
⎧⎪
y31 = ⎨
⎩⎪
1,
if x1 = 1
0, otherwise
1,
if x2 = 1
0, otherwise
1,
if x3 = 1
0, otherwise
⎧⎪
y12 = ⎨
⎪⎩
⎧⎪
y22 = ⎨
⎪⎩
⎧⎪
y32 = ⎨
⎩⎪
1,
if x1 = 2
0, otherwise
1,
if x2 = 2
0, otherwise
1,
if x3 = 2
0, otherwise
⎧⎪
y13 = ⎨
⎪⎩
⎧⎪
y23 = ⎨
⎪⎩
⎧⎪
y33 = ⎨
⎩⎪
1,
if x1 = 3
0, otherwise
1,
if x2 = 3
0, otherwise
1,
if x3 = 3
0, otherwise
For example, look at the definitions of y11, y12, and y13. These definitions imply that
(y11, y12, y13) = (0, 0, 0)
if
x1 = 0
(y11, y12, y13) = (1, 0, 0)
if
x1 = 1
(y11, y12, y13) = (0, 1, 0)
if
x1 = 2
(y11, y12, y13) = (0, 0, 1)
if
x1 = 3
These four alternative values of x1 are the only possible values. Since these alternative values are
mutually exclusive alternatives, y1, y2, and y3 need to satisfy the constraints
y11 + y12 + y13 ≤ 1
y11, y12, y13 are binary
Selecting values of y11, y12, and y13 that satisfy these constraints is equivalent to selecting a value
of x1 that satisfies the constraints
x1 ≤ 3
x1 ≤ 0
x1 is integer
In just the same way, the other auxiliary binary variables need to satisfy the constraints
y21 + y22 + y23 ≤ 1
y31 + y32 + y33 ≤ 1
y21, y22, y23, y31, y32, y33 are binary
Selecting values of these variables that satisfy these constraints is equivalent to selecting values
of x2 and x3 that satisfy the constraints
x2 ≤ 3
x3 ≤ 3
x2 ≥ 0, x3 ≥ 0
x2, x3 are integers
CD S7-1-16
Therefore, we now can formulate a model for the Supersuds problem in terms of these auxiliary
binary variables by including the above constraints on these variables. We also need to add a
constraint that will ensure that the original constraint,
x1 + x2 + x3 = 5
still will hold. The key here is to note that the definitions of the auxiliary binary variables imply
that
x1 = y11 + 2y12 + 3y13
x2 = y21 + 2y22 + 3y23
x1 = y31 + 2y32 + 3y33
Therefore, the original constraint can be replaced by the constraint
y11 + 2y12 + 3y13 + y21 + 2y22 + 3y23 + y31 + 2y32 + 3y33 = 5
Finally, we come to the whole reason for bothering with all of this, namely, that the
auxiliary binary variables enable us to formulate a legitimate objective function. Using monetary
units of millions of dollars, the three profit columns of Table 3 respectively indicate that
Profit from product 1 = y11 + 3y12 + 3y13
Profit from product 2 = 2y22 + 3y23
Profit from product 3 = –y31 + 2y32 + 4y33
Therefore, adding these three profits, the total profit is
P = y11 + 3y12 + 3y13 + 2y22 + 3y23 – y31 + 2y32 + 4y33
Consequently, the complete BIP model for the Supersuds problem can be formulated on a
spreadsheet as shown in Figure 7, where the changing cells Solution (D11:F13) display the values
of the auxiliary binary variables. Clicking on the Solve button then provides the optimal solution
shown in these changing cells in the figure, namely,
y11 = 0
y12 = 1
y13 = 0
so
x1 = 2
(allocate 2 TV spots to product 1)
y21 = 0
y22 = 0
y23 = 0
so
x2 = 0
(allocate 0 TV spots to product 2)
y31 = 0
y32 = 0
y33 = 1
so
x3 = 3
(allocate 3 TV spots to product 3)
which yields a profit of P = 7 ($7 million), according to the objective cell TotalProfit (I13). The
number of TV spots being allocated to the respective products is shown in NumberOfSpots
(D18:F18).
CD S7-1-17
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
B
C
D
E
F
H
I
Supersuds Corp. Marketing Plan
Profit
($millions)
Number
of
Spots
Solution
Number
of
Spots
1
2
3
1
2
3
Total
Max Of One
Product 1
1
3
3
Product 2
0
2
3
Product 3
-1
2
4
Product 1
0
1
0
1
<=
1
Product 2
0
0
0
0
<=
1
Product 3
0
0
1
1
<=
1
2
0
3
Number of Spots
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
14
C
D
Total =SUM(D11:D13)
Total
Profit
($millions)
7
Total
Spots
5
=
Range Name
MaxOfOne
NumberOfSpots
Profit
RequiredSpots
Solution
Total
TotalProfit
TotalSpots
Solver Parameters
Set Objective Cell: TotalProfit
To: Max
By Changing Variable Cells:
Solution
Subject to the Constraints:
Solution = binary
Total <= MaxOfOne
TotalSpots = RequiredSpots
18
G
Required
Spots
5
Cells
D16:F16
D18:F18
D5:F7
I18
D11:F13
D14:F14
I13
G18
I
10
Total
11
Profit
12
($millions)
13 =SUMPRODUCT(Profit,Solution)
E
=SUM(E11:E13)
F
=SUM(F11:F13)
C
D
E
Number of Spots =SUMPRODUCT(C11:C13,D11:D13) =SUMPRODUCT(C11:C13,E11:E13)
G
16
Total
17
Spots
18 =SUM(NumberOfSpots)
Figure 7
A spreadsheet formulation of the BIP model for the Supersuds problem, where
the optimal solution obtained by Solver is given in Solution (D11:F13), which
yields NumberOfSpots (D18:F:18).
