EC611--(Ch 02) Optimization Techniques

EC611--Managerial Economics

Optimization Techniquesand New Management Tools

Dr. Savvas C Savvides, European University Cyprus

Managerial Economics DR. SAVVAS C SAVVIDES 1

Models and Data

Modela framework based on simplifying assumptionsit helps to organize our economic thinking based on a simplified picture of realityWe focus on key elements

Datathe economist’s link with the real world1. time series2. cross section


Real and Nominal Variables

Many economic variables are measured in money termsNominal values

measured in current pricesReal values

adjusted for price changes compared with a base yearmeasured in constant prices


Real & Nominal Values--Example

£125,000£27,000£2,500Land Prices (Hilton Park Area, Nicosia)

100.039.37.4Price Index (2000=100)

200319751960

(2,500*100) / 7.4 = 33,783 (125,000*7.4) / 100 = 9,250

£125,000£68,702£33,783Real Land Price (in 2000 prices)

£9,250£5,084£2,500Real Land Price (in 1960 prices)


Evidence in Economics

Evidence collected and produced from empirical observation and testing may allow us to accumulate support for a theory, or to reject it, or indicate points for further research and investigationScatter diagrams help us to test and validate economic theory with empirical realityEconometrics is a more sophisticated method that takes this task of empirically validating theory further using statistical techniques


Data & Scatter Diagrams

886.57

807.06

876.05

856.54

906.03

1055.52

1006.01

QuantityPriceYearPrice

Quantity

X (7.0, 80)

X

XX

X X (6.0, 100)

X

7.0

6.0

10080


Economic Models: An Example

Examples:1. Quantity of CDs demanded depend on

(or is a function of): f (Prices, income, preferences)

2. Revenues are a function of Sales:

f (Q)


0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

Q

TR

Expressing Economic Relationships

Equations: TR = 100Q - 10Q2

Tables:

Graphs:

Q 0 1 2 3 4 5 6TR 0 90 160 210 240 250 240

if Q=3 TR = 100(3) – 10(3)2 = 210e.g. if Q=1 TR = 100(1) – 10(31)2 = 90


25


AC = TC/Qe.g. for Q=3

AC = 180/3 =60

MC = ∆TC/∆Q

For ∆Q from 3 to 4:

MC = (240-180)/(4-3) =60 / 1 = 60

96606080140

-AC

240602020120

-MC

48052404180316021401200TCQ

Total, Average, & Marginal Cost


0

60

120

180

240

0 1 2 3 4Q

TC ($)

0

60

120

0 1 2 3 4 Q

AC, MC ($)MC

AC

Total, Average, & Marginal Cost

TC


Q TR TC Profit0 0 20 -201 90 140 -502 160 160 03 210 180 304 240 240 05 250 480 -230

Profit Maximization

Profit = TR - TC


Profit Maximization

0

60

120

180

240

300

0 1 2 3 4 5Q

($)

-60

-30

0

30

60

TR

TC

Profit


Slope of a Line

Slope between A & B

∆P/∆Q = -5 / +5 = - 1


Slope of a Line

Price

Quantity


Slope of Non-Linear Relationships

A B

Slope of TR at A is positive:Slope of tangency at pt. A

TR

Total Revenue

Quantity

Slope of TR at B is negativeSlope of tangency at pt. B


Concept of the Derivative (1)

For example, if TR = Y and Q =X, the derivative of Y with respect to X is equal to

the ∆Y w.r.t. X, as the ∆X approaches zero.

0l i mX

d Y Yd X X∆ →

∆=

∆

Optimization analysis can be conducted much more efficiently using differential calculus.This relies on the concept of the derivative, which resembles the concept of the margin.


Concept of the Derivative (2)

∆Y = f(X+∆X) – f(X) Divide both sides by ∆X ∆Y/ ∆X = [f(X+∆X) – f(X] / ∆X

Let’s expand on the right hand side. Since Y depends on X, Y = f ( X ) ∆Y = ∆X ∆X = ∆X (tautology). Add & subtract X on RHS. ∆X = (X+∆X) – (X)

Substituting the RHS of the last expression in the derivative expression, we get

dY/ dX = [f(X+∆X) – f(X] / ∆X


The Derivative – An Example

If Y = X2 dY/ dX = [(X+∆X)2 – X2 ] / ∆X

dY/ dX = [ X2+ 2X * ∆X) + (∆X)2 - X2 ] / ∆X

dY/ dX = [ (2X * ∆X) + (∆X)2 ] / ∆X

dY/ dX = [ (2X * ∆X)/ ∆X ] + [(∆X)2 / ∆X]

Cancelling the ∆X terms dY/ dX = (2X + ∆X)

This says that at the limit, i.e., as ∆X 0, the whole expression will approach 2X (since ∆X=0)


Rules of DifferentiationConstant Function Rule: The derivative of a constant, Y = f(X) = a, is zero for all values of a (the constant).

( )Y f X a= = 0dYdX

=

Y

X

Y = 1010

Changes in X do not affect the value of Y. Horizontal lines have zero slope!


Rules of Differentiation

Power Function Rule: The derivative of a power function, where a and b are constants, is defined as follows.

( ) bY f X aX= = 1bdY b aXdX−= ⋅

Example: Y = 3X2

Derivative: dY/dX = 2 * 3X2-1

= 6X


0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

Q

TR

Power Function --Example

Equations: TR = 100Q - 10Q2

Tables:

Graphs:

Q 0 1 2 3 4 5 6TR 0 90 160 210 240 250 240

MR = dTR/dQ = 100 – 20Q

40

3

20

4

06080100MR

5210Q

MR

TR



Sum-and-Differences Rule: The derivative of the sum or difference of two functions U and V, is defined as follows.

