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Managerial Economics
Dr Nihal Hennayake
What is Microeconomics and Macroeconomics ? •Ragnor Frisch : Micro means “ Small” and Macro means “Large”Microeconomics deals with the study of individual behaviour.
• It deals with the equilibrium of an individual consumer, producer, firm or industry. Macroeconomics on the other hand, deals with economy wide aggregates.
• Determination of National Income Output, Employment• Changes in Aggregate economic activity, known as Business Cycles• Changes in general price level , known as inflation, deflation• Policy measures to correct disequilibrium in the economy, Monetary policy and Fiscal policy
What is Managerial Economics?
“Managerial Economics is economics applied in decision making. It is a special branch of economics bridging the gap between abstract theory and managerial practice” – Willian Warren Haynes, V.L. Mote, Samuel Paul
“Integration of economic theory with business practice for the purpose of facilitating decisionmaking and forward planning” Milton H. Spencer
“Managerial economics is the study of the allocation of scarce resources available to a firm or other unit of management among the activities of that unit” Willian Warren Haynes, V.L. Mote, Samuel Paul
“ Price theory in the service of business executives is known as Managerial economics” Donald Stevenson Watson
BUSINESS ADMINISTRATION
DECISION PROBLEMS
MANAGERIAL ECONOMICS : INTEGRATION OF ECONOMIC
THEORY AND METHODOLOGY WITH TOOLS AND TECHNICS BORROWED FROM OTHER DECIPLINES
OPTIMAL SOLUTIONS TO BUSINESS PROBLEMS
TRADITIONAL ECONOMICS : THEORY AND METHODOLOGY
DECISION SCIENCES : TOOLS AND TECHNICS
Nature, Scope and Significance of Managerial Economics:
Managerial Economics – Business Economics Managerial Economics is ‘Pragmatic’ Managerial Economics is ‘Normative’ Universal applicability The roots of Managerial Economics came from Micro Economics Relation of Managerial Economics to Economic Theory is much like that of Engineering to Physics or Medicine to Biology. It is the relation of applied field to basic fundamental discipline
Core content of Managerial Economics :
Demand Analysis and forecasting of demand Production decisions (InputOutput Decisions) Cost Analysis (Output Cost relations) Price – Output Decisions Profit Analysis Investment Decisions
1. Demand Analysis : Meaning of demand : No. of units of a commodity that
customers are willing to buy at a given price under a set of conditions.
Demand function : Qd = f (P, Y, Pr W)
Demand Schedule : A list of prices and quantitives and the list is so arranged that at each price the corresponding amount is the quantity purchased at that priceDemand curve : Slops down words from left to right.
Law of demand : inverse relation between price and quantity Exceptions to the law of demand : Giffens paradox
Price expectations
Elasticity : Measure of responsiveness Qd = f (P, Y, Pr W)
E = percentage change in DV/ percentage change in IV
Concepts of price, income, and cross elasticity
Price Elasticity :
Ep = Percentage change in QD/Percentage change in P
Types of price elasticity : 1. Perfectly elastic demand Ep = ∞ 2. Elastic demand Ep > 13. Inelastic demand Ep < 14. Unit elastic demand Ep = 15. Perfectly inelastic demand Ep = 0
Elasticity and expenditure : If demand is elastic a given fall in price causes a relatively larger increase in the total expenditure.
P TR when demand is elastic.↓ ↑ P TR when demand is inelastic.↓ ↓ P TR remains same when demand is Unit elastic.↓ ↑
Measurement of elasticity : Point and Arc elasticity Elasticity when demand is linear
Determinants of elasticity : (1) Number and closeness of its substitutes, (2) the commodity’s importance in buyers’ budgets, (3) the number of its uses.
Other Elasticity Concepts Income elasticity Cross elasticity
Functions of a Managerial Economists:
The main function of a manager is decision making and
managerial Economics helps in taking rational decisions.
The need for decision making arises only when there are more
alternatives courses of action.
Steps in decision making :
Defining the problem
Identifying alternative courses of action
Collection of data and analyzing the data
Evaluation of alternatives
Selecting the best alternative
Implementing the decision
Follow up of the action
Specific functions to be performed by a managerial Economist :
1.Production scheduling
2.Sales forecasting
3.Market research
4.Economic analysis of competing companies
5.Pricing problems of industry
6. Investment appraisal
7.Security analysis
8.Advice on foreign exchange management
9.Advice on trade
10.Environmental forecasting
OPTIMIZATION
Managerial economics is concerned with the ways in which managers should make decisions in order to maximize the effectiveness or performance of the organizations they manage. To understand how this can be done we must understand the basic optimization techniques.
Functional relationships:
relationships can be expressed by graphs:
This form can be expressed in an equation:
Q = f ( P )Though useful, it does not tell us how Q responds to P,
but this equation do.
