Ch 7 LINEAR PROGRAMMING
Linear programming1 is a mathematical
technique that enables a decision maker to
arrive at the optimal solution to problems
involving the allocation of scarce resources.
Typically, many economic and technical
problems involve maximization or minimization
of a certain objective subject to some
restrictions.
1During
World War II US army began to formulate certain
linear optimization problems. Their solutions were called plans
or programs. Today important application areas include airline
crew scheduling, shipping or telecommunication networks, oil
refining and blending, and stock and bond portfolio selection.
1
The History of Linear Programming
Leonid Kantorovich
George Dantzig
2
Programming problems, in general, are
concerned with the use or allocation of
scarce resources - labor, materials,
machines, and capital - in the ”best”
possible manner so that costs are
minimized or profits maximized.
In using the term ”best possible” it is
implied that some choice or set of
alternative courses of actions is
available for making the decision.
3
Typical Applications of Linear
Programming
1. A manufacturer wants to develop a
production schedule and inventory
policy that will satisfy sales demand in
future periods and same time minimize
the total production and inventory cost.
2. A financial analyst must select an
investment portfolio from a variety of
stock and bond investment alternatives.
He would like to establish the portfolio
that maximizes the return on
investment.
4
Typical Applications of Linear Programming
continued
3. A marketing manager wants to
determine how best to allocate a fixed
advertising budget among alternative
advertising media such as radio, TV,
newspaper, and magazines. The goal is
to maximize advertising effectiveness.
4. A company has warehouses in a number
of locations throughout the country. For
a set of customer demands for its
products, the company would like to
determine how much each warehouse
should ship to each customer so that the
total transportation costs are minimized.
5
Constructing Linear Programming
Next we list what is required in order
Models
to construct a linear programming
model:
1. Objective Function. There must be
an objective (or goal or target) the
firm or organization wants to
achieve. For example, maximize
dollar profits, minimize dollar cost,
maximize total number of expected
potential customers, minimize total
time used, and so forth.
6
Constructing Linear Programming Models
continued
2. Restrictions and Decisions. There must
be alternative courses of action or
decisions, one of which will achieve the
objective.
3. Linear Objective Function and Linear
Constraints. We must be able to express
the decision problem incorporating the
objective and restrictions on the
decisions using only linear equations
and linear inequalities. i.e., we must be
able to state the problem as a linear
programming model.
7
Economic Significance of Linearity
The simplifying assumption of linearity
causes some problems:
1. In profit-maximizing production
problem linearity of the objective
function implies constant profit rate
per unit as output increases; this
means
a) that the selling price is constant;
and
8
b) that average variable cost is constant;
the law of diminishing returns does not
influence the production process, and
input prices are constant
 perfect competition in output and input
markets
2. Linear resource constraints imply
constant combination of inputs;
this means constant returns to scale.
9
Summary of the Economic Implications
of the Linearity Assumption
Linear objective
function
Constant gross profit per unit
(GP = P - AVC = constant)
Price (P) is
constant
Constant returns
to variable inputs
Constant
input prices
Firm is a price taker
in the output market
10
The three basic steps in constructing
a linear programming model:
Step I
Identify the unknown variables to
be determined (decision
variables), and represent them in
terms of algebraic symbols.
11
Step II
Identify all the restrictions or
constraints in the problem and
express them as linear equations
or inequalities which are linear
functions of the unknown
variables.
12
Step III
Identify the objective or criterion
and represent it as a linear
function of the decision variables,
which is to be maximized or
minimized.
13
Example 1 Product-Mix Problem
The Handy-Dandy Company wishes
to schedule the production of a
kitchen appliance which requires
two resources – labor and material.
The company is considering three
different models of this appliance
and its engineering department has
furnished the following data:
14
Resources required Product/Model
C
B
A
to produce 1 unit
6
3
7
Labour (hours/unit)
5
4
4
Material (lbs./unit)
3
2
4
Profit ($/unit)
Resources
Available
150
200
The supply of raw materials is
restricted to 200 pounds per day. The
daily availability of manpower is 150
hours. Formulate a linear programming
model to determine the daily
production rate of the various models
of appliances in order to maximize the
total profit.
15
Step I
Identify the Decision Variables. The
unknown activities to be determined
are the daily rate of production for the
three models (A, B, C) in order to
maximize the total profit. Representing
them by algebraic symbols,
xA = daily production of model A
xB = daily production of model B
xC = daily production of model C
16
Step II
Identify the Constraints. In this
problem the constraints are the
limited availability of the two
resources (labor and material).
Model A requires 7 hours of labor for
each unit, and its production quantity
is xA. Hence, the requirement of
manpower for model A alone will be
7xA hours (assuming a linear
relationship).
17
Similarly, models B and C will
require 3xB and 6xC hours,
respectively. Thus, the total
requirement of labor will be
7xA + 3xB + 6xC, which should not
exceed the available 150 hours. So
the labor constraint becomes:
7xA + 3xB + 6xC  150
18
Similarly, the raw material constraint
is given by
4xA + 4xB + 5xC  200
In addition, we restrict the variables
to have non-negative values. This is
called the non-negativity constraint,
which the variables must satisfy
xA, xB and xC  0.
19
Step III
Identifying the Objective. The
objective is to maximize the total
profit from the sales. Assuming
that perfect market exists for the
product such that all that is
produced can be sold, the total
profit from sales becomes
Z = 4xA + 2xB + 3XC.
20
Thus, the linear programming model
for our product mix problem is:
Find numbers xA, xB, xC which will
maximize
Z = 4xA + 2xB + 3XC
subject to the constrains
7xA + 3xB + 6xC  150
4xA + 4xB + 5xC  200
xA  0, xB  0, xC  0
21
Exercise 1: Formulating an LP-Problem
Advertising Media Selection
An advertising company wishes to plan an advertising campaign in three different
media – television, radio, and magazines. The purpose of the advertising program is
to reach as many potential customers as possible. Result of a market study are given
below:
Cost of an
Advertising
unit, in $
TV,
TV,
day time prime time Radio
Magazines
40 000
75 000
30 000
15 000
# of potential
customers
400 000
reached/unit
900 000
500 000
200 000
# of women
customers
300 000
reached/unit
400 000
200 000
100 000
The company does not want to spend more than $800 000 on advertising. It further
requires that (1) at least 2 million exposures take place among women; (2)
advertising on TV be limited to $500 000; (3) at least 3 advertising units be bought
on day time TV, and two units during prime time; and (4) the number of advertising
units on radio and magazines should each be between 5 and 10.
22
Formulate as an LP-problem.
Solving Linear Programming
Problems

