Download MEN311S MANAGERIAL ECONOMICS

MANAGERIAL ECONOMICS
DEFINITION
• PROVIDES LINK BETWEEN
ECONOMIC THEORY AND
DECISION SCIENCES IN
ANALYSIS OF MANAGERIAL
DECISION MAKING.
• THE DIAGRAMME BELOW
SHOWS HOW MANAGERIAL
ECONMICS DIFFER FROM
MICROECONOMICS:
• MICROECONOMICS IS LAGERLY
DESCRIPTIVE, THAT IS, IT
ATTEMPTS TO DESCRIBE HOW
THE ECONOMY WORKS
WITHOUT INDICATING HOW IT
SHOULD OPERATE.
A LINK BETWEEN ECONOMIC THEORY AND THE
DECISION SCIENCE IN ANALYSIS OF MANAGERIAL
ECONOMICS
PROBLEM
FACED BY
DECISION
MAKERS IN
MANAGEMENT
ECONOMIC
THEORY
MANAGERIAL
ECONOMICS, WHICH
APPLIES AND EXTENDS
ECONOMICS AND THE
DECISION
SCIENCES
TO
SOLVE MANAGEMENT
PROBLEMS.
SOLUTIONS
TO DECISION
PROBLEMS
FACED BY
MANAGERS.
DECISION
SCIENCES
RELATIONSHIPS
• MANAGERIAL ECONOMICS IS LARGELY
PRESCRIPTIVE, THAT IS, IT ATTEMPTS TO
ESTABLISH RULES AND TECHNIQUES TO FULLFILL
SPECIFIC GOALS.
• MANAGERIAL ECONOMICS ON THE OTHER HAND
USES OPTIMIZATION TECHNIQUES, SUCH AS
DIFFERENTIAL CALCULUS AND MATHEMATICAL
PROGRAMMING, TO DETERMINE OPTIMAL
COURSES OF ACTION FOR DECISION MAKERS.
THE BASIC PROCESS OF DECISION
MAKING
 ESTABLISH OR IDENTIFY THE OBJECTIVES
 In making any decision, you as the decision maker
should determine what the organisation or individuals
objective. Unless you know what it is you trying to
achieve, there is no sensible way to make the decision.
 DEFINE THE PROBLEM
 One of the difficult part of decision making is to
determine exactly what the problem is. To solve shortcomings in business, you should identify the problem
brought about the short-comings otherwise, you as a
manager, you will have little chance of solving the
problem.
THE BASIC PROCESS OF DECISION
MAKING
IDENTIFY POSSIBLE SOLUTIONS
 Once the problem is defined, try to construct and
identify possible solutions, which includes,
effective production and marketing of its
products based on existing designs, as well as
redesigning its entire products line.
SELECT BEST POSSIBLE SOLUTION
 Having identified the set of alternative possible
solutions, evaluate each one and determine
which is best, give the objective of the
organisation.
THE BASIC PROCESS OF DECISION
MAKING
IMPLEMENT THE DECISION
This is of crucial importance stage, because
the particular solution which has been chosen
might positively or negatively affect the
business processes if the implementation
process was not followed properly or
effectively.
THEORY OF THE FIRM
• The firms main aim is to make profit, at its
maximum level. This is determined by looking
at the expected profit of the firm, which can
be expressed as an equation:
• The value of the firm equals:
• Present value of expected future profit
•
𝜋1
=
1+𝑖
+
𝜋2
(1+𝑖)2
+
𝜋𝑛
……..+
(1+𝑖)𝑛
THEORY OF THE FIRM
•
𝜋𝑡
𝑛
=σ𝑡=𝑖
(1+𝑖)𝑡
• Where 𝜋𝑡 is the expected profit in year t. ί is
the interest rate, and t goes from 1(next year)
to n ( the last year in the planning horizon)
• Please note that, this equation will be used in
all upcoming classes to determine profits and
losses/cost to company.
OPTIMIZATION TECHNIQUES
• Theses refers to the basic elements of differential
calculus, including the rules of differentiation and
use of derivative to maximize a function such as
profit or minimize a function such as cost.
• Differential tells us what changes will occur in
one variable called, the dependent variable when
a small marginal change is made in another
variable called, the independent variable.
OPTIMIZATION TECHNIQUES
• While we want to maximize the profit of our
firm or minimize the cost of production, such
maximization or minimization is often subject
to constraints such as, producing certain
amount to adhere to a contract or utilizing a
certain amount of labor in a union agreement.
