II
6I
INT RDCI T
LIEA PRGAMN
SION EDITIO
. .. .. .. R ST NSB RY TOC TO
Peter E. Hildebrand
Agricultural Economics
INTRODUCTION
TO
LINEAR
PROGRAMMING
EDITION
SECOND
INTRODUCTION
ALLYN
AND BACON,
BOSTON
TO
INC.
1963
LINEAR PROGRAMMING
SECOND EDITION
R. STANSBURY STOCKTON
Professor and Chairman, Product Management Area,
Graduate School of Business
Indiana University
Copyright 1963 by Allyn and
Bacon, Inc., 150 Tremont Street,
Boston. All rights reserved. No
part of this book may be repro
duced in any form, by mimeo
graph or any other means, with
out permission in writing from the
publisher.
Library of Congress Catalog Card
Number: 6319120
Printed in the United States of
America
PREFACE
DURING THE THREE YEARS SINCE THE FIRST PUBLICATION OF THIS
book linear programming has become an integral part of many
courses in Schools of Business. The main issue now is not whether
it should be taught but how much, where, and in what manner.
Consequently, the objective of this edition, like the original, is to
introduce students of business administration to linear program
ming methods.
The increasing mathematical competence of business students at
both undergraduate and graduate levels has made it possible to
raise our sights in many fields. Special courses and programs of
study which stress the application of modern mathematics and sta
tistics are no longer unusual. Such sources and programs are, how
ever, appropriate only for a minority. The educational objective
for most business students should be to develop a general adminis
trative understanding of programming methods. Introduction to
Linear Programming is designed to meet the need for suitable study
materials to accomplish this limited objective.
Every effort has been made to explain programming methods in
simple terms which are familiar to the business student. The ratio
of words to symbols is somewhat higher than in most programming
texts as a consequence. The desire to keep the text at the basic
level has made it necessary to exclude many interesting aspects of
linear programming. The student who wishes to develop special
skills in the field will find the selected bibliography helpful in select
ing advanced material in the field.
The descriptive material in Chapter 1 has been expanded in this
edition so that the reader has a better understanding of the types
of problems to which linear programming may be applied. The
chapter on the transportation or distribution method has been re
written and now includes Northwest Corner solutions, the MODI
VI PREFACE
method, and methods for handling degenerate solutions. In addi
tion, the number of exercises has been increased so that many new
problems are included at the end of each chapter.
I am especially indebted to Professor Rocco Carzo of The Penn
sylvania State University for his suggestions on the revision and to
Mrs. Karen Strom who has typed most of the manuscript.
R. Stansbury Stockton
CONTENTS
1
LINEAR PROGRAMMING AND THE
DECISION PROCESS 1
THE MANAGER AND LINEAR PROGRAMMING 1
AREAS OF APPLICATION 3
BASIC ASSUMPTIONS 5
LINEAR PROGRAMMING METHODS 6
COMPUTERS AND LINEAR PROGRAMMING 6
2
LINEARITY AND LINEAR EQUATIONS 8
LINEARITY AND PROBLEMSOLVING 8
SYMBOLIC EXPRESSIONS FOR RELATIONSHIPS
BETWEEN VARIABLES 9
MEANING OF LINEARITY 9
LINEAR EQUATIONS 12
DETERMINATION OF THE EQUATION OF A LINE 12
vii
CONTENTS
DETERMINATION OF INTERCEPTS 13
EXERCISES 16
3
THE GRAPHICAL METHOD 18
PROCEDURE 18
TECHNICAL PHASES OF THE PROBLEM 19
ECONOMIC PHASES OF THE PROBLEM 25
DETERMINATION OF THE OPTIMAL SOLUTION 27
THE HUMAN FACTOR 33
SUMMARY 34
REVIEW QUESTIONS 35
EXERCISES 36
4
THE SIMPLEX METHOD 40
GENERAL NATURE OF THE METHOD 40
COMPUTATION PROCEDURE 42
PRODUCTMIX EXAMPLE 42
SUMMARY 64
REVIEW QUESTIONS 66
EXERCISES 67
CONTENTS
5
THE TRANSPORTATION METHOD 69
GENERAL NATURE OF THE METHOD 69
STEP 1: FRAME THE PROBLEM 72
STEP 2: DEVELOP AN INITIAL SOLUTION 75
STEP 3: TEST THE SOLUTION FOR DEGENERACY 82
STEP 4: EVALUATE UNUSED CELLS 83
STEP 5: DEVELOP REVISED MATRIX 92
METHOD FOR HANDLING DEGENERATE SOLUTIONS 100
SUMMARY 105
REVIEW QUESTIONS 105
EXERCISES 105
SELECTED BIBLIOGRAPHY
I LINEAR PROGRAMMING AND
THE DECISION PROCESS1
1.1 THE MANAGER AND LINEAR PROGRAMMING
LINEAR PROGRAMMING is one of a number of recently developed
analytical techniques that have proved useful in solving certain
types of business problems. These quantitative methods of prob
lemsolving, like many employed in operations research, are
based upon mathematical and statistical concepts. Therefore,
they present a problem to the business executive who finds
it difficult to translate what appears to be mathematical "jibber
ish" into meaningful terms. Under such circumstances, the busi
nessman tends to equate his inability to understand with imprac
ticality and, as a consequence, to dismiss these new techniques
as a plot by mathematicians to confuse and befuddle business
men! There is evidently some truth in the adage that "People
1 Much of the material in this chapter is drawn from the author's article
"Linear Programming and Management," Business Horizons, Vol. 3, No. 2,
Summer, 1960.
2 LINEAR PROGRAMMING
would rather live with problems which they cannot solve than
with solutions to those problems which they cannot understand."
Publications in the field of management science con
tain an increasing number and variety of business applications of
these analytical methods, including linear programming. Fur
thermore, the number and scope of these applications can be ex
pected to increase in the future as variations and refinements in the
methods are developed. The question is how they can be ap
plied in a specific business.
Much of the credit for demonstrating the applicability of
linear programming methods to business problems belongs to
those people engaged in operations research. However, it is not
necessary to be in operations research to learn the fundamentals of
the subject. In fact, lack of understanding on the part of line man
agers appears to be a major factor limiting both present and future
applications of many potentially useful explicit methods of analy
sis.
Inasmuch as linear programming represents a type of
"model," one appropriate method of study would be to place it
within the broader framework of the managerial decisionmaking
process. Most statements of the latter are, adaptations of the so
called scientific method. One of the justifications for the study
of any quantitative method of analysis is that these methods illus
trate all steps or phases of the decisionmaking process on an
explicit basis. The assumption here is that the decisionmaking
skills developed through the use of explicit exercises will be ap
plicable to other problems, including those in which many factors
are essentially qualitative.
A second reason for studying linear programming is that
a general or "administrative" understanding of the method is a
necessary prerequisite of effective use of the technique within
any organization. Line managers as differentiated from staff
specialists or analysts play an important role in the initial and
final stages of problemsolving projects, that is, in the formula
tion of the problem and in the evaluation and application of
the findings. They must, as a consequence, be capable of effec
tive communication with any staff specialist, including the opera
tions analyst. A more specific objective in the study of linear pro
1.2 AREAS OF APPLICATION 3
gramming might, therefore, be stated as the development of
sufficient insight into the method to enable one to (1) recognize
problems that might be subjected to analysis by the method, (2)
assist the analyst in the initial stage of the investigation, (3)
evaluate and interpret the results intelligently, and (4) apply
the results with the confidence that comes only with some under
standing of the "whys" as well as the "whats" involved.
1.2 AREAS OF APPLICATION
Linear programming methods may chiefly be applied to
the general class of problems known as allocation problems. Econ
omists have 'traditionally defined such problems as those involv
ing the allocation of scarce resources among alternative ends ac
cording to some criterion. Scarce resources for the business firm
include capital, personnel, equipment, and materials. The various
products and/or services that constitute the output of the firm
represent alternative ends to which resources must be allocated.
The criterion or objective, on the basis of which allocation deci
sions are to be made, may be some form of profit maximization or
any other appropriate measure of desired performance.
Business managers always have been and always will be
confronted with allocation decisions. The methods of analysis
used to resolve these problems can and should be varied. For
example, decisions based upon judgment and intuition may be
satisfactory where the number of factors in the problem is lim
ited and their relationships are clear. More difficult problems
may require preliminary data collection, followed by the applica
tion of some formal method of analysis such as those character
istic of industrial engineering. The adequacy of these standard
techniques falls off rapidly, however, as the number of variables in
the problem increases. Linear programming is most appropriate
for complex allocation problems that cannot be handled satis
factorily with conventional analytical techniques.
Many types of allocation problems are found in business,
4 LINEAR PROGRAMMING
especially in the production or operations function. Some examples
to which linear programming methods have been successfully ap
plied are:
Determination of product mix. The types and quantities
of products to be manufactured during the next planning period
must be determined. The relative profitability of the items in the
product line varies. The planned product mix must take into
consideration expected demand, the capability and capacities of
production and distribution facilities, and management policies,
such as policy on products carried to "round out the line." Given
these limitations or restrictions on the mix, the solution should also
make maximum economic sense; that is, it should maximize
profits.
Blending or mixing problems. One or more products are
manufactured by mixing or blending various ingredients; for ex
ample, paint, cattle feed, petroleum products. Many different com
binations of these ingredients can result in end products that will
meet all technical specifications. Given the availability and relative
costs of the ingredients, which blend will result in minimum
material cost per unit of end product?
Production scheduling and inventory planning. Given
a seasonal demand and limited production facilities, what pro
duction schedule and planned inventory levels over the next plan
ning period will meet expected demand and also result in mini
mum cost?
Machine loading. A series of orders is to be processed
through a group of machines. The cost of processing each order
depends in part on the particular machine to which it is as
signed. Limited machine capacity precludes assignment of each
order to the lowestcost equipment. What allocations of available
capacity to the series of orders will result in minimum total proc
essing cost?
Shipping and physical distribution problems. Goods are to
be shipped from several production facilities having limited capac
ities to field warehouses that anticipate a given demand over the
next planning period. Transportation and/or production costs
vary among the alternative methods of supplying the warehouses.
What physical distribution pattern will be both within the capacity
1.3 BASIC ASSUMPTIONS 5
and demand restrictions and will at the same time minimize total
production and distribution costs over the planning period?
1.3 BASIC ASSUMPTIONS
All explicit analytical methods are based upon certain
assumptions. Just as one must understand certain accounting
conventions to use accounting statements intelligently, one must
understand the basic assumption of any quantitative analytical
tool if it is to be applied properly. The two central assumptions
in all linear programming methods are (1) linearity, and (2) cer
tainty. Linearity means that all problem relationships can be
expressed in the form of linear equations. The straightline
method of depreciation assumes that capital value decreases at
a linear rate. Breakeven charts assume that both variable costs
and revenue are linear functions; that is, that they increase in
direct proportion to output. The term certainty indicates that no
significant variations are expected in the numerical value for a
problem factor. Average cost, for example, may be used in mak
ing a decision even though actual cost may vary slightly from this
average. Statistical quality control, on the other hand, is based
upon the expected variations in a process within the control limits,
i.e., a degree of uncertainty exists. Linear programming is an ap
propriate analytical tool only for those complex allocation prob
lems that are characterized by both linearity and certainty.
An understanding of linear functions is so fundamental
to understanding linear programming that Chapter 2 is devoted
to a general review of this subject. The exact meaning and sig
nificance of linearity and certainty in a programming context will
become apparent in the explanations of the various methods. It
should be noted that, by reducing the mathematical complexity of
the methods, these assumptions make linear programming a good
starting point for the beginning student of quantitative methods of
analysis. The ultimate payoff, of course, is in solving real prob
lems. The usefulness of linear programming, like any other method
6 LINEAR PROGRAMMING
of analysis, is dependent in large part on the reasonableness of the
assumptions about the problem being studied.
1.4 LINEAR PROGRAMMING METHODS
The mathematical computational procedures of linear
programming depend in part on which of the several program
ming methods is adopted for a particular problem. The basic or
general case is called the Simplex method, since it is based upon
the simplex algorithm. Certain types of allocation problems may
be solved by special, less complex, versions of the Simplex method
known as the Graphical and Transportation methods. In addition,
there are a number of variations such as the socalled Index
method, which yields only approximate solutions but has mini
mum computational requirements. All methods are, in effect,
nothing more than effective search procedures that seek optimal
solutions to allocation problems in which there are more un
knowns than linear equations. In this sense, linear programming
may be viewed as systematic trial and error in which the most
promising shortcuts to a solution are indicated by mathematics
rather than intuition and elbow grease.
1.5 COMPUTERS AND LINEAR PROGRAMMING
Formulating a business problem for solution by any
linear programming method requires setting up a large number of
linear equations. Once the problem is framed, an initial solution is
determined. A set of operating rules is then applied to determine
if a better solution exists and, if so, to develop an improved solu
tion. The solution process is iterative; that is, the rules are ap
plied again and again until such time as an optimal solution is
found.
1.5 COMPUTERS AND LINEAR PROGRAMMING 7
The fact that the computational rules and procedures for
linear programming are so well defined makes this a natural ap
plication for modern electronic computers. Standard programs
are available for linear programming on nearly all computer hard
ware today so that the burdensome task of computer program
ming is less of an obstacle than in earlier years. Problems of lim
ited complexity, like those included in subsequent chapters, may
be solved efficiently on paper. However, as the number of vari
ables in the problem increases, the magnitude of the computational
task increases sharply. For this reason the use of expensive com
puter time can be justified in the analysis of most complex busi
ness problems.
2 LINEARITY AND LINEAR
EQUATIONS
2.1 LINEARITY AND PROBLEMSOLVING
THE CONCEPT of linearity is an abstraction that is frequently
used in the analysis of certain problems found in life. Many
personal decisions involve the assumption of linear relationships
even though we may not be conscious of these assumptions. From
a practical standpoint, as long as decisions made on the basis of an
informal, intuitive, and highly personalized analysis "work," there
is little to be gained by questioning the assumptions. In business,
however, problemsolving often requires formal methods of analy
sis. Many individuals may be involved both in the analysis of the
problem and in the application of the results. Where this is
the case, it is important that assumptions concerning the relation
ships between factors in the problem be made explicit.
Inasmuch as linearity is one of the basic assumptions, a
general knowledge of the concept is useful to the student of busi
ness. One should keep in mind, however, that the concept of
linearity has many applications above and beyond those found
in linear programming.
2.2 SYMBOLIC EXPRESSIONS 9
2.2 SYMBOLIC EXPRESSIONS FOR RELATIONSHIPS
BETWEEN VARIABLES
Suppose that observation of certain phenomena convinces
one that there is some relationship between factors X and Y.
