

Operations
Research (MTH601)
82
Formulate
this problem as a linear programming
model by defining each component of
the model separately
and
then
combining the components into a single
model.
Decision
Variables
The
decision confronting management in
this problem is how many
plates and mugs to produce.
As such,
there
are two decision variables
that represent the number of
plates and mugs to be
produced on a daily basis.
The
quantities
to be produced can be represented
symbolically as,
X1 =
the number of plates to
produce
X2 =
the number of mugs to
produce
The
Objective Function
The
objective of the company is to
Maximize total profit. The
company's profit is the sum
of the individual
profits
gained from each plate
and mug. As such, profits
from plates is determine by
multiplying the unit profit
for
each
plate, Rs. 4, by the number
of plates produced, X1.
Likewise, profit derived
from mugs is the unit
profit of a
mug,
Rs. 5, multiplied by the
number of mugs produced,
X2.
Thus, total profit, Z, can be
expressed mathematically
as
Maximize
Z = 4X1 +
5X2
where
Z
= total profit per
day
Rs
4X1 =
profit from plates
Rs
5X2 =
profit from mugs
By
placing the term Maximize in
front of the profit function, the
relationship expresses the
objective of the
firm
to Maximize total profit.
Model
Constraints
This
problem has two resources
used for production, which
are limited, labor and
clay. Production of
plates
and
mugs require both labor
and clay. For each
plate produce, one hour of
labor is required. Therefore,
the labor
used
for the production of plates
is 1X1 hours.
Similarly, each mug requires
two hours of labor; the
labor used for
the
production of mugs is 2X2 hours.
Thus, the labor used by
the company is the sum of
the individual amounts
of
labor
used for each
product.
1X1 +
2X2
However,
the amount of labor
represented "1X1 +
2X2" is limited to 40 hrs per
day, thus, the complete
labor
constraint
is
82
Operations
Research (MTH601)
83
1X1 +
2X2 < 40 hours
The
"less than or equal to ( < )"
inequality is employed instead of an
equality ( = ) because the
forty hours
of
labor is a maximum limitation
that can be used, but
not an amount that must be
used, but not an amount
that must
be
used. This allows the
company more flexibility in
that it is not restricted to
use the 40 hours exactly,
but whatever
amount
necessary to Maximize profit up to
and including forty hours.
This means that the
possibility of "idle or
excess
capacity" (i.e., the amount
under forty hours not
used) exists.
The
constraint for pottery clay is
formulated in the same way
as the labor constraint.
Since each plate
requires
four pounds of clay, the
amount of clay used daily
for the production of plates
is 4X1 pounds,
and since
each
mug requires three pounds of
clay, the amount of clay
used for mugs daily is
3X2.
Given that amount of clay
available
for production each day is
120 pounds, the material
constraint can be formulated
as
4X1 +
3X2 < 120 pounds
A
final restriction is that
the number of plates and
mugs produced be either zero
or a positive value, since it
would
be impossible to produce negative
items. These restrictions
are referred to as nonnegative
constraints and are
expressed
mathematically as
X1 > 0,
X2 > 0
The
complete linear programming
model for this problem can
now be summarized as
Maximize
Z
= Rs. 4X1 +
5X2
Subject
to
1X1 +
2X2 < 40
4X1 +
3X2 < 120
X1,
X2 > 0
The
solution of this model will
result in numerical values
for X1 and
X2, which will maximize total
profit,
Z.
As one possible solution,
consider X1 =
5 plates and X2
= 10 mugs.
First we will substitute this
hypothetical
solution
into each of the constraints in
order to make sure that
the solution does not
require more resources than
the
constraints
show are available.
1(5)
+ 2(10) < 40
25
< 40
4(5)
+ 3(10) < 120
50
< 120
and
4(5)
+ 3(10) < 120
50
< 120
Thus,
neither one of the
constraints is violated by this hypothetical
solution. As such, we say
the solution is
feasible
(i.e., it is possible). Substituting
these solution values in the
objective function gives Z = 4(5) +
5(10) = Rs.
70.
However, the maximum
profit.
Now
consider a solution of X1 = 10 plates and X2 = 20
mugs, This would result in a
profit of
Z
= Rs. 4(10) + 5 (20) = 40 +
100 = Rs. 140
83
Operations
Research (MTH601)
84
While
this is certainly a better solution in terms of
profit, it is also infeasible
(i.e., not possible) because
it
violates
the resource constraint or
labor:
1(10)
+ 2(20) < 40
50
< 40
Thus,
the solution to this problem
must both Maximize profit
and not violate the
constraints. The
actual
solution
to this model which achieves
this objective is X1 =
24 plates and X2 = 8
mugs, with a corresponding
profit
of
Rs. 136.
84
Table of Contents:

