Transcript Sensitivities in OptimalDesign

Computational Methods for Design
Lecture 5 - Design and Optimization Problems
John A. Burns
Center for Optimal Design And Control
Interdisciplinary Center for Applied Mathematics
Virginia Polytechnic Institute and State University
Blacksburg, Virginia 24061-0531
A Short Course in Applied Mathematics
2 February 2004 – 7 February 2004
N∞M∞T Series Two Course
Canisius College, Buffalo, NY
1D Model Problem
LET 1 < q <  and consider the boundary value problem
(S)
3
d 2 w( x) + 1  d w( x)  0, 0  x  q ,
8  dx

dx 2
w( 0 )  0,
w( q )  4
Given data wˆ (x ) , 0 < x < 1 the goal is to match wˆ (x ) by
solving the following
OPTIMAL DESIGN PROBLEM: Find the parameter 1 < q0 , to
minimize the cost function
1
(COST)
J ( q ) = F( w(,q ),q )  1 2  | w( x,q )  wˆ ( x) |2 dx
0
Model Problem #1
1
d J( q )  w(x,q )-wˆ (x),  (w(x,q )) dx

q
dq
0

s(x)  s(x ,q) 
w(x ,q ) 0  x 
q
q
The sensitivity equation for s(x, q ) = q w(x , q) in the
“physical” domain (q) = (0,q) is given by
(S)
2
d 2 s( x) + 3  d w( x ,q) d s( x)  0, 0  x  q ,
8  dx
 dx
dx 2
s( 0 )  0, s( q )   dxd w( x ) | q
x
Can be made “rigorous” by the method of mappings.
MORE ABOUT THIS NEAR THE END
Typical Cost Function
1
J (q ) = F( w(, q ), q)  1 2  | w( x, q )  wˆ ( x) |2 dx
0
WHERE w( x , q ) USUALLY SATISFIES A DIFFERENTIAL EQUATION
AND q IS A PARAMETER (OR VECTOR OF PARAMETERS)
CONTINUOUS
SENSITIVITY
THE CHAIN RULE PRODUCES
1
d
dq
J( q ) =
d
dq
F( w(, q ),q )   [ w( x, q )  wˆ ( x)]  [ q w( x, q )]dx
0
OR (Reality) USING NUMERICAL SOLUTIONS
d
dq
h
[J ( q )] =
d
dq
1
DISCRETE
SENSITIVITY
F( w (, q ),q )    w h( x, q )  wˆ ( x), q [ wh( x,q )] dx
h
0
Computing Gradients
TYPICAL APPROACHES TO COMPUTE
d Jh( q )
dq 

q =q0
(I) BY FINITE DIFFERENCES
hq
hq
q

J
(


)
J
( 0 )
0
d
h
J ( q0 )  

q
dq





(II) BY DISCRETE SENSITIVITIES
d
dq
h
[J (q0)] =
1
d
dq
h
F( w (,q0),q0)    wh( x,q0)  wˆ ( x), q [ wh( x,q0)] dx
0
Computing Gradients
FINITE DIFFERENCES
DISCRETE SENSITIVITIES
• REQUIRES 2 NON-LINEAR
SOLVES
• REQUIRES THE EXISTENCE OF THE
DISCRETE SENSITIVITY
• IF SHAPE IS A DESIGN
VARIABLE, FD REQUIRES 2
MESH GENERATIONS
• IF SHAPE IS A DESIGN VARIABLE,
THE DISCRETE SENSITIVITY LEADS TO
MESH DERIVATIVES COMPUTATIONS
WHAT IS THE “CONTINUOUS / HYBRID”
SENSITIVITY EQUATION METHOD? --- SEM
d
dq
1
[J (q0)]    wh( x,q0 )  wˆ ( x),[ q w ( x,q0)]
h
0
APPROXIMATE
h, k
 dx
A Sensitivity Equation Method
FOR q > 1 AND h=q/(N+1) CONSIDER (FORMAL)
w(x)
w h(x) = Finite Element Approximation
x
x=0
x=1
x=q
NUMERICAL APPROXIMATION
(S)
h
3
d 2 wh( x) + 1  d wh( x)  0, 0  x  q ,
8  dx

