Managerial Economics in a Global Economy

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Copyright  2007 by Oxford University Press, Inc.
Slide 1
1
Optimization Techniques
• Methods for maximizing or minimizing
an objective function
• Examples
– Consumers maximize utility by purchasing
an optimal combination of goods
– Firms maximize profit by producing and
selling an optimal quantity of goods
– Firms minimize their cost of production by
using an optimal combination of inputs
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Copyright  2007 by Oxford University Press, Inc.
Slide 2
Expressing Economic
Relationships
Equations:
Tables:
TR = 100Q - 10Q2
Q
TR
0
0
1
90
2
3
4
5
6
160 210 240 250 240
TR
300
250
Graphs:
200
150
100
50
0
0
1
2
3
4
5
6
7
Q
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Copyright  2007 by Oxford University Press, Inc.
Slide 3
Total, Average, and Marginal
Revenue
TR = PQ
AR = TR/Q
MR = TR/Q
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Q
0
1
2
3
4
5
6
TR
0
90
160
210
240
250
240
AR
90
80
70
60
50
40
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MR
90
70
50
30
10
-10
Slide 4
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Slide 5
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Copyright  2007 by Oxford University Press, Inc.
Slide 6
TR
300
250
Total Revenue
200
150
100
50
0
0
1
2
3
4
5
6
7
Q
AR, MR
120
100
Average and
Marginal Revenue
80
60
40
20
0
-20
0
1
2
3
4
5
6
7
-40
Q
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Slide 7
Total, Average, and
Marginal Cost
AC = TC/Q
MC = TC/Q
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Q
0
1
2
3
4
5
TC AC MC
20 140 140 120
160 80 20
180 60 20
240 60 60
480 96 240
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Slide 8
PowerPoint Slides Prepared by Robert F. Brooker, Ph.D.
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Slide 9
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Copyright  2007 by Oxford University Press, Inc.
Slide 10
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Slide 11
Geometric Relationships
• The slope of a tangent to a total curve
at a point is equal to the marginal value
at that point
• The slope of a ray from the origin to a
point on a total curve is equal to the
average value at that point
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Slide 12
Geometric Relationships
• A marginal value is positive, zero, and
negative, respectively, when a total
curve slopes upward, is horizontal, and
slopes downward
• A marginal value is above, equal to, and
below an average value, respectively,
when the slope of the average curve is
positive, zero, and negative
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Slide 13
Profit Maximization
Q
0
1
2
3
4
5
PowerPoint Slides Prepared by Robert F. Brooker, Ph.D.
TR
0
90
160
210
240
250
TC Profit
20
-20
140
-50
160
0
180
30
240
0
480 -230
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Slide 14
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Slide 15
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Slide 16
Steps in Optimization
• Define an objective mathematically as a
function of one or more choice variables
• Define one or more constraints on the
values of the objective function and/or
the choice variables
• Determine the values of the choice
variables that maximize or minimize the
objective function while satisfying all of
the constraints
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Slide 17
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Slide 18
New Management Tools
•
•
•
•
Benchmarking
Total Quality Management
Reengineering
The Learning Organization
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Slide 19
Other Management Tools
•
•
•
•
Broadbanding
Direct Business Model
Networking
Performance Management
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Slide 20
Other Management Tools
•
•
•
•
•
Pricing Power
Small-World Model
Strategic Development
Virtual Integration
Virtual Management
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Slide 21
Chapter 2 Appendix
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Slide 22
Concept of the Derivative
The derivative of Y with respect to X is
equal to the limit of the ratio Y/X as
X approaches zero.
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Slide 23
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Slide 24
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Slide 25
Rules of Differentiation
Constant Function Rule: The derivative
of a constant, Y = f(X) = a, is zero for all
values of a (the constant).
Y  f (X )  a
dY
0
dX
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Slide 26
Rules of Differentiation
Power Function Rule: The derivative of
a power function, where a and b are
constants, is defined as follows.
Y  f (X )  aX b
dY
 b  a X b 1
dX
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Slide 27
Rules of Differentiation
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 )
Y  U V
dY dU dV


dX dX dX
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Slide 28
Rules of Differentiation
Product Rule: The derivative of the
product of two functions, U and V, is
defined as follows.
U  g( X )
V  h( X )
Y  U V
dY
dV
dU
U
V
dX
dX
dX
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Slide 29
Rules of Differentiation
Quotient Rule: The derivative of the
ratio of two functions, U and V, is
defined as follows.
U  g( X )
dY

dX
V  h( X )

