(jg Unconstrained Optimization Equality Constrained Optimization Equality/ Inequality Constrained Optimization R Lusby (42111) KKT Conditions 2/40
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[PDF] A Karush-Kuhn-Tucker example - UBC Math
The first KKT condition says λ1 = y The second KKT condition then says x − 2yλ1 + λ3 = 2 − 3y2 + λ3 = 0, so 3y2 =2+ λ3 > 0, and λ3 = 0 Thus y = √2/3, and x = 2 − 2/3 = 4/3 Again all the KKT conditions are satisfied
[PDF] Karush-Kuhn-Tucker Conditions
(jg Unconstrained Optimization Equality Constrained Optimization Equality/ Inequality Constrained Optimization R Lusby (42111) KKT Conditions 2/40
[PDF] KKT example
The KKT conditions are usually not solved directly in the analysis of practical large nonlinear programming problems by software packages Iterative successive
[PDF] Chapter 11
Ch 11 - Optimization with Equality Constraints 14 11 4 Necessary KKT Conditions - Example Example: Let's minimize f(x) = 4(x – 1)2 + (y – 2)2 with constraints
[PDF] Karush-Kuhn-Tucker conditions
Today: • KKT conditions • Examples • Constrained and Lagrange forms • Uniqueness The Karush-Kuhn-Tucker conditions or KKT conditions are: • 0 ∈ ∂f(x)
[PDF] Karush-Kuhn-Tucker Conditions - CMU Statistics
Today: • KKT conditions • Examples • Constrained and Lagrange forms • Uniqueness The Karush-Kuhn-Tucker conditions or KKT conditions are: • 0 ∈ ∂
[PDF] Applications of Lagrangian: Kuhn Tucker Conditions
In the example we are using here, we know that the budget constraint will be binding but it is not clear if the ration constraint will be binding It depends on the size
[PDF] KKT Examples - MIT OpenCourseWare
1 oct 2007 · The KKT conditions are usually not solved directly in the analysis of practical large nonlinear programming problems by software packages
[PDF] Kuhn-Tucker Example
Kuhn-Tucker Example Consider the problem min f ( r x ) = (x 1 - 4) 2 + (x 2 - 4) 2 { }, such that g The Kuhn - Tucker conditions are : —L( r x ) = 0, ni ≥ 0, ni
[PDF] CONSTRAINED OPTIMIZATION
DEFINITION: The Lagrangian function for Problem P1 is defined as L(x,λ) = f(x) + Σj=1 ,m λj hj(x) The KARUSH-KUHN-TUCKER Conditions If the point
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Karush-Kuhn-Tucker Conditions
Richard Lusby
Department of Management Engineering
Technical University of Denmark
Today's Topics
(jgUnconstrained OptimizationEquality Constrained Optimization
Equality/Inequality Constrained Optimization
R Lusby (42111) KKT Conditions2/40
Unconstrained Optimization
R Lusby (42111) KKT Conditions3/40
Unconstrained Optimization
(jgProblem minimizef(x) subject to:x2RnFirst Order Necessary Conditions Ifxis a local minimizer off(x) andf(x) is continuously dierentiable in an open neighbourhood ofx, then rf(x) =0 That is,f(x) isstationa ryat xR Lusby (42111) KKT Conditions4/40Unconstrained Optimization
(jgSecond Order Necessary Conditions Ifxis a local minimizer off(x) andr2f(x) is continuously dierentiable in an open neighbourhood ofx, then rf(x) =0 r2f(x) is positivesemi denite Second Order Sucient Conditions
Suppose thatr2f(x) is continuously dierentiable in an open neighbourhood ofx. If the following two conditions are satised, thenx is a local minimum off(x). rf(x) =0 r2f(x) isp ositivedenite R Lusby (42111) KKT Conditions5/40
Equality Constrained Optimization
R Lusby (42111) KKT Conditions6/40
Equality Constrained Optimization
(jgProblem minimizef(x) subject to:hi(x) = 08i= 1;2;:::m x2RnR Lusby (42111) KKT Conditions7/40Equality Constrained Optimization
Consider the following example
(jgExample minimize2x21+x22 subject to:x1+x2= 1Let us rst consider the unconstrained caseDierentiate with respect tox1andx2
@f(x1;x2)@x1= 4x1 @f(x1;x2)@x2= 2x2These yield the solutionx1=x2= 0Doesnot satisfy the constraintR Lusby (42111) KKT Conditions8/40
Equality Constrained Optimization
Example Continued
(jgLet us penalize ourselves for not satisfying the constraintThis gives