dx 2
w h( 0 )  0 ,
w h( q )  4
DISCRETE STATE EQUATION
A Sensitivity Equation Method
(S)
h
(S)
h
3
d 2 wh ( x) + 1  d wh ( x)  0, 0  x  q ,
8  dx

dx 2
wh ( 0 )  0,
wh ( q )  4
2
d 2 s ( x) + 3  d w h( x ,q ) d s ( x)  0, 0  x  q ,
8  dx
 dx
dx 2
s ( 0 )  0, s ( q )   dxd wh( x ) | q
x
 IMPORTANT OBSERVATIONS
 The sensitivity equations are linear
 The sensitivity equation “solver” can be constructed
independently of the forward solver -- SENSE™
 When done correctly “mesh gradients” are not required
A Sensitivity Equation Method
(S)
h
2
d 2 s ( x) + 3  d w h( x ,q ) d s ( x)  0, 0  x  q ,
8  dx
 dx
dx 2
s ( 0 )  0, s ( q )   dxd w h( x ) | q
x
FOR q > 1 AND k = q/(M+1) CONSIDER (FORMAL)
s(x)= qw(x,q)
s h,k(x) = Finite Element Approximation of
(S)
h
x
x=0
x=1
x=q
2nd NUMERICAL APPROXIMATION
s



q
q
( x, )   w( x, )
 q

h,k
h,k
Convergence Issues
d
dq
1
[J (q )]    w ( x,q )  wˆ ( x),[ w( x,q )]
h
h

q
h,k
def
 dx  G ( x, q , h ,
k
)
0
THEOREM. The finite element scheme is asymptotically consistent.
lim
h 0
k0
IDEA:
d
dq
[Jh( q )] - G( x , q , h ,
When the error
d
dq
k
) 0
[J h(q )] - G ( x, q , h ,
k
) is small, then
a trust region method should (might?) converge.
R. G. Carter, “On the Global Convergence of Trust-Region Algorithms Using
Inexact Gradient Information”, SIAM J. Num. Anal., Vol 28 (1991), 251-265.
J. T. Borggaard, “The Sensitivity Equation Method for Optimal Design”, Ph.D.
Thesis, Virginia Tech, Blacksburg, VA, 1995.
J. T. Borggaard and J. A. Burns, “A PDE Sensitivity Equation Method for Optimal
Aerodynamic Design”, Journal of Computational Physics, Vol.136 (1997), 366-384.
Convergence Issues
N=16, M=32
Convergence Issues
THE CASE k = h is often used, but may not be “good enough”
d
dq
[J h(q )] - G ( x, q , h ,
h
)
NOT CONVERGENT
N=M=16
Tol = 0.00001 Tol = 0.0001 Total: 378.82
Time (secs) Cost Time Grad. Time
Iter
q
Grad. Norm
Step
0
1.2000
4.3998E+00
-3.6427E-02
0.1231
1
1.1636
3.1583E+03
1.4051E-03
31.3210
31.2697
0.0478
2
1.1650
3.0910E+03
-1.4339E-03
36.2310
36.1798
0.0480
3
1.1635
5.8909E+02
7.8372E-03
46.1160
46.0075
0.1043
4
1.1714
5.0139E+03
-8.8462E-04
45.3550
45.3006
0.0511
5
1.1705
2.9396E+03
-1.5052E-03
43.6720
43.6208
0.0470
6
1.1690
1.7238E+04
2.5880E-04
42.5810
42.5301
0.0468
7
1.1693
2.5888E+03
1.7342E-03
46.3470
46.2965
0.0472
8
1.1710
4.6995E+04
-9.4732E-05
44.1900
44.1396
0.0468
9
1.1709
1.5743E+02
0.0000E+00
42.8790
42.8265
0.0485
Timing Issues
THE CASE k = 2h offers flexibility and
convergence.
N=16,
Iter
0
1
2
3
4
5
6
7
8
d
dq
[J h(q )] - G ( x, q , h , 2h)
But, what about timings?
M=32
Tol = 0.00001 Tol = 0.0001 Total: 39.81
Time (secs) Cost Time Grad. Time
q
Grad. Norm
Step
1.2000
4.8489E+00
-3.2414E-02
0.1968
1.1676
2.0720E+01
4.0347E-01
34.9270
34.8053
0.0911
1.5711
4.4544E+00
3.7808E-01
1.2613
1.1234
0.1075
1.9491
6.9846E-02
-1.2442E-02
0.9941
0.8714
0.0925
1.9367
1.5779E-02
2.7472E-03
0.4190
0.2907
0.0938
1.9394
3.3558E-03
-5.8723E-04
0.4095
0.2845
0.0943
1.9389
7.3235E-04
1.2801E-04
0.4525
0.3135
0.1083
1.9390
1.5892E-04
-2.7785E-05
0.8602
0.6327
0.1968
1.9390
3.0703E-05
0.0000E+00
0.2914
0.1451
0.1083
Approximately 96 .6% of cpu time spent in function evaluations
Approximately 02 .4% of cpu time spent in gradient evaluations
Mathematics Impacts “Practically”
UNDERSTANDING THE PROPER
MATHEMATICAL FRAMEWORK CAN
BE EXPLOITED TO PRODUCE
BETTER SCIENTIFIC COMPUTING
TOOLS
 A REAL JET ENGINE WITH 20 DESIGN VARIABLES


