Readings
Readings
Chapter 7
Integer Linear Programming
BA 452 Lesson B.4 Binary Fixed Costs
1
Overview
Overview
BA 452 Lesson B.4 Binary Fixed Costs
2
Overview
Fixed Costs of Production are those production costs that are present
whenever production is positive. The simplest model of fixed costs in a linear
program restricts decision variables to binary (0 or 1).
Fixed Costs of Assignment are modeled like Fixed Costs of Production. The
simplest model of fixed costs restricts decision variables (the fraction of work
completed) to be binary.
Location Covering Problems are a special kind of Linear Programming problem
when outputs are fixed because the firm has only established customers, who
require coverage by service centers.
Worker Covering Problems are like Location Covering Problems, where
customers require scheduling workers so that at each time period is covered by
a prescribed minimal number of workers.
Transportation Problems with New Origins are Transportation Problems
extended so that new origins may be added, at a fixed cost. They choose the
best plant locations and how much to ship.
Transshipment Problems with New Nodes are Transshipment Problems
extended so that new transshipment nodes may be added, at a fixed cost. They
choose the best transshipment locations.
BA 452 Lesson B.4 Binary Fixed Costs
3
Tool Summary
Tool Summary
 Use binary variables to indicate whether an activity, such as a
production run, is undertaken.
 Write a multiple-choice constraint: The sum of two or more binary
variables equals 1, so any feasible solution choose one variable to
equal 1.
 Write a mutually-exclusive constraint: The sum of two or more binary
variables is at most 1, so any feasible solution chooses at most one
variable to equal 1. All variables could equal 0.
 Write a conditional constraint: An inequality constraint so that one
binary variable cannot equal unless certain other binary variables
also equal 1.
 Write a corequisite constraint: An equality constraint of binary
variables, so are either both 0 or both 1.
BA 452 Lesson B.4 Binary Fixed Costs
4
Fixed Cost of Production
Fixed Cost of Production
BA 452 Lesson B.4 Binary Fixed Costs
5
Fixed Cost of Production
Overview
Fixed Costs of Production are those production costs that
are present whenever production is positive. The simplest
way to model fixed costs in a linear program is to restrict
decision variables to be binary (0 or 1).
For example, suppose the cost of producing quantity x is
5x. On the one hand, if x can take on any non-negative
value (like x is the pounds of hamburger produced), then 5
is the constant unit cost of production. On the other hand, if
x can take on only binary values (like x is the number of
new books adopted), then cost is 0 if there is no production
and 5 if there is positive production, so 5 is the fixed cost of
production.
BA 452 Lesson B.4 Binary Fixed Costs
6
Fixed Cost of Production
Question: W. W. Norton & Company, the oldest and
largest publishing company wholly owned by its
employees, must decide which new textbooks to adopt
next year. The books considered are described along
with their expected three-year sales:
Book
Projected Sales
Business Calculus
20
Finite Math
30
General Statistics
15
Mathematical Statistics
10
Business Statistics
25
Finance
18
Financial Accounting
25
Managerial Accounting
50
English Literature
20
German
30
BA 452 Lesson B.4 Binary Fixed Costs
7
Fixed Cost of Production
Three individuals in the company can be assigned to
these projects, all of whom have varying amounts of
time available. John has 60 days, Susan has 52 days,
and Monica has 43 days. The days required by each
person to complete each project are showing in the
following table. For example, if the business calculus
book is published, it will require 30 days of John’s time
and 40 days of Susan’s time.
Book
Projected Sales
John
Susan
Monica
Business Calculus
20
30
40
X
Finite Math
30
16
24
X
General Statistics
15
24
X
30
Mathematical Statistics
10
20
X
24
Business Statistics
25
10
X
16
Finance
18
X
X
14
Financial Accounting
25
X
24
26
Managerial Accounting
50
X
28
30
English Literature
20
40
34
30
German
30
x
50
36
BA 452 Lesson B.4 Binary Fixed Costs
8
Fixed Cost of Production
Norton will not publish more than two statistics books
or more than one accounting book in a single year. In
addition, one of the math books (business calculus or
finite math) must be published, but not both. Which
books should be published, and what are the projected
sales? If Monica has 1 more day available, which
books should be published, and what are the projected
sales? Comment.
