On double one-parametric optimization problems and some applications
Paulo Mbunga
Mathematisch-Naturwissenschaftliche Fakultt II Institut fr Mathematik Humboldt-Universitt zu Berlin
24th May 1999
Abstract
This paper deals with a special class of quadratic optimization prob- lems with quadratic constraints, which we call double parametric. We try to solve these problems using Pathfollowing Methods with a special case of the Standard Embedding. The singularity theory developed by Jongen, Jonker and Twilt plays a great role in our investigations. In cases where a jump from one connected component to another one in the set of local minimizers and in the set of generalized critical points is not possible, we prove that the turning points are in negative direction. We survey results with respect to the choice of the starting point and make some proposals in order to overcome cases where jumps are not possible. We show the role of the Mangasarian-Fromovitz constraints qualication in the existence of the jumps and the solution of considered problems. Some examples of ap- plications to global quadratic optimization and multicriteria optimization are given.
Keywords: parametric optimization, pathfollowing methods, jumps, gen- eralized critical point, turning point in the negative sense
1 Introduction 2
1 Introduction
In this paper, we consider the following double quadratic optimization problems minf(x;xo)TD(x;xo)jx2Mig i = 1 2:
(Pi) M1 :=
(
x2Rn gj(x) = aTjx + bj 0 j2J gs+i(x) = xTAix + aTs+ix + bs+i 0 i2I
)
M2 :=
(
x2Rn gj(x) = aTjx + bj 0 j2J gs+1(x) = xTA1x + aTs+1x + bs+1 0
)
J = f1 sg I =f1 2 3g
where D is a positive-denite matrix, A1 a negative semi-denite matrix, A2 and A3positive semi-denite matrices, aja vector inRnand xo2Rna specially chosen point. According to the properties of the matrices Ai,i = 1 ::: 3, the functions gs+i are the respective quadratic concave and convex constraints (for i = 2 3).
Let
P :=fx2RnjaTjx + bj0 (j = 1 s)g
G1:=x2Rnjgs+1(x) = xTA1x + aTs+1x + bs+1 0 G2:=fx2Rnjgs+2(x) = xTA2x + aTs+2x + bs+20g G3:=fx2Rnjgs+3(x) = xTA2x + aTs+3x + bs+30g: The feasible sets M1and M2 can also be given by
M1 = P\\3
i=1Gi
M2 = P\G1:
We notice that the sets M1 and M2 are not necessarily convex (see Fig. 1).
Remark 1.
The kind of problems Pi,i = 1 2, was considered e.g. in 13].Let us introduce some conditions, that we shall need later. First we assume:
(C1)
P is a convex polyhedronThen the sets M1 and M2 are compact. The sets
H1 = fx2RnjDxgs+3(x) = 2A3x + as+3= 0g H2 = fx2RnjDxgs+1(x) = 2A1x + as+1= 0g represent the a ne subspaces with the dimension
n;rank(A3) and n;rank(A1), respectively.
1 Introduction 3
H1 xo
M1 gs+3 gs+2
P
gs+1
(a)
H1 xo
M1 M1
P
gs+2
gs+1
gs+3
H2 (b)
xo M2
P
gs+1
(c)
H2 xo
M2 M2
P
gs+1
(d) Figure 1: possible structure of M1 and M2
Starting situation
One of the important issues we have to deal with is the choice of the starting point xo. We shall see later that solving the problems (Pi) i = 1 2 successfully will depend on the choice of this starting point. In this paper we suggest to choose the starting point in the following way (see Fig. 1):
Problem P1:
C2(a) xo2int(P\G1\G2)n(H1G3):
Problem P2:
C2(b) xo 2 intP n(G1H2): Furthermore, we claim that the observed constraints must have full dimension(cf. 13]):
Mi = cl intMi i = 1 2:
(C3)
With the condition (C3) we want to exclude cases where M1@(P \\3
i=1Gi) M2@(P\G1):
Let us recall now the well-known concept of embedding (cf. e.g. Allgower 1], Guddat et al. 5], Dentcheva 3], Gfrerer et al. 10]) and propose the following embeddings in order to solve (Pi) i = 1 2:
Let xo2P be arbitrarily chosen and dene
minff(x t)jx2M1(t)g t2(;1 1]
P
1(t) :
2 Theoretical Background 4
where
f(x t) := (x;xo)TD(x;xo)
M1(t) := fx2Rnjgj(x t)0 (j = 1 s + 3)g gj(x t) := gj(x) j = 1 s + 2
gs+3(x t) := gs+3(x) + (t;1)gs+3(xo) and
minff(x t)jx2M2(t)g t2(;1 1]
P
2(t) : where
f(x t) := (x;xo)TD(x;xo)
M2(t) := fx2Rnjgj(x t)0 (j = 1 s + 1)g gj(x t) := gj(x) j = 1 s
gs+1(x t) := gs+1(x) + (t;1)gs+1(xo)
In Section 2 we present the necessary background for one-parametric problems, of which Pi(t) is a particular case. In Section 3 we study Pi(t) and in section 4 we analyze under which perturbations on Pi(t) we obtain a parametric problem PD(t) belonging to the generic class F of Jongen, Jonker and Twilt 15]. In Section 5 we present some applications.