One method of expressing this relationship would be to describe
it with words, that is, with a verbal model. For example, "it
appears that the value of Y generally depends upon the value of
X." Provided that a very general statement of this type is suf
ficient, words are an acceptable tool to use in describing the rela
tionship between the variables. However, as the number of
variables increases and their interrelationships become more com
plex, words become cumbersome and inadequate tools for model
building. Fortunately, this is a problem that mathematicians and
statisticians solved long ago. The use of symbolic expressions
is a second and more precise method for expressing relation
ships. Thus, the mathematician would simply write: Y = F (X).
These statements may be read as "the value of Y is a function of
(depends on) the value of X" or more simply as "Y is a function
of X." By convention, X is called the independent variable and
Y the dependent variable, since the value of the latter depends
on the value of the former.
Although the expression Y= F(X) represents a notational
improvement over the verbal description, it remains very general
in the sense that we still do not know the exact nature of the rela
tionship between the two variables. In solving business problems,
for example, it is not too helpful to know only that "profits depend
on sales."
2.3 MEANING OF LINEARITY
Linearity represents a special case of the relationship Y =
F(X). The relationship may be defined as linear if, for all
10 LINEARITY AND LINEAR EQUATIONS
possible values of X and Y, a given change in the value of X
produces a constant change in the value of Y.
2.31 Nongraphical Interpretation. Consider the table of values
given below:
TABLE 2.1
Change Change
X inX Y in Y
3 7
2 +1 4 +3
1 +1 1 +3
0 +1 2 +3
1 +1 5 +3
2 +1 8 +3
3 +1 11 +3
In this example, Y = F(X) is linear since a given change
in X (always + 1) produces a constant change in the value of
Y (always + 3).
Now consider this second table:
TABLE 2.2
Change Change
X in X Y in Y
0 0
1 +1 1 +1
2 +1 4 +3
3 +1 9 +5
4 +1 16 +7
2.3 MEANING OF LINEARITY 11
This relationship between X and Y is nonlinear, since the
change in X (always + 1) produces varying changes in Y.
2.32 Graphical Interpretation. Plotting the data from Table 2.1
on a graph (Figure 2.1) reveals that all of the points lie on a
straight line. The graph of a linear equation is always a straight
line. The points of Table 2.2 form a curve rather than a straight
line, thus indicating that it is a nonlinear relationship.
Y
17 
16 
15 
14
13 
12
11
10
9
8
7
6 / Example 2
5
4
3
2
J 4 3
Example
Figure 2.1
12 LINEARITY AND LINEAR EQUATIONS
2.4 LINEAR EQUATIONS
The general expression of a linear function of one inde
pendent variable (slope intercept form) is F(X) =a + b(X),
usually written Y =a + b(X) or Y = a + bX, where:
Y is the dependent variable
X is the independent variable
a is a numerical constant called the intercept
b is a numerical constant called the slope
The intercept (a) is the point where the line crosses the Y
axis, that is, it is the value for Y when X is zero. In Table 2.1
this occurred at 2, that is, a = 2. The slope (b) of a linear func
tion (straight line) is the amount of change in Y caused by a
unit (1) change in X. In Table 2.1 each unit change in X
produced a change of + 3 in the value for Y; thus, b = 3. The
equation of the function described in Table 2.1 is, therefore,
Y = 2 + 3 (X) or Y= 2 + 3X.
2.5 DETERMINATION OF THE EQUATION OF A LINE
Occasionally it is necessary to find the equation of a line
based upon limited information. This can be done provided that
the coordinates are known for at least two points on the line.
For example, suppose that at point 1, X = X1 and Y = Y1, and
that, at point 2, X = X2 and Y = Y2 (where X1, X2, Y1, and
Y2 are specific values for X and Y). The change in Y from point
1 to point 2 would be (Y2 Y1) and the change in X between
in Y by the total change in X between the points would give us the
change in Y per unit change in X, that is, the slope of the line,
Y2 Y1
b = (2.1)
X2 X1
2.6 DETERMINATION OF INTERCEPTS 13
EXAMPLE. Suppose that the following points are known for a
linear function:
At point 1, X1 = 1 and YI = 5
At point 2, X2 = 3 and Y2 = 11
Using Eq. (21),
SY Y1 115 6 3 3
X2 X1 3 1 2 1
The general equation for a line is Y = a + b(X). Since
the specific value for (b) is now known, the expression for this
particular function may now be written as Y = a + 3(X) or Y =
a + 3X.
To find the value of the intercept (a), substitute the values
for X and Y at one of the known points. For example, using
point 1, where X1 = 1 and Y1 = 5,
Y=a+3(X)
5=a+3(1)
a=2
The complete expression may now be written at
Y= 2+3(X).
2.6 DETERMINATION OF INTERCEPTS
In business one is frequently confronted with the problem
of evaluating alternative courses of action, and interest frequently
centers on the special case where the consequences of two or
more alternatives are equal.
2.61 Graphical Determination of Intercepts. Consider the fol
lowing linear equations, which have been plotted in Figure 2.2:
14 LINEARITY AND LINEAR EQUATIONS
Y, = 2 + X1
Y2 = 2 X2
Y3 = 10 1/2 X3
(2.2)
(2.3)
(2.4)
Note that Eq. (2.4) has a negative slope; that is, the
value of Y decreases as the value of X increases. This is the type
of line most frequently encountered in the graphical method.
14 2
13
12"
11
10
9
8
Y
7"
6
5 3
4 A
3
2
1
0 1
0 1 2 3 4 5 6 7 8 9 10 11
X
Figure 2.2
It is evident from the graph that the Y values for Eqs.
(lines) (2.2) and (2.3) are equal (Y1 = Y2 = 4) when X1 and
X2 are both equal to 2. Similarly, Eqs. (2.3) and (2.4) "cross"
at X = 4 and Y = 8. The intersection of Eqs. (2.2) and (2.4)
is somewhat more difficult to determine precisely. Perhaps the
best answer that could be given at this point is that their point
2.6 DETERMINATION OF INTERCEPTS 15
of intersection is somewhere between an X value of 5 and 6 and
a Y value of 7 and 8. Graphs typically do not produce precise
results. If more accurate answers are required, one must resort to
other methods of analysis.
2.62 Algebraic Solution for Intercepts. When two lines "cross"
one another, this means that at that particular point (X value)
the Y values are equal. At all other X values, of course, the cor
responding Y values would be different. If we are interested
only in the point of intersection (crossover), we can take ad
vantage of this temporary equality.
At point A:
Y1 =2+X
Y2 = 2X
But we also know Y1 = Y2 (at this special point only).
Substitution gives us:
Y, = Y2
2 + X = 2X
2 =2XX
2 = X (value of X, and X2 at intersection)
The value of Y that results from a value of X = 2 in both
equations is:
Y1 =2+X Y2 =2(X)
=2+2 =2(2)
=4 =4
At point B:
Y2 = Y3, Y2 = 2X
2X = 10 2X =2(4)
=8
2X + 10 X
2 Thus: Y3 = 10 2
X =10 = 10
X=4 =8
Develop your own solution for point C (the proper values
are X = 5 1/3 and Y = 7 1/3).
16 LINEARITY AND LINEAR EQUATIONS
2.7 EXERCISES
1. The following values for X and Y are given:
X Y
9 60
10 57
11 54
12 51
15 42
a. Is this a linear function? Why or why not?
b. What equation describes this relationship?
c. Determine the X and Y intercepts.
2. Three linear equations are given as follows:
Y=464X
5
Y= X
Y=7+ X
a. Plot these equations on graph paper. Approximate their in
tercepts from your plot.
b. Determine the three intercepts directly from the equations;
that is, give an algebraic solution.
3. Mr. John Franklin is a salesman who drives his own car on com
pany business. His employer reimburses him for such travel at the
rate of 9 cents per mile. Franklin estimates that his fixed costs
per year, such as taxes, insurance, and depreciation, are $513.
The direct or variable costs, such as gas, oil, and lubrication, aver
age about 3.6 cents per mile.
a. Draw a breakeven chart (total annual cost vs. miles) and de
termine the number of miles Franklin must drive each year
on company business to break even on total automobile
expenses.
b. Determine his breakeven point algebraically by finding the
point of intersection for the equations that describe his "in
2.7 EXERCISES 17
come" from driving the automobile and his total annual
expense of owning and operating the automobile.
4. The Saveio Company is anticipating an order for a machined
part. The size of the order has not yet been specified by the cus
tomer. Three manufacturing alternatives exist for producing this
part. Method No. 1 will involve tooling costs of $180 and a direct
cost per unit of $2.50. Method No. 2 involves more extensive tool
ing, having a cost of $390, but would reduce direct cost to $1.30
per unit. Method No. 3 involves no tooling and utilizes direct labor
only. The expected cost per unit using this alternative would be
$3.90.
a. Determine the total cost per order and the cost per unit for
each of the methods using the following order sizes: 60, 120,
200, 300.
b. Plot the total costs obtained from (a) on a graph. Deter
mine from your graph the approximate order sizes at which
the costs for the alternatives are equal.
c. Write a linear equation which describes the total cost per
order for each manufacturing method. Use these equations
to determine the exact order size at which total cost is equal.
d. What advantage does the use of equations in the analysis of
this problem have over the tabular approach used in (a)
above? Explain.
3 > THE GRAPHICAL METHOD
THE GRAPHICAL method of linear programming is limited in
application to certain types of elementary problems. Most
business problems worth subjecting to formal analysis within a
linear programming format would probably be far too complex
for the graphical method. It is, however, the simplest method and
as such is a useful starting point in any discussion of linear pro
gramming fundamentals.
3.1 PROCEDURE
It is essential that a logical and systematic procedure be
used for problemsolving so that maximum results may be accom
plished with minimum expenditures of time and effort. The sug
gested procedure for the graphical method is as follows:
1. Frame the problem.
a. Determine the restrictions or constraints.
(1) Make numerical computations required. (Only
two points need to be determined for linear
restrictions.
3.2 TECHNICAL PHASES OF THE PROBLEM 19
(2) Determine polygon of technical feasibility
through graphical display of linear restrictions.
b. Select an appropriate objective function.
(1) Measure of effectiveness should be consistent
with higher order objectives.
(2) The function must be linear.
2. Solve for the optimum solution, using either:
a. Direct graphical method.
b. Algebraic method using basic solutions.
3. Modify the solution to take account of those factors
in the problem that are not included in the quantitative
portion of the analysis.1
The meaning and significance of this procedure are best
demonstrated by example. The remainder of this chapter will,
therefore, be devoted to a sample problem.
THE BAKER'S DECISION
The owner of a small bakery that specializes in cookies is
concerned with the kinds and quantities of cookies to be pre
pared for sale tomorrow. Let us assume that there are only two
kinds of cookies sugar cookies and iced cookies from which
he may select his offering to the buying public.
3.2 TECHNICAL PHASES OF THE PROBLEM
(DETERMINATION OF RESTRICTIONS)
Like any production system, the bakery has limited re
sources available to create end products. Since we are concerned
with what to produce tomorrow, the bakery's resources are rela
tively fixed. There is little opportunity, for example, to in
1 Technically, this is not part of the procedure for linear programming. It is
included here to emphasize the decisionmaking approach.
20 THE GRAPHICAL METHOD
crease overnight either the size of the work force or the equip
ment available. As a consequence, the baker's shortrun capa
bility to produce is restricted by the amount of material, labor,
equipment, and other resources available to him during the period
of time over which the decision will be effective. Any product
mix decision made by the bakery manager must, therefore, be
technically or physically feasible, in the sense that it is possible
to accomplish the task with the available resources.
Let us assume that the following production resources are
available:
Cookie mix 120 pounds
Icing mix 32 pounds
Baking equipment continuous oven with capacity of
120 dozen per day
Bakery labor 15 hours.
In order to analyze the problem, we need to know exactly
what "scarce resources" are required to produce the alternative
products or "ends." The recipes provide this information. For
example, one dozen iced cookies require 1.0 pounds of cookie
mix and 0.4 pounds of icing mix. One dozen sugar cookies con
sume 0.6 pounds of cookie mix and no icing mix.2 Through
experience the baker knows that 0.10 hours of bakery labor are
required per dozen sugar cookies. The same requirement for
iced cookies is 0.15 per dozen.3
3.21 Determination of Cookie Mix Restriction. At this point we
can begin to limit the number of alternatives available to the
decisionmaker. For example, should he elect to produce only
sugar cookies, he has enough cookie mix on hand to produce
only 200 dozen (120 0.6). In view of his limited cookie
mix resource, any decision to produce more than that quantity of
sugar cookies is not a technically feasible alternative. Similarly,
it is not technically feasible to produce more than 120 dozen
2 Note the similarity of the recipe to the bill of material and operations
sheet in metalworking plants.
3 Note that all data for the problem are given as exact numbers rather than
as probability distributions. Certainty is assumed.
3.2 TECHNICAL PHASES OF THE PROBLEM 21
iced cookies (120 1.0) if only iced cookies are produced.
These two figures, then, represent the maximum possible produc
tion of each product if produced exclusively. It is also possible,
of course, to produce many combinations of these two products.
Only those combinations that require a total of 120 pounds or
less of cookie mix are, however, technically feasible. Several poten
tial combinations are shown in Table 3.1.
TABLE 3.1
Dozens Cookie Mix
Combination Sugar Iced Sugar Iced Total
1 200 0 120 pounds 0 pounds 120 pounds
2 150 30 90 30 120 "
3 100 60 60 60 120 "
4 50 90 30 90 120 "
5 0 120 0 120 120 "
It is apparent that for every 50 dozen sugar cookies the
baker "gives up" in his product mix, enough cookie mix is "re
leased" to produce 30 dozen iced cookies. This is not surprising,
since the ratio of this ingredient in the two products is 3 to 5
(0.6 to 1.0). Another way of stating this relationship is that
the exchange rate or physical rate of substitution of these two
products in terms of a particular mutually required resource is
3 to 5.
Suppose, now, that we relate the information developed
about the availability of cookie mix as a constraint on the de
cision and the rate of substitution between sugar and iced cookies
relative to this constraint, and display it on a graph. The five
points from Table 3.1 are shown in Figure 3.1.
These points all fall on the line determined by the two
"exclusive maximums" (Qs = 0, Q, = 120) and (200, 0) deter
mined earlier. Thus, if these two intercepts can be determined and
22 THE GRAPHICAL METHOD
rate of substitution remains constant, the restriction line can be
drawn between them.4 This restriction or constraint is simply
a graphical display of the ways in which the cookie mix
resource could be allocated. Thus, any of the following combina
tions are feasible (50, 60) (100, 50) (150, 25) because they lie
between the origin and the restriction and would consume less
than 120 pounds of cookie mix." Points that fall outside the
shaded area (50, 120) (100, 90) (150, 60) are not feasible de
cisions in terms of the mix available for tomorrow's production.