PREVIOUS ENGINEERING DESIGN METHODOLOGY
REQUIRED 8400 CPU HRS ~ 1 YEAR
USING A HYBRID SEM DEVELOPED AT VA TECH AS
IMPLEMENTED BY AEROSOFT IN SENSE™ REDUCED THE
DESIGN CYCLE TIME FROM ...
8400 CPU HRS ~ 1 YEAR TO 480 CPU HRS ~ 3 WEEKS
NEW MATHEMATICS WAS THE ENABLING TECHNOLOGY
Special Structure of SE’s
(DE)
d
x(t )  qx (t )
dt
x(0)  5
(SE)
d
s (t )  q  s (t )  x(t )
dt
s(0)  0
d
x(t )  qx (t )
dt
x(0)  5
(DE)
FIRST: SOLVE (DE)
(SE)
d
s (t )  q  s (t )  x5(etqt)
dt
x(t )  5e
qt
s(0)  0
SECOND: SOLVE (SE)
General Comments

THERE ARE MANY VARIATIONS THAT CAN IMPROVE
THE BASIC IDEA




COMBINING AUTOMATIC DIFFERENTIATION AND SEM
SMOOTHING AND GRADIENT PROJECTIONS
ADAPTIVE GRID GENERATION
THE ORDER OF THINGS MATTER


DIFFERENTIATE-THEN-APPROXIMATE
DERIVE SENSITIVITY EQUATION BEFORE MAPPING TO A
“COMPUTATIONAL DOMAIN”
– DOES NOT REQUIRE MESH DERIVATIVES
– REQUIRES A MORE SOPHISTICATED MATHEMATICAL
FRAMEWORK
– NEEDS A “DIFFERENT THEORY”
J. A. Burns and L. G. Stanley, “A Note on the Use of Transformations in Sensitivity
Computations for Elliptic Systems”, Journal of Mathematical & Computer Modeling,
Vol. 33, pp. 101-114, 2001.
MODEL PROBLEM #2
LET 1 < q <  and consider the boundary value problem
(S)
2
d

w( x)  f ( x ), 0  x 
dx 2
where
w( 0 )  0,
q,
w( q )  0
0, 0  x  1
f(x)=
1, 1  x  
OPTIMAL DESIGN PROBLEM: Find the parameter 0 < q0 < 1, to minimize
the cost function
1
(COST)
J ( q ) = F( w(,q ),q )  1 2  | w( x,q )  wˆ ( x) |2 dx
0
DERIVE THE SENSITIVITY EQUATION
MODEL PROBLEM #2

q

  d 2

 
w( x, q)   f ( x)  0
 q
q  dx2

2
d

w( x, q)  f ( x)
dx 2

d2
 2  w( x, q)   0
dx  q

w(0)  0

q
w(0, q)  0
s(0, q)  0
d2
 2 s ( x, q )  0
dx
w(q)  0

q
w(q, q)  0


w(q, q) 
w(q, q)  0
x
q


w(q, q)   w(q, q)
q
x
MODEL PROBLEM #2
(S)
2
d

w( x)  f ( x ), 0  x 
dx 2
q,
w( 0 )  0,
w( q )  0
The sensitivity equation for s(x, q ) = q w(x , q) in the
“physical” domain (q) = (0,q) is given by
(S)
2
d

s( x)  0, 0  x  q , s( 0 )  0, s( q )   dxd w( x ) | q
x
dx 2
APPROXIMATIONS and CHANGE OF VARIABLES
(METHOD OF MAPPINGS)
 = T(x,q) = x/q
0
1
(q) = (0,q)
q
x
1
0
 = (0,1)