( )U g X= ( )V h X=

dY dU dVdX dX dX

= ±

Y U V= ±



Product Rule: The derivative of the product of two functions U and V, is defined as follows.

( )U g X= ( )V h X=

dY dV dUU VdX dX dX

= +

Y U V= ⋅



Quotient Rule: The derivative of the ratio of two functions U and V, is defined as follows.

( )U g X= ( )V h X= UYV

=

( ) ( )2

dU dVV UdY dX dXdX V

−=



Chain Rule: The derivative of a function that is a function of X is defined as follows.

( )U g X=( )Y f U=

dY dY dUdX dU dX

= ⋅


Optimization With Calculus (1)Optimization often requires finding the max. or the min. of a function (e.g. maxTR, minTC, or maxΠ)

Find X such that dY/dX = 0. This means that the curve of the function has zero slope

Example: Given that TR = 100Q – 10Q2

d(TR) / dQ = 100 – 20Q Setting dTR/dQ =0,

we get 0 =100 – 20Q 20Q = 100 Q* = 5

Therefore, Total Revenues are maximized at Q* = 5

To find the optimum Price, we go to the demand equation from which the TR function derived:

P = 100 – 10Q P* = 100 – 10 (5) = 50


0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

Q

TR

Equation: TR = 100Q - 10Q2

MR = dTR/dQ = 100 – 20Q = 0

20Q = 100

Q = 5

MR

TR

Optimization With Calculus (2)


Optimization With Calculus (2)To distinguish between a max and a min, we use the second derivative.

Second derivative rules:

If d2Y/dX2 > 0 (positive), then X is a minimum.

If d2Y/dX2 < 0 (negative), then X is a maximum.In the example, we found d(TR) / dQ = 100 – 20Q

d2(TR)/dQ2 = - 20 (negative)Therefore, we know that the TR function is at a maximum (“top of the hill”) at Q = 5


Multivariate Optimization

Multivariate functions:TR = f (Sales, Advertising, prices, …)

TC = f ( wages, interest, raw materials, …)

Demand = f (price, income, P of substitutes, …)

To optimize a function that has more than one independent variables, we use the partial derivative.


Multivariate Optimization (2)

The Partial Derivative:

The partial derivative (indicated by ∂) is used in order to isolate the marginal effect of each one of the independent variables.

The same rules of differentiation apply, except that when we differentiate the dependent variable w.r.t. one variable, we hold all other variables constant.


Partial Derivative--Example

Suppose that Profits (π) are a function of the sales of products X and Y as follows:

π = f (X, Y) = 80X – 2X2 – XY – 3Y2 + 100Y

To find the partial derivative of Π w.r.t X, we hold Y constant (i.e. ∆Y =0) to get:

∂π / ∂X = 80 – 4X – Y

To find the partial derivative of Π w.r.t Y, we hold X constant (i.e. ∆X =0) to get:

∂π / ∂Y = 100 – X – 6Y


Max or Min Multivariate FunctionsExample (cont)

To max or min a multivariate function, we set each partial derivative equal to zero and solve the resulting simultaneous equations:

∂ π / ∂ X = 80 – 4X – Y = 0

∂ π / ∂ Y = 100 – X – 6Y = 0

To solve these simultaneous equations, we multiply the 1st by (-6) and the 2nd by (-1) to get:

- 480 + 24X +6Y = 0

100 – X – 6Y = 0

- 380 + 23X = 0

Therefore, X = 380 / 23 = 16.52


Max or Min Multivariate FunctionsExample (cont)

Substituting X = 16.52 into the first equation, we find the value of Y:

80 – 4 (16.52) – Y = 0

80 – 66.08 – Y = 0

Y = 13.92

Thus, the firm maximize Profits when it sells 13.92 unit of Y and 16.52 units of X. Thus:

π = 80X – 2X2 – XY – 3Y2 + 100Y

π = 80(16.52) – 2(16.52)2 – 16.52 * 13.92 –3(13.92)2 + 100(13.92)

π = 1,356.52


Constrained OptimizationSo far, we dealt with unconstrained optimization

However, in most real life situations, firms are faced with a series of constraints (budget, capacity, lack of raw materials, etc).

In these cases, we need to optimize (max or min) the objective function (profits, revenues, costs, market share, etc) subject to the constraints faced by the firm.

We have two methods to solve constrained optimization problems:

1. Substitution Method (used for simple functions)

2. Lagrangian Method (used for complex functions)


New Management ToolsBenchmarking: finding out what processes or techniques “excellent” firms use and adopt & adapt

Total Quality Management: the constant improvements in product quality and processes to deliver consistently superior service and value to customers

Reengineering: seeks to completely reorganize the firm (processes, departments, entire firm). Radically redesigning processes to achieve significant gains in speed, quality, service, profitability

The Learning Organization: continuous learning both on the individual level as well as on the collective level.

It is based on five ingredients: a new mental model -achieve personal mastery – develop system thinking –develop shared vision – strive for team learning


Other Management ToolsBroad Banding: eliminating multiple layers of salary levels, and increasing labor flexibility

Direct Business Model: dealing directly with the consumer, eliminating distributors and saving on time and costs (e.g, Dell )

Networking: the formation of strategic alliances to increase the synergies and capitalize on individual competences

Pricing Power: being able to increase prices faster than costs thus increasing profits

Small-World Model: large firms may gain efficiency by simulating the operation of small firms by breaking up the process in smaller scale and linking the units or individuals through organizational systems

Virtual Integration: the blurring of traditional boundaries between manufacturer and its suppliers and manufacturer and customer supply chain management

Virtual Management: the simulation of the production process and consumer behavior using computer models

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EC611--(Ch 02) Optimization Techniques