120
o
90
60
a
30
U
50 100 150 200
Sugar Cookie Production (Qs) (in dozens)
Figure 3.1
4 If the rate of substitution were not constant, the constraint would not be a
linear one. This would not fit one of the basic assumptions of linear pro
gramming.
5 The equation of the line is Q, = 120 0.6 Qs = 120 3
1 5
Note that the coefficient of the last term represents the rate of a~bsttittion
and is, therefore, the slope of the line.
3.2 TECHNICAL PHASES OF THE PROBLEM 23
3.22 Determination of Additional Restrictions. We are now
ready .to determine the additional technical constraints on the
decision. The 15 direct labor hours available, for example, might
be allocated to the production of 100 dozen iced cookies (15 
0.15), 150 dozen sugar cookies (15 0.1), or any combina
tion of the two that requires 15 or less direct labor hours. A
plot of this information on the graph (Figure 3.2) indicates that
this constraint is more restrictive than that for cookie mix.6 As
I" 120
0
100(
S50
S Area of Technical Feasibility
U (Labor and Cookie Mix)'/
0
0 50 100 150 200
Sugar Cookie Production (Qs) (in dozens)
Figure 3.2
a consequence, the area of technical feasibility, in terms of both
cookie mix and direct labor, becomes the area defined by the
latter resource alone. Expressed another way, the baker has
more than enough cookie mix to make any of the possible com
6 The equation of the direct labor constraint is
.1 2
QI = 100 Qs = 100 Q
.15The rate of substitution is 2 to 3 for the labor resource.
The rate of substitution is 2 to 3 for the labor resource.
24 THE GRAPHICAL METHOD
binations permitted by the scarce resource, direct labor. For this
reason, we can ignore the cookie mix in our further analysis; it
is not a strategic or key technical factor in this particular problem.
If, however, the baker were concerned with deciding how much
cookie mix to order tomorrow a different problem the
amount of cookie mix on hand would obviously be an important
factor.
3.23 The Technical Feasibility Polygon. Figure 3.3 shows the
effect of all technical constraints on the problem. Note that icing
120 G
o 100 F
.5
B Icing Mix
80
D
60
0 20 40 60 80 100 120 140 160
Sugar Cookie Production (Qs) (in dozens)
Figure 3.3
mix is a constraint on the production of iced cookies only. There
is no rate of substitution between the two product ends relative to
this specialized resource. Giving up the production of some
iced cookies, for example, in order to free some of the icing mix
resource for the production of sugar cookies has no practical
technical meaning, since the latter do not utilize this resource.
On the other hand, the rate of substitution for the ovens is one
3.3 ECONOMIC PHASES OF THE PROBLEM 25
to one: the ovens can produce one dozen sugar cookies for every
dozen iced cookies given up, and vice versa.
The net effect of technical constraints is to limit the num
ber of alternative solutions available to the decision maker. The
baker, however, still has a very large number of alternatives from
which he must choose. Any combination within the area of the
technical feasibility polygon is a physically feasible decision for
tomorrow's product mix.
Having thus determined the effect of the physical factors
in the problem, we can now examine the economic phases of the
problem.
3.3 ECONOMIC PHASES OF THE PROBLEM
One of the decisions that must be made in the socalled
observation stage of the scientific method is to determine an ob
jective function, or measure of effectiveness. Since a bakery is
an economic institution, it would be reasonable to assume that
the highest order economic objective of the firm would be maxi
mization of return on investment. Figure 3.4 indicates the first
few stages of the approach to this objective.
Let us assume that (1) all of the technically feasible alter
natives (product quantities) are realistic from the standpoint of
a sales forecast, and (2) investment is fixed in the short run, that
is, no changes in total investment can be made for tomorrow.7
The objective, then, might be restated as maximizing the total
dollar contribution to fixed costs (overhead expenses) and profit.8
Suppose that the selling price per dozen iced cookies is
70 and the direct material and labor costs associated with
their production is 50 per dozen; the corresponding data for plain
sugar cookies might be 60 and 450 respectively. Every dozen iced
7 If any of the feasible production quantities exceed the sales forecast, one
might use the expected sales quantities as additional constraints on the
decision.
8 From Figure 3.4.
26 THE GRAPHICAL METHOD
Figure 3.4
cookies sold "contributes" 20 to fixed costs (at all quantities
below the breakeven point) or to profit (for all output over the
breakeven point). In like manner, the "contribution" of sugar
cookies is 150 per dozen. Knowing this, our bakery manager can
now proceed to determine the most economically desirable of the
many feasible alternatives. This is the optimal economic solution.
It would be useful at this point to state the objective func
tion in terms that will allow explicit evaluation of various solu
tions that may be indicated. The objective is to maximize the
total dollar contribution to overhead and profit. This can be
expressed in symbolic form as follows:
Return on Investment:
= Investment Turnover times Profit Margin or IT X PM.
SR P (SR TC) SR (VC + FC)
I SR I I
= (SR VC) FC
I
3.4 OPTIMAL SOLUTION 27
If costs and investment are fixed in the short run, then
maximizing total dollar contribution (sales revenue minus variable
costs) becomes the shortrun equivalent of maximizing return on
investment.
TC = Q X CI + Qs X Cs
where:
TC = Total dollar contribution
Qi = Production of iced cookies (dozens)
Ci = Contribution of iced cookies (per dozen)
Qs = Production of sugar cookies (dozens)
Cs = Contribution of sugar cookies (per dozen)
3.4 DETERMINATION OF THE OPTIMAL SOLUTION
(3.1)
In framing the problem we have developed one equation
and two independent variables (Qs and QI). This complicates
the solution process because there are a very large number of
values for Qs and QI that will "satisfy" the single equation. All
of the combinations in Table 3.2, for example, produce a total
contribution of $6.00.
TABLE 3.2
Production Contribution
Mix Iced Sugar Iced Sugar Total
1 30 0 $6.00 $0.00 $6.00
2 24 8 4.80 1.20 6.00
3 18 16 3.60 2.40 6.00
4 6 32 1.20 4.80 6.00
5 0 40 0.00 6.00 6.00
28 THE GRAPHICAL METHOD
These are "solutions" for the equation, but not for the
problem. We are interested in finding those values for Qs and QI
that (1) maximize the objective function (total dollar contribu
tion in this case) and (2) remain within the technical restric
tions imposed by the limited resources.
One method of determining the optimal solution would be
trial and error. By inserting different values for the quantity of
sugar and iced cookies to be produced in Eq. (3.1), we might
eventually find those values that represent the optimal produc
tion program. An alternative would be to use one of the solution
methods characteristic of the graphical method of linear program
ming. These two methods make it possible to determine the op
timal solution with the least possible computation. Linear pro
gramming may, in fact, be described as an effective search pro
cedure for determining optimal solutions to types of problems in
which there are more unknowns than linear equations.
Before proceeding with the specifics of these solution meth
ods, it might be interesting to attempt an intuitive decision. We
might reason, for example, that since the contribution of one
product is higher than that of the other, an optimal solution
would be (1) to produce as many iced cookies as possible, and
(2) to use any resources not required for the production of iced
cookies to produce sugar cookies. Such a decision makes sense,
but does it make maximum sense? The graphical methods de
scribed in the following pages should provide us with a basis for
evaluation of this intuitive decision rule.
3.41 Direct Graphical Solution. If the values of Table 3.2 are
plotted on a graph (Figure 3.5), they trace a straight line of
the equation, Y = 30 (X). This is not surprising, since Eq.
(3.1) is a linear function. Furthermore, the slope of the line is
the ratio of contributions for the products (.15 .20 = /). Any
point on this line, then, represents a solution that is feasible and
results in a total contribution of $6.00. Our objective, however, is
to maximize the total contribution, so these do not appear to be
optimal solutions.
In Figure 3.6, the lines of "equal contribution" have
been drawn in for $12.00, $18.00 and $21.00. Notice that the
3.4 OPTIMAL SOLUTION
100
A )
0
80
60
040
0
0
Sugar Cookie Production (Qs) (in dozens)
Figure 3.5
"equal contribution" lines are parallel since their slopes are equal.
It is necessary to use a dashed line for a portion of the $18.00
line since some of those combinations represent solutions that are
not technically feasible. Were we to continue drawing parallel lines
further out from the origin, the solutions outside the area of
technical feasibility would increase. Finally, one line would just
intersect point D (see $21.00 line). The optimal solution indi
cated at this point is 60 dozen each of the sugar and iced cookies;
the corresponding profit is $21.00. We also know that both the
direct labor and oven capacity will be fully utilized while there
will be extra capacity (slack) in terms of the other resources 
icing and cookie mixes. Table 3.3 gives this information in tabu
lar form.
3.42 Algebraic Solution. A second and more direct method of
solving graphical linear programming problems is to work ex
clusively with those solutions represented by the points on the
polygon of technical feasibility. As might be expected, the optimal
30 THE GRAPHICAL METHOD
TABLE 3.3
Cookie Mix Icing Mix Labor Oven
(pounds) (pounds) (hours) (dozen per day)
Sugar 36 0 6.0 60
Iced 60 24 9.0 60
Total 96 24 15.0 120
Capacity 120 32 15.0 120
"Slack" 24 8 0 0
solution will normally be represented by one of these points
(B, C, D, or E in Figure 3.6).9 The coordinates of these
points may be approximated directly from the graph or computed
algebraically using the equations for the various constraints and
the method suggested in Sec. 2.6 for determining the point of
intersection for two linear functions. These coordinates are shown
in Table 3.4.
TABLE 3.4
Point Iced Sugar
A 0 dozen 0 dozen
B 80 0 "
C 80 30 "
D 60 60 "
E 0 120 "
The analysis may conveniently be started at the origin
solution (point A in Figure 3.6).'0 This is a technically feas
ible solution even though it makes no economic sense. Total
contribution Eq. (3.1) would be equal to zero. The intuitive de
9 In the special case when the slope of the "equal contribution" lines is the
same as that of one of the constraints, there will be a range of optimal
solutions.
t1 The reason for starting at the origin will become more apparent when the
simplex method is introduced.
3.4 OPTIMAL SOLUTION
100 
= 80
a40
8 20
0 20 40 60 80 100 120 140
Sugar Cookie Production (Qs) (in dozens)
Figure 3.6
cision rule developed in Sec. 3.4 would indicate that as many
dozens of iced cookies as possible should now be introduced
into the solution. The visual analogy would be to move from
point A to point B (Figure 3.6). The value of the objective
function at this point would be $16.00 (80 X .20 + 0 X .15).
This would require all of the icing mix resource, but a substan
tial amount of slack, or unused capacity, would remain for labor
and the ovens. Our intuitive rule would indicate, therefore, that
we begin to introduce quantities of the next most profitable prod
uct into the solution, that is, we move in the direction of point
C. Each dozen of sugar cookies adds 15 to the objective function.
At point C, however, we encounter the direct labor constraint.
This presumably represents the optimal solution by our intuitive
evaluation, inasmuch as we are maximizing production of our
most profitable product and are producing as much as possible
of the second most profitable product with the remaining capacity.
Total dollar contribution at point C is 80 X .20 + 30 X .15 =
$20.50. It is known from the direct graphical analysis (Sec. 3.41)
32 THE GRAPHICAL METHOD
that this is not the true optimal solution. Why did the intuitive
answer fail?
The answer to this question lies in the relative rates of
substitution and profitability in the problem. Moving from point
C to point D on the graph implies that the baker "trades"
units of labor capacity between the products. Since iced cookies
require .15 labor hours per dozen and sugar cookies require only
.1 hours per dozen, each dozen iced cookies may be "traded" for
one and onehalf dozen sugar cookies. This makes technical
sense, but what about the economic sense? A dozen iced cookies
contribute 206, but 1V2 dozen sugar cookies contribute 22.50
(15 X 1.5). Changing the product mix from point C to D is,
therefore, a profitable move, as shown below:
Value of objective function at point C .__  $20.50
Added contribution from sugar cookies
(30 dozen @ 150)  $4.50
Loss of contribution iced cookies
(20 dozen @ 20)  $4.00
Net added contribution  .50
Value of objective function at point D  $21.00
Would it also be profitable to move from point D to E, to
give up oven capacity for iced cookies in order to produce more
sugar cookies? The rate of substitution for the oven resource is
one to one. Such a change would result in a 50 decrease (200 
155) in the value of the objective function for each dozen "traded
off" in the solution. Therefore, point D represents the optimum
economic solution to our problem as given.'1
11 It is also possible to solve the problem by introducing the maximum num
ber of sugar cookies into the solution first. As long as desirable "trades"
can be made (as measured by the objective function) the solution is being
improved. In more complex problems the number of trials required to reach
the optimum solution is ordinarily reduced by introducing the products in
order of their desirability at each stage. Desirability has, in effect, been
defined in this case as contribution per dozen. Other definitions may be more
appropriate, as will be illustrated in Chapter 4.
3.5 THE HUMAN FACTOR 33
Before reaching an economic conclusion, the baker should
review and reevaluate two factors. One is the realism of the
assumptions made thus far in his analysis. Are all of the rela
tionships really linear? If not, is the assumption of linearity so
severe an abstraction as to negate the results of the analysis? A
second factor is the treatment of the data employed. Are they
really certainties or only probabilities?12
3.5 THE HUMAN FACTOR
At this point the baker has developed a model that ana
lyzes explicitly the technical and economic phases of the prob
lem. What about the human phases? Is the optimum economic
solution a practical one in terms of the people involved? Sup
pose, for example, that the baker and his helpers enjoy making
iced cookies more than plain sugar cookies. This means that the
overall objective function should be the maximization of some
combination of dollar contribution and enjoyment. Developing
an explicit statement of this function is extremely difficult, if not
impossible. We can, however, take this intangible factor into
account in the final stages of our analysis, as shown in Table 3.5.
TABLE 3.5
Production Dollar Total
Point Iced Sugar Contribution Enjoyment Satisfaction
A 0 dozen 0 dozen $ 0.00 Nil Nil
B 80 0 16.00 Maximum ?
C 80 30 20.50 Great ?
D 60 60 21.00 Moderate ?
E 0 120 18.00 Nil ?
12 Presumably, he should have asked and answered these questions in the
negative before he began his analysis. Good practice dictates that one
review the nature of all basic assumptions after the analysis as well.