BA 452 Lesson B.4 Binary Fixed Costs
9
Fixed Cost of Production
i
Book
1
Business Calculus
2
Finite Math
3
General Statistics
4
Mathematical Statistics
5
Business Statistics
6
Finance
7
Financial Accounting
8
Managerial Accounting
9
English Literature
10
German
BA 452 Lesson B.4 Binary Fixed Costs
10
Fixed Cost of Production
Max
s.t.
20x1
30x1
40x1
x1
i
Book
Projected Sales
John
Susan
Monica
1
Business Calculus
20
30
40
X
2
Finite Math
30
16
24
X
3
General Statistics
15
24
X
30
4
Mathematical Statistics
10
20
X
24
5
Business Statistics
25
10
X
16
6
Finance
18
X
X
14
7
Financial Accounting
25
X
24
26
8
Managerial Accounting
50
X
28
30
9
English Literature
20
40
34
30
10
German
30
x
50
36
+
30x2
+
15x3
+
10x4
+
25x5
+
+
16x2
24x2
+
24x3
+
20x4
+
10x5
30x3
x3
+
+
24x4
x4
+
+
16x5
x5
+
+
+
18x6
14x6
+
+
+
25x7
+
50x8
+
20x9
+
+
+
40x9
34x9
30x9
24x7
26x7
+
+
28x8
30x8
x7
+
x8
+
+
+
x2
BA 452 Lesson B.4 Binary Fixed Costs
30x10
50x10
36x10





=
60
52
43
2
1
1
John
Susan
Monica
No. of Stat Books
Account Book
Math Book
11
Fixed Cost of Production
Max
s.t.
20x1
30x1
40x1
+
30x2
+
15x3
+
10x4
+
25x5
+
+
16x2
24x2
+
24x3
+
20x4
+
10x5
30x3
x3
x1
+
+
+
24x4
x4
+
+
16x5
x5
+
+
18x6
+
25x7
+
50x8
+
20x9
+
30x10
14x6
+
+
24x7
26x7
+
+
28x8
30x8
+
+
+
40x9
34x9
30x9
+
+
50x10
36x10
x7
+
x8
x2
BA 452 Lesson B.4 Binary Fixed Costs





=
60
52
43
2
1
1
John
Susan
Monica
No. of Stat Books
Account Book
Math Book
12
Fixed Cost of Production
i
Book
1
Business Calculus
2
Finite Math
3
General Statistics
4
Mathematical Statistics
5
Business Statistics
6
Finance
7
Financial Accounting
8
Managerial Accounting
9
English Literature
10
German
Publish finite math, business statistics, and financial
accounting. Projected sales are 80,000.
BA 452 Lesson B.4 Binary Fixed Costs
13
Fixed Cost of Production
i
Book
1
Business Calculus
2
Finite Math
3
General Statistics
4
Mathematical Statistics
5
Business Statistics
6
Finance
7
Financial Accounting
8
Managerial Accounting
9
English Literature
10
German
Publish finite math, business statistics, and financial
accounting. Projected sales are 80,000.
BA 452 Lesson B.4 Binary Fixed Costs
14
Fixed Cost of Production
If Monica is available 1 more
day (44 days total), optimal
projected sales are now
98,000. A big (discrete) gain
for a small increase in a
constraint.
BA 452 Lesson B.4 Binary Fixed Costs
15
Fixed Cost of Assignment
Fixed Cost of Assignment
BA 452 Lesson B.4 Binary Fixed Costs
16
Fixed Cost of Assignment
Overview
Fixed Costs of Assignment are modeled like Fixed Costs of
Production. The simplest model of fixed costs restricts
decision variables to be binary.
