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Optimization Techniques Methods for maximizing or minimizing an objective function Examples...
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Transcript of Optimization Techniques Methods for maximizing or minimizing an objective function Examples...
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
0
50
100
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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
Total, Average, andMarginal Cost
Q TC AC MC0 20 - -1 140 140 1202 160 80 203 180 60 204 240 60 605 480 96 240
AC = TC/Q
MC = TC/Q
Total, Average, andMarginal Cost
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120
180
240
0 1 2 3 4Q
TC ($)
0
60
120
0 1 2 3 4 Q
AC, MC ($)AC
MC
Profit Maximization
Q TR TC Profit0 0 20 -201 90 140 -502 160 160 03 210 180 304 240 240 05 250 480 -230
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
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.
0limX
dY Y
dX X
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
0dY
dX
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
1bdYb aX
dX
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
dY dU dV
dX dX dX
Y U V
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
dY dV dUU V
dX dX dX
Y U V
Rules of Differentiation
Quotient Rule: The derivative of the ratio of two functions U and V, is defined as follows.
( )U g X ( )V h X UY
V
2
dU dVV UdY dX dXdX V
Rules of Differentiation
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 dU
dX dU dX
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.
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.
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
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
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
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)
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
Quadratic Formula
• Write the equation in the following form:aX2 + bX + c = 0
• The solutions have the following form:2b b 4ac
2a
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
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
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
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
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
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
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
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
New Management Tools
• Benchmarking
• Total Quality Management
• Reengineering
• The Learning Organization
Other Management Tools
• Broadbanding
• Direct Business Model
• Networking
• Pricing Power
• Small-World Model
• Virtual Integration
• Virtual Management