Algebra review
Functions. A certain variable Y is said to be a function of X if the relationship between
those variables is such that there is one and only one value of Y for each value of X.
Functions are often referred to as (mathematical) models. The notation generally used for
a linear function is:
Y f( X) => f( X) m. X b
Simple linear equations. Most of the problems in this course can be framed and solved
using simple linear equations that relate one variable (e.g., profits) to one other variable
(e.g., unit sales volume). These equations take the (familiar) slope-intercept form of a
simple linear equation presented in most introductory algebra courses:
Y m. X b
where Y and X are the dependent and independent variables, respectively; and the
parameters m and b are referred to as the slope coefficient and intercept term,
respectively.
Graph of simple linear function. As will become clear during this course, it is often
helpful to draw (or at least think about) the graph of a particular function used in solving
a problem. As an example, managerial accountants often want to know what a company's
"break-even point" is in terms of its unit sales volume (i.e., the accountants want to know
how many units of the company's product must be sold in order for the company not to
lose money). Consider the following example of the graph of a linear profit function
where it can easily be seen that: (1) if unit sales voume is 0, then profits equal -1,000; (2)
if unit sales voume is 250, then profits equal 0; and (3) if unit sales voume is 1,000, then
profits equal 3,000.
f( X) 4. X 1000
Profit as function of unit sales volume
4000
3000
Profits
f( X) 2000
1000
0 250 500 750 1000
1000
X
Unit sales volume
Solving equations. Decision-making problems posed in this course are generally limited
to those with one equation and one unknown variable where the algebraic solution is
relatively straightforward. This implies that an understanding of basic algebra is
necessary in this course.
Basic decision-making algorithm
In this course we will use a basic algorithm (i.e., set of procedures) for solving business
decision problems using accounting information:
(1) Decision model. Frame the decision within a linear decision model (e.g., profit
equation or function) based on the optimization objective; and consider whether
the model captures all factors relevant to achieving the optimization objective.
(2) Decision variable. Identify the decision variable the value of which must be
chosen in order to achieve some specific objective (e.g., maximize profits or
minimize costs through choice of unit sales/production volume).
(3) Parameter estimates. Identify or estimate decision model parameters.
(4) Constraints. Identify the nondecision variable value(s) which constrain the
decision to be made (noting that a unique solution to the problem is possible only
if there is no more than one unknown variable value in each equation in the
problem).
(5) Solution. Substitute the parameter estimates and constraint(s) into the decision
model and solve for the unknown value of the decision variable.
Constrained linear optimization models
In this course, we will deal primarily with constrained linear optimization models of the
previously discussed simple linear equations in two variables--one with a known value
and the other of unknown value. As an example of the way in which the basic decision-
making algorithm will be used in this course, consider again the problem of determining
how many units a company must produce and sell in order to break-even (i.e., to achieve
zero profits).
(1) Decision model. Since our objective is to achieve zero profits, the decision
model used in solving this problem is the basic linear profit equation referred to
earlier:
Profit ( Price Variable_cost). X Fixed_cost
To simplify this example for ease of exposition, this equation can be rewritten as
P ( CM). X FC
(The definition of the terms in these profit equations will be thorougly explained
in the first week of this course, so do not be concerned if they are not familiar to
you at this point.) Note that this profit equation is a simple linear equation that
relates the profit variable (P) to the X variable which represents unit sales volume
through the slope-coefficient parameter CM and the intercept term parameter FC.
(2) Decision variable. Since the problem is to find the unit sales volume necessary to
achieve zero profits, the decision variable that must be chosen to achieve that
objective is (naturally) X--the variable representing unit sales volume.
(3) Parameter estimates. In this course, problems will either require parameters to be
estimated or will be given in the problem. For this example, I will use the
parameters used in the profit function graphed on p. 1 of this material, and
substitute it into the decision model.
P 4. X 1000
(4) Constraints. Since X (unit sales volume) is our decision variable, we must
constrain the value of the one other variable in the equation, P (profits), to a single
value if we are to arrive at a unique solution of how many units must be sold to
achieve zero profits. Since we are interested in zero (break-even) profits, this
implies that P must be set to zero.
( P 0) 4. X 1000
(5) Solution. Soving the above equation (this is a constrained linear optimization
model) with one unknown:
0 4. X 1000
4. X 1000
1000
X
4
X 250
This solution tells us that the company must produce and sell 250 units in order to
achieve zero (break-even) profits. (Note that this solution corroborates what the graph of
this profit function told us on p. 1 of this material.)