Chapter 2-2.pdf

R. Zakhama
Structural Optimisation course
Chapter 2: Mathematical
Programming
Part 2: Constrained Optimisation

R. Zakhama 2
Chap2: Mathematical Programming
I. Introduction
 Feasible point: That satisfies all constraints.
 Feasible region: The set of values X of the design variables that satisfy all
the constraints.
X = {x∈Rn / gj(x) ≤ 0 j=1, ..., ng and hk (x) = 0 k=1, ..., ne}
(P) can be written as:
 Remarks:
- Generally, the solution x* is on the surface of the feasible region (if not,
the constraints are useless, ill-posed problem).
- From algorithmic point of view, one of the difficulties, is that we do not
know a priori what constraints will be active at the optimum.
The idea is to convert the problem to an unconstrained optimisation.
[ ]
where
1
0
)
(
1
0
)
(
s.t.
)
(
min
)
( 2
1
T
n
e
k
g
j
x
,...,x
,x
x
x
...n
k
x
h
...n
j
x
g
x
f
P =







=
=
=
≤
s.t.
)
(
min





∈ X
x
x
f
x

R. Zakhama 3
II. Feasible and Descent Directions
• Let F(x) = Set of feasible directions at x ∈ X.
• Let D(x) = Set of descent directions at x ∈ X.
s.t.
)
(
min





∈ X
x
x
f
x
Definition
A vector , is said to be a feasible direction at if
there exists δ1 > 0 such that for all α∈(0,δ1).
n
R
x∈
n
R
d ∈
X
∈
+ d
x α
0
≠
d
Definition
A vector , is said to be a descent direction at if
there exists δ2 > 0 such that for all α∈(0,δ2).
n
R
x∈
n
R
d ∈
)
(
)
( x
f
d
x
f <
+α
0
≠
d

R. Zakhama 4
g2
g1
x1
x2
s
s
FigII.1. : Feasible directions

R. Zakhama 5
s.t.
)
(
min





∈ X
x
x
f
x
Theorem
Let X be a nonempty set in Rn and be a local minimum
of f over X . Then,
X
∈
*
x
φ
=
∩ )
(
)
( *
*
x
x D
F
Proof.
Let be a local minimum.
By contradiction, assume that ∃ a nonzero .
∴∃ δ1 > 0 ∋ ∀ α ∈(0,δ1) and
∃ δ2 > 0 ∋ ∀ α ∈(0,δ2).
Hence, ∃ x ∈ B(x*,α)∩X ∋ f(x) < f(x*), for all α∈(0,min(δ1 ,δ2)).
)
(
)
( *
*
x
f
d
x
f <
+α
X
∈
*
x
)
(
)
( *
*
x
x
d D
F ∩
∈
X
∈
+ d
x α
*

R. Zakhama 6
• is a local minimum
• Consider any and assume
• Let such that  d is a descent
direction  d ∈ D(x)
• Let
• is a local minimum
• If F(x*) = Rn (every direction in Rn is locally feasible), x* ∈ X is a local
minimum
• Can we characterise F(x*) algebraically for a constrained optimisation
problem?
s.t.
)
(
min





∈ X
x
x
f
x
X
∈
*
x φ
=
∩
 )
(
)
( *
*
x
x D
F
X
∈
x 2
C
∈
f
n
R
d ∈ 0
)
( <
∇ d
x
f T
)
(
)
( x
f
d
x
f <
+
 α
)
(
}
0
)
(
:
{
)
(
~
x
d
x
f
d
x T
D
D ⊆
<
∇
=
X
∈
*
x φ
=
∩
 )
(
~
)
( *
*
x
x D
F
0
)
(
}
0
)
(
:
{ *
=
∇

=
<
∇
 x
f
d
x
f
d T
φ

R. Zakhama 7
• Assume f, gj ∈ C2 j=1…ng
• X = {x ∈ Rn : gj(x) ≤ 0 , j=1…ng}
• Active constraints : A(x) = {j : gj(x) = 0}
 is a local minimum
• This is only a necessary condition for a local minimum.
• Cannot be directly used for equality constrained problems
...
1
,
0
)
(
s.t.
)
(
min
j