Graphical Technique

First graph the constraints:
the solution set of the system is
that region (or set of ordered
pairs), which satisfies ALL the
constraints. This region is called
the feasible set
23
Solving Linear Programming Problems:
Graphical Technique continued
￭
Locate all the corner points of the
graph:
the coordinates of the corners will be
determined algebraically
It is important to note that the optima
is obtained at the boundary of the
solution set and furthermore at the
corner points.
For linear programs, it can be shown
that the optima will always be
obtained at corner points.
24
Solving Linear Programming Problems: Graphical
Technique continued
￭
Determine the optimal value:
test all the corner points to see
which yields the optimum value for
the objective function
Objective function
Feasible
set
Optimum
25
Example 2
Suppose a company produces two types of widgets, manual and electric.
Each requires in its manufacture the use of three machines; A, B, and C. A
manual widget requires the use of the machine A for 2 hours, machine B
for 1 hour, and machine C for 1 hour. An electric widget requires 1 hour
on A, 2 hours on B, and 1 hour on C. Furthermore, suppose the maximum
numbers of hours available per month for the use of machines A, B, and C
are 180, 160, and 100, respectively. The profit on a manual widget is $4
and on electric widget it is $6. See the table below for a summary of data.
If the company can sell all the widgets it can produce, how many of each
type should it make in order to maximize the monthly profit?
Manual Electric
A
B
C
profit
2
1
1
$4
1
2
1
$6
Hours
available
180
160
100
26
Example 2 continued
Step I
Identify decision variables:
x = number of manual widgets
y = number of electric widgets
Step II
Identify constraints:
2x + y  180
x + 2y  160
x + y  100
x0
y0
27
Example 2 continued
Step III
Define objective function:
max P = 4x + 6y
Solving P for y gives
y = -2/3 + P/3.
This defines a so-called ”family” of
parallel lines, isoprofit lines.
Each line gives all possible combinations
of x and y that yield the same profit.
28
Example 2 continued
29
Example 2 continued
30
Exercise 2: Solving an LP-problem:
A California vintner has available 660 lbs
of Cabernet Sauvignon (CS) grapes, 1860
lbs of Pinot Noir (PN) grapes, and 2100
lbs of Barbera (B) grapes. The vintner
makes a Pinot Noir (PN) wine, which
contains 20% CS, 60% PN, and 20% B
grapes and sells $3 a bottle, and a
Barbera (B) wine, which contains 10%
CS, 20% PN, and 70% B grapes and sells
for $2 a bottle. Assuming each bottle of
wine requires 3 lbs of grapes, determine
how many bottles of each type of wine
should be produced to maximize income.
31
Solving an LP-problem continued:

Algebraic Technique



The graphical method solving linear
programming problems can be used
for problems with two variables
(with some difficulty three)
However, for problems were the
number of variables might run into
hundreds or thousands, algebraic
techniques must be used
The simplex method, with the aid of
the computer, can solve these
problems
32
As with the graphical procedure, the
simplex method finds the optimal
corner-point solution of the set of
feasible solutions.
Regardless of the number of decision
variables and regardless of the
number of constraints, the simplex
method uses the key property of a
linear programming problem, which is:
33

