What is a limiting factor?
A limiting factor is a factor that prevents a
company from achieving the level of activity that it would like to. Uses the
contribution concept to solve the allocation of scarce resources.
Scarce resources maybe one or more
manufacturing inputs (material, labor, machine time) needed to make a product
and in short supply (the resource i.e.). Production can also be affected by
demand or sales.
Example 1:
Suppose A Ltd makes
two products, X and Y. Both products use the same machine and the same raw
material that are limited to 200 hours and $500 per week respectively.
Individual product details are as follows:
|
|
Product
X per unit
|
Product
Y per unit
|
|
Machine hours
|
5.0
|
2.5
|
|
Materials
|
$10
|
$5
|
|
Contribution
|
$20
|
$15
|
|
Maximum weekly demand
|
50
units
|
100
units
|
Required:
Identify the limiting
factor.
Example 2
Two products, alpha
and Gamma are made of Material X and require skilled labor in the production
process. The product details are as follows:
|
|
Alpha
$
|
|
Gamma
$
|
|
Selling price
|
10.00
|
|
15.00
|
|
Variable cost
|
6.00
|
|
7.50
|
|
Contribution
|
4.00
|
|
7.50
|
|
|
|
|
|
|
Material X required per unit
|
2
kg
|
|
4
kg
|
|
Skilled labor time required per
unit
|
1
hour
|
|
3
hours
|
The maximum demand per
week is 30 units of Alpha and 10 units of Gamma. The company can sell all the
Alphas and Gammas that it can make but there is a restriction on the
availability of both Material X and skilled labor. There are 150 kg of
material, and 45 hours of skilled labor, available per week.
Required:
Identify the limiting
factor.
Contribution per unit
of limiting factor
In order to decide
order of production, the contribution per unit of limiting factor will have to
be calculated. The following illustrate the steps:
A Ltd makes products X
and Y using the same machine and raw materials. Machine hours are limited to
200 hours and materials $500 per week respectively. Details:
|
|
Product
X per unit
|
Product
Y per unit
|
|
Machine hours
|
5.0
|
2.5
|
|
Materials
|
$10
|
$5
|
|
Contribution
|
$20
|
$15
|
As there are two
possible limiting factors, a test will be necessary if one or both are. You can
only apply these steps on one limiting factor situations. In this case, say
machine hours is the limiting factor. The following is then calculated:
|
Contribution per unit of limiting
factor
|
=
|
Contribution
per unit
|
|
Units
of limiting factor required per unit
|
|
Product X
|
$20
/ 5 hours
|
=
|
$4 per machine hour
|
|
Product Y
|
$15/2.5
hours
|
=
|
$6 per machines hour
|
Product Y has a higher
contribution per unit of limiting factor (machine hours) so it should be made
first. The balance of machine hours would then be assigned to Product X.
Optimal production plan
When limiting factors
are present, the objective is to maximize contribution (which in turn maximize profits) by producing
products in the order of the highest amount of contribution per unit of
limiting factor. The profit maximizing production mix is known as the optimal
production plan. Steps:
1)
Calculate the
contribution per unit of product.
2)
Calculate the
contribution per unit of limiting factor (scarce resource)
3)
Rank products
4)
All the scarce
resource to the highest ranked products.
5)
Once demand for
highest ranked products is satisfied, move on to the next highest ranking
product, repeat until the resource is used up.
Example 1
A company is able to
produce four products and is planning its production mix for the following
period. Relevant data is given below:
|
|
A
|
B
|
C
|
D
|
|
Selling price ($) per unit
|
19
|
25
|
40
|
50
|
|
Labor cost per unit ($)
|
6
|
12
|
18
|
24
|
|
Material cost per unit ($)
|
9
|
9
|
15
|
16
|
|
Maximum demand (units)
|
1,000
|
5,000
|
4,000
|
2,000
|
Labor is paid $6 per
hour and labor hours are limited to 12,000 hours in the period.
Required:
Determine the optimal production plan and
calculate the total contribution it earns for the company.
Example 2
The following data
relates to Products Able and Baker.
|
|
|
Product
|
|
|
|
|
Able
|
Baker
|
|
Direct materials per unit
|
|
$10
|
$30
|
|
Direct labor:
|
|
|
|
|
Grinding
|
$5 per hour
|
7
hours per unit
|
5
hours per unit
|
|
Finishing
|
$7.50 per hour
|
15
hours per unit
|
9
hours per unit
|
|
Selling price per unit
|
|
$206.50
|
$168.00
|
|
Budgeted production
|
|
1,200
units
|
600
units
|
|
Maximum sales for the period
|
|
1,500
units
|
800
units
|
Notes:
1)
No opening or closing
inventory anticipated
2)
The skilled labor used
for the grinding processes is highly specialized and in short supply, although
there is just sufficient to meet the budgeted production. However, it will not
be possible to increase the supply for the budget period.
Required:
Determine the optimal
production plan and calculate the total contribution it earns for the company.
Linear
programming
The optimal production
plan is only applicable to situations with only limiting factor. When there are
more than one limiting factor, then a technique known as linear programming is
used.
Formulating a linear programming problem
The steps involved are
as follows:
1)
Define the unknowns,
i.e. the variables that need to be determined.
2)
Formulate the
constraints, i.e. the limitations that must be placed on the variables.
3)
Formulate the
objective function that needs to be maximized or minimized.
Note that
non-negativity constraints will be needed to ensure that there are no negative
values.