=
≤ g
x
n
j
x
g
x
f
Lemma
For any x ∈X ,
)
(
)}
(
,
0
)
(
:
{
)
(
~
x
x
j
d
x
g
d
x T
j F
A
F ⊆
∈
<
∇
=
X
∈
*
x φ
=
∩
 )
(
~
)
(
~ *
*
x
x D
F

R. Zakhama 8
II. Optimality conditions
II.1 The K.K.T. necessary conditions ( Karush, Kuhn, Tucker)
 Equality constraints
• Let consider the following optimisation problem:
• We introduce the Lagrangian function:
• The necessary condition for the local minimiser is
and it must be a feasible point (i.e. constraints are satisfied).
• These are Karush-Kuhn-Tucker conditions
1
0
)
(
s.t.
)
(
min
)
(





=
= ...n
k
x
h
x
f
P
e
k
x
s
multiplier
Lagrangian
the
are
where
)
(
)
(
1
k
n
k
k
k λ
x
h
x
f
e

=
+
= λ
L
0
,
0
0
)
,
( *
*
=
∇
=
∇
⇔
=
∇ L
L
L λ
λ x
x

R. Zakhama 9
• These conditions, however, apply only at a regular point, that is at a point
where the gradients of the constraints are linearly independent.
s
multiplier
Lagrangian
the
are
where
)
(
)
(
1
k
n
k
k
k λ
X
h
X
f
e

=
+
= λ
L
KKT necessary conditions (First order)
If x* ∈ X is a local minimum and a regular point, then there exists a
unique vector λ* (= (λ1
*, … , λne
*)T) such that
tarity)
(complemen
...
1
R
ty)
(feasibili
...
1
0
)
(
y)
(optimalit
0
)
(
)
(
*
*
*
1
*
*
e
k
e
k
k
n
k
x
k
x
n
k
n
k
x
h
x
h
x
f
e
=
∀
∈
=
∀
=
=
∇
+
∇ 
=
λ
λ

R. Zakhama 10
• Example 1
0
1
)
(
s.t.
2
min
2
2
2
1
2
1
, 2
1
=
−
+
=
+
x
x
x
h
x
x
x
x
)
1
(
2 2
2
2
1
2
1 −
+
+
+
= x
x
x
x λ
L







=
+
−
=
−
=






=
=
+
=
+






=
∂
∂
=
∇

=
∇

1
)
)
2
(
1
(
)
1
(
)
2
(
1
1
0
)
(
0
2
1
0
2
2
0
0
0
)
,
(
2
2
2
1
2
1
*
*
λ
λ
λ
λ
λ
λ
λ
λ x
x
x
h
x
x
x
x
L
L
L
KKT
2
5
±
=
 λ
2 solutions satisfy the KKT conditions, but only one is a minimum: these
conditions are necessary but not sufficient to be a local minimum.

R. Zakhama 11
• Example 1
FigII.2. : Example 1

R. Zakhama 12
• Geometrical interpretation
• For single equality constraint: simple geometrical interpretation of
Lagrange optimality condition:
Meaning
Gradients parallel 
tangents parallel  h tangent to isolines
0
=
∂
∂
+
∂
∂
x
h
x
f
λ
x
h
x
f
∂
∂
∂
∂
//
f
h
x1
x2
∇h
∇f
FigII.3. : Geometrical interpretation 1

R. Zakhama 13
• Geometrical interpretation
f
h
∇h
∇f
f
h
∇h
∇f
f
h
∇h
∇f
f
h
∇h
∇f
maximum minimum
minimum no extremum

R. Zakhama 14
 Inequality constraints
• Let consider the following optimisation problem:
• To be able to apply the Lagrange multiplier method we first transform the
inequality constraints to equality constraints by adding slack variables.
• We can now form a Lagrangian function:
• The necessary condition for the local minimiser is
and it must be a feasible point (i.e. constraints are satisfied).
• These are Karush-Kuhn-Tucker conditions
2
,
1
0
)
(
s.t.
)
(
min
)
( C
g
f
...n
j
x
g
x
f
P j
g
j
x
∈