For example, suppose the cost of worker i performing the
fraction xij of job j is 5xij. On the one hand, if xij can take on
any non-negative value between 0 and 1 (like xij is the
fraction of the job performed), then 5 is the constant unit
cost of assignment (like the time cost of working). On the
other hand, if xij can take on only binary values, then cost is
0 if you are not assigned (xij = 0) and 5 if you are assigned
the entire job (xij = 1), so 5 is the fixed cost of assignment
(like the cost of commuting to a job across town).
BA 452 Lesson B.4 Binary Fixed Costs
17
Fixed Cost of Assignment
Question: 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
2
23
14
8
20
Tailor
3 4 5
20 21 18
X 12 10
11 X 9
20 18 21
BA 452 Lesson B.4 Binary Fixed Costs
18
Fixed Cost of Assignment
Formulate an integer program for determining the tailorgarment 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 particular problem can be formulated as either a
binary program or as an integer program. Any feasible
solution to the latter program is binary (0-1).
BA 452 Lesson B.4 Binary Fixed Costs
19
Fixed Cost of Assignment
Answer: Define the decision variables
xij = 1 if garment i is assigned to tailor j
= 0 otherwise.
 Find the number of decision variables
= [(number of garments)x(number of tailors)]
- (number of unacceptable assignments)
= [4x5] - 3 = 17
 Find the number of constraints. 1 for each garment and each tailor = 9.
BA 452 Lesson B.4 Binary Fixed Costs
20
Fixed Cost of Assignment

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
BA 452 Lesson B.4 Binary Fixed Costs
21
Fixed Cost of Assignment

Define the constraints of 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
BA 452 Lesson B.4 Binary Fixed Costs
22
Fixed Cost of Assignment

Define the constraints of 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
BA 452 Lesson B.4 Binary Fixed Costs
23
Fixed Cost of Assignment
Minimum time: 55 hours
Optimal assignments:
Garment
Wedding gown
Clown costume
Admiral's uniform
Bullfighter's outfit
BA 452 Lesson B.4 Binary Fixed Costs
1
19
11
12
X
Tailor
2 3 4 5
23 20 21 18
14 X 12 10
8 11 X 9
20 20 18 21
24
Location Covering
Location Covering
BA 452 Lesson B.4 Binary Fixed Costs
25
Location Covering
Overview
Location Covering Problems are a special kind of Linear
Programming problem when outputs are fixed because the
firm has only established customers. Commitments to
established customers require building service centers so
that each customer area is covered by a prescribed
minimal number of service centers. The objective is to
minimize the cost of building service centers. If each
center is equally costly, the objective reduces to minimizing
the number of service centers. The simplest way to model
the all-or-nothing decision to build or not build a serve
center is to restrict the fraction of the center built to be a
binary (0 or 1) decision variable.
BA 452 Lesson B.4 Binary Fixed Costs
26
Location Covering
Question: UPS is drawing up new zones for the location of
drop boxes for customers. The city has been divided into
the four zones shown below. You have targeted six
possible locations for drop boxes (numbered 1 through 6).
The list of which drop boxes could be reached easily from
each zone is listed below.
Zone
Can be Served by Drop Box Locations:
Downtown
Financial
1, 2, 5, 6
Downtown Legal
2, 4, 5
Retail South
1, 2, 4, 6
Retail North
3, 4, 5
BA 452 Lesson B.4 Binary Fixed Costs
27
Location Covering
Formulate and solve a model to provide the fewest dropbox locations yet make sure that each zone is covered by
at least two boxes.
BA 452 Lesson B.4 Binary Fixed Costs
28
Location Covering
Answer: Min x1 + x2 + x3 + x4 + x5 + x6
s.t.
x1 + x2 + x5 + x6 > 2
x2 + x4 + x5 > 2
x1 + x2 + x4 + x6 > 2
x3 + x4 + x5 > 2
BA 452 Lesson B.4 Binary Fixed Costs
29
Worker Covering
Worker Covering
BA 452 Lesson B.4 Binary Fixed Costs
30
Worker Covering
Overview
Worker Covering Problems are like Location Covering
Problems. Outputs are fixed because the firm has only
established customers. Commitments to established
customers require scheduling workers so that at each time
period customer needs are covered by a prescribed
minimal number of workers. The objective is to minimize
the cost of scheduling workers. If each worker is equally
costly, the objective reduces to minimizing the number of
workers scheduled. The simplest way to model the all-ornothing decision to schedule a worker is to restrict the
number of workers scheduled at each time period to be an
integer decision variable.