Q = 200 - 5 p
Marginal AnalysisThe marginal value of a dependent variable is
defined as the change in this dependent variable associated with a 1-unit change in a particular independent variable. e.g.
units marginal profit average profit 0 0 - 100 1 100 100 150 2 250 125 350 3 600 200 400 4 1000 250 350 5 1350 270 150 6 1500 250 50 7 1550 221 -50 8 1500 188 -100 9 1400 156
Total profit is maximized when marginal profit shifts from positive to negative.
Average Profit = Profit / Q
Slope of ray from the origin:– Profit / Q = average
profit
Maximizing average profit doesn’t maximize total profit
MAX
C
B
profits
Q
PROFITS
quantity
Marginal Profits = /Q Q1 is breakeven (zero profit)
maximum marginal profits occur at the inflection point (Q2)
Max average profit at Q3
Max total profit at Q4 where marginal profit is zero
So the best place to produce is where marginal profits = 0.
profits max
Q2
marginalprofits
Q
Q
averageprofits
Q3
Q4
(Figure 2.1)
Q1
Differential Calculus in Management
A function with one decision variable, X, can be written as: Y = f(X)
The marginal value of Y, with a small increase of X, is My = Y/X
For a very small change in X, the derivative is written:
dY/dX = limit Y/X X 0
Marginal = Slope = Derivative
The slope of line C-D is Y/X
The marginal at point C is Y/X
The slope at point C is Y over X
The derivative at point C is also this slope
X
C
DYY
X
Quick Differentiation Review
Constant Y = c dY/dX = 0 Y = 5
Functions dY/dX = 0
A Line Y = c • X dY/dX = c Y = 5X
dY/dX = 5
Power Y = cXb dY/dX = b•c•X b-1 Y = 5X2
Functions dY/dX = 10X
Name Function Derivative Example
Sum of Y = G(X) + H(X) dY/dX = dG/dX + dH/dX Functions
example Y = 5X + 5X2 dY/dX = 5 + 10X
Product of Y = G(X) • H(X)
Two Function dY/dX = (dG/dX)H + (dH/dX)G
example Y = (5X)(5X2 ) dY/dX = 5(5X2 ) + (10X)(5X) = 75X2
Quick Differentiation Review
Quotient of Two Y = G(X) / H(X) FunctionsdY/dX = (dG/dX)•H - (dH/dX)•G H2
Y = (5X) / (5X2) dY/dX = 5(5X2) -(10X)(5X) (5X2)2
= -25X2 / 25X4 = - X-2
Chain Rule Y = G [ H(X) ]dY/dX = (dG/dH)•(dH/dX) Y = (5 + 5X)2
dY/dX = 2(5 + 5X)1(5) = 50 + 50X
Quick Differentiation Review
y
dy/dx 10
10 20
20
0
Max of x
Slope = 0
value of x
Value of dy/dx which
Is the slope of y curve
Value of Dy/dx when y is max
x
x
the function
Y = -50 + 100X - 5X2
i.e.,
dYdX
= 100 - 10X
if dYdX
= 0
X = 10
i.e., Y is maximized when
the slope equals zero.
Note that this is not sufficient for maximization or minimization problems.
Min value of y
Max value of y
d2y/dx2 > 0 d2y/dx2 < 0
value of dy/dx
y
Dy/dx
x
x
Since dYdX
= 0 at two points, we need another
condition to distinguish between the maximum and
minimum points.
Look at the dYdX
curve
* at point 5 the curve is upward, i.e., its slope ( the
second derivative (the derivative of the derivative)) is
positive. Hence
d YdX
22
= > 0 ( minimum point )
* at point 10 the curve is downward, i.e., its slope is
negative. Hence
d YdX
22
= < 0 ( maximum point )
Optimization Rules Maximization conditions:
1 - dYdX
= 0
2 - d YdX
22
= < 0
Minimization conditions:
1 - dYdX
= 0
2 - d YdX
22
= > 0
Applications of Calculus in Managerial Economics
maximization problem:
A profit function might look like an arch, rising to a peak and then declining at even larger outputs. A firm might sell huge amounts at very low prices, but discover that profits are low or negative.
At the maximum, the slope of the profit function is zero. The first order condition for a maximum is that the derivative at that point is zero.
If = 50Q Q2, then d/dQ = 50 2∙Q, using the rules of differentiation. Hence, Q = 25 will maximize profits where 50 2Q = 0.
More Applications of Calculus
minimization problem: Cost minimization supposes that there is a least cost point to produce. An
average cost curve might have a Ushape. At the least cost point, the slope of the cost function is zero. The first order condition for a minimum is that the derivative at that point is zero.
If TC = 5Q2 – 60Q, then dC/dQ = 10Q 60. Hence, Q = 6 will minimize cost Where: 10Q 60 = 0.
Competitive Firm: Maximize Profits – where = TR - TC = P • Q - TC(Q)
– Use our first order condition:
– d/dQ = P - dTC/dQ = 0.
