Constrained NL Optimization (NLP)
Goals:
• Optimality conditions for linear constraints
• Lagrange multipliers and Lagrangian
• Optimality conditions for nonlinear constraints • Duality
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General Statement
A minimization problem with constraints is often written as:
min f (x)
subjectto gi(x)0, iE
gj(x)0, jI where f and g are C2.
Essential Difference Between LP and NLP
For solvable LP problems, optimal solution occurs at an extreme point.
For NLP problems, the optimal solution may not be at an extreme point or even on the boundary of the feasible set.
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Example 1: max P KL,
subject to 4K L 8,
Example 2:
max f(x)sinx,
K, L 0 3/21/2020
subject to 0 x .
Examples
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Case 1: Linear Equality Constraints Consider the optimization problem:
min f (x)
subjectto Axb, Amn
with rank(A) m n.
Idea: transform the constrained problem into an unconstrained optimization problem.
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A Motivating Example
Solve the optimization problem with a linear equality constraint:
min f(x)x2 2x x2 x2 4x 11233
subjectto x x 2x 2. 123
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A Motivating Example (cont’d)
The optimization problem can be made “unconstrained” as follows:
min f(x)2×2 3×2 4x x 2x 23232
using x 2x 2x. 123
Answer
A strict local minimizer to the unconstrained problem
is:
x2 1.5, x3 1. Thus, for the original problem,
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x (2.5, 1.5, 1), *
f x 1.5. *
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An Exercise
Solve the following optimization problem:
max f (x) x x x 123
subject to 123
x x x 1 aaa
123 withai 0,xi 0,i1,2,3.
Observation
Indeed, any optimization problem with linear equality constraints Ax=b
can be transformed into an unconstrained problem!
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Some Terms from Matrix Theory
To generalize, let’s recall the following notions.
The null space of a matrix Amn , with m n, is: NApn: Ap0.
T
The range space of A is defined as:
RAT qn:qAT,forsomem .
NARA. nT
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Null‐space Matrix Z: AZ=0
A matrix Z nr is a null-space matrix of A if any vector in N ( A) can be expressed as
a linear combination of the columns of Z.
If r nm, then Z is called a basis matrix for
the null space of A (as the columns of Z are linearly
independent!).
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Representation of the Null Space
Since the null space can be represented as
N(A) p: pZvforsomev , r
it follows that
N(A) R(Z).
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Comment
If Amn is a matrix of full (row) rank m,
then without loss of any generality, we can assume
AB N, Bmm fullrank,Nm(nm). As it can be directly checked,
Z B1Nn(nm). I
nm
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As a consequence, we have:
If x is any point satisfying Ax b, then all other points can be written as
x xZv, forsomevectorv Particular non-homogeneous solution
because x x N ( A).
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General homogenous solutions
Optimization with Linear Equality Constraints
min f (x) subjectto Axb
m i nr ( v ) f ( x Z v ) . v
The function is called the reduced function.
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Comment
If Z is a basis matrix for the null space of A, then
is a function of n‐ m variables, i.e., r = n ‐ m. In other words, the transformed unconstrained
problem involves less variables.
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Motivating Example revisited
min f(x)x2 2x x2 x2 4x 11233
subjectto x x 2x 2. 123
1 2 B1N
TakeZ1 0,forAB N
and x200.
1112 0 1 I
22 T
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Motivating Example revisited
2 1 2v xxZv01 01.
as before.
v2 0 0 1
min(v)2v2 3v2 4vv 2v 12121
Optimality Conditions
minf(x) min(v) f(xZv) subjectto Axb vr
Using the chain rule,
(v) ZT f (x Zv) ZT f (x) 2(v)ZT2 f(xZv)ZZT2 f(x)Z
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Necessary Conditions
Ifx isalocalminimizeroff over x:Axb
*
and Z is a null-space matrix for A, then
ZTf(x)0. (x:”stationarypoint”) *
ZT2 f (x )Z is positive semi-definite. *
That is, the reduced gradient is zero and the reduced Hessian matrix is positive semidefinite.
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Comments
•
A point at which the reduced gradient is zero is
•
called a stationary point.
The second‐order condition implies:
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pT2f(x)p0, pN(A). *
However, this condition does not require that 2 f (x ) be positive semidefinite!
*
Ax b, *
Sufficient Conditions
If x satisfies: *
ZTf(x)0,and *
ZT2f(x)Zispositivedefinite, *
where Z n(nm) is a basis matrix for N(A), then, x is a strict local minimizer of
*
fover x:Axb.
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Motivating Example revisited
min f(x)x2 2x x2 x2 4x 11233
subjectto x x 2x 2. 123
1 2 B1N TakeZ1 0 ,
I 0 1
forA B N 1 1 2 .