FUNCTIONAL RELATIONSHIP
• To understand this chapter, you must know
how economic relationships are expressed.
Frequently, the relationship between two or
more economic variables can be represented
by a table, graph or in a form of an equation.
For example: Q = f(p), where Q are numbers of
units sold and P represent the price per unit.
FUNCTIONAL RELATIONSHIP
• The equation above should be read as,
number of units sold is a function of price,
which means that, the number of units sold
depends on price.
• The number of units in this case, is the
dependent variable and price is the
independent variable.
FUNCTIONAL RELATIONSHIP
• GIVEN THAT, Q = 200 – 5p, calculate the
number of units sold, given that, the product
was sold at a price of $10.
• This is a very simple calculation as p is
replaced by $10 as the price and your answer
will be Q=200-5(10) = 150 units.
RELATIONSHIP BETWEEN OUTPUT AND
PROFIT
• STUDENTS WILL BE EXPECTED TO KNOW HOW
TO CALCULATE MARGINAL REVENUE AND
AVERAGE PROFIT.
OPTIMISATION TECHNIQUES
• DERIVATIVES
• Using ∆(called Delta) to denote change in the
independent variable can be expressed as ∆X, and a
change in the dependent variable can be expressed as
∆Y. thus, the marginal value of Y can be estimated by
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑌
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑋
=
Δ𝑦
Δ𝑥
• For Example, if a two – unit increase in X results in a
one unit increase in Y, ∆X = 2 and ∆Y = 1, this means
that, the marginal value of Y is about ½ that is, the
depended variable Y increases by about ½ if the
independent variable X increases by 1.
OPTIMISATION TECHNIQUES
• Unless the relationship between Y and X can be
represented as a straight line graphically, the value of
Δ𝑦
𝑖𝑠 𝑛𝑜𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
Δ𝑥
• Derivatives of Constant
• Managers wants to know how much to optimize the
performance of their organizations. If Y is some measure of
the organisational performance and X is a variable under a
particular managers control, he or she would like to know
the value of X that maximizes Y.
• If the dependent variable Y is a constant, its derivative with
respect to X is always zero. That is if Y=a ( where a is a
𝑑𝑦
constant) = 0.
𝑑𝑥
OPTIMIZATION TECHNIQUES
• Derivatives of power functions
• A power function can be expressed as
𝑌 = 𝑎𝑥 𝑏 , where a and b are constant, if the
relationship between X and Y is of this kind,
the derivative of Y with respect to X equals b
times. Students are expected to read further
on this and practice.
OPTIMIZATION TECHNIQUES
• EXAMPLE: a new company did a study to estimate the
effects of advertising expenditures on the sales of their
product and found that the relationships between
advertising expenditure and sales in two district were:
𝑆1 = 10 + 5𝐴1 and 𝑆2 = 12 + 4𝐴2 − 0.5𝐴2 2 .
• Where 𝑆1 is the sales of the product (in millions of dollars
per year) in the first district and 𝑆2 is its sales of the
product in the second district.
• Determine the amount of additional sales that an extra
dollar of advertising would generate in each district.
• To answer this question, the derivatives of sales with
respect to advertising must be calculated for each district.
OPTIMIZATION TECHNIQUES
𝑑𝑆1
𝑑𝐴1
= 5 − 3𝐴1
𝑑𝑆2
•
𝑑𝐴2
= 4 − 𝐴2
• THEREFORE:
this means, in each district, the
effective on sales of an extra dollar
of advertising expenses, depend
on the amount spend on
advertising.
OPTIMIZATION TECHNIQUES
• Suppose that $0.5 million was being spend on
advertising in the first district and $1 million was
being spend on advertising in the second district.
•
𝑑𝑆1
𝑑𝐴1
•
𝑑𝑆2
𝑑𝐴2
= 5 − 3 0.5 = 3.5
= 4 − 1 = 3, this means an extra dollar of
advertising generated an extra $3.50 of sales in
the first district and extra $3.00 sales in the
second district.
OPTIMIZATION TECHNIQUES
• RECOMMENDATIONS: its recommended that,
if the company wants to boost the total sales
of its product, more should be spend on
advertising in the first district and less in the
second district.
OPTIMIZATION TECHNIQUES
• MAXIMIZATION AND MINIMIZATION PROBLEMS.