34 THE GRAPHICAL METHOD
We cannot fill in all of the Total Satisfaction column until
the baker, using judgment, evaluates it for us. He might decide,
for example, that it was "worth" 500 per day to him to be
allowed to produce 80 dozen iced cookies and only 30 dozen sugar
cookies (point C). On the other hand, he might also decide that the
loss of an additional $4.50 required to avoid the production of any
sugar cookies (point B) was not acceptable to him. In this case
the optimal decision, considering all phases of the problem,
would be the product mix indicated by point C.
3.6 SUMMARY
The graphical method is capable of handling linear pro
gramming problems of only limited complexity and is, there
fore, a special rather than a general method. The most limiting
factor is the number of variables involved. A twodimensional
graphical display of relationships is restricted to two independent
variables.
The two basic assumptions common to all linear pro
gramming methods linearity and certainty are recognized
explicitly both in framing the problem and in solving it.
Except in special cases, the optimal solution will always
lie at one of the several points or vertices of the polygon of tech
nical feasibility. The search for these points is made more effective
by the visual display of the relationships among variables and con
straints.
The precision of solutions developed by purely graphical
solution methods is a function of the scale and accuracy of the
graphical display. The accuracy of solutions may be improved
through the use of algebraic methods. In such cases the graphi
cal portion of the analysis becomes of secondary importance.
Most of the technical and economic phases of a problem
may be expressed explicitly and treated directly in the analysis.
Many qualitative and intangible human factors cannot be handled
in this manner. The primary contribution of any quantitative
3.7 REVIEW QUESTIONS 35
method is in narrowing the judgment portion of decisionmaking,
not in eliminating it. It is important, therefore, to recognize that
the optimal solution from a linear programming model is not
necessarily an optimal overall solution to the problem. Mana
gerial judgment rather than mathematics must be used in large
part for the selection of the best alternative course of action
and, of course, for the implementation of the decision.
3.7 REVIEW QUESTIONS
1. What are restrictions or constraints? What is their
net effect on the number of alternatives open to the decision
maker?
2. Justify the use of maximization of total dollar contri
bution to fixed costs and profit as the objective function in the
quantitative portion of the analysis for the bakery problem.
3. Show specifically where the two basic assumptions of
linear programming linearity and certainty enter into the
analysis.
4. Why, except in special cases, does the solution always
lie at one of the points or vertices of the technical feasibility
polygon?
5. Under what circumstances would there be a range of
optimal solutions? How could the decisionmaker go about select
ing a final alternative from this optimal range?
6. Linear programming is sometimes described as effec
tive search procedure. Explain the meaning of this statement.
7. Explain briefly the purpose and content of Table 3.5.
Give specific reasons for including question marks in the final
column.
36 THE GRAPHICAL METHOD
3.8 EXERCISES
1. Determine the optimal product mix for the bakery
problem discussed in the chapter on the basis of the following
contributions:
a. Ci = 250 and Cs =150
b. CI = 300 and Cs = 200
2. The Zeus Company manufactures a deluxe and a
standard model. All production is to stock rather than to cus
tomer order.
At the present time a seller's market exists for the com
pany's products. Although the Zeus Company has made no at
tempt to charge prices other than those that they have charged
for several years, they are concerned with the profitability of their
product mix. The profit contribution of the deluxe model is
$3.00 per unit and that of the standard model is $1.00 per unit.
The manufacture of the product is relatively simple.
Both models require a machining operation, and the deluxe
model requires an additional painting operation. The same ma
chines are used for machining either model. Since all units are
placed in inventory before being sold, warehouse capacity as
well as equipment capacities must be considered in making prod
uct mix decisions.
The technical requirements and restrictions, on a monthly
basis, are as follows:
Resource Requirements
Machining Ware
(machine housing Painting
hours per (square feet (man hours
piece) per piece) per piece)
Standard 2 4 0
Deluxe 3 3 1
Total Capacities Available 24,000 36,000 6,000
3.8 EXERCISES 37
a. Determine the most profitable mix for next month.
b. Determine the best mix if the contributions of the two
products were reversed, if Cs = $3, CD = $1.
c. In general, what effects would changes in the con
tribution ratio have on the solution? Show specifically
where changes in the solution occur.
d. Under what circumstances would there'be a range of
equally profitable product mixes for The Zeus Com
pany? (Hint: This would be a very special case. A
very careful answer to Question (c) will help you.)
e. What are the linear characteristics of the problem?
f. In what respects is the human element considered
in this problem?
3. The Lotanoiz Corporation manufactures a "Holly
wood" automobile muffler in one size only. However, they pro
duce two models the "LongLife" and the "Economuf." The
"LongLife" is of heavier metal construction and also receives a
special alloy dipping before being painted.
Since the corporation is young and operates on a com
paratively small margin of profit, the ownermanagers are very
anxious to find the exact product mix that will yield maximum
return. At present, both types of mufflers are selling quite well
and neither model seems to have the better market potential.
Total Direct Man Time Required Time Required Contri
Hours Required per Pair for Dip per Pair bution
for Fabrication ping Operation for Painting per
Model per Unit (Hours) (Hours) Unit
Economuf 1.0 0.1 $0.80
LongLife 1.5 0.2 0.1 $0.90
Painting operations may be carried on 37.5 hours per
week. Two units are painted at a time. The dipping operation
38 THE GRAPHICAL METHOD
can be carried on 8 hours per day (5 days per week). Two units
may be dipped at a time. As both mufflers are the same size,
the shipping department can process up to 800 units weekly re
gardless of the model mix. A total of 900 manhours is available
per week in the fabricating department.
(a) Given the information provided above, what weekly
production schedule would you suggest?
(b) Which departments will be operating at capacity if
your schedule is adopted? How much idle capacity will
exist in those departments that are operating at less
than full capacity?
(c) What could the management of the Lotanoiz Cor
poration do to "relax" the various technical restrictions
in this problem, that is, to move the restrictions
further away from the origin? To which ones should
their initial efforts be directed? Why?
4. The Plummer Company produces two types of men's
billfolds. The higherpriced Chieftain line uses more first
grade materials and is virtually handmade. The lowerpriced
Warrior line is largely machinemade and uses a higher percent
age of standard grade materials. Each Chieftain unit contributes
$5.00 and each Warrior unit contributes $3.00 toward overhead
and profit.
The firm has been producing 600 of the Chieftain model
and 300 of the Warrior model every week. The most recent sales
forecast indicates that these figures represent maximum possible
sales during the next week as well. The company's labor force of
ten skilled longservice employees work 36hour weeks.
The company's suppliers have just notified them that due
to a temporary shortage of materials no shipments can be made
for at least one week. This means that the materials now avail
able, the sales forecast, and labor availability must be considered
in developing next week's schedule.
3.8 EXERCISES 39
The material
are as follows:
requirements and labor hour availability
Chieftain Warrior Available
First grade materials .4 sq ft .2 sq ft 256 sq ft
per unit per unit
Standard materials .3 sq ft .5 sq ft 276 sq ft
per unit per unit
Labor .5 hrs .2 hrs 360 hrs/week
per unit per unit
a. The sales manager has proposed that only Chieftain
models be produced next week because of its higher contribution.
(1) Frame the problem in a graphical format.
(2) Evaluate the sales manager's proposal.
b. The production engineer suggests that any reduction
of work hours should be held to an absolute minimum. He pro
poses that the schedule adopted should be that one which will
maximize the number of employees' hours to be worked during
the week.
(1) Evaluate this proposal.
4 THE SIMPLEX METHOD
4.1 GENERAL NATURE OF THE METHOD
ANY PROBLEM that fits the two basic assumptions of linear pro
gramming can be solved by the simplex method. It is, in fact,
the general method, as differentiated from the graphical, trans
portation, and other special methods. This broad capability is
gained, from the student's viewpoint at least, only at the expense
of a considerable increase in mathematical complexity. The term
"simplex" is somewhat misleading; it is actually the most diffi
cult method to comprehend. The reader who fully understands
the graphical method, however, should not find it too difficult.
The basic problem in linear programming is to find the
particular set of variables that satisfies all constraints and maxi
mizes or minimizes the value of the objective function. The
orderly solution of problems using the simplex format requires the
performance of a series of carefully defined steps. The purpose
of the procedure is to produce the desired result with minimum
computation effort.
The major advantage of the graphical method is that the
visual display of relationships enables the analyst to see the points
4.1 GENERAL NATURE OF THE METHOD 41
on the polygon of technical feasibility. Since the optimal solution
ordinarily lies at one of these vertices, the search procedure may
be limited to an analysis of these points. Theoretically, a three
dimensional display might be useful in problems involving three
variables. Constraints in this case would become planes, as
would the objective function, and the optimal solution would
represent one of the many intersections generated by these
planes.
Most linear programming problems that lend themselves to
analysis by the simplex method involve many more than three
variables. It would be most difficult to visualize a fivedimen
sional problem, let alone a twentydimensional problem. Never
theless, it is useful in understanding the rationale of the simplex
method to attempt certain analogies. One such analogy is to
compare the computation procedure to that of developing a plan
for climbing a mountain. The objective may be stated simply:
to reach the summit (maximize the objective function). There
would be, presumably, an infinite number of routes that might
be taken by the climbers. If the mountain were high, it might
be necessary to accomplish the total task in stages. Intuitive
reasoning would indicate that as long as each of these camps is
higher up the mountain than the previous one, the party will pro
gress toward the summit. Thus, many alternative routes can be
eliminated even though they are feasible.' Starting from the base
(origin), an effective climbing procedure (computation proced
ure) would reach the summit (optimal solution) with a minimum
number of camps (steps or iterations).
Mountains, of course, have only three rather than five or
twenty dimensions. At the same time, the simplex procedure can
and does "reach the summit" of linear programming problems
having many variables.
1 For example, a climbing plan that establishes camp 1 at 10,000 feet, camp
2 at 5,000 feet, camp 3 at 8,000 feet, and camp 4 at 2,000 feet, is feasible
but ridiculous in terms of the objective.
42 THE SIMPLEX METHOD
4.2 COMPUTATION PROCEDURE
1. Frame the problem.
a. Select the pertinent variables and constraints.
b. Express the relationship between all variables and
constraints in explicit form, that is, in equations.
c. Determine the objective function or measure of
effectiveness.
2. Develop an initial feasible solution.
a. Technically, any basic feasible solution is accept
able.
b. For practical purposes, the origin solution is usu
ally selected.
3. Evaluate the alternative variables that might be
brought into the solution. Several methods of evalua
tion may be used, including:
a. Net effect per unit.
b. Net total effect.
4. Select one of the variables and determine the number
of units of each variable represented by the revised
solution.
5. Make the necessary adjustments to express the new
rates of substitution between those variables in the
solution and all variables.
6. Repeat Steps 3, 4, and 5 until an analysis at Step 3
reveals that no additional changes of a favorable nature
can be made.
4.3 PRODUCTMIX EXAMPLE
The productmix problem represents an elementary prob
lem capable of analysis by the simplex as well as by the graphi
4.3 PRODUCTMIX EXAMPLE 43
cal method. Let us, therefore, rework the problem using the sim
plex format.
4.31 Step 1. The relationship between variables and constraints
must be expressed in equations rather than by the visual pre
sentation acceptable in the graphical method.
The total demand for the direct labor resource is deter
mined by the number of each type of cookie in the solution at any
point. Each dozen iced cookies introduced into the product mix
will require the allocation of .15 direct labor hours. Similarly,
each dozen sugar cookies produced requires .10 labor hours.
Furthermore, only those combinations that require a total of 15
or less hours are feasible solutions, as shown below:
Labor hours (iced) + Labor hours (sugar) Total
labor available
This is an inequality; the two sides are not always equal
to one another. It can be converted into an equation by adding
unused, idle, or "slack" capacity (resources) to the left side,
as follows:
Labor hours (iced) + Labor hours (sugar) + Labor
hours (slack) = Total labor hours available
The two sides of this expression will always be equal. It
is necessary, however, to make the expression more specific for
it to be useful in the simplex analysis:
Labor hours (iced) = .15 Qi, where Qi = dozens of
iced cookies
Labor hours (sugar) = .10 Qs, where Qs = dozens of
sugar cookies
Thus:
.15 QI + .10 Qs + Labor hours (slack) = 15 (4.1)
For computation purposes it is necessary to express the
slack capacity in the same general form as the real variables. A use
ful convention is to assume that fictitious products are produced
with those capacities not required by the real variables in the solu
44 THE SIMPLEX METHOD
tion. For example, suppose that some imaginary cookie (type
L) can be produced with the slack labor capacity. The symbol
QL would represent the number of dozens produced. If each
dozen L cookies is assumed to require 1.0 hours of labor capacity
and no other resources, then 1.0 QL represents the demand on the
labor resource to produce QL dozens.2 The supply and demand
equation for the labor resource may now be written as:
.15 Q, + .10 Qs + 1.0 QL = 15 (4.2)
This equation means that the sum of the labor require
ments for iced, sugar, and L cookies is always equal to 15 hours.
The requirement for L cookies is a mathematical fact but a fiction
in reality, since it actually represents that portion of labor
capacity not required for the real variables.
Let O cookies represent the slack variable for unused
oven capacity and M cookies the corresponding slack variable for
icing mix.3 The complete set of relationships between all five
variables two real and three slack and the labor constraint
may now be expressed as follows:
.15 Qi + .10 Qs + 1.0 QL + 0.0 Qo + 0.0 Q, = 15
(4.3)
where:
Qx = dozens of iced cookies
real variables
Qs = dozens of sugar cookies
QL = dozens of L cookies
Qo = dozens of O cookies slack variables
QM = dozens of M cookies
2 The assumption of 1.0 labor hour is based on computation convenience.
Technically, any positive number would be an acceptable assumption. Note
also that slack variables are assumed to require only the resource for which
they represent idle capacity. Thus, the "recipe" for L cookies calls for 1.0
hour of labor capacity and zero icing mix and oven capacity.
3 Using the procedure suggested in footnote 2, the "recipe" for 0 cookies
calls for 1.0 unit of oven capacity per dozen. Making M cookies requires 1.0
pound of icing mix per dozen.
4.3 PRODUCTMIX EXAMPLE 45
Eq. (4.3) differs from Eq. (4.2) only in that the relation
ship between the two slack variables representing oven and icing
mix resources have been added to the expression. Their coeffic
ients are zero since their production requires no labor capacity.
The addition of these two terms does not add to the "practical"
meaning of the equation but it does satisfy one of the basic
requirements of the simplex method, that all relationships between
variables and restrictions must be stated explicitly and com
pletely.