=
≤
s
multiplier
Lagrangian
the
are
where
)
)
(
(
)
(
1
2
k
n
j
j
j
j
g
s
x
g
x
f µ
µ

=
+
+
=
L
0
,
0
,
0
0
)
,
,
( *
*
*
=
∇
=
∇
=
∇
⇔
=
∇ L
L
L
L s
x
s
x µ
µ
g
j
j
j n
j
s
s
x
g ...
1
able,
slack vari
a
is
,
0
)
( 2
=
=
+

R. Zakhama 15
• The KKT can be written as:
• What is the sign of µ?
,...,
1
0
1
n
i
x
g
x
f
x
g
n
j i
j
j
i
i
=
=
∂
∂
+
∂
∂
=
∂
∂

=
µ
L
0 g
j
j
j
n
j
s
g ,...,
1
2
=
=
+
=
∂
∂
µ
L
,...,
1
0
2 g
j
j
j
n
j
s
s
=
=
=
∂
∂
µ
L
We have
Feasible direction
Descent direction
Since x* is a local minimum
1

=
∇
−
=
∇
g
n
j
j
x
j
x g
f µ
g
j
x
T
j
x
T
n
j
g
d
g
d
d ,...,
1
0
0
/ =
≥
∇
−

≤
∇
∃

0
≤
∇
 f
dT

=
∇
−
=
∇

g
n
j
x
T
j
x
T
T
g
d
f
d
d
1
by
g
Multiplyin µ
φ
=
∩
 )
(
~
)
(
~ *
*
x
x D
F
0
≥
 j
µ

R. Zakhama 16
s
multiplier
Lagrangian
the
are
where
)
(
)
(
1
k
n
k
k
k λ
X
h
X
f
e

=
+
= λ
L
If x* is a local minimum of problem (P) and a regular point, then there
exists a unique vector µ* (= (µ1
*, … , µng
*)T) such that
tarity)
(complemen
...
1
0
...
1
0
ty)
(feasibili
...
1
0
)
(
y)
(optimalit
0
)
(
)
(
*
*
*
2
*
*
*
1
*
*





=
∀
≥
=
∀
=
=
∀
=
+
=
∇
+
∇ 
=
g
j
g
j
j
g
j
j
j
n
j
x
j
x
n
j
n
j
s
n
j
s
x
g
x
g
x
f
g
µ
µ
µ

R. Zakhama 17
• Graphical interpretation of KKT conditions
( ) g
T
j n
k
j
d
g ≤
=
≤
∇ …
1
0
( ) 0
≤
∇ d
f
T
1
g
∇
2
g
∇
g2
g1
x1
x2
∇f
-∇f
1
g
∇
− 2
g
∇
−
Feasible cone
• Feasible direction:
• Descent direction:
• x* is a local optimum: -∇x f (x*) form an
obtuse angle (> 90 °) with each feasible
direction (we can not decrease f
without violating the feasible domain).
• negative gradient (descent direction) lies
in cone spanned by positive constraint
gradients.
• no descent direction exists within the
cone of feasible directions.
• -∇x f (x*) is a linear combination, with
coefficients µj > 0 of ∇xgj (x*), for
any gj active at x*

R. Zakhama 18
• Graphical interpretation of KKT necessary conditions
f3 > f2 > f1
No active constraints at x*,
g(x)≤0
g(x)=0
f3 > f2 > f1
x* is not a minimiser, μ<0
g(x)≤0
g(x)=0
-∇g(x*)
f3 > f2 > f1
x* is a minimiser, μ>0
g(x)=0
-∇g(x*)
g(x)≤0

R. Zakhama 19
 Equality and Inequality constraints
[ ]
where
1
0
)
(
1
0
)
(
s.t.
)
(
min
)
( 2
1
T
n
e
k
g
j
x
,...,x
,x
x
x
...n
k
x
h
...n
j
x
g
x
f
P =







