Lecture 5

Necessary Conditions for

Finite‐variable
Constrained Minimization
ME260 Indian Institute of Science
S t r u c t u r a l O p t i m i z a t i o n : S i z e , S h a p e , a n d To p o l o g y
G. K. Ananthasuresh
P r o f e s s o r, M e c h a n i c a l E n g i n e e r i n g , I n d i a n I n s t i t u t e o f S c i e n c e , B e n g a l u r u

suresh@iisc.ac.in

1
Outline of the lecture
How do constraints influence the ability to minimize the objective
function?
The concept of Lagrange multipliers
Feasible space
Active and inactive constraints
Necessary conditions
What we will learn:
Constrained optimum lies on the boundary of the feasible space.
Conditions for constrained local minimum; constraint qualification
Sensitivity of the constrained optimum to small changes in constraints.
Physical meaning of Lagrange multipliers.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 2
Two variables and an equality constraint
Min f  x1 x2 (The intent here is to maximize the
x1 ,x2 product of two numbers such that their
Subject to sum is equal to 1)

x1  x2  1 Min f   x1 (1  x1 )
x1

One easy way to solve this or

problem is to eliminate either
x1 or x2 by expressing in terms Min f  (1  x2 ) x2
x2
of the other using the equality
constraint and make this an But this kind of explicit elimination
unconstrained problem in one of a variable may not always be
variable. possible. What do we do then?

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 3

Eliminate a variable in the first order
approximation
Min f (x1 , x2 ) Let the optimum be: (x1* , x2* )
x1 ,x2 x1  x1  x1*
Let the first order perturbations be:
Subject to x2  x2  x2*
Now, consider first‐order approximations of
h(x1, x2 )  0 the objective function and the constraint.

f f
f (x1 , x2 )  f (x , x ) 
*
1
*
2 x1  x2
x1 x* ,x *
x2 x1* ,x2*
1 2

h h h h
h(x1 , x2 )  h(x1* , x2* )  x1  x2  x1  x2  0
x1 x1* ,x2*
x2 x1* ,x2*
x1 x1* ,x2*
x2 x1* ,x2*

 h   h 
 x2    x1   
 x1 x1* ,x2*   x2 x1* ,x2* 

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 4

After eliminating one variable’s first‐order
perturbation in terms of the other…
 h   h 
With x2    x1   
 x1   x2 
 x1* , x2*   x1* , x2* 
 h 
 x1 
f f  x 
f ( x1 , x2 )  f ( x1* , x2* )  x1   1 x1* , x2*   f ( x* , x* )  f x1  
h
x1
1 2
x1 x1* , x2*
x2 x1* , x2*
 h  x1 x1* , x2*
x1 x1* , x2*
 
 x2 x* , x* 
 1 2 

 f   f   h  Because the
 
 x2      0 first order
where    
x1* , x2* 
 x1 x1* ,x2*   x1 x1* ,x2*  derivative
 h  should be
   f   h 
 x2  zero for
 x1* , x2* 
    0 any x
 x2 x1* ,x2*   x2 x1* ,x2*  1

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 5

Some generality and a new concept
Two things are remarkable about the two equations we got: (i) they are
similar and (ii) they both give a similar expression for the constant,  .

 f   h 
    0  f   f 
 x1 x1* ,x2*   x1  
x
  
 x2
x1* ,x2*
 1 x1 ,x2 
* * * *
x1 ,x2
 f   h   
 h   h 
    0    
 x2 x1* ,x2*   x2 x1* ,x2*  x
 1 x1 ,x2 
* *  x2 * * 
x1 ,x2

 This is called the Lagrange multiplier corresponding to the equality constraint.