Similar expressions representing the relationship between
all five variables and the oven and icing mix capacities may now
be derived as follows:
Oven capacity (iced) + Oven capacity (sugar) 5 Total
oven capacity
This inequality can be converted into an equation by add
ing a third term representing that portion of total capacity not
required by the two real products. The total oven capacity is
given as 120 dozen per day. The equation thus becomes:
Oven capacity (iced) + Oven capacity (sugar) + Oven
capacity (slack) = 120
Each dozen iced and sugar cookies will require 1.0 unit of
the total oven capacity. The O cookies, the variable representing
slack oven capacity, are assumed, for computation convenience,
to require 1.0 unit of oven capacity and no other resources:
1.0 QI +1.0 Qs + 1.0 Qo = 120
Finally, the addition of the two additional slack variables
(L and M cookies) to the equation will represent the complete
set of relationships between all variables and the oven capacity
constraint:
1.0 QI + 1.0 Qs + 0.0 Q, + 1.0 Qo + 0.0 Q, = 120
(4.4)
Note that the coefficients of QL and QM are zero in
46 THE SIMPLEX METHOD
this equation, since neither of these variables requires any oven
capacity.
Only one real and one slack variable make a demand on
the icing mix resource iced cookies and M cookies (the slack
variable for icing mix). The recipe for iced cookies requires
.4 pounds per dozen, and M cookies, by definition require 1.0
pound per dozen. The amount of icing mix on hand is equal to
32 pounds. Thus, the complete set of relationships would be:
.4 Qx + 0.0 Qs + 0.0 QL + 0.0 Qo + 1.0 QM = 32 (4.5)
The coefficients of Qs, QL, and Qo are zero, since their
production would not require any of the icing mix resource.
It is more convenient for computation purposes to express
the relationships between variables and constraints in the form of
a table than to retain the expressions in their more familiar equa
tion form. The general format for such a presentation is given in
Table 4.1.
TABLE 4.1
Variables Qi Qs QL Qo QM Capacity
Labor .15 .10 1 0 0 15
Oven 1.0 1.0 0 1 0 120
Icing Mix .4 .0 0 0 1 32
The columns in the table represent the recipes for the
various products (variables) in the problem. For example, read
ing down the column headed by the symbol Qi reproduces the
requirements for iced cookies, .15 labor hours, 1.0 unit of oven
capacity, and .4 pounds of icing mix per dozen. Similarly, each
dozen sugar cookies requires .10 labor hours, 1.0 unit of oven
capacity, and zero icing mix. The requirements for all slack vari
ables are unique in that, by definition, they require 1.0 unit of
the capacity they represent and nothing more. Just as sugar
4.3 PRODUCTMIX EXAMPLE 47
cookies require no icing mix, L cookies (the variable that represents
slack labor capacity) require no oven capacity or icing mix. In
like manner, O cookies require only oven capacity and M cookies
require icing mix but no labor or oven capacity.
Equations (4.3), (4.4), and (4.5) may be reproduced from
Table 4.1 by multiplying each of the series of coefficients in each
row by the appropriate Q, which appears at the head of each col
umn. For example, the first row produces
(.15) Q + (.10) Qs + (1) QL + (0) Qo + (0) QM = 15.
Similar interpretation of the second row shows that this row repro
duces Eq. (4.4) and that the third row represents Eq. (4.5).
The objective function, as such, does not appear in these
tables. The general form of this equation was formulated in
the graphical example (Sec. 3.3) as:
Qix Ci+ Qs x Cs = TC
where:
Ci = contribution per dozen iced cookies
Cs = contribution per dozen sugar cookies
TC = total contribution to fixed costs and profit
The requirements of the simplex method have, however,
made it necessary to introduce three additional variables into
the problem. Thus, a complete statement of the objective
function must include the "contribution" of the slack as well
as of the real variables, as follows:
QI X CI + Qs XCs + QL X CL +Qo X Co
+ QM X CM = TC
What values should be assigned to CL, Co, and CM, that
is what addition to the total dollar contribution results from the
addition of one dozen L, O, or M cookies to the product mix?
Since these are fictitious products introduced to represent slack
capacity, their economic contribution should represent the value
of slack capacity. By implication the contribution of slack re
sources in the graphical method was zero; unused capacity resulted
48 THE SIMPLEX METHOD
in no additional income or costs. In accordance with the need
for explicit relationships in the simplex method, assume that
CL, Co, and CM are zero. Thus:
Qi X 200 + Qs X 15( + QL x 0 + Qo x O0 +
QM X 0 = TC
(4.6)
This information concerning the relative economic values
of the variables can now be added to the computation table.
By convention these values are placed above the corresponding
Q term at the head of each column, as in Table 4.2.
TABLE 4.2
Contribution 20 150 00 0 0
Variables Qi Qs QL Qo QM Capacity
Labor .15 .10 1 0 0 15
Oven 1.0 1.0 0 1 0 120
Icing Mix .4 0.0 0 0 1 32
The information presented in Table 4.2 can be summarized
as follows:
a. There are five variables in the problem. Only two of
these (iced and sugar cookies) are real variables. The
three slack variables (L, 0, and M cookies) are ficti
tious products introduced into the problem because of
the necessity for explicit relationships. In reality, these
slack variables represent unused or idle capacity in the
labor, oven, and icing mix resources.
b. The numbers in the columns below each variable indi
cate the resources required to produce one unit of each
variable.
c. The numbers in the rows show the relationship be
tween each variable in the problem and each resource
of limited capacity (restrictions or constraints).
4.3 PRODUCTMIX EXAMPLE 49
d. The economic values above each variable indicate the
contribution to the objective function or measure of
effectiveness that results from the introduction of one
unit of each variable into the solution. In the bakery
problem these values represent the contribution to profit
and fixed costs that results from the production of one
dozen of each type of cookie. Only real products have
positive contributions. Slack variables normally have
zero values since they represent fictitious products that
cost nothing to produce and have no market value.
4.32 Step 2 (Explanation). The computation procedure for the
simplex method is actually a method of hunting for better and
better solutions until the optimal solution is discovered. Theoreti
cally, the solution of a simplex problem may be started at any
one of the several basic solutions,4 but for practical reasons the
origin solution should be chosen as the initial solution since it
is the easiest to visualize and to formulate.
The origin solution is represented by point A in Figure
3.6 (page 31). It is a technically feasible solution which, in
the graphical analysis, was interpreted to mean that no cookie
production takes place. Because this solution made no economic
sense, the succeeding stages of the graphical analysis proceeded
to move or change the product mix to those represented by other
points on the feasibility polygon.
The meaning of an origin solution in the simplex method is
somewhat different than in the graphical method. No real varia
bles are produced. This means that all resources are idle insofar
as iced and sugar cookies are concerned. Remember, however,
that slack variables were defined as that portion of each resource
not required for the production of real products. Therefore,
the origin solution in the simplex method represents a solution
that calls for the production of slack variables only. This makes
no more economic sense than the origin solution in the graphical
4 A basic solution is defined as a solution at one of the points or vertices of
the polygon of technical feasibility. Points A, B, C, D, and E in Figure 36
represent the basic solutions for this problem.
50 THE SIMPLEX METHOD
method, but it does provide a useful starting point from which
better solutions may be developed.
In the bakery example, the origin solution indicates that
no iced or sugar cookies are to be produced and that all resources
are to be allocated to the production of slack products (L, O,
and M cookies). What quantities of each type are in the prod
uct mix at this point? Reference to Table 4.2 shows that 15
hours of direct labor are available. The only slack product that
requires this resource is L cookies, so that the entire 15 hours
may be devoted to their production. The first coefficient under
QL in the table indicates that one hour of direct labor is required
per dozen L cookies. The number of L cookies in the origin solu
tion is, therefore, 15 dozen (15 1 = 15). Similar reasoning ap
plied to the other slack variables leads to the conclusion that the
120dozen oven capacity should be used to produce 120 dozen O
cookies and that the 32 pounds of icing mix should be allocated
to the production of 32 dozen M cookies. In summary, the
product mix indicated by the origin solution in the simplex
method would be:
Q0 = 0, Qs = 0, QL = 15, Qo = 120, QM = 32
For computation purposes the origin solution must be
shown in tabular form, as in Table 4.3.
TABLE 4.3
Contribution 200 15i 00 00 0 Production
Variables Qi Qs QL Qo QM
Of QL .15 .10 1 0 0 15 Dozen L
cookies
00 Qo 1.0 1.0 0 1 0 120 Dozen O
cookies
00 Qm1 .4 0.0 0 0 1 32 Dozen M
cookies
4.3 PRODUCTMIX EXAMPLE 51
Note that this table differs from Table 4.2 in two respects.
First, the variables presently in the solution are shown on the
left side of the table together with their economic values. The
fact that these are all slack variables identifies the solution as
that of the origin. Secondly, the title of the final column in the
table has been changed from "Capacity" to "Production." Of
course, the production of 15 dozen L cookies, 120 dozen O
cookies, and 32 dozen M cookies actually represents idle capacity
in the labor, oven, and icingmix resources. The adoption of
this fiction was necessary to meet the needs of the simplex com
putation framework. It is, however, an assumption that tries the
patience of the reader who finds it difficult to describe "idle
capacity" by any other term.
4.33 Steps 3 and 4 (Explanation and Analysis). Of the five
variables in the baking problem, only two iced cookies and
sugar cookies were not included in the product mix repre
sented by the origin solution. Would the introduction of either of
these products into the solution improve the economic perform
ance of the system? Answering this question is the objective of
Step 3 in the simplex computation procedure.
The production of each dozen iced cookies would add a
20 contribution to fixed costs and overhead. The data for this
product (see column of coefficients under Qi in Table 4.2) shows
that the production of each dozen iced cookies requires the allo
cation of .15 labor hours, 1 unit of oven capacity, and .4 pounds
of icing mix. In the special language of the simplex method
(see Table 4.3), the introduction of one dozen iced cookies into
the solution would require that some portion of the variables in
the existing solution be "given up." The specific products and
amounts that would have to be withdrawn for each dozen iced
cookies introduced would be .15 dozen L cookies, 1 dozen O
cookies, and .4 dozen M cookies. Since the economic contribu
tion of these slack products by definition is zero, this exchange of
real products for slack variables would appear to be a worthwhile
one. Similarly, the substitution of sugar cookies for slack vari
ables would also be a favorable change. It is necessary, therefore,
to make a detailed analysis of the consequences of these two
52 THE SIMPLEX METHOD
alternative changes in the product mix to determine which is most
favorable.
Alternative No. 1 Add iced cookies to product mix.
(1) Add: 1 dozen iced cookies having an economic value of
200 per dozen. Gross gain per unit = 200.
(2) Less: value of products given up to permit production of
each dozen iced cookies.
.15 dozen L cookies @ 0 per dozen = .15 X O0 = 00
1.0 dozen O cookies @ O0 per dozen = 00
.4 dozen M cookies @ 0 per dozen = 00
Contribution loss per dozen = 00
(3) The net change in contribution for each dozen iced cookies
introduced is the difference between the value of the variable
introduced and the variables given up to make this possible
(204 0 = 20 per gain per dozen).
Alternative No. 2 Add sugar cookies to product mix.
(1) Gross gain = 15 per dozen.
(2) Value of products displaced in product mix by each dozen
sugar cookies introduced.
1.0 dozen O cookies @ O0 = 0O
.10 dozen L cookies @ O0 = 0O
Loss of contribution = 0O
(3) Net gain per dozen sugar cookies introduced= 150 0 = 150
One criterion that may be used to select the alternative
variable to be introduced at any stage of a simplex problem is
that of net effect per unit on the measure of effectiveness or
objective function. At this stage of the bakery example, this
criterion would indicate that iced rather than sugar cookies
should be brought into the product mix since they have a higher
net contribution to profit and overhead per dozen.5
This same conclusion was reached in the early stages of the graphical
method analysis on the basis of their contribution per dozen. The simplex
may seem like "the hard way" in this and other respects. Remember, how
ever, that the simplex is also the only way of solving problems of this type
that involve more than two variables.
4.3 PRODUCTMIX EXAMPLE 53
A second criterion that may be used in selecting from
among the alternative variables that might be introduced is that
of net total effect on the objective function. This is defined as the
product of the net effect per unit and the number of units that
could be introduced at that stage of the solution.6 Since the net
contributions per dozen have already been determined, it is now
necessary to determine the number of dozens of iced and sugar
cookies that might be introduced into the origin solution.
The number of units that might be introduced is deter
mined, as in the graphical example, by the restrictions.7
A. If iced cookies are introduced (move from A toward B in
Figure 3.3):
1. L cookies (labor capacity) must be given up.
a. Each dozen iced cookies displaces .15 dozen L cookies.
b. The limit or restriction for L cookies is 15 dozen.
c. The maximum number of iced cookies that may be
introduced in terms of this restriction is, therefore,
100 dozen (point F in Figure 3.3).
2. O cookies (idle oven capacity) must also be given up.
a. Each dozen iced cookies displaces 1.0 dozen O
cookies.
b. The restriction on O cookies is 120 dozen.
c. The maximum number of iced cookies that may be
introduced in terms of this restriction is 120 (point
G in Figure 3.3).
3. In addition, M cookies (idle icing mix) must also be
given up.
a. The rate of substitution or exchange between iced
cookies and M cookies is 1.0 to .4, each one dozen
6 A significant advantage of the total effect criterion, especially when hand
computation methods are being used, is that it should produce an optimum
solution in the fewest total number of tables. In this particular problem it
will require only three tables rather than the four needed by the per unit
criterion.
7 The computations which follow are based on the data in Table 4.3 (page
50). The graphical interpretations are drawn from Fig. 3.3 (page 24).
54 THE SIMPLEX METHOD
iced cookies displaces .4 dozen M cookies.
b. The restriction on M cookies is 32 dozen.
c. Limit on iced cookies corresponding to this restric
tion is 80 (point B in Figure 3.3).
4. The most limiting restriction for iced cookies is M
cookies. Therefore, the number of iced cookies that
could be introduced at this stage is limited to 80 dozen.
(Move to point B in Figure 3.3 (page 24). Solutions F
and G are not technically feasible.)
5. If 80 dozen iced cookies are brought into the solution,
what quantity of the slack products that were represented
in the origin solution will be displaced? The coefficients
of Table 4.3 show that .15 dozen L cookies (actually .15
hours of labor capacity) must be given up for each dozen
iced cookies introduced. Thus, (80 X .15 = 12) dozen
are displaced. Since the total dozens of L cookies in the
origin solution was 15, 3 dozen remain.8
QL (second stage) = 15 80 X .15 = 3 dozen L cookies
Similar reasoning indicates that the values for Qo and QM
at the second stage would be:
Qo (second stage) = 120 80 X 1.0 = 40 (point G 
point B)
Qu1 (second stage) = 32 80 X .4 = 0 (out of solution
at point B; no idle icing mix would be available for the
production of M cookies).