=
=
=
≤
s
multiplier
Lagrangian
the
are
where
)
(
)
(
1
k
n
k
k
k λ
X
h
X
f
e

=
+
= λ
L
If x* is a local minimum of problem (P) and a regular point, then there
exists a unique vector λ* (= (λ1
*, … , λne
*)T) and a unique vector
µ* (= (µ1
*, … , µng
*)T) such that
y)
(optimalit
0
)
(
)
(
)
( *
1
*
*
1
*
*
=
∇
+
∇
+
∇ 
 =
=
x
g
x
h
x
f j
n
j
x
j
k
n
k
x
k
x
g
e
µ
λ
tarity)
(complemen
,...,
1
0
,...,
1
,...,
1
0
ty)
(feasibili
,...,
1
0
)
(
,...,
1
0
)
(
*
*
*
*
2
*
*
*







=
∀
≥
=
∀
∈
=
∀
=





=
∀
=
+
=
∀
=
g
j
e
k
g
j
j
g
j
j
e
k
n
j
n
k
R
λ
n
j
s
n
j
s
x
g
n
k
x
h
µ
µ

R. Zakhama 20
II.2 KKT Sufficient optimality conditions
s
multiplier
Lagrangian
the
are
where
)
(
)
(
1
k
n
k
k
k λ
X
h
X
f
e

=
+
= λ
L
KKT sufficient conditions (Second order)
If there exists x* ∈ X, λ* ∈ Rne , µ* ∈ R+
ng and s* ∈ Rng such that
The KKT necessary conditions are satisfied
and
For all d ≠ 0 such that
then x* is a strict local minimum of (P)
e
T
k n
k
d
x
h ,...,
1
,
0
)
( *
=
=
∇
0
)
,
,
,
( *
*
*
*
2
≥
∇ d
s
x
L
d x
T
µ
λ
0
and
)
(
,
0
)
( *
>
∈
=
∇ j
T
j x
j
d
x
g µ
A
0
and
)
(

}
,...,
1
{
,
0
)
( *
=
∈
≤
∇ j
g
T
j x
n
j
d
x
g µ
A

R. Zakhama 21
 Example 2
• Problem:
• Write the KKT necessary conditions and find a solution of this problem by
discussing the different cases. Check if that solution is a local minimum.
0
1
)
(
0
9
)
(
s.t.
)
(
min
2
1
2
2
2
2
1
1
2
2
1
≤
−
+
=
≤
−
+
=
+
=
x
x
x
g
x
x
x
g
x
x
x
f

R. Zakhama 22
 Example 2
• f , g1 , g2 are differentiable.
• The Lagrangian function can be written as:
• The KKT conditions are:
( ) ( ) 0
)
(
)
(
)
( 2
2
2
2
2
1
1
1 =
+
+
+
+
= s
x
g
s
x
g
x
f µ
µ
L















∈
≥
=
=
=
+
−
+
=
+
−
+






=






+






+












∈
≥
=
∇
R
s
s
s
s
s
x
x
s
x
x
x
x
x
R
s
s
s
x
L
2
1
2
1
2
2
1
1
2
2
2
1
2
1
2
2
2
1
2
2
1
1
1
2
1
2
1
,
(6)
0
,
(5)
0
(4)
0
(3)
0
1
(2)
0
9
(1)
0
0
1
1
2
2
1
2
,
0
,
0
)
,
,
(
µ
µ
µ
µ
µ
µ
µ
µ
µ

R. Zakhama 23
 Example 2
• To solve this optimisation problem, we have to study the different possible
cases (it is not a general method to solve an optimisation problem).
• If µ1 = µ2 = 0, (4) and (5) are satisfied but not (1)
• If µ1 ≠ 0 and µ2 = 0, (1)  2x1+2µ1x1 = 0
if x1 ≠ 0  µ1 = -1 which does not satisfy (6)
 x1 = 0
(4)  s1 = 0 since µ1 ≠ 0  (2) is simplified to x2
2 – 9 =0  x2 = ± 3
if x2 = 3 , (1)  1+6µ1 = 0 does not satisfy (6)
if x2 = -3 , (1)  1-6µ1 = 0  µ
µ
µ
µ1 = 1/6
• The resolution of KKT necessary conditions give
x1
* = 0, x2
* = -3, µ
µ
µ
µ1
* = 1/6, µ
µ
µ
µ2
* = 0
•  ∇x
2L is positive definite  x* is a local minimum