2 Theoretical Background
We consider the general one-parametric problem:
minff(x t)jx2M(t)g t2R P(t) :
where
M(t) = fx2Rnjhi(x t) = 0 i2I gj(x t)0 j2Jg I = f1 mg m < n J =f1 sg
and f hi gj2Cq(RnRR) q 2.
Furthermore, we introduce the following notations:
P
gc := f(x t)2RnRj x is a g.c. point of P(t)g
P
stat := f(x t)2RnRj x is a stationary point of P(t)g
Ploc := f(x t)2RnRj x is a local minimizer of P(t)g H := (h1 hm)T G := (g1 gs)T:
Note 1.
For the denition of a g.c. point we refer to 15], see also 5].2 Theoretical Background 5
xo M1 gs+3
gs+2
P
gs+1
gs+3(x 0) = 0
(a)
xo gs+3(x 0) = 0
M1 M1
P
gs+2
gs+1
gs+3
(b)
xo
gs+3(x 0) = 0 M2
P
gs+1
(c)
xo
gs+3(x 0) = 0
M2 M2
P
gs+1
(d) Figure 2: possible structure of M1(t) and M2(t)
The LinearIndependence Constraint Qualication (LICQ)is said to hold at x2M(t) if the vectors
Dxhi(x t) i2I Dgj(x t) j2Jo(x t) are linearly independent (Jo(x t) :=fj2Jjgj(x t) = 0g).
TheMangasarian-Fromovitz Constraint Qualication (MFCQ)is satised at x2M(t) if:
MF1 fDhi(x t) i2Igare linearly independent,
MF2 there exists a vector 2Rnsatisfying:
Dhi(x t) = 0 i2I Dgj(x t) < 0 j2Jo(x t):
Next we cite our short characterization of the classF introduced by Jongen, Jonker and Twilt 15] from 2:5 in 5] . In 15] the local structure ofPgc is completely described if (f H G) belongs to a Cs3-open and dense subset F of C3(Rn+1 R)1+m+s, where Cs3denotes the strong (or Whitney-)Cs3-topology (see
2 Theoretical Background 6
also 5]). A typical baseneighbourhood of this topology with f 2Ck(Rn+1 R) and 2C+(Rn+1 R) is given by
Bk(f) :=
(
g2Ck(Rn+1 R)
j@g(x);@f(x)j< (x)
8x2Rn+1 8 withjjk
)
where
C+(Rn+1 R) :=f :Rn+1;!Rj is continuous and (x) > 0 8x2Rn+1g: If (f H G)2F , thenPgccan be divided into 5 types.
Type
1:
A point z = (x t)2Pgc is of Type 1 if the following conditions are satised:There exists i j2R i2I j2Jo(z) satisfying
0
@Dxf +X
i2I iDxhi+ X
j2Jo(z) jDxgj
1
A
jz =
z
= 0 (1)
(ND1) LICQ is satised at x2M(t),
therefore i j 2Ri2I j2Jo(z) are uniquely dened, (ND2) j6= 0 j2Jo(z),
(ND3) Dx2L(x t) jT(z) is nonsingular, where D2xL is the Hessian of the Lagrangian
L(x t) = f(x t) +X
i2I hi(x t) + X
j2Jo(z) gj(x t) and the uniquely determined numbers i j are taken from (1).
Furthermore,
T(z) =f2RnjDxhi(z) = 0 Dxgj(z) = 0 j2Jo(z)g
is the tangent space at z. Dx2L(z)jT(z)represents VTD2xLV , where V is a matrix whose columns form a basis of T(z).