A linear programming problem always
has an optimal solution occurring at a
corner-point solution
Simplex method begins with a feasible
solution and tests whether or not it is
optimum.
If not optimum, the method proceeds to
a better solution.
34
Solution of a maximum-type linear
programming problem by the
simplex algorithm involves the
following steps:
1. Adding slack variables to convert
the inequalities into equations
In the case of less-than-or-equal-to
constraints, slack variables are used
to increase the left-hand side to
equal the right-hand side limits of
the constraint conditions.
35
Example:
max Z = 3x1 + x2
st.
2x1 + x2  8
2x1 + 3x2  12
adding slacks and rewriting the object
row:
2x1 + x2 + s1
=8
2x1 + 3x2
+ s2
= 12
-3x1 – x2
+Z =0
where x1, x2, s1 and s2 are non-negative.
36
2x1 + x2 + s1
=8
2x1 + 3x2
+ s2
= 12
-3x1 – x2
+Z=0
2. Setting up the initial simplex
tableau
Form an augmented coefficient
matrix:
x1
x2
s1
s2
Z
s1  2 1 1 0 0 8 


s2  2
3 0 1 0 12
Z  3  1 0 0 1 0 
37
3. Finding an initial feasible solution
Simplex algorithm starts always from
origin (if possible). In our case it
means that the initial feasible solution
is:
x1 = 0, x2 = 0, s1 = 8, s2 = 12.
Simplex method proceeds from this
corner point to an adjacent corner
point. Such corner points are called
basic feasible solutions.
38
4. Introducing basic and nonbasic
variables as the natural way to
express basic feasible solutions
For any basic feasible solution
(B.F.S.), the variables held zero are
called nonbasic variables and the
other are called basic variables.
Moving form one B.F.S. to another
means that one basic variable (we
call it a departing or exiting variable)
becomes nonbasic and a nonbasic
variable (entering variable) becomes
a basic variable.
39
40
5. Choosing the proper pivot element
to advance the solution and
maintain the non-negativity of all
variables
How to decide which variable to
make basic and which nonbasic (in
other words in which direction to
move from the current B.F.S.)?
41
The natural direction is the one in which
the value of the objective function
increases the most:
If in Z = 3x1 + x2, x1 is allowed to
become basic, x2 remains at 0 and Z =
3x1; thus for each one-unit increase in
x1, Z increases by 3 units. On the other
hand, if x2 is allowed to become basic, Z
will increase only by one unit if x2 is
increased by one unit. This is one way
to determine the pivot column.
42
Choosing the proper pivot element continued
Another way to determine the pivot
column is to examine the bottom
row of the simplex tableau:
x1
x2
s1
s1  2
1 1
s2  2
3 0
Z  3  1 0
s2
Z
0
1
0
0 8
0 12
1 0 
indicators
entering
variable
43
The bottom row entries to the left of
the vertical line are called indicators.
We choose the column with the most
negative indicator as the pivot column.
Having chosen the pivot column, we
must now determine the pivot row in
order to know which element in our
simplex tableau is the pivoting
element.
44
For that purpose we divide the right
hand side entries of the simplex tableau
by corresponding entries of the pivot
column.
Of the resulting quotients we choose
the smallest (minimum quotient). So the
pivot element is the intersection of the
pivot column and pivot row.
45
The pivot element is the intersection of
the pivot column and pivot row:
departing
variable
x1
x2
s1
s1  2
1 1
s2  2
3 0
Z  3  1 0
s2
Z
Quotients
0
1
0
0 8
0 12
1 0 
8 ÷ 2 = 4 (smaller)
12 ÷ 2 = 6
entering variable (most negative indicator)
46
Since x1 and s2 will be the basic
variables in our new B.F.S, it would
be convenient to change our
previous tableau by elementary row
operations into a form where the
values of x1, s2, and Z can be read
off with ease just as in the initial
basic solution.
47
To do this we want to find a matrix
which is equivalent to the tableau above
but which has the form:
x1
x2
s1
s2
Z
1
0


0
?
?
?
?
?
?
0
1
0
0
0
1
?

?
?

where the question marks represent
numbers to be determined.
48
6. Pivoting, which is done column wise
We must transform the tableau to an
equivalent matrix that has a 1 at the
place of the pivot element (pivot entry)
and 0’s elsewhere in the column x1.
x1
departing
variable
s1
s2
Z
 2*

 2
 3

x2
s1
s2
1 1
3 0
1 0
0
1
0
Z
0
0
1
8

12 
0 
entering variable
49
By elementary row operations, we have:
x1
R1
R2
R3
½ R1
 2
 2


 3
Z
s2
s1
x2
1
1
0
0
3
0
1
0
1
0
0
1
1
1
 1
0
2
2

0 1
3
 2
 3  1 0 0

s1 s2
x2
x1
-2R1 + R2
3R1 + R3
1

0
0

1
2
1
2
0
0
1
8 
12 

0 

4 

12 
0 

Z
0
0
2
1
1
0
1
2
3
2
0
1
4 

4 
12 

50
7. Continuing and recognizing when
the algorithm terminates.
General termination rule: all values in
the indicator row are non-negative.
51
We have a new simplex tableau:
x1
x1  1

s 2 0
Z 0
x2
1
2
s1
1
2
s2
Z
0
0
2
1
1
0
1
2
3
2
0
1
4

4
12 
indicators
All values in the indicator row are non-negative.
Hence, we have found the optimal solution for
the problem.
The solution is: x1= 4, x2= 0 and Z= 12
52