optimization unconstrained and constrained calculus part ii
TRANSCRIPT
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OptimizationOptimizationunconstrained and unconstrained and
constrained constrained
Calculus part II
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Setting-Up Optimization Setting-Up Optimization
ProblemsProblems
• Define the agent’s goal: objective function and identify the agent’s choice (control) variables
• Identify restrictions (if any) on the agent’s choices (constraints). If no constraints exist, then we have unconstrained minimization or maximization problems.
If constraints exist, what type? Equality Constraints (Lagrangian)Inequality Constraints (Linear Programming)
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Mathematically,Optimize y = f(x1, x2, . . . ,xn)
subject to (s.t.)gj (x1, x2, . . . ,xn) ≤ bj
or= bj j = 1, 2, . . ., m.
or
≥ bj
y = f(x1, x2, . . . ,xn) → objective function
x1, x2, . . . ,xn → set of decision variables (n)optimize → either maximize or minimize
gi(x1, x2, . . . ,xn) → constraints (m)
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Constraints refer to
• restrictions on resources• legal constraints• environmental constraints• behavioral constraints
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Review of DerivativesReview of Derivatives• y=f(x): First-order condition:
• Second-order condition: • Constant function:• Power function:• Sum of functions:
• Product rule:
• Quotient rule:
• Chain rule:
)(' xfdx
dy
)(''2
2
xfdx
yd
axfy )( 0)(' xfbaxxfy )( 1)(' bbaxxf
)()( xgxfy )(')(' xgxfdxdy
)()( xgxfy )(')()()(' xgxfxgxfdxdy
)(
)(
xg
xfy
2)(
)(')()()('
xg
xgxfxgxfdxdy
))(( xgfy )('))((' xgxgfdxdy
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Unconstrainted UnivariateUnconstrainted UnivariateMaximization Problems: max Maximization Problems: max
ff((xx))• Solution:
• Derive First Order Condition (FOC): f’(x)=0• Check Second Order Condition (SOC): f’’(x)<0
• Local vs. global: If more than one point satisfy both FOC and SOC, evaluate the objective function at each point to identify the maximum.
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ExampleExamplePROFIT = -40 + 140Q – 10Q2
Find Q that maximizes profit
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ExampleExamplePROFIT = -40 + 140Q – 10Q2
Find Q that maximizes profit
140 – 20Q = set 0
Q = 7
- 20 < 0
max profit occurs at Q = 7max profit = -40 + 140(7) – 10(7)2
max profit = $450
dQ
dPROFIT
2
2
dQ
PROFITd
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Minimization Problems: Minimization Problems: Min Min ff((xx))• Solution:
• Derive First Order Condition (FOC): f’(x)=0• Check Second Order Condition (SOC): f’’(x)>0
• Local vs. global: If more than one point satisfy both FOC and SOC, evaluate the objective function at each point to identify the minimum.
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ExampleExampleCOST = 15 - .04Q + .00008Q2
Find Q that minimizes cost
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ExampleExample
COST = 15 - .04Q + .00008Q2
Find Q that minimizes cost
-.04 + .00016Q = set 0
Q = 250
.00016 > 0
Minimize cost at Q = 250min cost = $10
dQ
dCOST
2
2
dQ
COSTd
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Unconstrained Unconstrained Multivariate OptimizationMultivariate Optimization
0),(g2
2
zxx
yxx
• Max
• FOC:
• SOC:
),( zxgy
),( zxgxy
x ),( zxgz
yz
0),(g zz2
2
zxz
y
0),(),(),(),(22
2
2
2
2
zxzxzxzx xz
yzxy
z
y
x
y
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ExampleExample
2122
2121 681010014060 QQQQQQPROFIT
Find Q1 and Q2 that maximize Profit
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ExampleExamplePROFIT is a function of the output of two products
(e.g.heating oil and gasoline)Q1 Q2
Solve Simultaneously Q1 = 5.77 units Q2 = 4.08 units
0616100 122
setQQdQ
dPROFIT
100166
140620
21
21
2122
2121 681010014060 QQQQQQPROFIT
0620140 211
setQQdQ
dPROFIT
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Second-Order Second-Order ConditionsConditions
202
1
2
dQ
PROFITd 621
2
dQdQ
PROFITd
1622
2
dQ
PROFITd
02
21
2
22
2
21
2
dQdQ
PROFITd
dQ
PROFITd
dQ
PROFITd
(-20)(-16) – (-6)2 > 0
320 – 36 > 0
we have maximized profit.
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Constrained Constrained OptimizationOptimization
• Solution: Lagrangian Multiplier Method• Maximize y = f(x1, x2, x3, …, xn)
• s.t. g(x1, x2, x3, …, xn) = b• Solution:
• Set up Lagrangian:
• FOC:
.),...,,(),...,,(),,...,,( 212121 bxxxgxxxfxxxL nnn
0),...,2,1(),,...,2,1(
0),,...,2,1(
...
0),,...,2,1(1
bxnxxgxnxxL
xnxxL
xnxxL
xn
x
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Lagrangian MultiplierLagrangian Multiplier
• Interpretation of Lagrangian Multiplier λ: the shadow value of the constrained resource.
o If the constrained resource increases by 1 unit, the objective function will change by λ units.
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ExampleExampleMaximize Profit =
subject to (s.t.) 20Q1 + 40Q2 = 200 Could solve by direct substitution
Note that 20Q1 = 200 – 40Q2 → Q1 = 10 – 2Q2
Maximize Profit =
2122
2121 681010014060 QQQQQQ
2222
2)222 )10(68210(10100)210(14060 QQQQQQ
units 56.5
units 22.2
1
2
Q
Q
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Lagrangian Multiplier MethodLagrangian Multiplier Method
)2004020(
681010014060L
Function Lagrangian Formulate
21
2122
2121PROFIT
QQQQQQ
0set )2004020(L
0set 40616100L
0set 20620140L
.,, are ariablesDecision v
.2004002 as long asfunction profit theMaximizes L Maximizing
2profit
122
profit
211
profit
21
21profit
QQd
d
QQdQ
d
QQdQ
d
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.constraint theof valuein the changeunit one a from resulting
function objective theof valuein the change themeasures
.774
units 22.2 and units 56.5en Wh
before asanswer Same
units 22.2
units 56.5
20040 20 also
180434
or
.1661001240280 Therefore,
21
2
1
21
21
2121
Q
Q
QQQQ