Think about what the Lagrange multiplier physically means…

It is the negative of the ratio of the rate of change of objective
function to the rate of change of the constraint with respect to either
variable.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 6
The concept of the Lagrangian
Min f (x1 , x2 )
x1 ,x2 An alternative formulation…
Subject to Min L  f (x1 , x2 )   h(x1, x2 )
h(x1 , x2 )  0 x1 ,x2
Both have the
same
necessary L is called
conditions!
 f   h 
the
    0 Lagrangian.
 x1 x1* ,x2*   x1 x1* ,x2* 
 f   h 
    0
 x2 x1* ,x2*   x2 x1* ,x2* 

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 7

General problem in two variables and
one equality constraint
Min f (x1 , x2 )
x1 ,x2
 Min L  f (x1 , x2 )   h(x1, x2 )
x1 ,x2
Subject to
h(x1 , x2 )  0
f h
 0 Three variables (x1 , x2 ,  )
conditions
Necessary

x1 x1 And three

f h equations!
 0
x2 x2 We are fine.
h( x1 , x2 )  0

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 8

n variables and m equality constraints
Min f (x1 , x2 ,, xn ) Min f (x)

Short form
x1 ,x2 ,,xn x
Subject to Subject to
h1 (x1 , x2 ,, xn )  0
h(x)  0
h2 (x1 , x2 ,, xn )  0
  x1   h1 (x1 , x2 ,, xn ) 
   
hm (x1 , x2 ,, xn )  0 
x
x2   h2 (x1 , x2 ,, xn ) 
 and h   
     
Can there be more constraints than variables?  xn   hm (x1 , x2 ,, xn ) 
   
mn No; feasible values may not exist. It is over‐constrained.
mn Some discrete feasible values may exist; but cannot do minimization.
mn This must be true in order to do minimization of the objective function.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 9
Feasible space; partitioning variables
Let us partition n variables in
If there are m constraints (note: m < to s (solution or dependent)
n), we can choose only (n-m) variables and d (decision or
variables freely because m variables independent variables).
can be found using the m equality
constraints.  x1   s1 
   
So, we search in the (n-m)‐  x2   s2 
dimensional feasible space.      
   
Feasible space is the reduced space  xm   sm   s 
that satisfies the constraints. x      
 xm1   d1   d 
When we say it is (n-m)‐dimensional,  xm2   d2 
we mean that m variables are    
somehow eliminated using m      
 xn   dnm 
equality constraints.    

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 10

Taylor series again
 f (x *
n
)
f (x)  f (x )  
*
(xi  xi* )  O(2)
i1 xi

 f (x * )  f T (x * ) x * Approximated to first order.

As per
 f (x * )   s f T (x * ) s*   d f T (x * ) d* partitioned
variables.
For the necessary condition, we want the first order terms go
to zero; then, the function value does not change in the
vicinity of the minimum up to first order.
But we know that perturbation in s variables cannot be
independent of those in d variables. So… (next slide)
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 11
Taylor series for equality constraints
n h j (x * )
h j (x)  h j (x * )  
i1 xi
(xi  xi* ) O(2)
j  1,2,,m
As per
 h j (x * )  h j T (x * ) x * partitioned
variables.
0
 h j (x * )   s h j T (x * ) s*   d h j T (x * ) d*  0
Because x* is Because x
feasible; i.e., should remain
it satisfies   s h j T (x * ) s*  d h j T (x * ) d*  0 feasible after
the perturbation
constraints. from x*.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 12
Eliminating s‐perturbations
s h j (x ) s  d h j (x ) d  0 j  1,2,,m
T * * T * *

 s h T ( x* )  s *   d h T ( x * )  d *  0 Compact notation for all constraints.

Note the sizes of the quantities.

m m m  (n  m) m 1
m 1 (n  m)  1
1
  s   s h (x )   d hT (x * ) d*
* T *

Remember that we can take the inverse of the m x m matrix here only
if it is not singular. This implies that the gradients of the constraints
should be linearly independent. This is known as constraint
qualification.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 13
Reduced gradient Should be zero for
the necessary
f (x)  f (x * )  s f T (x * ) s*   d f T (x * ) d* condition for a
minimum.
1
With  s   s h (x )   d hT (x * ) d*
* T *

We get  s f T (x * ) s*  d f T (x * )d*  0
1
 s f (x )  s h (x )   d hT (x * ) d*   d f T (x * )d* = 0
T * T *

 T * T *
1
 s f (x ) s h (x )  d hT (x * )   d f T (x * ) d* = 0 
Think of f as z that depends
Reduced gradient in the z(d)
p=n‐m space. only the d variables
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 14
 Reduced gradient is
The multipliers, again zero is the necessary
condition.