CD S7-1-18
REVIEW QUESTIONS
1. What is the distinction between a binary decision variable and an auxiliary binary variable?
2. What is meant by mutually exclusive products?
3. How can an auxiliary binary variable be defined in terms of whether to allow the production
of a certain product?
4. How does an either-or constraint arise in Example 2?
5. When two individual constraints are paired together as either-or constraints, how can an
auxiliary binary variable be defined in terms of which one of these individual constraints is
chosen as the one that must hold?
6. How does restriction 1 for Example 3 relate to the restriction imposed in Example 1?
7. After introducing auxiliary binary variables for Example 3, what constraint on these variables
forces choosing at most two of the possible new products as the ones that can be produced?
8. When using the natural (integer) decision variables (x1, x2, x3) defined for Example 4, why is
it not possible to write a legitimate objective function in terms of these decision variables to
fit integer programming?
9. What are the groups of mutually exclusive alternatives that arise when introducing the
auxiliary binary variables for Example 4?
Glossary
auxiliary binary variable A binary variable that is introduced into the model, not to represent a
yes-or-no decision, but simply to help formulate the model as a (pure or mixed) BIP problem.
either-or constraints A pair of constraints such that either one can be chosen to be observed and
then the other one would be ignored.
CD S7-1-19
Problems
To the left of the problems (or their parts), we have inserted an E* whenever Excel should be
used (unless your instructor give you contrary instructions).
E* 7s.1. The research and Development Division of the Progressing Company has been developing four
possible new product lines. Management must now make a decision as to which of these four
products actually will be produced and at what levels. Therefore, a management science study has
been requested to find the most profitable product mix.
A substantial cost is associated with beginning the production of any product, as given in the first
row of the following table. Management’s objective is to find the product mix that maximizes the
total profit (total net revenue minus start-up costs).
Start-up cost
Marginal revenue
1
$50,000
70
Product
2
3
$40,000
$70,000
60
90
4
$60,000
80
Let the continuous decision variables x1, x2, x3, and x4 be the total number of units produced
of products 1, 2, 3, and 4, respectively. Management has imposed the following policy constraints
on these variables:
1. No more than two of the products can be produced.
2. Either product 3 or 4 can be produced only if either product 1 or 2 is produced.
3. Either 5x1
+
3x2
+
or
4x1 + 6x2 + 3x3 + 5x4 ≤ 6,000
6x3
+
4x4
≤
6,000
Use auxiliary binary variables to formulate and solve a mixed BIP model on a spreadsheet for this
problem.
E* 7s.2. The Toys-R-4-U Company has developed two new toys for possible inclusion in its product line
for the upcoming Christmas season. Setting up the production facilities to begin production would
cost $50,000 for toy 1 and $80,000 for toy 2. Once these costs are covered, the toys would
generate a unit profit of $10 for toy 1 and $15 for toy 2.
The company has two factories that are capable of producing these toys. However, to avoid
doubling the start-up costs, just one factory would be used, where the choice would be based on
maximizing profit. For administrative reasons, the same factory would be used for both new toys
if both are produced.
Toy 1 can be produced at a rate of 50 per hour in factory 1 and 40 per hour in factory 2. Toy 2
can be produced at the rate of 40 per hour in factory 1 and 25 per hour in factory 2. Factories 1
and 2, respectively, have 500 hours and 700 hours of production time available before Christmas
that could be used to produce these toys.
It is not known whether these two toys would be continued after Christmas. Therefore, the
problem is to determine how many units (if any) of each new toy should be produced before
Christmas to maximize the total profit. Formulate and solve a mixed BIP model on a spreadsheet
for this problem.
CD S7-1-20
E* 7s.3. Reconsider the Fly-Right Airplane Co. problem introduced in Problem 7.7. A more detailed
analysis of the various cost and revenue factors now has revealed that the potential profit from
producing airplanes for each customer cannot be expressed simply in terms of a start-up cost and
a fixed marginal net revenue per airplane produced. Instead, the profits are given by the following
table.
Airplanes Produced
0
1
2
3
4
5
Customer 1
0
–$1
2
4
Profit (Millions)
Customer 2
0
$1
5
Customer 3
0
$1
3
5
6
7
Use auxiliary binary variables to formulate and solve a BIP model on a spreadsheet for this new
version of the problem.
E* 7s.4. Reconsider Problem 3.6, where the management of the Omega Manufacturing Company is
considering devoting excess production capacity to one or more of three products. (See the Partial
Answers to Selected Problems in Appendix B in the back of the book for the optimal solution for
Problem 3.6.) Management now has decided to add the restriction that no more than two of the
three prospective products should be produced. Use auxiliary binary variables to formulate and
solve a mixed BIP model on a spreadsheet for this new version of the problem.
E* 7s.5. Consider the following algebraic form of an integer programming model:
Maximize Profit = 4x12 – x13 + 10x22 – x24
subject to
x1 + x2 ≤ 3
and
x1 ≥ 0
x2 ≥ 0
x1 and x2 are integers
E*
a. Reformulate this model in algebraic form as a pure BIP model with six binary
variables.
b. Display and solve this model on a spreadsheet.
c. Reexpress the optimal solution obtained in part b in terms of the variables, x1 and x2,
for the original model.