METHODS OF MAPPINGS
w( x, q)  z(T ( x, q), q)  z( , q)
d2
 2 w( x)  f ( x)
dx
=T(x,q)
d2 
 2 [ q w( x)]  0
dx
s(M ( , q))  r ( , q)
(q)
d2
 2 z ( )  g ( , q)
d
d2 
 2 [ q z ( )]  q g ( , q)
d
d2

r ( )  0
2
d

“SOLVE”
wh( x)
[

q
z h ( )
h
w] ( x )
h
[ q w ] (  )
x=M(,q)
[

q
h
z ] ( )
r h(  )
MODEL PROBLEM #2
Map (0,q) to (0,1) by  = T(x,q) = x/q and note that the
inverse mapping M( ,q) = q maps (0,1) to (0, q).
Define
z( ,q) = w(M( ,q), q) = w(q , q) - transformed state
p( , q) = q z( ,q) - sensitivity of the transformed state
and
r ( , q) = s(M( ,q), q) = s(q, q) - transformed sensitivity.
M(S)
2
d

z( )  g(  , q ), 0    1 ,
dx 2

0,


g( , q )=
 -q2,


0  
1
q
z( 0 )  0,
1
q
  1
z( 1 )  0
MODEL PROBLEM #2
M(S)
M(S)
2
 d p( )  2g(  , q )  1 (  ),
q
dx 2
2
d

r( )  0, 0    1,
dx 2
p( 0 )  0, p( 1 )  0
1 d
r( 0 )  0, r( 1 )  ( q )
z(  ) |
 1
d
To compute s(x, q) one has two choices
Solve M( S) for r( , q) and transform back to get
(1)
s(x, q) = r( , q) = r(T(x,q), q) = r(x/q , q)
Solve M(S) for p( , q) and transform back to get
(2)
s(x, q) = p(x/q , q) -  z(x/q , q)[ M (x/q , q)]-1[qM (x/q , q)]
MESH DERIVATIVE
MODEL PROBLEM #2
FOR q > 1 AND h=q/(N+1) CONSIDER (FORMAL)
w(x)
w h(x) = Finite Element Approximation
x
x=0
x=1
x=q
NUMERICAL APPROXIMATION
2 h
d

w ( x)  f ( x ), 0  x  q ,
dx 2
(S)
h
(S)
h
w h( 0 )  0,
wh( q )  0
2
d

s( x)  0, 0  x  q , s( 0 )  0, s( q )   dxd w h( x ) | q
x
dx 2
MODEL PROBLEM #2
(S)
h
2
d

s( x)  0, 0  x  q , s( 0 )  0, s( q )   dxd w h( x ) | q
x
dx 2
FOR q > 1 AND k = q/(M+1) CONSIDER (FORMAL)
s(x)= qw(x,q)
s h,k(x) = Finite Element Approximation of
(S)
h
x
x=0
x=1
x=q
2nd NUMERICAL APPROXIMATION


h,k

q
q

s ( x, ) 
w( x, )
 q

h,k
MODEL PROBLEM #2
? WHAT HAPPENS ?
Linear Finite Elements
q = 1.5
q = 1.5
w(x ,q )
T
z( ,q )
MODEL PROBLEM #2
H1 - ERROR FOR w(x ,q )
MODEL PROBLEM #2
0
N = 03
-0.05
N = 05
N = 09
-0.1
-0.15
-0.2
-0.25
-0.3
-0.35
s(x ,q )
0
M by (1)
0.1
0.2
0.3
0.4
0.5