V dU
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dX
 
 U dV
V
U
Y
V
dX

2
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Slide 30
Rules of Differentiation
Chain Rule: The derivative of a function
that is a function of X is defined as follows.
Y  f (U )
U  g( X )
dY dY dU


dX dU dX
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Slide 31
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Slide 32
Optimization with Calculus
Find X such that dY/dX = 0
Second derivative rules:
If d2Y/dX2 > 0, then X is a minimum.
If d2Y/dX2 < 0, then X is a maximum.
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Slide 33
Univariate Optimization
Given objective function Y = f(X)
Find X such that dY/dX = 0
Second derivative rules:
If d2Y/dX2 > 0, then X is a minimum.
If d2Y/dX2 < 0, then X is a maximum.
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Slide 34
Example 1
• Given the following total revenue (TR)
function, determine the quantity of
output (Q) that will maximize total
revenue:
• TR = 100Q – 10Q2
• dTR/dQ = 100 – 20Q = 0
• Q* = 5 and d2TR/dQ2 = -20 < 0
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Slide 35
Example 2
• Given the following total revenue (TR)
function, determine the quantity of
output (Q) that will maximize total
revenue:
• TR = 45Q – 0.5Q2
• dTR/dQ = 45 – Q = 0
• Q* = 45 and d2TR/dQ2 = -1 < 0
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Slide 36
Example 3
• Given the following marginal cost
function (MC), determine the quantity of
output that will minimize MC:
• MC = 3Q2 – 16Q + 57
• dMC/dQ = 6Q - 16 = 0
• Q* = 2.67 and d2MC/dQ2 = 6 > 0
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Slide 37
Example 4
• Given
– TR = 45Q – 0.5Q2
– TC = Q3 – 8Q2 + 57Q + 2
• Determine Q that maximizes profit (π):
– π = 45Q – 0.5Q2 – (Q3 – 8Q2 + 57Q + 2)
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Slide 38
Example 4: Solution
• Method 1
– dπ/dQ = 45 – Q - 3Q2 + 16Q – 57 = 0
– -12 + 15Q - 3Q2 = 0
• Method 2
– MR = dTR/dQ = 45 – Q
– MC = dTC/dQ = 3Q2 - 16Q + 57
– Set MR = MC: 45 – Q = 3Q2 - 16Q + 57
• Use quadratic formula: Q* = 4
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Slide 39
Quadratic Formula
• Write the equation in the following form:
aX2 + bX + c = 0
• The solutions have the following form:
 b  b  4ac
2a
2
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Slide 40
Multivariate Optimization
• Objective function Y = f(X1, X2, ...,Xk)
• Find all Xi such that ∂Y/∂Xi = 0
• Partial derivative:
– ∂Y/∂Xi = dY/dXi while all Xj (where j ≠ i) are
held constant
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Slide 41
Example 5
• Determine the values of X and Y that
maximize the following profit function:
– π = 80X – 2X2 – XY – 3Y2 + 100Y
• Solution
– ∂π/∂X = 80 – 4X – Y = 0
– ∂π/∂Y = -X – 6Y + 100 = 0
– Solve simultaneously
– X = 16.52 and Y = 13.92
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Slide 42
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Slide 43
Constrained Optimization
• Substitution Method
– Substitute constraints into the objective
function and then maximize the objective
function
• Lagrangian Method
– Form the Lagrangian function by adding
the Lagrangian variables and constraints to
the objective function and then maximize
the Lagrangian function
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Slide 44
Example 6
• Use the substitution method to
maximize the following profit function:
– π = 80X – 2X2 – XY – 3Y2 + 100Y
• Subject to the following constraint:
– X + Y = 12
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Slide 45
Example 6: Solution
• Substitute X = 12 – Y into profit:
– π = 80(12 – Y) – 2(12 – Y)2 – (12 – Y)Y – 3Y2 + 100Y
– π = – 4Y2 + 56Y + 672
• Solve as univariate function:
– dπ/dY = – 8Y + 56 = 0
– Y = 7 and X = 5
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Slide 46
Example 7
• Use the Lagrangian method to
maximize the following profit function:
– π = 80X – 2X2 – XY – 3Y2 + 100Y
• Subject to the following constraint:
– X + Y = 12
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Slide 47
Example 7: Solution
• Form the Lagrangian function
– L = 80X – 2X2 – XY – 3Y2 + 100Y + (X + Y – 12)
• Find the partial derivatives and solve
simultaneously
– dL/dX = 80 – 4X –Y +  = 0
– dL/dY = – X – 6Y + 100 +  = 0
– dL/d = X + Y – 12 = 0
• Solution: X = 5, Y = 7, and  = -53
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Slide 48
Interpretation of the
Lagrangian Multiplier, 
• Lambda, , is the derivative of the
optimal value of the objective function
with respect to the constraint
– In Example 7,  = -53, so a one-unit
increase in the value of the constraint (from
-12 to -11) will cause profit to decrease by
approximately 53 units
– Actual decrease is 66.5 units
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Slide 49