A point of Type 1 is a nondegenerate critical point. The set Pgc is the closure of the set of all points of Type 1, the points of the Types 2;5 constitute a discrete subset ofPgc.
The points of the Types 2;5 represent basic degeneracies:
Type
2:
- violation of (ND2)Type
3:
- violation of (ND3)Type
4:
- violation of (ND1) andjIj+jJo(z)j;1 < nType
5:
- violation of (ND1) andjIj+jJo(z)j= n + 12 Theoretical Background 7
For each of these ve types Figure 3 illustrates the local structure of Pgc in the neighbourhood of stationary points. The full curves stand for the curve of stationary points z = (x t), and the dotted curve represents the curve of g.c.
points which are not stationary points. LetPgc 2 f1 5gbe the set of g.c. points of Type . The classF is dened by
F:=
(
(f H G)2C3(RnRR)1+m+sjPgc 5
=1
Pgc
)
: Recall thatF is Cs3 open and dense in C3(RnR R)1+m+s(cf. 15]).
The algorithm PATH III
This algorithm computes a numerical description of a compact connected com- ponent inPgc, i.e., in particular, it nds a discretization of the interval tA tB] tA<
0 < tB(not necessarily tA tB] 0 1]) and corresponding g.c. points starting at (xo 0)2Pgc. The algorithm is based on the active index set strategy and is a so-called predictor-corrector scheme if the active index set is constant. A Newton corrector is used.
The main point of the approach consists in the computation of the new index sets for the possible continuations.
We note that we do not have any numerical di culties walking around turning points of the Types 3 or 4.
More precisely: if there exists a PC2-path connecting (xo 0) and a point (x 1), then we obtain, in a nite number of predictor and corrector steps, a point lying in the radius of convergence of the Newton method for x with respect to the problem P(1). Since PATH III is not successful in nding a point (x 1)2Pgc in general, we propose to jump from one connected component in clPloc and
P
gc, respectively, to another one. For more information we refer the reader to Guddat et al. 5] and PAFO 11].
The algorithm JUMP I
This algorithm works in the set clPloc. Starting at the known local minimizer xoat to= 0, a connected component in clPlocfor increasing t will be numeric- ally described by using PATH III. Depending on the appearance of a singularity, a direction of descent will be computed. Using a feasible direction method, a local minimizer on another connected component inPlocwill be calculated and PATH III starts again. We have to take into account that we have no proposals for jumps in any case. Jumps are possible if there occurs a turning point of Type 2 or a point of Type 3. Let t be near t t < t, and let xm(t) and xs(t) be the local minimizer and a point ofPstatnPloc, respectively. Then, as t tends to t, the vector
u(t) := xs(t);xm(t)
kxs(t);xm(t)k
2 Theoretical Background 8
stat
gc
MFCQ holds (k)
z
MFCQ violated (l)
z
MFCQ violated (m) Type 5 z
J0(z)6= (g)
z
J0(z)6= (h) z
J0(z) = (i)
z
J0(z) = (j) Type 4 z
(e)
z
(f) Type 3 z
(a) z
(b) z
(c) z
(d) Type 2 z
Type 1 z
t
Figure 3: 5 Types of Jongen, Jonker and Twilt
3 Properties of the double parametric embedding 9
tends to a tangential vector, say u, which is a direction of (higher order) descent.
Hence, for t near t t < t, the vector xs(t);xm(t) provides an approximately tangential direction of descent (see Figure 4)
A g.c. point of Type 4 is a quadratic turning point and, when passing z along clPloc, the local minimizer switches into a local maximizer. We have the following cases for t < tand t close to t:
Case
I:
The value of f decreases.Case
II:
The value of f increases.In Case I it is possible to jump to another brach of local minimizers. In fact, since the feasible set M(t) is compact, we compute a point onPgc beyond the turning point, say (xmax(t) t) with t < tclose to t. The point xmax(t) is a local maximizer for P(t) and we can start at xmax(t) with a descent method in order to nd a local minimizer. In Case II, there is no proposal for possible jumps (see Fig. 4).
If a point of Type 5 appears, we also do not know a jump in case the MFCQ is violated. Such a situation is characterized by the fact that the connected com- ponent in the feasible set shrinks to one point and becomes empty for increasing t.
The algorithm JUMP II
The algorith Jump II works inPgc and it is useful for the computation of as many connected components as possible using PATH III and jumps in Pgc. For more details, we refer the reader to Guddat et al. 5] for the pathfollowing algorithms in the Sections 4 and 5.
