• TABLE OF CONTENTS
HIDE
 Front Cover
 Half Title
 Title Page
 Preface
 Table of Contents
 Linear programming and the decision...
 Linearity and linear equations
 The graphical method
 The simplex method
 The transportation method
 Bibliography






Group Title: The Allyn and Bacon series in quantitative methods for business and economics
Title: Introduction to linear programming
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 Material Information
Title: Introduction to linear programming
Series Title: The Allyn and Bacon series in quantitative methods for business and economics
Physical Description: ix, 112 p. : illus. ; 22 cm.
Language: English
Creator: Stockton, Robert Stansbury
Publisher: Allyn and Bacon
Place of Publication: Boston
Publication Date: 1963
Copyright Date: 1963
Edition: 2d ed.
 Subjects
Subject: Linear programming   ( lcsh )
Business mathematics   ( lcsh )
Programmation linéaire   ( rvm )
Mathématiques financières   ( rvm )
Genre: bibliography   ( marcgt )
non-fiction   ( marcgt )
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Table of Contents
    Front Cover
        Front Cover
    Half Title
        Page i
    Title Page
        Page ii
        Page iii
        Page iv
    Preface
        Page v
        Page vi
    Table of Contents
        Page vii
        Page viii
        Page ix
    Linear programming and the decision process
        Page 1
        Page 2
        Page 3
        Page 4
        Page 5
        Page 6
        Page 7
    Linearity and linear equations
        Page 8
        Page 9
        Page 10
        Page 11
        Page 12
        Page 13
        Page 14
        Page 15
        Page 16
        Page 17
    The graphical method
        Page 18
        Page 19
        Page 20
        Page 21
        Page 22
        Page 23
        Page 24
        Page 25
        Page 26
        Page 27
        Page 28
        Page 29
        Page 30
        Page 31
        Page 32
        Page 33
        Page 34
        Page 35
        Page 36
        Page 37
        Page 38
        Page 39
    The simplex method
        Page 40
        Page 41
        Page 42
        Page 43
        Page 44
        Page 45
        Page 46
        Page 47
        Page 48
        Page 49
        Page 50
        Page 51
        Page 52
        Page 53
        Page 54
        Page 55
        Page 56
        Page 57
        Page 58
        Page 59
        Page 60
        Page 61
        Page 62
        Page 63
        Page 64
        Page 65
        Page 66
        Page 67
        Page 68
    The transportation method
        Page 69
        Page 70
        Page 71
        Page 72
        Page 73
        Page 74
        Page 75
        Page 76
        Page 77
        Page 78
        Page 79
        Page 80
        Page 81
        Page 82
        Page 83
        Page 84
        Page 85
        Page 86
        Page 87
        Page 88
        Page 89
        Page 90
        Page 91
        Page 92
        Page 93
        Page 94
        Page 95
        Page 96
        Page 97
        Page 98
        Page 99
        Page 100
        Page 101
        Page 102
        Page 103
        Page 104
        Page 105
        Page 106
        Page 107
        Page 108
        Page 109
        Page 110
        Page 111
    Bibliography
        Page 112
Full Text


















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: 63-19120

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 PROBLEM-SOLVING 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

PRODUCT-MIX 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-
lem-solving, 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 decision-making
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 decision-making process on an
explicit basis. The assumption here is that the decision-making
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 problem-solving 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 lowest-cost 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 straight-line
method of depreciation assumes that capital value decreases at
a linear rate. Break-even 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 so-called 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 PROBLEM-SOLVING


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, problem-solving 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 inter-relationships 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. (2-1),

SY -Y1- 11-5 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 =2X-X
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=46--4X
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 break-even 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 break-even 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 problem-solving 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 decision-making 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 short-run 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
decision-maker. 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 metal-working 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 so-called
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 break-even point) or to profit (for all output over the
break-even 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 short-run 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 one-half 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 re-evaluate 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
over-all 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 two-dimensional
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 decision-making,
not in eliminating it. It is important, therefore, to recognize that
the optimal solution from a linear programming model is not
necessarily an optimal over-all 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 decision-maker 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 "Long-Life" and the "Economuf." The
"Long-Life" 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 owner-managers 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
Long-Life 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 man-hours 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 higher-priced Chieftain line uses more first
grade materials and is virtually hand-made. The lower-priced
Warrior line is largely machine-made 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 long-service employees work 36-hour 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 five-dimen-
sional problem, let alone a twenty-dimensional 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 PRODUCT-MIX EXAMPLE


