The purpose of this exercise is to explore a condition that implies all of the existence results for Lagrange multipliers that we have proved in this section for the case of inequality constraints only. Consider the problem in
The purpose of this exercise is to explore a condition that implies all of the existence results for Lagrange multipliers that we have proved in this section for the case of inequality constraints only. Consider the problem in
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.4: Linear Transformations
Problem 34EQ
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part D E
![3.3.5 (Constraint Qualifications for Inequality Constraints)
The purpose of this exercise is to explore a condition that implies all of the
existence results for Lagrange multipliers that we have proved in this section for
the case of inequality constraints only. Consider the problem
minimize f(æ)
subject to g;(x) < 0,
j = 1,... ,r.
For a feasible x, let F(x) be the set of all feasible directions at x defined by
F(x) = {d|d 0, and for some ā > 0, g(x + ad) < 0 for all a E [0,a]}
and denote by F(x) the closure of F(x). Let x* be a local minimum. Show that:
(a)
Vf(x*)'d >0,
Vde F(x*).
(b) If we have F(x*) = V (x*), where
V (a*) = {d|V9;(x*")'d < 0, V j € A(x*)},
then there exists a Lagrange multiplier vector u* > 0 such that
Vf(x*)+ H;V9; (x*) = 0,
H; = 0,
V j ¢ A(x*).
%3D
j=1
Hint: Use part (a) and Farkas' Lemma (Prop. 3.3.13).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F162c6403-ce91-41f8-a781-2ad0c6649c2f%2F42f1834e-6a0f-45e8-b732-6ecdca60b677%2F0t8oho_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3.3.5 (Constraint Qualifications for Inequality Constraints)
The purpose of this exercise is to explore a condition that implies all of the
existence results for Lagrange multipliers that we have proved in this section for
the case of inequality constraints only. Consider the problem
minimize f(æ)
subject to g;(x) < 0,
j = 1,... ,r.
For a feasible x, let F(x) be the set of all feasible directions at x defined by
F(x) = {d|d 0, and for some ā > 0, g(x + ad) < 0 for all a E [0,a]}
and denote by F(x) the closure of F(x). Let x* be a local minimum. Show that:
(a)
Vf(x*)'d >0,
Vde F(x*).
(b) If we have F(x*) = V (x*), where
V (a*) = {d|V9;(x*")'d < 0, V j € A(x*)},
then there exists a Lagrange multiplier vector u* > 0 such that
Vf(x*)+ H;V9; (x*) = 0,
H; = 0,
V j ¢ A(x*).
%3D
j=1
Hint: Use part (a) and Farkas' Lemma (Prop. 3.3.13).
![(c) The condition of part (b) holds if any one of the following conditions holds:
1. The functions g; are linear.
2. (Arrow, Hurwitz, Uzawa [AHU61]) There exists a vector d such that
Vg;(x*)'d < 0,
V jE A(x*).
Hint: Let d satisfy Vg;(x*)'d < 0 for all j E A(x*), and let d, = yd+ (1–
y)d. From the mean value theorem, we have for some e E [0, 1]
9j(x* + ad,) = 9;(x*)+ aVg;(x* + cad, )'dy
<a(^Vg;(x* + €ad, )'d + (1 – y)V9; (x* + cad, )'d).
For a fixed y, there exists an ā > 0 such that the right-hand side above is
nonpositive for a € [0,a). Thus d, e F(x*), for all y and lim,-o d, = d.
Hence d e F(x*).
3. (Slater [Sla50]) The functions g; are convex and there exists a vector a
satisfying
9;(T) < 0,
Vje A(x*).
Hint: Let d = x – x* and use condition 2.
4. The gradients Vg,(x*), j E A(x*), are linearly independent. Hint: Let d
be such that Vg;(x*)'d = -1 for all j e A(x*) and use condition 2.
%3D
(d) Consider the two-dimensional problem with two inequality constraints where
f(x1, x2) = xı + x2,
J -xị + x2
4.
-xị + x2 if x1 < 0,
if xi 2 0,
91 (X1, x2) = {
if x1 > 0,
if x1 < 0.
J xi - x2
92(X1, 22)
X, - X2
(0,0) is a local minimum, but each of the conditions 1-4
of part (c) is violated and there exists no Lagrange multiplier vector.
Show that x*
(e) Consider the equality constrained problem minh(x)=0 f(x), and suppose
that x* is a local minimum such that Vf(x*) # 0. Repeat part (d) for the
case of the equivalent inequality constrained problem
min f(x).
||h(x)||2<0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F162c6403-ce91-41f8-a781-2ad0c6649c2f%2F42f1834e-6a0f-45e8-b732-6ecdca60b677%2F9pih1in_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(c) The condition of part (b) holds if any one of the following conditions holds:
1. The functions g; are linear.
2. (Arrow, Hurwitz, Uzawa [AHU61]) There exists a vector d such that
Vg;(x*)'d < 0,
V jE A(x*).
Hint: Let d satisfy Vg;(x*)'d < 0 for all j E A(x*), and let d, = yd+ (1–
y)d. From the mean value theorem, we have for some e E [0, 1]
9j(x* + ad,) = 9;(x*)+ aVg;(x* + cad, )'dy
<a(^Vg;(x* + €ad, )'d + (1 – y)V9; (x* + cad, )'d).
For a fixed y, there exists an ā > 0 such that the right-hand side above is
nonpositive for a € [0,a). Thus d, e F(x*), for all y and lim,-o d, = d.
Hence d e F(x*).
3. (Slater [Sla50]) The functions g; are convex and there exists a vector a
satisfying
9;(T) < 0,
Vje A(x*).
Hint: Let d = x – x* and use condition 2.
4. The gradients Vg,(x*), j E A(x*), are linearly independent. Hint: Let d
be such that Vg;(x*)'d = -1 for all j e A(x*) and use condition 2.
%3D
(d) Consider the two-dimensional problem with two inequality constraints where
f(x1, x2) = xı + x2,
J -xị + x2
4.
-xị + x2 if x1 < 0,
if xi 2 0,
91 (X1, x2) = {
if x1 > 0,
if x1 < 0.
J xi - x2
92(X1, 22)
X, - X2
(0,0) is a local minimum, but each of the conditions 1-4
of part (c) is violated and there exists no Lagrange multiplier vector.
Show that x*
(e) Consider the equality constrained problem minh(x)=0 f(x), and suppose
that x* is a local minimum such that Vf(x*) # 0. Repeat part (d) for the
case of the equivalent inequality constrained problem
min f(x).
||h(x)||2<0
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