Therefore, if 80 dozen iced cookies were to be introduced,
the variables in the second solution and their quantities
would be QT = 80, Q1, = 3, and Qo = 40.
6. The total net change in the objective function Eq. (4.6)
corresponding to this change in variables would be as
follows:
a. The new value of the objective function (at point B),
TC = 80 X 20 + 0 X 15 + 3 X O0 + 40 X 0 + 0 X
0 = $16.00
8 Note that 3 hours of labor capacity would be sufficient to produce 20
dozen iced cookies (3 .15). This is the difference between points F and B
in Figure 3.3.
4.3 PRODUCTMIX EXAMPLE 55
b. Less the previous value of the objective function (ori
gin), TC = 0 X 200 + 0 X 150 + 15 X O0 + 120 X
0 + 32 X O0 = $0.00
c. Net change = $16.00 0.00 = $16.00
B. If sugar rather than iced cookies were to be introduced (move
from A toward point E in Figure 3.3):
1. L cookies must be allocated to production of sugar
cookies.
a. Limit on sugar cookies would be 150 dozen (15.10).
2. O cookies must also be given up.
a. Limit is 120 (120 1.0).
3. M cookies must also be given up.
a. Limit = infinity (32 0), since sugar cookies require
no icing mix.
4. Most limiting restriction is O cookies.
5. If Qs = 120, then slack production would be:
Qo = 120 120 X 1.0 = 0 (Now out of solution)
QL =15 .10X 120 = 3
QM = 32 120 X 0.0 = 32
6. Net change in objective function:
a. New value = $18.00 (point E)
b. Previous value = $0.00 (point A)
c. Net change = $18.00
C. Conclusion: Sugar cookies should be introduced since the
net total contribution of this variable at this stage of the analy
sis is greater ($18.00) than that of iced cookies ($16.00).
D. The variables in the second stage solution and the quantities of
each would be (from Step B5 above):
Qs = 120, QL = 3, Q_1 = 32.
Both Q, and Qo are, of course, still in the problem but are
not in the solution, since they are zero at this stage of the
problem.
56 THE SIMPLEX METHOD
The initial purpose of Step 3 in the simplex procedure
is to determine if the introduction of any of the problem vari
ables not in the existing solution (origin solution, in this case)
would have a favorable effect on the objective function. The
method of analysis used to evaluate this condition is to deter
mine the net effect per dozen of iced and sugar cookies. Both
have favorable effects in this case because the only products dis
placed are slack variables.
The method suggested here for selecting the variable to be
introduced is the use of a net total effect criterion. This requires
that the maximum number of units that could be introduced be
determined for both iced and sugar cookies.
The net change in contribution was found to be 80 dozen
X 200 per dozen or $16.00 for iced cookies, and 120 dozen X
15 per dozen or $18.00 for sugar cookies.
It was concluded from the analysis of Step 3 that 120 dozen
sugar cookies should be added to the product mix represented
by the existing (origin) solution. As a consequence, it was then
necessary to determine the number of dozens of the slack
products (actually, amounts of the various resources) that were
displaced by the introduction of the sugar cookies and the
amounts remaining in the solution.
The graphical analogy of these computations is "moving"
from point A to point E. At point E, 120 sugar cookies and
zero iced cookies are being produced, the oven capacity is fully
utilized, all of the icing mix capacity remains idle, and only 12
of the 15 labor hours are required. The simplex states that:
Qs = 120
Q, =0
Qo =0
Q = 32
Q, = 3
4.34 Step 5. It is necessary to develop a table to represent the
"solution" at each stage of the search procedure. Some of the
necessary information to construct a second table for the bakery
4.3 PRODUCTMIX EXAMPLE 57
example was developed in the later stages of Steps 3 and 4.
This information is shown in Table 4.4.
TABLE 4.4
Contribution 20 15b 0 0 0
Variables Qx Qs QL Qo QM Production
00 QL 3 Dozen L
cookies
150 Qs 120 Dozen Sugar
cookies
0 QM 32 Dozen M
cookies
Note that the middle row heading has changed, since sugar
cookies have displaced O cookies in the origin solution. The
production figures have also changed, due to the change in
product mix adopted in Step 4.
The most trying computation task associated with the
simplex method is the determination of the new coefficients for
Table 4.4. Most of the coefficients in Table 4.3 are no longer
applicable since they represented the rates of substitution be
tween all possible variables and the variables in the solution
at that time. The new set of coefficients may be determined
from the former set. From Table 4.3:
S= (.10L) + (1.00) + (0.0 M)
This expression represents the original recipe for sugar
cookies. It was very useful in the graphical method in that it
indicated what resources and amounts must be allocated to the
production of sugar cookies as the solution (product mix) was
moved from point A to point E. It was also a useful expression
in Steps 3 and 4 of the simpler method inasmuch as it showed
58 THE SIMPLEX METHOD
exactly what quantities of the variables in the solution (all slacks
at the origin) had to be given up to produce each dozen sugar
cookies.
Having arrived at a secondstage solution in the simplex
method, the mathematical equivalent of point E, Table 4.4 does an
adequate job of describing the product mix (QL = 3, Qs = 120,
and QM = 32) in its present form. What is needed now is a new
set of coefficients, which describe the exchange rates or rates of
substitution between the new set of variables in the solution (L, S,
and M) and all variables in the problem (1, S, L, 0, and M).
Solving the original sugar cookie recipe for 0 produces
the following expression:
S= S (.10 L) (0.0 M) (4.7)
By substituting the righthand side of Eq. (4.7) for O
wherever the latter is found in the "recipes" applicable to the
origin solution, a new set of coefficients will be found that are
relevant to the secondstage solution.
For example, from Table 4.3:
I = .15 L + 1.0 (0) + .4 M (original coefficients)
= .15 L+ 1.0(S .10L0.0M) +.4M
= .05 L+ 1.S+.4 M (4.8)
L =1.0 L+0(0)+0(M)
= 1.0 L+0(S.10 L0.0M)+0(M)
=1.0 L+0(S) +0(M) (4.9)
S = .10L+1.00+0.0M
= .10L+ 1.0(S.10L0.0M) +0.0M
=0.0 L+I.OS+O.OM (4.10)
M= 0.0 (L) +0.0(0) + 1.0 (M)
=0.0 L +0.0(S .10 L0.0 M) + 1.OM
=0.0 L+O.OS+ .OM (4.11)
4.3 PRODUCTMIX EXAMPLE 59
The coefficients of Eqs. (4.7) through (4.11) may now
be transferred to the appropriate columns of the table describ
ing the solution at the second stage (Table 4.5).
TABLE 4.5
Contribution 20 15 0 0 0
Variables QI Qs QL Qo QM Production
0 QL .05 0 1 .10 0 3 Dozen L
cookies
15 Qs 1 1 0 1 0 120 Dozen sugar
cookies
00 QM .4 0 0 0 1 32 Dozen M
cookies
The solution described is that defined by point E in Fig
ures 3.3 and 3.6. The only "real" production is 120 dozen
sugar cookies. In the graphical method it was acceptable to indi
cate that the product mix represented by this point did not require
all of the labor or icing mix capacities. The amount of this slack
capacity in the labor resource was defined as an amount equal
to the production requirement for 30 dozen sugar cookies
(point H point E). In the simplex method, this slack is defined
as the capacity required to produce 3 dozen L cookies. The
latter may be converted back to the former through their ratio
of labor requirements as follows:
One dozen L cookies require 1.0 labor hours
One dozen sugar cookies require .10 labor hours
The slack labor capacity at point E = 3 dozen L cookies
or 30 dozen sugar cookies (3 .10= 30).
The point to be stressed once again is that the use of slack
variables does not change the essential structure of the problem.
The meaning of the coefficients in Table 4.5 may be in
terpreted as follows:
60 THE SIMPLEX METHOD
A. The introduction of each dozen O cookies into the solu
tion indicated by Table 4.5 would require:
1. Giving up one dozen sugar cookies because of
limited oven capacity.
2. This reduction in the number of sugar cookies
will release labor capacity equal to .10 dozen L
cookies. Thus, the negative coefficient of .10
means that .10 dozen L cookies would be added
to the product mix.
3. There is no change in the output of M cookies
(QM) since the coefficient is zero.9
B. The introduction of each dozen iced cookies requires
giving up:
1. One dozen sugar cookies.
2. An additional .05 dozen L cookies.
3. Fourtenths (.4) dozen M cookies.
C. Introduction of L or M cookies would have no effect
on the solution. Present units in the solution would
simply be exchanged on a oneforone basis for new
units. Similar reasoning may be applied to the in
troduction of sugar cookies.
4.35 Step 3, Second Stage. The evaluation process for Table
4.5 (using total effect criterion) may be facilitated by the use
of a tabular presentation, as shown in Table 4.6. This table
represents a tabular summary of the computations required in
Steps 3 and 4. The general method for making these computa
tions is the same as that given under these two steps for the
firststage solution. For that reason, they will only be sum
marized here:
A. Enter the gross gain for each variable.
B. Determine the loss of contribution due to the fact that
a portion of the variables now in the solution must
" Note the similarity of reasoning (but differences in answers) between the
substitution of O cookies for sugar cookies here and the substitution of
sugar cookies for 0 cookies in connection with Table 43. Reintroduction of
O cookies would, in effect, return the solution to that at the origin.
TABLE 4.6
Computation Q1 Qs QL Qo QM
Gross 20 150 0 00 0
A. gainperozen 20 15
Loss per
dozen L .05 x 04= 04 Ox 0= 0 1 X 04 = 0 .10x 04= 04 0x 04=04
B. S 1 x 15 = 15 1 x 154 = 15 0 x 154 = 0 1 x 154 =154 0 x 15 = 04
M .4 X 0 = 0 X = 0 0 0 = 00X x 04= 0 1 X 00 = O
Total 154 154 04 154 04
O
C
Net Change i
C NetChange 20 15 = +5 15 15 = 04 0 04=0 0 15 =154 04 0= 04
per dozen
Number
D. of dozens L 3+ .05 = 60 mi
X
S 120+ 1=120
M 32 .4= 80
Total Change Mr
E. in Objective
Function 60 X 54 = $3.00
O
62 THE SIMPLEX METHOD
be given up in order to introduce other products.
(Multiply coefficients for each variable by contribu
tion of product given up.)
C. Find net change per unit by subtracting total of
Computation B from Computation A for each vari
able. (Note that only iced cookies have a positive
contribution at this stage of the analysis. Reintro
ducing O cookies would decrease the value of the
objective function by 150 per dozen and return the
solution to the origin.)
D. For all products having a favorable net effect per
dozen (iced cookies only in this case), determine the
most limiting restriction. (This is labor capacity in
the case of iced cookies.)
E. Determine the total effect on the objective function for
all variables that were carried through Step D.
It is not essential that a table such as 4.6 be constructed
for each iteration. Once the reader understands the reasons for and
the logic behind the computations involved, several shortcuts
should be evident. For example, Computations C and D can be
performed mentally from the data in Table 4.5.10
4.36 Step 4. The conclusion implied by Table 4.6 is that 60
dozen iced cookies should be brought into the solution. As a
consequence, the number of sugar cookies will be reduced to 60
dozen (120 1 X 60 60), the number of M cookies to 8 dozen
(32 .4 X 60 = 8), and L cookies will be reduced to zero (3 
60 X .05 = 0). The corresponding value of the objective func
tion (Eq. [4.6]) is now:
60 X 20+ + 60 X 150 + 0 X 01 + 0 X 0 + 8 X O0 = $21.00
10 Other evaluation procedures and notational methods will, no doubt, sug
gest themselves as the reader's experience with the simplex method increases.
The important thing is to keep things straight rather than to follow the
methods suggested here.
4.3 PRODUCTMIX EXAMPLE 63
4.37 Step 5. Iced, sugar, and M cookies (idle icing mix) are
now in the solution.11 This means that another new set of rela
tionships (coefficients) must be determined for the table that
describes this solution. Since iced cookies replaced L cookies
in the product mix, from Table 4.5:
I = (.05)L + (1)S+ (.4)M
L = I S .4M = 201 20S 8M
.05
0 = .10(L) + 1.0(S) + O.O(M)
= .10(201 205 8M) + 1.0(S) + 0(M)
= 2(1) + 2(S) + .8(M) + 1(S) + 0(M)
=2(1) + 3(5) + .8(M)
I = .05(201 20S 8M) + 1(S) + .4(M)
= 1(1) 1(S) .4(M) + 1(S) + .4(M)
= 1(1) + 0(S) +0(M)
The new rates of substitution, together with the other
information developed in Step 4, may now be used to construct
the table representing the third stage.
TABLE 4.7
Contribution 200 150 0 00 0
Variables QI Qs QL Qo QM Production
20 Qi 1 0 20 2 0 60 Dozen iced
cookies
150 Qs 0 1 20 3 0 60 Dozen sugar
cookies
0 Q:M 0 0 8 .8 1 8 Dozen M
cookies
11 This product mix is that represented by point D in Figure 3.3.
64 THE SIMPLEX METHOD
4.38 Step 3. Although the reader "knows" that the thirdstage
solution is optimal, the final step in the simplex method requires
"proof" of this optimality; i.e., no variable has a favorable net
effect per unit on the measure of effectiveness. The evaluation
procedure, shown in Table 4.8, is based on the information
given in Table 4.7.
Since no variable has a favorable effect on the objective
function (note net change per dozen), it is not necessary to carry
the computations further. The optimal product mix is, therefore,
60 dozen each of sugar and iced cookies. Technically, 8 dozen M
cookies are also in the optimal mix but, since they represent a
fictitious product, they are not a "real" portion of the answer.
Actually, this means that 8 pounds of icing mix will remain on
hand at the end of the day.
4.4 SUMMARY
The simplex is the general method of linear programming.
Any allocation problem that involves linearity and certainty can
be solved through the use of this method.
Any linear programming method may be likened to an
effective search procedure. The purpose of the search is to find
that set of values for all variables that is technically feasible
and that optimizes an appropriate objective function. The term
"effective" is used in the sense that an optimal solution is
obtained with the least possible computation effort.
The simplex computation procedure is more complex
than those of special methods such as the graphical and trans
portation methods because the absence of simplifying factors and
assumptions makes it necessary to rely entirely on mathematical
expressions.