=
∇
3
1
0
0
3
7
2
xL
(KKT sufficient conditions)

R. Zakhama 24
X
 Example 2
FigII.7. : Example 2

R. Zakhama 25
III. Constrained optimisation methods
III.1 Feasible directions method
• The method is intended for problems with general inequality constraints.
• The iterative solutions are moving along the boundary.
• The method relies on determining a search direction and performing 1-D
minimisation along the specified direction,
• We seek a direction that will reduce the objective function without violating
the constraints for a finite move along the direction:
k
k
k
k
d
x
x α
+
=
+1
• A direction which reduces the objective function is called a usable
direction,
• A direction which does not cause violation of the active constraints
upon move is a feasible direction,
• Direction satisfying both conditions is a usable feasible direction.
0
)
( ≤
∇ k
T
k
d
x
f
)
(
,
0
)
( k
k
T
k
j x
j
d
x
g A
∈
≤
∇

R. Zakhama 26
• Zoutendijk’s basic idea is at each stage of iteration to determine a vector dk
that will be both a feasible direction and a descent direction. This is
accomplished numerically by finding a normalised direction vector dk and a
scalar parameter θ < 0 such that:
• This subproblem is linear and can be solved using the simplex algorithm,
• If θ < 0, we have found a usable feasible.
• If θ = 0, it can be shown that the KKT conditions are satisfied.







≤
≤
−
∈
≤
∇
≤
∇
1
1
)
(
,
)
(
)
(
s.t.
min
)
(
,
k
i
k
k
T
k
j
k
T
k
d
d
x
j
d
x
g
d
x
f
L
k
A
θ
θ
θ
θ

R. Zakhama 27
(a) Initialise: x0 and k = 0
(b) Solve the linear problem (L) to find dk and θ
(c) If θ < 0,
• Find the maximum step such that will be in the feasible domain
• Find αk along dk such that
•
• k = k+1, go to (b)
(d) Else if θ ≥ 0, the iteration terminates.
Output: x* = xk , a stationary point.
∞
=
>
≥
=
=
+
=
α
α
α
α
α
α
set
exists,
0
no
if
}
0
and
,...,
1
,
0
)
(
:
min{ g
k
k
j n
j
d
x
g
α k
k
d
x α
+
d
α
x
x k
k
k
k
+
=
+1
[ ]
)
(
min
arg
,
0
k
k
k
d
x
f α
α
α
α
+
=
∈

R. Zakhama 28
III.2 Penalty function methods
• Consider the problem:
where X ∈ Rn
• Idea:
• Transform constrained problem into unconstrained one.
• Solve a sequence of unconstrained optimisation problems.
• Define a function,
• Solve an equivalent unconstrained problem:
X
∈
x
x
f
x
s.t.
)
(
min



∉
∞
+
∈
=
X
X
x
x
x
if
if
0
)
(
ψ
)
(
)
(
min x
x
f
x
ψ
+

R. Zakhama 29
• The function ψ(x) is not a practical approach.
• Replace ψ(x) by a sequence of continuous non-negative functions that
approach ψ(x).
• Penalty function ψ :
- Exterior penalty functions.
- Interior penalty functions (or barrier functions).