L(x1;x2;1) = 2x21+x22+1(1x1x2)This is known as theLagrangian of the p roblem Try to adjust the value1so we use just the right amount of resource1= 0!get solutionx1=x2= 0;1x1x2= 1
1= 1!get solutionx1=14
;x2=12 ;1x1x2=141= 2!get solutionx1=12
;x2= 1;1x1x2=12 1=43 !get solutionx1=13 ;x2=23 ;1x1x2= 0R Lusby (42111) KKT Conditions9/40Equality Constrained Optimization
Generally Speaking
(jgGiven the following Non-Linear ProgramProblem
minimizef(x) subject to:hi(x) = 08i= 1;2;:::m x2RnA solution can be found using theLagrangianL(x;) =f(x) +mX
i=1 i(0hi(x))R Lusby (42111) KKT Conditions10/40Equality Constrained Optimization
Why isL(x;) interesting?(jgAssumexminimizes the following minimizef(x) subject to:hi(x) = 08i= 1;2;:::m x2RnThe following two cases are possible:1The vectorsrh1(x);rh2(x);:::;rhm(x) are linearly dependent2There exists a vectorsuch that
@L(x;)@x1=@L(x;)@x2=@L(x;)@x3=;:::;=@L(x;)@xn= 0 @L(x;)@1=@L(x;)@
2=@L(x;)@
3=;:::;=@L(x;)@
m= 0R Lusby (42111) KKT Conditions11/40
Case 1: Example
(jgExample minimizex1+x2+x23 subject to:x1= 1 x21+x22= 1The minimum is achieved atx1= 1;x2= 0;x3= 0The Lagrangian is:
L(x1;x2;x3;1;2) =x1+x2+x23+1(1x1) +2(1x21x22)Observe that: @L(1;0;0;1;2)@x2= 181;2Observerh1(1;0;0) =h1 0 0i
andrh2(1;0;0) =h2 0 0i
R Lusby (42111) KKT Conditions12/40
Case 2: Example
(jgExample minimize2x21+x22 subject to:x1+x2= 1The Lagrangian is: L(x1;x2;1) = 2x21+x22+1(1x1x2)Solve for the following: @L(x1;x2;1@x1) = 4x11= 0 @L(x1;x2;1@x2) = 2x21= 0 @L(x1;x2;1)@= 1x1x2= 0R Lusby (42111) KKT Conditions13/40Case 2: Example continued
(jgSolving this system of equations yieldsx1=13 ;x2=23 ;1=43Is this a minimum or a maximum?
R Lusby (42111) KKT Conditions14/40
Graphically
(jgx 1x 2x1+x2= 11
1R Lusby (42111) KKT Conditions15/40
Graphically
(jgx 1x 2x1+x2= 11
1x 1=13x2=23rf(x) =λrh(x)R Lusby (42111) KKT Conditions15/40
Geometric Interpretation
(jgConsider the gradients offandhat the optimal pointThey must point in the same direction, though they may have
dierent lengths rf(x) =rh(x)Along with feasibility ofx, is the conditionrL(x;) = 0From the example, atx1=13 ;x2=23 ;1=43 rf(x1;x2) =h4x12x2i
=h 4343
i rh1(x1;x2) =h 1 1i
R Lusby (42111) KKT Conditions16/40
Geometric Interpretation
(jgrf(x) points in the direction of steepest ascentrf(x) points in the direction of steepest descentIn two dimensions:
I rf(xo) is perpendicular toa level curve of f I rhi(xo) is perpendicular tothe level curve hi(xo) = 0R Lusby (42111) KKT Conditions17/40Equality, Inequality Constrained Optimization
R Lusby (42111) KKT Conditions18/40
Inequality Constraints
What happens if we now include inequality constraints? (jgGeneral Problem maximizef(x) subject to:gi(x)0( i)8i2I h j(x) = 0( j)8i2JGiven a feasible solutionxo, the set ofbinding constrain tsis: I=fi:gi(xo) = 0gR Lusby (42111) KKT Conditions19/40The Lagrangian
(jgL(x;;) =f(x) +mX i=1 i(0gi(x)) +kX j=1 j(0hj(x))R Lusby (42111) KKT Conditions20/40Inequality Constrained Optimization
(jgAssumexmaximizes the following maximizef(x) subject to:gi(x)0( i)8i2I h j(x) = 0( j)8i2JThe following two cases are possible:1rh1(x);:::;rhk(x);rg1(x);:::;rgm(x) are linearly dependent2There exist vectorsandsuch that
rf(x)kX j=1 jrhj(x)mX i=1 irgi(x) = 0 igi(x) = 00R Lusby (42111) KKT Conditions21/40
Inequality Constrained Optimization
(jgThese conditions are known as the Karush-Kuhn-Tucker ConditionsWe look for candidate solutionsxfor which we can ndandSolve these equations using complementary slackness
At optimality some constraints will be binding and some will be slackSlack constraints will have a correspondingiof zeroBinding constraints can be treated using the Lagrangian
R Lusby (42111) KKT Conditions22/40
Constraint qualications
(jgKKT constraint qualication rgi(xo) fori2Iare linearly independentSlater constraint qualication gi(x) fori2Iare convex functionsA non boundary point exists:gi(x)<0 fori2IR Lusby (42111) KKT Conditions23/40