– Decision Rule: P = MC.a function of Q
Max = 100Q - Q2
First order = 100 -2Q = 0 implies
Q = 50 and;
= 2,500
Second Order Condition: one variable If the second derivative is negative,
then it’s a maximum
Max = 100Q - Q2
First derivative
100 -2Q = 0
second derivative is: -2 implies
Q =50 is a MAX
Max= 50 + 5X2
First derivative
10X = 0
second derivative is: 10 implies
Q = 10 is a MIN
Problem 1 Problem 2.
e.g.;
Y = -1 + 9X - 6X2 + X3
first condition
dYdX
= 9 - 12X + 3X2 = 0
Quadratic Function
Y = aX2 + bX + c
X = b b ac
a
2 42
a = 3
b = -12
c = 9
X = ( ) ( )12 122 4 9 3
6 = 2 1
therefore
Y = 0 at
X = 3 or X = 1
the second condition
d YdX
22
= -12 + 6X
at X = 3
d YdX
22
= -12 + 6(3) = 6 >0 ( minimum point)
at X = 1
d YdX
22
= -12 + 6(1) = - 6 <0 (maximum point)
Partial Differentiation
Economic relationships usually involve several independent variables.
A partial derivative is like a controlled experiment- it holds the “other” variables constant
Suppose price is increased, holding the disposable income of the economy constant as in
Q = f (P, I )
then Q/P holds income constant.
Sales are a function of advertising in newspapers and magazines ( X, Y)
Max S = 200X + 100Y -10X2 -20Y2 +20XY
Differentiate with respect to X and Y and set equal to zero.
S/X = 200 - 20X + 20Y= 0
S/Y = 100 - 40Y + 20X = 0
solve for X & Y and Sales
2 equations & 2 unknowns
200 - 20X + 20Y= 0100 - 40Y + 20X = 0
Adding them, the -20X and +20X cancel, so we get 300 - 20Y = 0, or Y =15
Plug into one of them: 200 - 20X + 300 = 0, hence X = 25To find Sales, plug into equation: S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250
PARTIAL DIFFERENTIATION AND MAXIMIZATION OF
MULTIVARIATE FUNCTIONS.
= f (Q1 , Q2 )
To know the marginal effect of Q1 on we hold Q2 constant, and
vice versa.
In order to do that we use partial derivative of with respect to
Q1 denoted by Q1
( treating Q2 as constant )
e.g.;
= -20 + 100Q1 + 80Q2 - 10Q12 - 10Q2
2 - 5Q1Q2;
to find the partial derivative of with respect to Q1 we treat Q2
as constant; hence
Q1
= 100 - 20Q1 - 5Q2; (1)
therefore
Q2
= 80 - 20Q2 - 5Q1; (2)
setting both partial derivatives equal to zero and solve
simultaneously
100 - 20Q1 - 5Q2 =0
80 - 20Q2 - 5Q1 =0 multiply by -4 and add
________________
- 220 + 75Q2 = 0
hence
Q2 = 2.933
substitute for Q2 at any of the eq. 1
100 - 20Q1 - 14.665; hence
Q1 = 4.267.
i.e.,
profit is maximized when the firm produces 4.267 of Q1 and 2.933 of Q2.
CONSTRAINED OPTIMIZATION
We assume that the firm can freely produce 4.267 of Q1 and 2.933
of Q2. Quite often this may not be the case.
e.g.
Minimize TC = 4Q12 + 5Q2
2 - Q1Q2;
subject to:
Q1 + Q2 = 30 The constraint function
Solution:
The lagrangian multiplier:
Steps:
1 - set the constraint function to zero
2 - form the lagrangian function by adding the constraint function
after multiplication with an unknown factor to the original
function.
3 - take the partial derivatives and set them equal to zero
4 - solve the resulting equations simultaneously
step 1:
30 - Q1 - Q2 = 0
step 2:
L = 4Q12 + 5Q2
2 - Q1Q2 + ( 30 - Q1 - Q2)
step 3:
LQ1
= 8Q1 - Q2 -
2QL
= -Q1 + 10Q2 -
L
= -Q1 - Q2 + 30
8Q1 - Q2 - = 0 (1)
-Q1 + 10Q2 - =0 (2)
-Q1 - Q2 + 30 =0 (3)
step 4
multiply eq(2) by -1 and subtract from eq(1)
9Q1 - 11Q2 = 0 (4)
multiply (3) by 9 and add to eq(4)
-9Q1 - 9Q2 + 270 = 0
9Q1 - 11Q2 = 0
____________________
-20Q2 +270 = 0
Q2 = 270/20 = 13.5
substituting in eq (3) Q1 = 16.5
the values of Q1 and Q2 that minimizes TC are 16.5 and 13.5
respectively.
substituting Q1 and Q2 in eq(1) or eq(2) we find that
= 118.5
the interpretation of
measures the change in TC if the constraint is to be relaxed by one
unit.
i.e., TC will increase ( has a positive sign ) by 118.5 if the constraint
becomes 29 or 31.