Example 1
A company makes two
products, X and Y, and wishes to maximize profit. Information on X and Y is as
follows:
|
|
Product
X
|
|
Product
Y
|
|
Material kg per unit
|
1
|
|
1
|
|
Labor hours per unit
|
5
|
|
10
|
|
|
$
|
|
$
|
|
Selling price per unit
|
80
|
|
100
|
|
Variable cost per unit
|
50
|
|
50
|
|
Contribution per unit
|
30
|
|
50
|
The company can sell
any number of product X, but expects the maximum annual demand for Y to be
1,500 units. Labor is limited to 20,000 hours and materials to 3,000 kg p.a.
Required:
Using the information
given, formulate the linear programming problem.
Example 2
A builder has purchased 21,000 m² of land on
which it is planned to build two types of dwelling, detached and town houses,
within an overall budget of $2.1 million. A detached costs $35,000 to build and
requires 600 m² of land. A town house costs $60,000 to build and requires 300
m² of land. To comply with local planning regulations, not more than 40
buildings may be constructed on this land, but there must be at least 5 of each
type. From past experience the builder estimates the contribution on the
detached house to be about $10,000 and on the town house to be about $6,000.
Contribution is to be maximized.
Required:
Using the information
given, formulate the linear programming problem.
It is possible to
solve to solve linear programming problems with two variables by drawing a
graph of the constraint and the objective function. Steps involved:
1)
Define the unknowns,
i.e. the variables that need to be determined.
2)
Formulate the
constraints, i.e. the limitations that must be placed on the variables.
3)
Formulate the
objective function (that needs to be maximized or minimized).
4)
Graph the constraints
and objective functions.
5)
Determine the optimal
solution to the problem by reading the graph.
Example 1
The constraints and
the objective function of a linear programming problem concerning how many
units of products X and Y to be produced and sold each year have been
formulated to be as follows:
Constraints
|
Materials
|
x
+ y
|
≤
|
3,000
|
|
Labor hours
|
5x
+ 10y
|
≤
|
20,000
|
|
Maximum sales
|
y
|
≤
|
1,500
|
|
Non-negativity
|
x
|
≥
|
0
|
|
|
y
|
≥
|
0
|
The objective function
is to maximize 30X + 50y where:
|
x
|
=
|
number of units of X produced and
sold each year
|
|
y
|
=
|
number of units of Y produced and
sold each year
|
Required:
Determine the optimal
solution to this linear programming problem using a graphical approach.
Example 2
The constraints and
the objective function of a linear programming problem concerning how many
detached houses and town houses to be built have been formulated to be as
follows:
Constraints
|
Area
|
600x
+ 300y
|
≤
|
21,000
|
|
Costs
|
35x
+ 60y
|
≤
|
2,100 (in $000)
|
|
Number
|
x
+ y
|
≤
|
40
|
|
Minimum
|
x
|
≥
|
5
|
|
|
y
|
≥
|
5
|
|
Non-negativity
|
x
|
≥
|
0
|
|
|
y
|
≥
|
0
|
The objective function
is to maximize 10X + 6y where:
|
x
|
=
|
number of detached houses
|
|
y
|
=
|
number of town houses
|
Required:
Determine the optimal
solution to this linear programming problem using a graphical approach. (Note:
do not draw the iso-contribution line in order to determine the optimal
solution.)
Using equations to
solve linear programming problems
Equations can be used
to determine where the two lines cross.
For e.g., in example 1,
we established that the optimal solution was at Point C using the graphical
method. Point C represents the point at which the sales constraint intersects
the labor constraint.
|
Labor constraint
|
5x
+ 10y
|
=
|
20,000
|
(1)
|
|
Materials constraint
|
x
+ y
|
=
|
3,000
|
(2)
|
The basic method is to
eliminate one of the two unknowns between the equations. This is achieved by
adding or subtracting the equations. This is known as solving simultaneous
equations.
Example 1
Solve the following
simultaneous equations.
|
5x
+ 10y
|
=
|
20,000
|
(1)
|
|
x
+ y
|
=
|
3,000
|
(2)
|
Example 2
Suppose that the
optimal solution to a linear programming problem is at the intersection of the
following two constraint lines:
|
2x
+ 3y
|
=
|
1,700
|
(1)
|
|
5x
+ 2y
|
=
|
2,600
|
(2)
|
Required:
Determine the values
of x and y using simultaneous equations.
Establishing the optimal solution using simultaneous equations
Using simultaneous
equations should be done with caution:
·
It is not recommended
that you solve sets of simultaneous equations until you have determined
graphically which constraints are effective in determining the optimal
solution.
·
Solutions using
graphical method relies on graphs being drawn clearly so accurate readings may
be taken.
·
If it is not easy to
read the coordinates of the required points from a graph then the simultaneous
equations provides a reliable check.
Example 3
Alfred Ltd is
preparing its plan for the coming month. It manufactures two products, the
Flaktrap and the Saptrap. Details are as follows:
|
|
Product
|
|
|
|
|
Flaktrap
|
Saptrap
|
Price/wage rate
|
|
Selling price ($)
|
125
|
165
|
|
|
Raw materials (kg)
|
6
|
4
|
$5/kg
|
|
Labor hours
|
|
|
|
|
Skilled
|
10
|
10
|
$3/hour
|
|
Semi-skilled
|
5
|
25
|
$3/hour
|
The company's variable OAR is $1 per labor hour (for both
skilled and semi skilled labor). The supply of skilled labor is limited to
2,000 hours per month and the supply of semi-skilled labor is limited to 2,500
hours per month.
At the selling prices
indicated, maximum demand for Flaxtraps is expected to be 150 units per month
and the maximum demand for Saptraps is expected to be 80 units per month.
Required:
Use simultaneous
equations to solve the linear programming problem given above.
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