• Having determined how to find the derivative of Y
with respect to X, we now take up the way to
determine the value of x that maximises or
minimizes Y.
• Given Y = -50+100X-5𝑥 2
»
𝑑𝑦
𝑑𝑥
= 10 − 10𝑥
» therefore., if this derivative equals zero,
» 100-10x = 0
• X=10
OPTIMIZATION TECHNIQUES
• GIVEN y = -1+9x-6𝑥 2 + 𝑥 3
• Find the values of output that maximizes or
minimizes profit.
•
•
𝑑𝑦
𝑑𝑥
= 9 − 12𝑥 + 3𝑥 2 = 0
𝑑2 𝑦
𝑑𝑥 2
= 12 + 6𝑥
• If x=1
𝑑2𝑦
𝑑𝑥 2
= −12 + 6 1 = −6
OPTIMIZATION TECHNIQUES
• Since the second derivative is negative, profit
is maximized.
• If x=3,
𝑑2𝑦
𝑑𝑥 2
= −12 + 6 3 = 6
• Since the second derivative is positive, profit is
minimum, when output equals 3 million units.
CONSTRAINED OPTIMIZATION
• Managers of firms and other organisations
generally face constraints that limit the
options available to them. A production
manager may want to minimize his or her
firms costs but may not be permitted to
produce less than is required.
• Suppose that, a company produces two
products and that its total cost equals
CONSTRAINED OPTIMIZATION
• TC =4𝑄1 2 + 5𝑄2 2 + 𝑄1 + 𝑄2
• Where 𝑄1 , equals its output per hour of the
first product and 𝑄2 equals its output per hour
of the second product. Because of the
commitments to customers, the number
produced of both products combined cannot
be less than 30 per hour.
2
2
• TC=TC =4𝑄1 + 5𝑄2 + 𝑄1 + 𝑄2
CONSTRAINED OPTIMIZATION
• 𝑄1 + 𝑄2 = 30
• SOLVE FOR 𝑄1, 𝑄1 = 30 − 𝑄2
2
2
• THEREFORE; TC = 4(30 − 𝑄2 ) + 5𝑄2 − (30 −
𝑄2 )𝑄2
• = 4(900 − 60𝑄2 2 ) + 5𝑄2 2 − 30𝑄2 − 𝑄2 2
2
• TC = 3600 − 270𝑄2 + 10𝑄2
• Find the value of 𝑄2 that minimizes TC, by
obtaining the derivative of TC with respect to 𝑄2
and we must set it equals to zero.
CONSTRAINED OPTIMIZATION
•
𝑑𝑇𝐶
𝑑𝑄2
= −270 + 20𝑄2 = 0
• 20𝑄2 = 270
• 𝑄2 = 13.5
• To ensure that, this is a minimum, we obtain the second
derivative
•
𝑑 2 𝑇𝐶
𝑑𝑄2
=
20, 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑓𝑖𝑛𝑑 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚.
CONSTRAINED OPTIMIZATION
• Find the value of 𝑄1 that minimizes total cost
• 𝑄1 + 𝑄2 = 30
• 𝑄1 = 30 − 𝑄2
• = 30 − 13.5
• =16.5
• This means, in order for the company to minimize the
total cost, it should produce 16.5 units of the first
product and 13.5 units of the second product per hour.
• What is the actual Total Cost?
PRICE ELASTICITY OF DEMAND
• Price Elasticity of demand is defined as the
percentage change in quantity demanded
resulting from a 1 percent change in price.
• Example: given the hypothetical demand function
for school uniforms
• 𝑄 = −700𝑝 + 200𝐼 − 500𝑆 + 0.01𝐴
• Assuming that per capital disposable Income(I) is
$13 000, the average price of software(S) is $400,
and advertising expenditure(A) is $50 million.
PRICE ELASTICITY OF DEMAND
•
•
•
•
Calculate price elasticity of demand
𝑄 = 2900000 − 700𝑝
Where p=$3000
𝑄 = 2900000 − 700 3000
=800 000
– Evaluate the partial derivative given 2 900 000 –
700p.
–
𝜕𝑄
𝜕𝑝
= −700
PRICE ELASTICITY OF DEMAND
• To obtain the price elasticity of demand, we
𝑑𝑄
multiply by P/Q
𝑑𝑝
−700 3000Τ800 000
•
• =-2.62 as the price elasticity of demand.
• What does it tell us?