0.6
0.7
0.8
0.9
q [z( ,q)] = p( ,q )
1
MODEL PROBLEM #2
0.7
0.6
 [z(x/q , q)]
0.5
 [z h(x/q , q)]
Finite Element Approximation
of the Spatial Derivative
0.4
0.3
0.2
0.1
0
-0.1
-0.2
(2)
0
0.5
1
1.5
s(x, q) = p(x/q , q) -  z(x/q , q)[ M (x/q , q)]-1[q M (x/q , q)]
MODEL PROBLEM #2
s(x ,q )
M by (2)
r( ,q )
THE HYBRID CONTINUOUS SENSITIVITY METHOD
1D Interface Problem
ELLIPTIC PROCESS MODEL - 2 MATERIALS
x
1
=2
q
=1

 κ1  1,
κ(x)=κ(x , q ) = 
κ  2,

 2
0  x< q
q
x 1
0
(S)


d 
d
(x, q ) w(x)=0, 0  x 1

dx 
dx

w( 0 )  0, w(1 )  1
w( q - )=w( q + )
CONTINUITY

 d w( q  )  ( q  ) d w( q  )
( q )

 

dx
dx

 

w(,q ) H1(0,1)
1D Interface Problem
OPTIMAL DESIGN PROBLEM: Find the parameter 0 < q0 < 1, to minimize
the cost function
1
(COST)
J ( q) = F( w(, q ), q )  1 2  | w( x,q )  wˆ ( x) |2 dx
0
1
d J( q )  w(x,q )-wˆ (x),  (w(x,q )) dx

q
dq
0
s(x) 
s(x ,q ) 

w(x ,q ) 0  x  1
q
OR ...
d J( q )  w(  , q ) - wˆ (  ), s(  , q )
L2
dq
1D Interface Problem
THE SOLUTION AND SENSITIVITY IS GIVEN BY
 ( 2 )x,
 q 1
w(x,q )=
1
( q  1 )(x  1 )  1,

(  2 )x,
 ( q  1 )2

w(x,q )=
 1 )(x  1 ),
q
(
 ( q  1 )2
HOW SMOOTH IS
s(x, q ) = q w(x , q) ?
s( · , q )  H1() ?
0  x q
q
 x 1
0 xq
q  x 1
1D Interface Problem
0.3
1
PLOT OF w(x, q) AT q = .5
PLOT OF SENSITIVITY s(x,q) AT q = .5
0.9
0.2
0.8
0.1
0.7
0
0.6
-0.1
0.5
0.4
-0.2
0.3
-0.3
0.2
-0.4
0.1
0
-0.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
s( · , q )  H1()
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1D Interface Problem

HOWEVER, THE SENSITIVITY EQUATION IS GIVEN BY THE
BOUNDARY VALUE PROBLEM
(S)
(CS)
(JS)





d 
d
(x) s(x)=0, 0  x 1

dx 
dx

[ ( q  )
s( 0 )  s( 1 )  0
d
d
s( q  )]  [ κ( q  ) s( q  )]
dx
dx
 s( q  ) - s( q  )    d u( q  ) - d u( q  )
 dx



dx
HOW DID WE DERIVE THIS SYSTEM?
WHAT DO WE MEAN BY A SOLUTION?
CAN THIS BE MADE RIGOROUS?
Formal Derivation
LET
w(x,q
w (x,q ),
 1
) = 

q
w (x, ),
 2
0 < x <q ,
q < x < 1.

T
w( x, q)  w1 ( x, q ) w2 ( x, q )  H L1 (0, q )  H 1 ( q ,1)
TAKE THE TOTAL DERIVATIVE OF
w1 ( q , q )  w2 ( q , q )




w1 ( q , q ) 
w1 ( q , q ) 
w2 ( q , q ) 
w2 ( q , q )
x
q
x
q
[




w ( q, q ) w ( q , q )]   [ w ( q , q )  w (q ,q )]
q 2
q 1
x 2
x 1
[




w( q+, q ) w( q-, q )]   [ w(q+, q )  w( q-, q )]
q
q
x
x
Formal Derivation
JUMP
s( q ) - s( q  )   d w( q  ) - d w( q  )
 dx



dx
LIKEWISE ...
CONTINUITY
[ ( q  )
d
d
s( q  )]  [ κ( q  ) s( q  )]
dx
dx
s(, q )   w(, q )  H1(0,1)
q
?
(S)

d 
d

(x) s(x)=0, 0  x 1 in [W (q)]’