BA 452 Lesson B.4 Binary Fixed Costs
31
Worker Covering
Question: Amazon.com is open 24 hours a day. The
number of phone operators need in each four hour period
of a day is listed below.
Period
Operators Needed
10 p.m. to 2 a.m.
8
2 a.m. to 6 a.m.
4
6 a.m. to 10 a.m.
7
10 a.m. to 2 p.m.
12
2 p.m. to 6 p.m.
10
6 p.m. to 10 p.m.
15
BA 452 Lesson B.4 Binary Fixed Costs
32
Worker Covering
Suppose operators work for eight consecutive hours.
Formulate and solve the company’s problem of determining
how many operators should be scheduled to begin working
in each period in order to minimize the number of cashiers
needed? (Hint: Workers can work from 6 p.m. to 2 a.m.)
BA 452 Lesson B.4 Binary Fixed Costs
33
Worker Covering
Answer: Define the decision variables
TNP = the number of operators who begin working at 10 p.m.
TWA = the number of operators who begin working at 2 a.m.
SXA = the number of operators who begin working at 6 a.m.
TNA = the number of operators who begin working at 10 a.m.
TWP = the number of operators who begin working at 2 p.m.
SXP = the number of operators who begin working at 6 p.m.
Min TNP + TWA + SXA + TNA + TWP + SXP
s.t.TNP + TWA > 4
TWA + SXA > 7
SXA + TNA > 12
TNA + TWP > 10
TWP + SXP > 15
SXP + TNP > 8, all variables > 0
BA 452 Lesson B.4 Binary Fixed Costs
34
Worker Covering
BA 452 Lesson B.4 Binary Fixed Costs
35
Transportation with New Origins
Transportation with New Origins
BA 452 Lesson B.4 Binary Fixed Costs
36
Transportation with New Origins
Transportation Problems with New Origins are
Transportation Problems extended so that new origins may
be added, at a fixed cost. Also called Distribution System
Design Problems, they choose the best plant locations and
determine how much to ship from each plant. The simplest
way to model those fixed costs in a linear program is with
binary (0 or 1) variables.
For example, consider a Transportation Problem extended
by allowing the possibility of developing a new origin.
Suppose the fixed development cost is 5 and the new
origin’s supply capacity is 7. That is a linear program with a
supply of 7Y at the new origin and added cost term of 5Y,
where Y = 0 indicates the new origin is not developed and
Y = 1 indicates the new origin is developed.
BA 452 Lesson B.4 Binary Fixed Costs
37
Transportation with New Origins
Question: Linksys operates a plant to produce its wireless
routers in St. Louis with an annual capacity of 30,000 units.
Product is shipped to regional distribution centers in
Boston, Atlanta, and Houston. Because of an anticipated
increase in demand, Linksys plans to increase 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 annual capacity for the
four proposed plants are as follows:
Proposed Plant
Annual Fixed Cost
Annual Capacity
Detroit
$175,000
10,000
Toledo
$300,000
20,000
Denver
$375,000
30,000
Kansas City
$500,000
40,000
BA 452 Lesson B.4 Binary Fixed Costs
38
Transportation with New Origins
The company’s long-range planning group forecasts of the
anticipated annual demand at the distribution centers are
as follows:
Distribution Center
Annual Demand
Boston
30,000
Atlanta
20,000
Houston
20,000
BA 452 Lesson B.4 Binary Fixed Costs
39
Transportation with New Origins
The shipping cost per unit from each plant to each
distribution center is as follows:
Plant\Distribution
Boston
Atlanta
Houston
Detroit
$5
$2
$3
Toledo
$4
$3
$4
Denver
$9
$7
$5
Kansas City
$10
$4
$2
St. Louis
$8
$4
$3
Formulate and solve the problem of minimizing the cost of
meeting all demands.