The mathematics involved in the simplex method requires
the use of equations rather than inequalities. To meet this re
quirement it is necessary to express unused or idle resources in
the form of slack variables. In the bakery example, these
TABLE 4.8
Computation QI Qs QL Qo Q3
A. Gross 20 15 0 0 0
gain per dozen 20 1500
I 1 x 200 = 200
S 0x 15 = o0
M OX 00 = 0O
0 x200 = 0
1 x 150 = 150
OX 0 = 0
20 x 200 = $4.00 2 x 200 = 400
20 x 150 = $3.00 3 x 150 = 450
8 x 0O = 0o .8 x 0o = 0o
0 x 200 = 0 '1
0 x 15i = o0 0
1 x 0o = 0 C
0

Total 200 15 $1.00 50 0
m
00 0o = 0o
U
r
m
m
Loss per
dozen
C. Net change
oer dozen
20 200 = 0 15 154 = 0 0 $1.00 = $1.00 0 50 = 54
__
L
I
66 THE SIMPLEX METHOD
slack variables represented fictitious products that were, by defini
tion, produced with all resources not required by real products in
the problem.
It is customary to start a simplex analysis at the origin
solution since this is the simplest of the many possible solutions
to formulate. Subsequent steps in the procedure involve bring
ing into the solution those variables that are shown to have a
favorable effect on the measure of effectiveness. When all such
favorable alternatives have been exhausted, the optimal solution
has been reached and the analysis completed.
As with any quantitative method of analysis, the optimal
solution to a simplex analysis is optimal only in the sense that
all pertinent factors in the problem have been treated explicitly
in the analysis. Qualitative factors may require modification of
this solution before a final decision can be made.
Most business problems that justify analysis within a sim
plex framework involve many variables and constraints. The use
of a computer is, therefore, frequently required for economical
computation. The most important steps in the problemsolving
process from a managerial viewpoint are, therefore, those of fram
ing the problem and evaluating the results of the analysis.
4.5 REVIEW QUESTIONS
1. Why is the simplex method the general method of
linear programming?
2. Explain why all relationships between variables and
restrictions must be stated in equation form in this method.
3. What are slack variables? Why are they necessary
in the simplex method? What effect do they have on the "an
swers" to problems analyzed by this method?
4. What is meant by a basic solution? The "origin" solu
tion? Why is the latter normally used as the initial solution?
4.6 EXERCISES 67
5. What do the coefficients within a simplex table repre
sent? Why is it necessary to compute a new set for each table
in the analysis?
6. What criteria may be used in determining which vari
able to bring into the solution at each stage of the computation
procedure? What specific advantage does the total effect criterion
have?
7. How does one know when an optimal solution has been
reached?
4.6 EXERCISES
1. Rework the bakery problem. Include cookie mix in your analysis
and use contributions of 250 for iced and 150 for sugar cookies.
a. Write the basic equations.
b. Develop the origin solution.
c. Determine which real product to introduce.
d. Develop the second stage solution table.
e. Carry your anaylsis to the optimum solution.
f. Explain the meaning of your final solution.
2. Use the simplex method to frame and solve the Zeus Company
problem in Section 3.8. Follow Steps (a) to (f) in Problem 1
above.
3. Solve the Lotanoiz Corporation problem in Section 3.8, using
the simplex format.
4. Determine the optimal product mix for the Plummer Company
(Section 3.8) by use of the simplex method.
5. The bakery manager is considering the introduction' of a larger
size sugar cookie ("Sugar Kings") into his product line. Several
experimental batches have been produced and sold. Data col
lected during these runs indicate that: (a) 1.2 pounds of cookie
mix and .12 hours of labor are required per dozen, and (b) direct
costs should be 65t per dozen. Limited market experience sug
gests that 95 per dozen would be an appropriate initial price.
68 THE SIMPLEX METHOD
a. Given these new facts and assuming no change in the other
factors in the bakery problem:
(1) Write the set of linear equations which describes this
threeproduct case.
(2) Develop an origin solution.
(3) Carry the problem to an optimum solution.
(4) Frame the problem using a threedimensional display. Ex
plain your analysis by means of this diagram.
b. Suppose that the baker decides: (a) to limit his initial pro
duction of "Sugar Kings" to 20 dozen per day, and (b) to use
the remaining resources for iced and regular sugar cookies.
(1) Determine the optimal product mix and expected contri
bution per day.
5 THE TRANSPORTATION
METHOD
5.1 GENERAL NATURE OF THE METHOD
COMPLEX ALLOCATION problems having certain characteris
tics may be solved by a special, highly simplified version of the
simplex referred to as the transportation, distribution, or stepping
stone method of linear programming. It is especially appropri
ate for sourcetodestination situations such as the transportation
of goods from plants to distribution facilities. The same solution
framework, however, may be applied to a wide variety of prob
lems. Thus, the characteristics of the problem itself rather
than its institutional or functional setting determine whether or not
this method is applicable.
The key characteristic of such problems is homogeneity.
All rates of substitution between variables must be one to one.
Such a condition was illustrated in the bakery example by the
relationship between the real products and the oven resource;
one dozen iced cookies could be substituted for a dozen sugar
70 THE TRANSPORTATION METHOD
cookies in terms of oven capacity and vice versa. Since the
other rates of substitution in the product mix example were not
onetoone, the key characteristic of the transportation method
was not met by that problem.
Like all linear programming methods, the transportation
method is an iterative process. After an initial solution is form
ulated, the computational procedure provides an effective man
ner for developing improved solutions until the optimum is
reached. The nature and meaning of the steps in this procedure
will be demonstrated in the following sample problem.
THE McCLAIN CONSTRUCTION COMPANY
The McClain Company will have four major construction
projects under way next month, an unusually high level of activ
ity for the firm. As a consequence, the volume of materials that
must be moved from the local supply yard exceeds the capacity
of the company's truck fleet. Rather than purchase additional
trucks and hire new drivers, the company management has de
cided to contract with local haulers for as much of the total
hauling task as is justified on an economic basis.
The expected hauling requirements, expressed in truck
loads, for the construction sites during the next month are as
follows:
Requirements
Site (in loads)
No. 1 60
No. 2 90
No. 3 75
No. 4 65
TOTAL 290 loads
5.1 GENERAL NATURE OF METHOD 71
Three local trucking firms have submitted bids indicating
their price per load from the local supply yard to each site
and the maximum number of loads they would be willing to
contract for. This information is summarized as follows:
Cost per load
Site Firm A Firm B Firm C
No. 1 $6 $3 $6
No. 2 $6 $6 $5
No. 3 $7 $4 $4
No. 4 $7 $5 $3
Maximum 100 loads 80 loads 70 loads
Capacity
Expected costs and capacity of the McClain Company's
truck fleet for the same period are as follows:
Cost per load
Site McClain Co.
No. 1 $5
No. 2 $3
No. 3 $5
No. 4 $6
Maximum Capacity 60 loads
Since substantial differences exist in the relative costs of
the various assignments that might be made, the McClain Com
pany must now decide which hauling requirements will be as
signed to company trucks and which to each of the trucking firms.
72 THE TRANSPORTATION METHOD
5.2 STEP 1: FRAME THE PROBLEM
The technical phases of the problem should be used to
develop the basic computation framework or matrix required in
the transportation method.
The total hauling capacity (sources) available to the com
pany is the sum of the individual capacities.
Capacity (McClain) 60
+Capacity (Firm A) +100
+Capacity (Firm B) + 80
+ Capacity (Firm C) + 70
= TOTAL CAPACITY = 310 LOADS (5.1)
The total material requirements of the various construction
sites (destinations) may also be expressed in equation form.
Requirements (No. 1) + Requirements (No. 2) + Require
ments (No. 3) + Requirements (No. 4) = Total Material
Requirements
60 + 90 + 75 + 65 = 290 loads (5.2)
Eqs. (5.1) and (5.2) indicate that the total hauling
capacity available exceeds hauling requirements for the period.
The transportation method, like the simplex, requires that slack be
explicitly recognized. One way of doing this in this particular
problem would be to create an artificial or fictitious site requiring
20 loads of hauling capacity.1 The addition of this slack site to
Eq. (5.2) results in the following expression:
60 + 90 + 75 + 65 + 20 = 310 loads (5.3)
Eqs. (5.1) and (5.3) may now be displayed as the technical or
rim requirements of the problem, as shown in Table 5.1.
1 Occasionally, problems will be encountered where requirements exceed
capacity. In such cases, slack would represent unfilled orders.
5.2 STEP 1: FRAME THE PROBLEM 73
The technical matrix of a transportation problem serves
the same basic purpose as the graphical display of the graphical
method and tables of coefficients used in the simplex method.
All frame the problem in such a manner as to facilitate the
search for better and better solutions.
The 20 cells within Table 5.1 represent the various source
todestination assignments that are technically feasible. For ex
ample, the 100load capacity of Firm A may be used, in whole or
TABLE 5.1
Sources of Hauling Capacity
Firm Material
Site M A B C Requirements
No. 1 60
No. 2 90
No. 3 75
No. 4 65
Slack 20
Hauling
Capacity 60 100 80 70 310
74 THE TRANSPORTATION METHOD
in part, to meet the needs of any of the four sites. A portion of
it (up to 20 units) could also be allocated to meet the slack re
quirement.2 Any combination of assignments for Firm A would
be acceptable as long as the total equaled exactly 100 units.
Similarly, the 60unit requirement of Site No. 1 may be met by
any combination of assignments among the various hauling re
sources as long as the total equals 60.
As in the graphical and simplex methods, the "best" or
optimal solution to a transportation problem (a) is feasible, that
is, it satisfies all rim requirements; and (b) maximizes or mini
mizes an appropriate objective function.
The economic data pertinent to the decision are those costs
associated with the allocation of a particular source of hauling
capacity to each site or destination. For example, each load
hauled to Site No. 1 by Firm A will cost $6; by Firm B, $3; by
Firm C, $6; and by company trucks, $5. Considered alone, the
lowest cost source for this site would be Firm B at a total cost of
$180 (60 x $3). Considerations at other sites may, however,
call for supplying Site No. 1 from another source if overall cost
is to be minimized.
The technical and economic data are summarized in tabu
lar form in Table 5.2. This matrix indicates in explicit form the
basic alternatives (cells), the costs associated with each alter
native, and the rim requirements. The fact that the problem is
framed within the transportation method also indicates, implicitly,
that the rates of substitution are all onetoone.3
The final task in framing the problem is to state an
explicit objective function. Assuming that the quality of the
service provided by the various truck fleets is equal, the mini
mization of total hauling costs for the month would be a rea
sonable criterion.4
2 An allocation of 20 units of Firm A's capacity to the slack site would
mean, in effect, that only 80 units would be contracted for.
3 The homogeneity assumption in this problem means that one "load" is the
same whether it is hauled in company or contractor trucks.
4 Technically, the assumptions relative to service are not required. If some
explicit cost could be assigned to poor service, this cost could be added to
the prices within each cell and a purecostminimization criterion adopted.
5.3 STEP 2: INITIAL SOLUTION 75
TABLE 5.2
Sources of Hauling Capacity
Firm Material
Site M A B C Requirements
$5 $6 $3 $6
No. 1 60
$3 $6 $6 $5
No. 2 90
$5 $7 $4 $4
No. 3 75
$6 $7 $5 $3
No. 4 65
$0 $0 $0 $0
Slack 20
Hauling
Capacity 60 100 80 70 310
5.3 STEP 2: DEVELOP AN INITIAL SOLUTION
In the transportation method, unlike the simplex, the
initial solution need not be the origin solution. The only require
ment of an initial solution in the transportation method is that it
be technically feasible; that is, it must not violate any of the
rim requirements. This means that the amount of subsequent
76 THE TRANSPORTATION METHOD
computational work is, in part, a function of the starting position
selected.
There are several standard procedures for developing ini
tial solutions. Two appropriate ones for the beginning student are
the Northwest Corner Rule and the North to South Row Rule, a
simple inspection method. Other more sophisticated methods
can be used to advantage once the basic structure of the trans
portation method has been mastered.
5.31 Inspection Method for Initial Solution. The inspection
method used here is a simple one which will be referred to as
the North to South Row Rule. This rule operates as follows:
A. Starting with the first or north row, fill the require
ments of each row, in order, using the lowest cost
assignment available within the limits imposed by
previous allocations.
B. After all row requirements have been met, add across
each row and down each column to insure that all rim
requirements have been met.
The North to South Row Rule will be applied to the McClain
problem in the remainder of this section.
Inspection of Table 5.3 shows that the lowest cost capac
ity for the 60load requirement at Site No. 1 is Firm B. This
TABLE 5.3
Sources of Hauling Capacity
Material
Site M A B C Requirements
$5 $6 $3 $6
No. 1 I6 60
60
Hauling
Capacity 60 100 80 70 310
5.3 STEP 2: INITIAL SOLUTION 77
firm has 80 loads available so 60 loads are assigned as shown
in cell 1B. Only 20 loads of Firm B's capacity are now available
to other sites.
For Site No. 2 the McClain fleet represents the lowest
cost resource but only 60 loads are available (Table 5.4). The
additional 30 loads must, therefore, be secured from Firm C,
which represents the next lowest cost capacity. Once these assign
ments have been recorded, the requirements for both Sites No. 1
and No. 2 have been met.
TABLE 5.4
Sources of Hauling Capacity
Material
Site M A B C Requirements
(in loads)
$5 $6 $3 $6
No. 1 0 60
60
$3 $6 $6 $5
No. 2 0 90
60 30
Hauling
Capacity 60 100 80 70 310
Table 5.5 shows two lowest cost sources for Site No. 3
Firms B and C. However, a portion of their capacities have
already been allocated. Also, the McClain fleet has been fully
assigned. Therefore, the most economical assignment which can
be made at this point is the one shown in Table 5.5.6
5 A feasible alternative would be to secure all 75 loads from Firm A. Such
an allocation would not, however, follow our rule of using the lowest cost
assignment available.
78 THE TRANSPORTATION METHOD
TABLE 5.5
Sources of Hauling Capacity
Material
Site M A B C Requirements
(in loads)
$5 $6 $3 $6
No. 1 60
60
$3 $6 $6 $5
No. 2 60 / 0 90
S60 30
$5 $7 $4 $4
No. 3 75
15 20 40
Hauling
Capacity 60 100 80 70 310
The initial solution may now be completed by assigning
65 loads of hauling capacity to Site No. 4 and 20 loads to the
slack site. Table 5.6 shows that, as a result of previous alloca
tions, it is necessary to select Firm A even though it has the high
est cost. Further analysis in later stages will enable adjustments in
this and other assignments to be made in the event they are war
ranted.