R. Zakhama 30
 Exterior penalty functions
• Constrained problem:
• Define,
• Unconstrained exterior penalty transformation:
• Solve for increasing sequence {rk} such that:
• Let
• Ideally,
[ ] 
 =
=
+
=
e
g n
l
l
n
j
j x
h
x
g
x
P
1
2
2
1
)
(
)}
(
,
0
max{
)
(
1
0
)
(
1
0
)
(
s.t.
)
(
min
...n
l
x
h
...n
j
x
g
x
f
e
l
g
j
R
x n
=
=
=
≤
∈
)
(
)
(
)
,
(
min x
rP
x
f
r
x
q
n
R
x
+
=
∈
+∞
→
<
<
<
+∞
→
k
k
r
r
r
r
k
2
1
lim
,
⋯
)
,
(
min
arg k
R
x
k
r
x
q
x n
∈
=
+∞
→
→ }
{
as
}
{ * k
k
r
x
x

R. Zakhama 31
rk
q(x,rk)
FigIII.1. : Inequality constrained problem in 1-D

R. Zakhama 32
(1) Initialise: x0 , r0 , ε > 0 and k = 0
(2) while
(a)
(b)
(c) k = k+1
endwhile
Output: x* = xk
( ) ε
>
− )
(
)
,
( k
k
k
x
f
r
x
q
)
,
(
min
arg
1 k
R
x
k
r
x
q
x n
∈
+
=
)
10
1
example
(for
1
<
<
=
+
c
c
r
r k
k

R. Zakhama 33
 Interior penalty functions
• Typically applicable to inequality constrained problems:
• Some interior penalty (Barrier) functions:
• Unconstrained exterior penalty transformation:
• Solve for increasing sequence {rk}.
[ ]
)
(
ln
)
(
or
)
(
1
)
(
1
1

 =
=
−
−
=
−
=
g
g n
j
j
n
j j
x
g
x
B
x
g
x
B
1
0
)
(
s.t.
)
(
min
...n
j
x
g
x
f
g
j
R
x n
=
≤
∈
)
(
1
)
(
)
,
(
min x
B
r
x
f
r
x
q
n
R
x
+
=
∈

R. Zakhama 34
 Interior penalty functions
1/rk
q(x,rk)
FigIII.2. : Inequality constrained problem in 1-D

R. Zakhama 35
III.3 Augmented Lagrangian (multiplier) methods
• Consider the problem (equality constraints):
• The Lagrangian function:
• The KKT necessary conditions:
• Augmented Lagrangian function:
• Note that the optimal value of λl are not known.
• If all λl = 0, then the method is identical to the Exterior Penalty Function.
• If, on the other hand, we have optimal values λl
* , then min T is the answer
for any finite value of r.

 =
=
+
+
=
e
e n
l
l
n
l
l
l x
h
r
x
h
x
f
r
x
T
1
2
1
)
(
)
(
)
(
)
,
,
( λ
λ
1
0
)
(
s.t.
)
(
min
...n
l
x
h
x
f
e
l
R
x n
=
=
∈

=
+
=
e
n
l
l
l x
h
x
f
x
L
1
)
(
)
(
)
,
( λ
λ
0
)
(
,
0
1
*
=
=
∇
+
∇
=
∇ 
=
x
h
h
f l
n
l
l
x
l
x
x
e
λ
L

R. Zakhama 36
• First order KKT conditions for the augmented Lagrangian function:
• The exact KKT conditions for the considered problem:
• Comparing the two last equations, we expect that:
• Hestenes (Hestenes, 1969) suggested using the following update for λl :
• Easy to extend the augmented Lagrangian for the inequality constraints
(g(x) ≤ 0 ) since we can write:
( ) 0
2
1
=
∇
+
+
∇
=
∇ 
=
e
n
l
l
x
l
l
x
x h
rh
f
T λ
0
1
*
=
∇
+
∇
=
∇ 
=
e
n
l
l
x
l
x
x h
f λ
L
*
2 l
l
l rh λ
λ →
+
)
(
2
1 k
l
k
l
k
l x
rh
+
=
+
λ
λ
0
)
( 2
=
+ s
x
g

R. Zakhama 37
(1) Input: r, ε > 0
(2) Initialise: x0 and k = 0
(3) while
(a)
(b)
(c) k = k+1
endwhile
Output: x* = xk
( ) ε
λ >
− )
(
)
,
,
( k
k
k
x
f
r
x
T
)
,
,
(
min
arg
1
r
x
T
x k
R
x
k
n
λ
∈
+
=
)
(
2
1 k
k
k
x
rh
+
=
+
λ
λ