3 Properties of the double parametric embedding
The rst theorem includes basic properties of the embeddings Pi(t), i = 1 2.
Theorem 1.
It holds:(E1) xo is a global minimizer forPi(0). (E2) Pi(1)Pi,
(E3) Mi(t1)Mi(t2) for t1t2,t1 t22(;1 1], (E4) Mi(t)6=and compact , for allt2(;1 1],
(E5) i(t)6= 8t2(;1 1] where i(t)denotes the set of all global minim- izers ofPi(t).
3 Properties of the double parametric embedding 10
Proof. Without loss of generality we consider the problem P1(t). (E1) and (E2) are obvious.
(E3) Let t1t2, t1 t22(;1 1]. If x2M1(t2), then gs+3(x) + (t2;1)gs+3(xo)0.
We have gs+3(x) + (t1;1)gs+3(xo)gs+3(x) + (t2;1)gs+3(xo), since gs+3(xo) > 0. So, x2M1(t1).
(E4) follows from (E3) and Mi6=. M1(t) is compact, since P is compact and the setT3i=1Giclosed.
(E5) follows from (E4).
For the problems Pi(t), i = 1 2 let us assume:
(C4) (f g1 gs+3)2F and (f g1 gs+1)2F respectively :
F is the class of Jongen, Jonker and Twilt.
Remark 2.
In this case the problemsPi(t),i = 1 2are regular in the sense of Jongen, Jonker and Twilt (i.e., JJT-regular).One important issue is to determine the kind of singularities which may appear. Theorem 2 gives us an answer.
Denition 1.
A point(x t)2Mi(t)Ri = 1 2is a vertex if xis a vertex of the polyhedron P.Theorem 2 (Singularity Theorem).
Assume that (C1), (C2), (C3), (C4) hold and letIko(x t) :=fji2f1 sgjgji(x t) = aTjix + bji = 0 i = 1 kg (2)
be the index-set of active linear inequalities of the problems Pi(t), i 2 f1 2g. Then the following assertions are true:
A) ForP(t) :=P1(t)
For any z = (x t)2clPlocnP1gcand t2(;1 1], the following holds:
(i) z2P2gcP3gcP4gcP5gc.
(ii) (xo 0)2P2gc\Ploc and (xo 0) is no turning point.
(iii) Each regular vertex zwithJo(z) =fs + 3gIko(z), is a point of Type 5and the MFCQ is fullled.
(iv) If zis no vertex of P and Jo(z) = fs + 3gIko(z), or Jo(z) =fs + 2gfs + 3gIko(x t), then
z2P2gcP3gcP5gc and the MFCQ is fullled.
(v) Ifs + 12Jo(z), then it holds:
(a) If the MFCQ is not satised at z =) z2P4gcP5gc, (b) If the MFCQ is fullled at z =) z2P2gcP3gcP5gc.
3 Properties of the double parametric embedding 11
t x
x
t (~xmax(t) t)
Jump (xmin(t) t)
f decreases
x x
t (~xmax(t) t)
(xmin(t) t)
f increases
Figure 4: Jump in Points of Type 4
B) ForP(t) :=P2(t)
For any z = (x t)2clPlocnP1gcand t2(;1 1], the following holds:
(i) z2P2gcP3gcP4gcP5gc.
(ii) (xo 0)2P2gc\Ploc and (xo 0) is no turning point.
(iii) If(x t)is a regular vertex withJo(z) =fs + 1gIko(z), thenz2P5gc. (iv) If zis no vertex of withJo(z) =fs + 1gIko(x t), then it holds:
(a) If the MFCQ is violated =) z2P4gcP5gc, (b) If the MFCQ is satised=) z2P2gcP3gcP5gc.
C) If z2clPloc\P4gc, then sign
0
@
;DtXp j=1 jgj(z)
1
A< 0:
Remark 3. C)
means that the jump is impossible.Proof. A) P(t) :=P1(t) (i) Follows from (ii);(v).
(ii) From the choice of xoin (V2) we have:
(a)
gs+3(xo t)
8
>
<
>
:
< 0 if t < 0
= 0 if t = 0
> 0 if t > 0 (b) gj(xo t) = gj(xo)6= 0 j6= s + 3,
(c) it follows from (a) and (b) that Jo(xo 0) =fs + 3g.