Where we used the new symbol for

Lagrange multipliers
appear again.
Compare with the two‐ Notice that both equations
variable case on slide 5 of have the same form; one is
this lecture. gradient w.r.t. to d and the
Same story here! other w.r.t. s.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 15
The Lagrangian

With

f m h j
   j 0 i  1,2,,n
xi j 1 xi
is a row vector. We have n scalar equations here.

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 16

Necessary conditions for equality‐
constrained minimization problem
Min f (x)
x

Subject to
h(x)  0

Variables: n+m
Equations: n+m
So, we are fine.

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 17

Geometric interpretation
This means that the gradient
of the objective function is a
f m h j linear combination of the
   j 0 i  1,2,,n gradients of the constraints.
xi j 1 xi

x2 “Contours” or “level‐set curves” of f(x1,x2)

f increases
Here, at the constrained
minimum, the gradients
h of f an h are in the same
direction.
f h( x1 , x2 )  0

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 18

With inequality constraints
Min f (x)
x

Subject to
h(x)  0 p inequality constraints.

g(x)  0 gk (x)  0 k  1,2,, p

Inequality constraints may be active or inactive at the minimum point.

g0 g0
Active constraints should Inequality constraints may
be treated just like simply be ignored because
equality constraints. they don not play any role.

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 19

Complementarity conditions
How do we know if an inequality constraint is active or not?
We don’t.
So, we express it in the form of equations!
This is an interesting set
k gk  0 k  1,2,, p of equations:
Either multiplier is zero
or the constraint is zero.
They are called
Lagrange multiplier complementarity
corresponding to kth conditions.
inequality constraint
Both may also be zero under
special situations.

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 20

Necessary conditions for constrained
minimization with equalities and inequalities.
Min f (x)
x

Subject to
h(x)  0
g(x)  0
Variables: n+m+p
Equations: n+m+p
So, we are fine.
.
But we are not done yet.
The Lagrange multipliers of inequality constraints are
restricted in sign. Let us discuss why.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 21
Inequality constraint; simplest problem
f ( x)

Constrained
local
minimum

x
Feasible region
For x* to be a local minimum.

For x* to continue to satisfy

the inequality even after
perturbation.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 22
Minimizability vs. feasibility

But necessary condition requires:

Multiply both sides by

It is a simple but a good  0

explanation for the non‐
negativity of the Lagrange So, this has to be non‐negative.
multipliers of inequality That is,
constraints.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 23
Necessary condition for general
constrained minimization problem
Min f (x)
x

Subject to
h(x)  0
g(x)  0
Variables: n+m+p
Equations: n+m+p
Number of inequalities: 2p
The first condition follows from the Lagrangian.

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 24

Necessary conditions: KKT conditions
Min f (x)
x

Subject to
h(x)  0
g(x)  0
Karush‐Kuhn‐Tucker conditions.

Independently done at Princeton University; later Kuhn dug up Karush’s

work and gave credit that is due to him. A rare and admirable gesture.

Had done this in his master’s thesis at University of Chicago before Kuhn and Tucker.

ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 25

A caveat: constraint qualification
KKT conditions are applicable as “necessary
conditions” only if the constraints qualification is
satisfied.
Constraint qualification requires that the gradients of
the equality constraints and active inequality
constraints be linearly independent at the optimum.
◦ See slide 13.
◦ One can construct special example where a point is a
minimum but KKT conditions are not satisfied.
◦ How can “necessary” conditions be not satisfied?
◦ It is because at such special points “constraint qualification” is not
satisfied. So, KKT conditions are not applicable.
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 26
The end note
Two variables and one equality constraint
for finite‐variable constrained optimization

The concept of Lagrange multiplier and the Lagrangian

Feasible space
Reduced gradient with equality constraints
Necessary conditions

Lagrange multipliers

Constraint qualification

Inequality constraint and the implication of

the sign of the Lagrange multiplier
Complementarity conditions

Karush‐Kuhn‐Tucker necessary conditions

Thanks
ME 260 / G. K. Ananthasuresh, IISc Structural Optimization: Size, Shape, and Topology 27