BA 452 Lesson B.4 Binary Fixed Costs
40
Transportation with New Origins
Answer:
Define binary variables for plant construction,
Y1 = 1 if a plan is constructed in Detroit; 0, if not
Y2 = 1 if a plan is constructed in Toledo; 0, if not
Y3 = 1 if a plan is constructed in Denver; 0, if not
Y4 = 1 if a plan is constructed in Kansas City; 0, if not
Define shipment variables just as in transportation
problems,
Xij = the units shipped (in thousands) from plant i (i = 1, 2,
3, 4, 5) to distribution center j (j = 1, 2, 3) each year.
BA 452 Lesson B.4 Binary Fixed Costs
41
Transportation with New Origins
The objective is minimize total cost.
From cost data
Plant\Distribution
Boston
Atlanta
Houston
Detroit
$5
$2
$3
Toledo
$4
$3
$4
Denver
$9
$7
$5
Kansas City
$10
$4
$2
St. Louis
$8
$4
$3
shipping costs (in thousands of dollars) are
5X11 + 2X12 + 3X13+ 4X21 + 3X22 + 4X23 + 9X31 + 7X32
+ 5X33 + 10X41 + 4X42 + 2X43 + 8X51 + 4X52 + 3X53
BA 452 Lesson B.4 Binary Fixed Costs
42
Transportation with New Origins
From cost data
Proposed Plant
Annual Fixed Cost
Annual Capacity
Detroit
$175,000
10,000
Toledo
$300,000
20,000
Denver
$375,000
30,000
Kansas City
$500,000
40,000
plant construction costs (in thousands of dollars) are
175Y1 + 300Y2 + 375Y3 + 500Y4
Hence, the objective to minimize total costs is
Min 5X11 + 2X12 + 3X13+ 4X21 + 3X22 + 4X23 + 9X31 +
7X32 + 5X33 + 10X41 + 4X42 + 2X43 + 8X51 + 4X52 +
3X53 + 175Y1 + 300Y2 + 375Y3 + 500Y4
BA 452 Lesson B.4 Binary Fixed Costs
43
Transportation with New Origins
From capacity data
Proposed Plant
Annual Fixed Cost
Annual Capacity
Detroit
$175,000
10,000
Toledo
$300,000
20,000
Denver
$375,000
30,000
Kansas City
$500,000
40,000
Detroit capacity constraint is X11 + X12 + X13 < 10Y1
Toledo capacity constraint is X21 + X22 + X23 < 20Y2
Denver capacity constraint is X31 + X32 + X33 < 30Y3
Kansas City capacity constraint is X41 + X42 + X43 < 40Y4
And St. Louis capacity constraint is X51 + X52 + X53 < 30
BA 452 Lesson B.4 Binary Fixed Costs
44
Transportation with New Origins
From demand data
Distribution Center
Annual Demand
Boston
30,000
Atlanta
20,000
Houston
20,000
Boston demand constraint is
X11 + X21 + X31 + X41 + X51 = 30
Atlanta demand constraint is
X12 + X22 + X32 + X42 + X52 = 20
Houston demand constraint is
X13 + X23 + X33 + X43 + X53 = 20
Non-negativity constraints complete the linear
programming formulation.
BA 452 Lesson B.4 Binary Fixed Costs
45
Transportation with New Origins
From demand data
Distribution Center
Annual Demand
Boston
30,000
Atlanta
20,000
Houston
20,000
Boston demand constraint is
X11 + X21 + X31 + X41 + X51 = 30
Atlanta demand constraint is
X12 + X22 + X32 + X42 + X52 = 20
Houston demand constraint is
X13 + X23 + X33 + X43 + X53 = 20
Non-negativity constraints complete the linear
programming formulation.
BA 452 Lesson B.4 Binary Fixed Costs
46
Transportation with New Origins
The Management Scientist solves this mixed integer linear
program of 4 binary variables Yj and 15 continuous
variables Xij, and 8 constraints.
BA 452 Lesson B.4 Binary Fixed Costs
47
Transportation with New Origins
BA 452 Lesson B.4 Binary Fixed Costs
48
Transportation with New Origins
The Management Scientist
solves this mixed integer
linear program of 4 binary
variables Yj and 15
continuous variables Xij, and
8 constraints.
