Differential Equations

1st Order Equations

Advertising Awareness

Page 1
Problem 1
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Problems 2-3

Differential equations are widely used to describe a variety of dynamic processes in Economic Sciences, Business and Marketing. Below we consider how an advertising campaign can be simulated using differential equations.

Imagine that a company has developed a new product or service. The marketing strategy of the company involves aggressive advertising. To construct the simple model we introduce two variables:

  • The quantity \(q\left( t \right)\) represents advertising activity that is described by spending rate, for example, by the amount of dollars (euros, pounds) the company spends for advertising per week;
  • The quantity \(A\left( t \right)\) represents awareness of the target group of potential consumers about the new goods or services.

Thus, we will consider the market niche as a black box (Figure \(1\)). The advertising activity \(q\left( t \right)\) is the input variable, and awareness of consumers \(A\left( t \right)\) is the output variable that measures response of the system to the advertising campaign.

The market niche as a black box

Figure 1.

A simple model of such type was proposed in \(1962\) and is called the advertising model of Nerlove and Arrow (the N-A model). This model relates advertising activity \(q\left( t \right)\) and awareness of consumers \(A\left( t \right)\) and is given by the differential equation:

\[\frac{{dA}}{{dt}} = bq\left( t \right) – kA,\]

where \(b\) is a constant describing advertising effectiveness, \(k\) is a constant corresponding to decay (or forgetting) rate.

The given equation contains two terms in the right side. The first term \(bq\left( t \right)\) provides the linear growth of awareness of consumers as a result of advertising. The second term \(-kA\) reflects the opposite process, i.e. forgetting about the product. In the first approximation, we can assume that the forgetting rate is proportional to the current level of awareness \(A.\)

This equation is a Linear Differential Equation of First Order. It’s convenient to rewrite it in the standard form:

\[\frac{{dA}}{{dt}} + kA = bq\left( t \right).\]

The integrating factor is the exponential function:

\[u\left( t \right) = {e^{\int {kdt} }} = {e^{kt}}.\]

Therefore the general solution of the given differential equation is given by

\[{A\left( t \right) }={ \frac{{b\int {{e^{kt}}q\left( t \right)dt} + C}}{{{e^{kt}}}}.}\]

As usual, the constant of integration \(C\) can be found from the initial condition \(A\left( {{t_0}} \right) = {A_0}.\)

In the examples below we consider how awareness of customers \(A\left( t \right)\) varies for different advertising modes.

Solved Problems

Click on problem description to see solution.

 Example 1

Management of a company decided to advertise a new product permanently during the year. The advertising budget is $12,000. The coefficients \(k\) and \(b\) are \(k = {\large\frac{1}{4}\normalsize},\;b = 25.\) Construct and solve the differential equation describing the number of people \(A\left( t \right)\) aware of this product.

 Example 2

Under the conditions of the previous problem \(1,\) figure out how will the number of potential buyers vary by the end of the year if the advertising budget is spent evenly for the first \(6\) months?

 Example 3

Investigate the dynamics of awareness \(A\left( t \right)\) for the case of Linear Advertising Function \(q\left( t \right).\) Use the same data as in Examples \(1,2.\)

Example 1.

Management of a company decided to advertise a new product permanently during the year. The advertising budget is $12,000. The coefficients \(k\) and \(b\) are \(k = {\large\frac{1}{4}\normalsize},\;b = 25.\) Construct and solve the differential equation describing the number of people \(A\left( t \right)\) aware of this product.

Solution.

The equation of dynamics of \(A\left( t \right)\) is written in the form

\[\frac{{dA}}{{dt}} + kA = bq\left( t \right).\]

We assume that the time \(t\) is measured in months. By the condition of the problem, the advertising costs are constant during the year, so every month they are equal to:

\[{{q_0} = \frac{{12,000}}{{12}} }={ 1,000\,\frac{$}{\text{month}}.}\]

Substituting the known data leads to the following differential equation:

\[\frac{{dA}}{{dt}} + \frac{A}{4} = 25,000.\]

The integrating factor for this equation has the form:

\[u\left( t \right) = {e^{\int {{\large\frac{1}{4}\normalsize}dt} }} = {e^{\large\frac{t}{4}\normalsize}}.\]

The general solution is written as

\[
{A\left( t \right) }={ \frac{{25,000\int {{e^{\large\frac{t}{4}\normalsize}}dt} + C}}{{{e^{\large\frac{t}{4}\normalsize}}}} }
= {\frac{{\frac{{25,000}}{{\frac{1}{4}}}{e^{\large\frac{t}{4}\normalsize}} + C}}{{{e^{\large\frac{t}{4}\normalsize}}}} }
= {100,000 + C{e^{ – \large\frac{t}{4}\normalsize}}.}
\]

We determine the constant of integration \(C\) from the initial condition \(A\left( {t = 0} \right) = 0.\) Hence, \(C = -100,000.\) Then the particular solution is expressed by the formula

\[{A\left( t \right) }={ 100,000\left( {1 – {e^{ – \large\frac{t}{4}\normalsize}}} \right).}\]
The number of potential buyers depending on time

Figure 2.

The graph of this function is shown in Figure \(2\). Thus in case of permanent advertising, the number of potential buyers aware of the product grows non-linearly approaching the maximum value

\[{A_{\max }} = \frac{{b{q_0}}}{k} = 100000.\]
Page 1
Problem 1
Page 2
Problems 2-3