3 Properties of the double parametric embedding 12
x Dxgs+3(x t) Dxgs+1(x t)
gs+3(x t) = 0 M(t) gs+1(x t)
Figure 5: Violation of the MFCQ
Since xo is a global minimizer of P1(t), t 0 we also have that xo is a g.c. point. Next, we verify that Dxgs+3(xo) 6= 0, since xo 62 H1, jJo(xo 0)j = 1, and Dx2L(xo 0) = D. Consequently the LICQ is satised and the rst conclusion follows from (C4). The latter statement follows from the fact that D2xL(xo 0)=T(xo 0) is positive- denite and the arguments described in (5:2:20) 5].
(iii) Since zis a regular vertex, we havejJo(z)j=jIko(z)j+ 1 = n + 1. Con- sequently, we obtain the rst statement. Using the convexity of gj j2 Jo(x t) we have
Dxgj(x t)((x t);(x t))gj(x t):
With our choice of ^x2int M1M1(t) we obtain gj(^x t) < 0. We set := (^x t);(x t): Here is an MFCQ-vector. For a given we have Dxgj(x t) < 0: Hence the MFCQ is satised.
(iv) Readily the same as (iii). Of course, ifjJo(z)j= n + 1
(jIko(z)j= n;1) and Jo(z) =fs + 2gfs + 3gjIko(z)j, then z is a point of Type 5 and the MFCQ is satised.
(v) (a) follows from Theorem 2.5.4 in 5] and (b) is obvious.
B) P(t) := P2(t) For (i), (ii) and (iv) we proceed as in
A)
For (iii) we also proceed as in
A)
. Here the MFCQ will not hold in general (see Fig. 5).C) In view of (5.2.31) and (5.2.32) in 5], it su ces indeed to check that sign
0
@
;DtXp j=1 jgj(z)
1
A< 0:
3.1 Illustrative examples 13
For this purpose we have Dt
0
@
p
X
j=1 jgj(z)
1
A= u^jg^j(xo) ^j = s + 1 s + 3:
Since z2P4gc, the MFCQ is not satised, we have ^j > 0. From g^j(xo) > 0 we obtain:
Dt
0
@
p
X
j=1ujgj(z)
1
A> 0:
Hence = sign(;DtL(z)) < 0. This completes the proof.
Remark 4.
Letz = (x t)2Pgc be a degenerate vertex and lets + 32Jo(z)(for the Problem P1) or s + 12Jo(z)( for the Problem P2), then it holds that(f G)62F:
From the practical point of view, we follow a path inPgc, starting at a point of Type 1.
Theorem 3.
Assume that (C4)for Pi(t),i = 1 2. Then it holds:(xo to)2P1gc 8to< 0:
Proof. Follows immediately from the fact that gs+3(xo to) < 0, Dxf(xo to) = 0 and Dx2f(xo to) = 2D for all to< 0.
3.1 Illustrative examples
Example 3.1.
minff(x1 x2) := (x1;x1o)2+ (x2;x2o)2j(x1 x2)2Mg M := f(x1 x2)2R2jgj(x1 x2)0 j = 1 7g g1(x1 x2) := x1;6x2+ 8
g2(x1 x2) := 3x1;x2;27 g3(x1 x2) := x1+ 3x2;29 g4(x1 x2) :=;6x1+ x2+ 22
g5(x1 x2) :=;(x1;7)2;(x2;4:5)2+ 9 g6(x1 x2) := (x1;7)2+ (x2;5)2;16 g7(x1 x2) := (x1;3)2+ (x2;5:5)2;9
x1o= 10:5 x2o= 5
g7(x1 x2 t) := g7(x1 x2) + (t;1)g7(x1o x2o)
3.1 Illustrative examples 14
0 1 2 3 4 5 6 7 8 9 10 11 01
23 45 67 89
xo
g1 g2
g3 g4
g7
g6 g5 t = 0
g7(x 0) = 0
Figure 6: Feasible Set for the Example 3.1
0 1 2 3 4 5 6 7 8 9 10 11 01
23 45 67 89
xo
g1
g2 g3
g4
g5
Figure 7: Feasible Set for the Example 3.2
3.2 On the role of the MFCQ 15 Example 3.2.
minff(x1 x2) := (x1;x1o)2+ (x2;x2o)2j(x1 x2)2Mg M :=f(x1 x2)2R2jgj(x1 x2)0 j = 1 7g g1(x1 x2) := x1;6x2+ 8
g2(x1 x2) := 3x1;x2;27 g3(x1 x2) := x1+ 3x2;29 g4(x1 x2) :=;6x1+ x2+ 22
g5(x1 x2) :=;(x1;10)2;(x2;4)2+ 9 x1o= 10 x2o= 5
g5(x1 x2 t) := g5(x1 x2) + (t;1)g5(x1o x2o)
Remark 5.
