CHAPTER 8

CHAPTER 8:

Linear Programming

Linear Programming (LP)

• LP is widely used mathematical modeling
technique designed to helps manager in
planning and decision making relative to
resource allocation.

• A few example of problems in which LP
has been successfully applied in
operations management are;

– Selecting the product mix in a factory to make
best use of available firm’s machine-hour and
labor-hour for maximum profit.

– Picking blends of raw materials in feed mills to
produce finished feed combinations at
minimum cost.

• The graphical procedure is useful only when
there are two decision variable in the problems.
When there are more than two variables, it is not
possible to plot the solution on a two-
dimensional graph and we must turn to more
complex approaches is called simplex method.

Requirements of LP Problems.

• LP Problems seek to maximize or minimize an
objective, usually profit or cost. We refer to this
property as the objective function of an LP problem.

• The present of restriction or constraints limit
(resources) the degree to which the objective can
be obtained.

• There must be alternative (course of action)
available. If there were no alternative, we would not
need LP.

• The objective and constraint in LP problems must
be expressed in terms of linear equation or
inequalities.

Basic Assumptions of LP

• We assume that conditions of certainty exist, do not
change during the period being studied.

• We also assume that proportionality exists in the
objective and constraint.

• Technical assumption deals with additivity, meaning
that the total of all activities equals the sum of the
individual activities.

• We make the divisibility assumption that solutions
need not be in whole numbers (integers).

• We assume that all answers or variables are non-
negative.

Formulating LP Problems

• A key principle of LP is that interactions exist
between variables.

• The feasible region is the overlapping area of
constraints that satisfies all of the restrictions on
resources. Recall that the theory of LP states the
optimal solution will lie at a corner point of feasible
region.

• The iso-profit method is the first method we
introduce for finding the optimal solution and

• A second approach is the corner point method.
Both methods were shown in the solution of the
example below.

Example:

Flair Furniture Company

• The company produces inexpensive tables and chairs. The
production process for each is similar in that both require a
certain number of hours of carpentry work and a certain
number of labor hours in the painting & varnishing
department. Each table takes 4 hours of carpentry and 2
hours in the painting & varnishing. Each chair requires 3
hours in carpentry and 1 hour in painting & varnishing.
During the next production period (a week), 240 hours of
carpentry time are available and 100 hours in painting &
varnishing time are available. Each table sold yields a profit
of RM 7.00; each chair produced is sold for a RM5.00 profit.

• Flair Furniture’s problem is to determine the best possible
combination of tables and chairs to manufacture in order to
reach the maximum profit.

Solution:

• Let us introduce some simple notation for use in the
objective function and constraints:
X1 = number of tables to be produced
X2 = number of chairs to be produced

• Create the LP objective function in term of X1 and X2.
The objective function is maximize profit; let say
Z = RM7 X1+ RM5 X2.

• The amount of a resource used is to be less than or
equal to (≤) the amount of resource available.

Hours required to produce

1 unit

Department Table (X1) Chair (X2) Available
Hours
week this
240
Carpentry 4 3 100
1
Painting & 2
RM5
Varnishing

Profit per unit RM7

From the table,
4 X1 + 3 X2 ≤240 (hours of carpentry time).

Similarly, the second constraint is a as follow: Painting &
varnishing time used is ≤ painting & varnishing time
available.

2 X1 + 1 X2 ≤100 (hours of painting & varnishing)

X1 and X2 ≥0 (No. of tables and chairs produced
greater than or equal to zero)

• Graphical Representation of constraint:
• To find the optimal solution to a linear

programming problem, we must first
identify a set, or region of feasible
solutions.
• First step: Plot the constraints graph.
• Second step: Find feasible solution or
feasible region.
• Third step: Find the optimum solution

• Method 1: Using Iso-profit line solution
method

We start the technique by letting profits equal some
arbitrary but small dollar amount. For the Flair Furniture
problem we may choose a profit of RM210. This is a
profit level that can be obtained easily without violating
either of the two constraints. The objective function can
be written as RM210 = 7 X1+ 5 X2. This expression is
just the equation of a line; we call it an iso-profit line.

• It represents all combinations of (X1; X2) that would yield
a total profit of RM210. Obviously, the profit line for
RM210 does not produce the highest possible profit for
the firm.

• We now have two clues as to how to find the optimal
solution to the original problem. We can draw a series of
parallel lines (by carefully moving our ruler in a plane
parallel to the first profit line) until the profit line touches
the last feasible region corner point.



• Method 2: Corner point solution method

This technique is simpler than the iso-profit line
approach, but it involves looking at the profit at every
point of the feasible region. The mathematical theory
behind LP states that an optimal solution to any problem
(that is, the values of X1, X2 that yield the maximum
profit) will lie at a corner point or extreme point of the
feasible region. Hence, it is only necessary to find the
values of the variables at each corner; the maximum
profit or optimal solution will lie at one (or more) of them.

• To find the( X1, X2 ), values producing the maximum profit, we
find the coordinates of each corner point and test their profit
levels as shown in the Figure 7.3 and calculation below:

• Point 1:(X1 = 0, X2= 0) profit= RM7(0) + RM5(0)= RM0
• Point 2:(X1 = 0, X2= 80)profit= RM7(0) + RM5(80)= RM400
• Point 3: (X1 = 50, X2= 0)profit= RM7(50) + RM5(0)= RM350
• Point 4: (X1 = 30, X2= 40)profit = RM7(30) + RM5(40)= RM410