BA 452 Lesson B.4 Binary Fixed Costs
49
Transportation with New Origins
All variables at the optimum are
zero except: X42 = 20, X43 =
20, X51 = 30, and Y4 = 1.
So, the Kansas City plant
should be built;
20,000 units should be shipped
from Kansas City to Atlanta;
20,000 units should be shipped
from Kansas City to Houston;
and 30,000 units should be
shipped from St. Louis to
Boston.
BA 452 Lesson B.4 Binary Fixed Costs
50
Transshipment with New Nodes
Transshipment with New Nodes
BA 452 Lesson B.4 Binary Fixed Costs
51
Transshipment with New Nodes
Overview
Transshipment Problems with New Transshipment Nodes
are Transshipment Problems extended so that new
transshipment nodes may be added, at a fixed cost. They
choose the best transshipment locations and determine
how much to ship through each location. The simplest way
to model those fixed costs in a linear program is with binary
(0 or 1) variables.
BA 452 Lesson B.4 Binary Fixed Costs
52
Transshipment with New Nodes
Question: Zeron Industries supplies three firms (Zrox, Hewes, Rockrite)
with customized shelving for its offices. Zeron orders shelving from the
same two manufacturers, Arnold Manufacturers and Supershelf, Inc.
Currently, weekly demands by the users are 50 for Zrox, 60 for Hewes,
and 40 for Rockrite. Both Arnold and Supershelf can supply up to 75
units to its customers.
Zeron currently ships from its Northside facilities, but it can develop
Southside facilities for a fixed cost of 8. Unit costs from the
manufacturers to the suppliers are:
Zeron N Zeron S
Arnold
5
8
Supershelf
7
4
The costs to install the shelving at the various locations are:
Zrox Hewes Rockrite
Zeron N
1
5
8
Zeron S
3
4
4
BA 452 Lesson B.4 Binary Fixed Costs
53
Transshipment with New Nodes
75
ARNOLD
Arnold
5
Zeron
N
8
75
4
50
Hewes
HEWES
60
RockRite
40
1
5
8
3
7
Super
Shelf
Zrox
Zeron
WASH
BURN
S
4
4
BA 452 Lesson B.4 Binary Fixed Costs
54
Transshipment with New Nodes

Define decision variables:
xij = amount shipped from manufacturer i to supplier j
xjk = amount shipped from supplier j to customer k
where
i = 1 (Arnold), 2 (Supershelf)
j = 3 (Zeron N), 4 (Zeron S)
k = 5 (Zrox), 6 (Hewes), 7 (Rockrite)
y = 1 if Zeron S is developed, y = 0 if Zeron S is not
developed
BA 452 Lesson B.4 Binary Fixed Costs
55
Transshipment with New Nodes









Define objective function: Minimize total shipping costs plus setup
costs.
Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37 + 3x45 + 4x46 +
4x47 + 8y
Constrain amount out of Arnold:
x13 + x14 < 75
Constrain amount out of Supershelf: x23 + x24 < 75
Constrain amount through Zeron N: x13 + x23 - x35 - x36 - x37 = 0
Constrain amount through Zeron S: x14 + x24 - x45 - x46 - x47 = 0
Constrain amount into Zrox:
x35 + x45 = 50
Constrain amount into Hewes:
x36 + x46 = 60
Constrain amount into Rockrite:
x37 + x47 = 40
Setup indicator for Zeron S:
x45 + x46 + x47 < 150y
(The first 4 constraints imply x45 + x46 + x47 < 150, so the setup indicator
constraint “x45 + x46 + x47 < 150y” means, at an optimum, y = 1 if any
material x45 or x46 or x47 transships through Zeron S, and y = 0 if no
material x45 or x46 or x47 transships through Zeron S.)
BA 452 Lesson B.4 Binary Fixed Costs
56
BA 452
Quantitative Analysis
End of Lesson B.4
BA 452 Lesson B.4 Binary Fixed Costs
57