The Figures 8, 9 and 10 show the dependence of the solution on the dierent starting points. If we start with the point xo = (10:5 5), we will not be successful, but starting withxo= (9:55 2:90)and xo= (4:5 7:5)leads us to the solution. Figure 11 shows the solution of Example 3.2.3.2 On the role of the MFCQ
In this section we give an answer to the questionunder which conditions the path- following methods are successfulwith a specially chosen starting point, when we assume that (C4) holds. Under the assumption that the MFCQ is satised at each x2Mi(t), points of Type 4 are excluded. Hence, the case II in Section 2 may not appear. We ask for a condition on the original problems (Pi) i = 1 2.
We discuss the so-called Enlarged Mangasarian-Fromovitz Constraint Qualic- ation (briey EnMFCQ) introduced by Gfrerer et al. 10].
Denition 2 (EnMFCQ).
The EnMFCQ is satised forP1 if, for all x2P (polyhedron), there exists a vector 2Rnwithgj(x) + Dgj(x) < 0 j2J+(x) :=fj2J jgj(x) 0 j6= s + 2g Dxgs+2(x) < 0 if gs+2(x) = 0:
The EnMFCQ is satised forP2 if, for allx2P, there exists a vector 2Rn with:
gj(x) + Dgj(x) < 0 j2J+(x) :=fj2Jjgj(x) 0g: The answer to the question above is given by the following theorem:
Theorem 4.
Assume the EnMFCQ is satised forPi i = 1 2. Then the MFCQ is satised for allx2Mi(t) i = 1 23.2 On the role of the MFCQ 16
9.4 9.6 9.8 10.0 10.2 10.4 10.6
2.5 3.0 3.5 4.0 4.5 5.0
x1
x2
type 2 type 3 type 4 type 5
9.4 9.6 9.8 10.0 10.2 10.4 10.6
2.5 3.0 3.5 4.0 4.5 5.0
x1
x2
-0.5 -0.4 -0.3 -0.2 -0.1 -0.0 0.1 0.2
9.4 9.6 9.8 10.0 10.2 10.4 10.6
t
x1
type 2 type 3 type 4 type 5
-0.5 -0.4 -0.3 -0.2 -0.1 -0.0 0.1 0.2
9.4 9.6 9.8 10.0 10.2 10.4 10.6
t
x1
-0.5 -0.4 -0.3 -0.2 -0.1 -0.0 0.1 0.2
9.4 9.6 9.8 10.0 10.2 10.4 10.6
t
x1
Figure 8: Example 1 in (x1 x2) and (t x1)-space
-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2
7.49 7.50 7.51 7.52 7.53 7.54 7.55 7.56
t
x2
-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2
7.49 7.50 7.51 7.52 7.53 7.54 7.55 7.56
t
x2
5.15 5.20 5.25 5.30 5.35 5.40 5.45 5.50
7.49 7.50 7.51 7.52 7.53 7.54 7.55 7.56
x1
x2
5.15 5.20 5.25 5.30 5.35 5.40 5.45 5.50
7.49 7.50 7.51 7.52 7.53 7.54 7.55 7.56
x1
x2
Figure 9: Example 1 in (x1 x2) and (t x1)-space
3.2 On the role of the MFCQ 17
4.20 4.25 4.30 4.35 4.40 4.45 4.50
2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85
x1
x2
4.20 4.25 4.30 4.35 4.40 4.45 4.50
2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85
x1
x2
-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2
2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85
t
x2
-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2
2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85
t
x2
Figure 10: Example 1 in (x1 x2) and (t x1)-space
7.5 8.0 8.5 9.0 9.5 10.0
4.5 5.0 5.5 6.0 6.5 7.0 7.5
x1
x2
type 2 type 3 type 4 type 5
7.5 8.0 8.5 9.0 9.5 10.0
4.5 5.0 5.5 6.0 6.5 7.0 7.5
x1
x2
7.5 8.0 8.5 9.0 9.5 10.0
4.5 5.0 5.5 6.0 6.5 7.0 7.5
x1
x2
-0.5 0.0 0.5 1.0 1.5 2.0
7.5 8.0 8.5 9.0 9.5 10.0
t
x1
type 2 type 3 type 4 type 5
-0.5 0.0 0.5 1.0 1.5 2.0
7.5 8.0 8.5 9.0 9.5 10.0