5.3 STEP 2: INITIAL SOLUTION 79
TABLE 5.6
Sources of Hauling Capacity
Material
Site M A B C Requirements
(in loads)
$5 $6 $3 $6
No. 1 / 60
/ 60
$3 $6 $6 $5
No. 2 / 90
S60 30
$5 $7 $4 $4
No. 3 75
15 20 40
$6 $7 $5 $3
No. 4 65
65
$0 $0 $0 $0
Slack 0 20
Hauling
Capacity 60 100 80 70 310
Initial Solution: North to South Row Rule
The meaning of the initial solution in Table 5.6 may be
interpreted as follows:
(1) Company trucks will haul 60 loads to Site No. 2.
80 THE TRANSPORTATION METHOD
(2) Only 80 loads of the 100 load capacity offered by
Firm A will be contracted for. Of the units hired,
15 will haul to Site No. 3 and 65 to Site No. 4.
(3) Sixty loads of Firm B's capacity will be assigned to
Site No. 1 and 20 to Site No. 3.
(4) Thirty loads of Firm C's capacity will be assigned to
Site No. 2 and 40 to Site No. 3.
This solution is feasible in the sense that none of the rim
requirements have been violated. This is the only necessary con
dition for an initial solution; it need not make much economic
sense. The North to South Row Rule for initial solutions did
make use of cost data but on a row by row rather than an overall
basis.
The cost of this solution (objective function) may be
determined by summing the costs for the various sites.
Site No. 1: $3 X60 =$ 180
Site No. 2: $3 X 60 + $5 X 30 = 330
Site No. 3: $7 X 15 + $4 20 + $4 X 40 = 345
Site No. 4: $7 X 65 = 455
Total Cost (Initial Solution) $1310
5.32 The Northwest Corner Rule for Initial Solution. There is
nothing special about the northwest corner of a matrix. This
socalled "rule" is simply a convenient convention which may be
stated as follows:
A. Starting at the upper left hand or northwest corner,
fill the requirements of each row, in order, from the
columns, in order.
B. Check for compliance with rim requirements.
This rule has been applied to the McClain problem in
Table 5.7. The 60load requirement of the first row is met with
the capacity of the first (M) column. Move down to the second
5.3 STEP 2: INITIAL SOLUTION 81
TABLE 5.7
Sources of Hauling Capacity
Material
Site M A B C Requirements
No. 1 / 60
60
No. 2 / 90
90
No. 3 /10/5 75
No. 4 5 65
115 50
Slack / 0 20
Hauling
Capacity 60 100 80 70 310
Initial Solution: Northwest Corner Method
row. This requirement is met with 90 units from the second
(A) column. The remaining 10 units are applied toward the 75
requirement of the third row. An additional 65 units must be
secured from the third (B) column to fill Row 3. The 15 units
left in Column B are applied to Row 4. The capacity of Column
C is then used to meet the remaining requirement of Row 4
and the slack site row.6 Adding rows and columns indicates that
this is a feasible solution.
Initial solutions based upon the Northwest Corner method
can always be recognized by their stairstep appearance. Be
6 It is unusual when only one column is needed to fill one row as in Cell
1M. When this occurs, it is a sign of degeneracy, a condition to be dis
cussed in the next section.
82 THE TRANSPORTATION METHOD
cause the method is concerned exclusively with developing this
pattern, it ignores costs. For example, the McClain trucks would
have been assigned to Site No. 1 no matter what cost appeared in
cell 1M. The cost data were omitted from the cells of Table 5.7
to illustrate this point.
When properly applied, both the North to South Row in
spection method and the Northwest Corner Rule will produce
feasible initial solutions. Because the latter ignores costs, more
computations are usually required to optimize the problem.7
Both are acceptable methods, so that the basis for selecting be
tween them is essentially one of personal preference.
5.4 STEP 3: TEST THE SOLUTION FOR DEGENERACY
Degeneracy is a special condition which can arise in any
type of linear programming problem. Since it is most likely to
occur even in elementary problems when using the transportation
method, the beginning student should be able: (a) to recognize
the condition, and (b) to devise measures for dealing with it.
Cells within which assignments have been made represent
the variables currently in a particular solution. For example, in
the initial solution of Table 5.6, cells 1B, 2M, and all others
with entries are used cells. The unused cells (lM, 2A, . .)
are those variables in the problem which are not yet in the
solution.
In order to carry out the computation procedure it is
necessary that the number of used cells within the matrix conform
to the Rim Minus One Rule.
Used Cells = No. of Rows + No. of Columns 1 (5.4)
There are five rows and four columns in the McClain problem.
Inspection of Table 5.6 reveals that there are eight used cells.
This solution is ready for Step 4. The initial solution of Table
5.7 is degenerate since this matrix contains only seven used cells.
7 The cost of the initial NW Corner solution is $1395 versus $1310 for the
inspection solution. The more "fat" in the initial solution, the more iterations
are required.
5.5 STEP 4: EVALUATE UNUSED CELLS 83
The measures which must be taken on this matrix will be more
readily understood if they are discussed in a later section.8
5.5 STEP 4: EVALUATE UNUSED CELLS
The evaluation process of Step 4 tests each unused cell for
its effect on the objective function. If an unused cell has a favor
able effect (can reduce cost in this case), it is brought into the
solution (becomes a used cell). Each cell handled in this fash
ion must displace a former used cell in order to comply with the
Rim Minus One rule. Two alternative procedures for evaluating
unused cells will be presented; the SteppingStone and MODI
methods.
5.51 The SteppingStone Method. What would happen if one
load were tentatively assigned to cell 2B as in Table 5.8? Note
TABLE 5.8
Sources of Hauling Capacity
Material
Site M A B C Requirements
No. 1
60
No. 2 6 3 90
,6'0 ^ 1 , 30
No. 3 0
No. 4
Slack
Hauling
Capacity 80
8 See Section 5.10.
84 THE TRANSPORTATION METHOD
that such an assignment would result in violations of the rim re
quirements for both Site 2 and Firm B. This means that, if cell
2B is to be utilized, compensating adjustments must be made in
other assignments. For example, if the McClain fleet were to haul
59 rather than 60 loads to Site No. 2, the change would become
feasible (59 + 1 + 30 = 90). An alternative would be to decrease
the loads destined for Site No. 2 which are carried by the trucks of
Firm C (60 + 1 + 29 = 90). To make the proposed change feas
ible in terms of Firm B, the number of loads to be hauled to Site
No. 1 might be decreased to 59 (59 + 1 + 20 = 80), or the loads
to Site No. 3 decreased by 1 (60 + 1 + 19 = 80).
For reasons that will become apparent later, suppose
that the decision is made to decrease cell 2C to 29 and cell 3B
to 19 loads. Has the full solution now been restored to a state of
technical feasibility? An inspection of Table 5.9 suggests that it
has not. Specifically, the change has produced secondorder dif
TABLE 5.9
Sources of Hauling Capacity
Material
Site M A B C Requirements
No. 1 60 60
No. 2 0 1 90
60 1 29
No. 3 75
No. 4 65
65
Slack 20
20
Hauling
Capacity 60 100 80 70 310
5.5 STEP 4: EVALUATE UNUSED CELLS 85
ficulties (side effects) within Row 3 and Column C. Site No. 3
now has only 74 loads assigned, rather than the required 75,
and Firm C has been reduced to 69 loads. The only appropriate
remedy would be to allocate one additional load to cell 3C (41
loads). Once this change has been made, the solution is again
feasible. Thus, the first step in evaluating unused cells using the
SteppingStone method is to find the chain of assignment changes
(evaluation route) necessary to make the proposal consistent with
the rim requirements.
Once these changes have been determined, we determine
their effect on the objective function (cost consequences in this
problem). The net results of the cell 2B proposal are sum
marized:
One load was added to cell 2B (0 1) and to cell 3C
(40 41). These changes would add $10 ($6 + $4) to the total
hauling cost. At the same time, units were tentatively subtracted
from cell 2C (30 29) and from cell 3B (20 19), with a
corresponding decrease in hauling cost of $9 ($5 + $4). Thus,
the net economic effect of these changes would be to increase the
total cost of the solution by $1 for each load shifted in this manner.
Since the objective is to reduce total hauling cost, the introduction
of cell 2B into the solution would not represent an improvement
in the assignment mix.
B C
$6 $5
2
+1 load 1 load
$4 $4
3
Load +1 load
86 THE TRANSPORTATION METHOD
The mechanics of determining evaluation routes is facili
tated by following these rules:
(1) Only one unused cell may be evaluated at a time.
(2) Other than the unused cell being evaluated, only used
cells may be part of an evaluation route.9
(3) There should always be only one unique route for
each cell.10
Generally the least difficult routes to visualize are those where
all adjustments in assignments can be made within four adjacent
cells, as for cell 2B in the sample problem. Routes forming a rec
tangle, three corners of which are used cells, are also easy to iden
tify. Some routes, however, involve more than four cells and
appear to be geometrical equivalents of the "calf path."
An additional illustration should be helpful in developing
the meaning of these rules. From Table 5.6, the evaluation of
cell 1C would proceed as follows. The addition of one load to
this cell would require that one unit be subtracted from a used
cell both in Row 1 and in Column C. The latter adjustment might
be made in either cell 2C or 3C, while the former can be made
only in cell 1B since this is the only used cell in Row 1. Reduc
tion of cell 1B by one unit means that one load must now be
added to some used cell in Column B. Cell 3B represents the
only possibility. The addition of one unit to cell 3B would then
require subtraction of one unit from some used cell in Row 3,
either 3A or 3C. Using the latter will complete the cycle in
that this will also satisfy the requirement that one unit be de
ducted from a used cell in Column C. The value of cell 1C would
be +$3 (+ 6 3 + 4 4). This, of course, is not a favorable
move.
9 Early descriptions of this method referred to unused cells as "water" and
to used cells as "stones." Thus, only "stones" could be used in evaluating
"water" cells; hence the term "SteppingStone method" arose.
1( As long as the number of used cells equals rimminusone, this will be true.
When two or more feasible routes are found, more assignments have been
made than necessary, and the solution can be simplified. Where no route is
available, degeneracy is indicated.
5.5 STEP 4: EVALUATE UNUSED CELLS 87
Other cells that are relatively easy to evaluate because
the evaluation route follows the three filled corners of a rec
tangle format, are indicated below:
Unused Cell Route Economic Effect
3M 3M, 3C, 2C, 2M +$3 per unit
SC SC, SA, 3A, 3C +$3 per unit
SB SB, SA, 3A, 3B +$3 per unit
4B 4B, 4A, 3A, 3B +$1 per unit
2A 2A, 2C, 3C, 3A $2 per unit
4C 4C, 4A, 3A, 3C $1 per unit
The evaluation routes for the other cells are somewhat
more difficult. The route for cell 1M, for example involves
1M, 1B, 3B, 3C, 2C, and 2M, and has an economic effect of
+$4. Evaluation of cell SM involves SM, 2M, 2C, 3C, 3A, and
SA. Each unit added to this cell will increase the total cost of
the assignment mix by $5. Obviously, this transfer does not
represent a favorable change.
For convenience and ease of analysis it is conventional to
note the net economic effect for each unused cell within the
cell, as shown in Table 5.10. Note that only two cells have
negative values (2A and 4C). These represent opportunities
to decrease the total cost of the hauling problem."
5.52 The MODI Method for Evaluating Unused Cells. The
Modified Distribution or MODI Method is an alternative to the
11 Since this is a cost minimization problem, a positive net economic effect
represents increased total cost. For profit maximization problems, where the
economic data within each cell represent profits rather than costs, positive
cell values would represent favorable changes on the measure of effectiveness.
88 THE TRANSPORTATION METHOD
TABLE 5.10
Cell Evaluations: First Stage Solution
Sources of Hauling Capacity
Material
Site M A B C Requirements
$5 $6 $3 $6
No. 1 60
+4 0 60 +3
$3 $6 $6 $5
No. 2 + 90
__ 60(32 +1 30
$5 $7 $4 $4
No. 3 + 75
+3 15 20 40
$6 $7 $5 $3
No. 4 65
+4 65 +1 Q
$0 $0 $0 $0
Slack 20
+5 20+3 +3
Hauling
Capacity 60 100 80 70 310
SteppingStone procedure for evaluating unused cells. It is some
what more expedient and proceeds as follows:
A. The first row is arbitrarily assigned a zero coefficient.
B. This coefficient together with the cost or profit entries
in the used cells are utilized to determine coefficients
for all other rows and the columns.
C. Each unused cell is evaluated by subtracting the sum
of the corresponding row and column coefficients
from the cell entry.
5.5 STEP 4: EVALUATE UNUSED CELLS 89
Steps (A) and (B) of the MODI Method for evaluating
unused cells are applied in Table 5.11 to the initial solution
of Table 5.6. Only the relevant portion of the matrix (cost
entries in used cells) is shown. The magnitude of the used cell
TABLE 5.11
Determination of Row and Column Coefficients: MODI Method
Ri K =O KA= KB =3 Kc = Kj
Kz~=1
assignments and the cost entries in unused cells are omitted
since they are not relevant to these steps.
First, a coefficient of zero is arbitrarily assigned to Row 1.12
12 Actually the arbitrary zero can be assigned to any row or any column.
The first row convention will be used here.
90 THE TRANSPORTATION METHOD
Next, the other row and column coefficients are determined using
Eq. (5.5).
Ri + K, = Cij (5.5)
where
Ri = coefficient of the ith row
Kj = coefficient of the jth column
Cij = cost or profit entry in used cell ij
For example, the entry in cell 1B is equal to the sum of the coeffi
cients for.Row 1 and Column B (RI +KB =C1B). Since both
R, and Cli are known, the equation can be solved for the unknown
column coefficient.
KB = C1B Ri = 3 0 = 3
Now this column coefficient can be used, together with the cell
entry in 3B, to determine the coefficient of Row 3.
Rs = C3B KB= 4 3 = 1
The remaining computations would be as follows:13
R4 =C4A KA = 7 6 = 1
KA = CA R3 = 7 1 = 6
Rs = CA KA = 0 6 = 6
Ko = C3 R3 = 4 1 = 3
R2 =C2c Kc =53=2
KM =C2M R2 =3 2 =1
Two important points about these computations should
be noted. First, the sequence in which the coefficients are de
termined is, with minor exceptions, significant. Second, there is
only one way in which each coefficient can be computed.
Step (C) of the MODI Method uses the row and column
coefficients together with the entries in unused cells as shown in
Table 5.12. Used cells are not relevant to this step and are omitted
from the table.
13 The reader should follow these computations using Table 5.11 and enter
each coefficient in the table as it is determined.