N
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T Sincef(x)2x 2, 2x , 2x 4 ,
123
ZTf(x)(0, 0)T atthefeasiblepoint *
x 2.5, 1.5, 1. *
In addition,
ZT2f(x)Z 0 2 01 0
1 1 02 0 01 2 * 201
4 4 positivedefinite 4 6
0 0 20 1
However, 2 f (x ) is not positive definite!!
*
Lagrange Multipliers
Fact : At a local minimizer x , *
f (x ) AT **
where*m iscalled”Lagrangemultipliers”.
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Proof
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Let Z be any n r null-space matrix for A. Then,vr,m s.t.
Zv 0. *
**
f(x)Zv AT ***
ZTZv 0, usingZT AT (AZ)T 0 *
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A Geometric Demonstration
aT x b
x*
f (x ) *
a
f (x)
x
isovalue contours
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Motivating Example revisited
min f(x)x2 2x x2 x2 4x 11233
subjectto x x 2x 2. 123
2x 2 1 1
2×2 1 usingthe1st-ordern.c. 2x 4 2
3
and x x 2x 2 the equality constraint
123
3, x 2.5, 1.5, 1 .
**
T
Lagrangian Function
For the problem: min f (x), subject to Ax b, Amn with rank(A) m,
we can define the Lagrangian function:
L(x,λ)f(x)m aTxbf(x)T(Axb)
note :
aiT stands for the ith row of A.
iii i1
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First‐Order Optimality Conditions
Lx , 0, **
because
L(x, )f(x)A ,
T L(x, )bAx.
x
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A Perturbed Problem
Assumethatf(x )min f(x), subjecttoAxb. *
Then, for the perturbed problem:
f ( x ) min f ( x), subject to Ax b .
Indeed, for small ,
f (x) f x x x f (x )
***
T f(x)xx A
***
f(x )T **
T
T
Exercise
Consider min f(x)x2 x2x2 2x x x4 8x 113 122 2
subjectto: 2x 5x x 3. 123
(a) Which of the following points are stationary: TT
T
(b) local minimizer? local maximizer? saddle point?
(0,0,2) ; 0,0,3 ; 1,0,1 .
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Case 2: Linear Inequality Constraints
Consider the optimization problem:
min f (x)
subjectto Axb, Amn.
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An Example
min f (x) 1 x2 1 x2 21 22
subjectto x 2x 2 12
x x 1 12
x 3. 1
x2
x
*
The optimal solution x* reduces to a solution of the equality constrained problem
with the active constraints.
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x1
More specifically,
The minimizer x* for the original problem is the same as the simplified problem:
min1x2 1×2 21 22
24 2
x* 5, 5,* 50.
subjectto x2x2.
For this example, x 2x 2 is an active constraint
12
Note :
at x* , while the other two are inactive.
12
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m i n f ( x )
subjectto Axb, Amn
Observations
m i n f ( x )
subjectto Axb, alltheactiveconstraints
ˆˆ
For the inactive constraints, we associate with zero Lagrange multipliers:
aTxb0, i1,,m. ii*i
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Remark
It may happen that none of the constraints in NL programming are active, i.e., constrained NL programming may be the same as unconstrained. A sharp difference with linear programming!
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22 minf(x)x1 x2
subject to
4x x 0,
12 2x x 0,
12
x 0, x 0.
12
12
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Necessary Conditions
At a local minimizer x , then s.t. **
f(x)AT,(or,ZTf(x)0), ***
* 0,
T Ax b0, and
**
ZT 2 f (x )Z is positive semi-definite, *
where Z is a null-space matrix for the matrix
ˆ
A of active constraints at x .
*
Comment 1
ˆ f(x)AT AT()
* * * active
because*i 0foranyinactiveithconstraint.
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Comment 2 (On the sign of Lagrange multipliers for the inequality constraints)
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By contradiction, find a stepsize s to decrease f at x : *
0 f(x s) f(x )f(x )T s. ***
On the other hand, for the active constraints
ˆˆ
atx , i.e., Ax b0, wewant
**
ˆˆˆˆˆ 0A x s bAx b As
Equivalently,
**
0
T
{f(x) s0andAs0 f(x)A, 0,
ˆˆ **
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Aˆ s
Clearly, the shaded area is empty only when these two vectors are in the same direction!
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f x *
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An Example
Check the necessary conditions at x (1, 0)T *
of the constrained minimization problem: 24
subject to
minx1 1.5 x2 0.5 x
1x x 0, 12
1x x 0, 12
1x x 0, 12
1x x 0. 12