t
x1
Figure 11: Example 2 in (x1 x2) and (t x1)-space
3.2 On the role of the MFCQ 18
Proof. Without loss of generality we consider the problem P1. The proof for P2 is readily the same as the proof for P1. Therefore, we omit it. Set (J1)o(x t) :=
fj2J jgj(x t) = 0 j6= s+3g. Then, for all j2Jo(x t) = (J1)o(x t)fs+3g, we have
gj(x t) = 0()
(gj(x) = 0 if j6= s + 3 gj(x) =;(t;1)gj(xo) if j = s + 3 Dxgj(x t) = Dxgj(x):
Let x2M1(t), j 2Jo(x t). In view of Denition 2, it su ces indeed to check that the given vector is also an MF-vector in (x t). For j 6= s + 3 we have Dxgj(x t) = Dxgj(x) <;gj(x) = 0 and for j = s + 3 we obtain
Dxgj(x t) = Dxgj(x) <;gj(x) = (t;1)gj(xo)0 8t2(;1 1]:
Hence is an MF-vector.
Remark 6.
(a) In case, the MFCQ is satised for allx2Mi(t) t2(;1 1] i = 1 2the setMi(t1)is homomorphic toMi(t2)for allt1 t220 1] i = 1 2. (b) Assume (C4) and that the MFCQ is satised for allx2Mi(t) t2(;1 1].Then we havePgc\P4gc=.
Remark 7.
Assumez2Pgcwithx2H1ands+32Jo(z)respectivelyx2H2 and s + 12Jo(z). Then it holds thatz2P4gcP5gc and the MFCQ is not satised.
According to the propositionsA)(i) (v) andCof Theorem 2 and the example in Fig. 5, we know that points of Type 4 and 5, where the MFCQ is not satised, may appear.
Denition 3 (Gomez 23]).
We assumePi(t),i = 1 2, to beJJT-regular with respect to(;1 1]and z = (x t)2Pgc. A g.c. point zis called a turning point in negative direction (positive direction) if there exists a neighbourhood V (z), such that8(x t)2Pgc\V (z)holds fortt (t t).The Figures 12 (a) and (b) show turning points in negativ direction and the Figures (c) and (d) those in positive direction.
Theorem 5.
AssumePi(t),i = 1 2, to beJJT-regular with respect to(;1 1]. Then every turning point without possibilities to jump is a turning point in negative direction.Proof. The proof relies on Theorem 1 in Gmez 23]. We will divide the proof into two steps. Without loss of generality we consider the problem P1.
Step 1 (Type 5).- Let Jo(z) = f1 pg and p = s + 3. We deduce from the
3.2 On the role of the MFCQ 19
(a) Type 4
z
(b) Type 5
z
(c) Type 5
z
(d) Type 4
z
Figure 12: 5 Turning points in negativ and positiv direction
properties of a point of Type 5 that the setfDxgj(z) j = 1 p;1gis linearly independent. Then we obtain a uniquely determined solution up= (up1 upp;1) of the system
Dxf(z) +pX;1
j=1 pjDxgj(z) = 0 with pj 6= 0, j = 1 p;1.
Next, consider the system
2
6
6
6
4
Dxf(x t) +Ppj=1;1 pjDxgj(x t) g1(x t)
gp;1...(x t)
3
7
7
7
5
= 0:
(3)
By the Implicit Function Theorem, there exists a uniquely determined function (xp(t) up(t)) : (t; t+ )!Rn+p;1
(4) such that
(a) xp(t) = x up(t) = up,
(b) for each t2(t; t+ ) the vector (xp(t) up(t) t) is a solution of (3).
Recall that (xp(t) t) is a g.c. point if the inequality gp(xp(t) t) = gs+3(xp(t) t)0 (5)
holds. Hence, the curve (xp(t) t) belongs toPgc either for t2(t t+ ) or for t2(t; t). It su ces to show that
dtgd s+3(xp(t) t) > 0
since gp becomes strictly positive, i.e., (xp(t) t) is not feasible.
We calculate this quantity:
dtgd p(xp(t) t) = Dxgp(x t)_xp(t) + Dtgp(x t):
(6)