The product-mix problem represents an elementary prob-
lem capable of analysis by the simplex as well as by the graphi-







4.3 PRODUCT-MIX 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 PRODUCT-MIX 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 PRODUCT-MIX 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 PRODUCT-MIX 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 3-6
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
120-dozen 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 PRODUCT-MIX 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 icing-mix 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 PRODUCT-MIX 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 PRODUCT-MIX 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 B-5 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 PRODUCT-MIX 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 second-stage 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 right-hand 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 second-stage solution.
For example, from Table 4.3:

I = .15 L + 1.0 (0) + .4 M (original coefficients)
= .15 L+ 1.0(S- .10L-0.0M) +.4M
= .05 L+ 1.S+.4 M (4.8)

L =1.0 L+0(0)+0(M)
= 1.0 L+0(S-.10 L-0.0M)+0(M)
=1.0 L+0(S) +0(M) (4.9)

S = .10L+1.00+0.0M
= .10L+ 1.0(S-.10L-0.0M) +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 L-0.0 M) + 1.OM
=0.0 L+O.OS+ .OM (4.11)







4.3 PRODUCT-MIX 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. Four-tenths (.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 one-for-one 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
first-stage 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 4-3. 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 PRODUCT-MIX 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 third-stage
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 problem-solving
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
three-product case.
(2) Develop an origin solution.
(3) Carry the problem to an optimum solution.
(4) Frame the problem using a three-dimensional 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 source-to-destination 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
one-to-one, 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-
to-destination assignments that are technically feasible. For ex-
ample, the 100-load 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 60-unit 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 one-to-one.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 pure-cost-minimization 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 60-load 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 1-B. 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
so-called "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 60-load 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 stair-step appearance. Be-

6 It is unusual when only one column is needed to fill one row as in Cell
1-M. 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 1-M. 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 1-B, 2-M, and all others
with entries are used cells. The unused cells (l-M, 2-A, . .)
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 Stepping-Stone and MODI
methods.

5.51 The Stepping-Stone Method. What would happen if one
load were tentatively assigned to cell 2-B 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
2-B 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 2-C to 29 and cell 3-B
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 second-order 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 3-C (41
loads). Once this change has been made, the solution is again
feasible. Thus, the first step in evaluating unused cells using the
Stepping-Stone 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 2-B proposal are sum-
marized:


One load was added to cell 2-B (0 1) and to cell 3-C
(40 41). These changes would add $10 ($6 + $4) to the total
hauling cost. At the same time, units were tentatively subtracted
from cell 2-C (30-- 29) and from cell 3-B (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 2-B 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 2-B 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 1-C 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 2-C or 3-C, while the former can be made
only in cell 1-B since this is the only used cell in Row 1. Reduc-
tion of cell 1-B by one unit means that one load must now be
added to some used cell in Column B. Cell 3-B represents the
only possibility. The addition of one unit to cell 3-B would then
require subtraction of one unit from some used cell in Row 3,
either 3-A or 3-C. 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 1-C 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 "Stepping-Stone method" arose.
1( As long as the number of used cells equals rim-minus-one, 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

3-M 3-M, 3-C, 2-C, 2-M +$3 per unit
S-C S-C, S-A, 3-A, 3-C +$3 per unit
S-B S-B, S-A, 3-A, 3-B +$3 per unit
4-B 4-B, 4-A, 3-A, 3-B +$1 per unit
2-A 2-A, 2-C, 3-C, 3-A -$2 per unit
4-C 4-C, 4-A, 3-A, 3-C -$1 per unit



The evaluation routes for the other cells are somewhat
more difficult. The route for cell 1-M, for example involves
1-M, 1-B, 3-B, 3-C, 2-C, and 2-M, and has an economic effect of
+$4. Evaluation of cell S-M involves S-M, 2-M, 2-C, 3-C, 3-A, and
S-A. 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 (2-A and 4-C). 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



Stepping-Stone 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 1-B 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 3-B, 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 =5-3=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.




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