R. Zakhama 38
IV. Problems
IV.1 Problem 1
In this exercise we reconsider the optimisation problem of a truss structure with 2 bars for
a minimum weight (figure IV.1).
We want to find the traversal sections X1 and X2 such that the structure weight Z is
minimised.
Bars must:
- resist under static loads.
- the displacement δy of node A must not exceed a limit δ0.
L et L/cosα
α
α
α : bares length
E : Young modulus
X1 et X2 : transversal bares sections
δx , δy : node displacements A
F : load applied at A
ρ : density
Z : total masse
σ0 and -σ0 : elasticity limit in tension and in
compression, respectively.
FigIV.1. : Problem 1

R. Zakhama 39
IV. Problems
IV.1 Problem 1
The problem formulation can be written as:
We take (1)
(P)






+
= 1
2
cos
min
2
1
X
α
L
LX
ρ
Z
,X
X
s.t.
E
L
0
0
σ
δ =

R. Zakhama 40
IV. Problems
IV.1 Problem 1
1- From the expression of equation (1), what does correspond physically the displacement
limit δ0.
2- We consider the following variables replacements:
and
We pose
Write the new formulation (P’) of the optimisation problem (P) with respect Y1 and Y2.
3- Demonstrate if the constraints g1 , g2 and g3 are satisfied then the constraints g4 and g5
are satisfied too and can be eliminated (we say g1 , g2 and g3 dominate g4 and g5).
4- Write the necessary KKT conditions for the new problem (P’).
5- Knowing that the only constraint g3 is active at the optimum, solve the new problem
(P’) and deduce the optimal section values X1 and X2 as well as the Lagrangian multipliers
values.
1
0
1
2
X
F
Y
σ
=
2
0
2
3
X
F
Y
σ
=
0
σ
ρLF
c =

R. Zakhama 41
IV. Problems
IV.2 Problem 2
Let consider the following problem:
Where s and h are positive values
1- Write the KKT necessary conditions for the given problem.
2- Knowing that the only constraint g1 and g5 are active at the optimum, compute the
optimum values x1 , x2 and the Lagrangian multipliers values (Assuming that h = 1200
and s = 100).
3- Compute the values of the objective and the Lagrangian function at the optimum.
( ) ( )
[ ]















>
≤
−
+
≡
≤
−
≡
≥
−
≡
≤
+
−
≡
≤
+
−
≡
≥
−
⋅
≡
−
+
+
+
=
0
,
0
1
150
,
0
1
750
0
1
10
,
0
1
01
.
0
0
1
13
.
0
,
0
1
10
12
.
2
s.t.
2
min
2
1
1
6
2
5
2
4
1
3
1
2
7
2
2
1
1
2
2
1
2
1
2
1
x
x
s
x
g
x
g
x
g
x
s
g
x
h
g
x
x
g
x
x
s
x
h
s
x
f
x
π
π
π

R. Zakhama 42
IV. Problems
IV.3 Problem 3
Given the problem:
1- Solve by the exterior penalty function for r1 = 0.1, r2 = 1.0. Show graphically q(X,r) as a
function of X= X1 = X2 for the values of r.
2- Solve by the interior penalty function for r1 = 10.0, r2 =100.0. Show graphically q(X,r) as
a function of X= X1 = X2 for the values of r.
3- Solve by the method of feasible directions. Calculate two iterations cycles, starting at the
initial point {X1}T = {7,4}. Show graphically the directions of move in the space of
X1 and X2.
0
9
)
5
(
)
5
(
g
s.t.
min
2
2
2
1
2
1
, 2
1
≤
−
−
+
−
≡
+
=
X
X
X
X
Z
X
X

R. Zakhama 43
Sources
Presentation based on material from:
• Haftka & Gürdal (1994): Elements of Structural Optimization
• van Keulen: TUD Optimization course
• Etman (2006): EM course Engineering Optimization

Chapter 2-2.pdf

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