dx 
dx

WEAKEST FORM OF THE ELLIPTIC PROBLEM
Sensitivity Computations
Petrov-Galerkin FE Method (Burns, Lin & STanley - 03)
2D Sensitivity Computations
Petrov-Galerkin FE Method (Burns, Lin & STanley - 03)
!! WORKS IN 2D !!
2D Sensitivity Computations
Petrov-Galerkin FE Method (Burns, Lin & STanley - 03)
!! WORKS IN 2D !!
FOR COMPLEX GEOMETRY
WHAT ABOUT
NUMERICAL METHODS
Numerical Methods
d
x(t )  f (t , x(t ) )
dt
(IVP)
x(t k 1 )  x(t k )
t
 f (t k , x(t k ))
x(t0 )  x0  Rn
FORWARD DIFFERENCE
x(t k1 )  x(t k )  tf (t k , x(t k ))
x0
t
t0
t1
t2
tk
t k 1
Explicit Euler
x0
t
t0
t1
t2
tk
t k 1
h  t , ti  t0  ih
x(t k1 )  x(t k )  tf (t k , x(t k ))
xk 1  xk  h  f (tk , xk )
Implicit Euler Method
d
x(t )  f (t , x(t ) )
dt
x0
t
t
t1
t2
0
d
x(t k 1 )  f (t k 1 , x(t k 1 ))
dt
  (  h)
x(t k 1  
h )  x(t k 1 )
h
 f (t k 1 , x(t k 1 ))
BACKWARD DIFFERENCE
x(t k )  x(t k 1 )
 f (t k 1 , x(t k 1 ))
h
x(t k1 )  x(t k )  hf (t k1 , x(t k1 ))
tk
t k 1
Implicit Euler
x0
t
t0
t1
t2
tk
t k 1
h  t , ti  t0  ih
x(t k1 )  x(t k )  hf (t k1 , x(t k1 ))
xk 1  xk  h  f (tk 1 , xk 1 )
Numerical Methods Matter
d
3
x(t )  q[ x(t )]
dt
x(0)  x0
d
x(t , q )  qx (t , q )
dt
x(0, q)  x0
x(t )  x(t , q)
DIFFERENTIATE THE EQUATION WITH RESPECT TO q
 d


q[ x(t , q)]3
 x(t , q)   
q  dt
q

(
)



2
 q   3[ x(t , q)]  x(t , q)   [ x(t , q)]3
q


Numerical Methods Matter



 d

2
3


x
(
t
,
q
)

q

3
[
x
(
t
,
q
)]

x
(
t
,
q
)

[
x
(
t
,
q
)]




q  dt

q



 d


2
3
x
(
t
,
q
)

q

3
[
x
(
t
,
q
)]

x
(
t
,
q
)

[
x
(
t
,
q
)]


q  dt
q

(
)
INTERCHANGE THE ORDER OF DIFFERENTIATION

d 

2
 s(tx,(tq, q))   q  3[ x(t , q)]  s(tx,(tq, q))  [ x(t , q)]3
dt  q
q

(
)
Numerical Methods Matter

d 

2
 s(tx,(tq, q))   q  3[ x(t , q)]  s(tx,(tq, q))  [ x(t , q)]3
dt  q
q

(
)
d
3
x(t )  q[ x(t )]
dt
(
x(0)  x0
)
d
s (t )  q  3[ x(t )]2  s(t )  [ x(t )]3
dt
x(t , q) 
x0
2q[ x0 ] t  1
2
s (t , q ) 
s(0)  0
 [ x0 ] t
3
(2q[ x ] t  1)
2
0
3/ 2
SOME RUNS
Numerical Methods Matter
x0  4, q  1
2
h
q[ x 0 ]2
2
h
q[ x 0 ]2
SOLUTION
Numerical Methods Matter
FORWARD EULER
xk1  xk  h  q [ x k ]  x k (1  h  q [ xk ] )
3
2
h
q[ x 0 ]2
2
x k 1  0
 x 0

 x 0
2
h
q[ x 0 ]2
x k 1
2
h
q[ x 0 ]2
x k 1  
Numerical Methods Matter
x k1  x k  h  q [ x k1 ]
3
BACKWARD EULER
x k1 (1  h  q[ x k1 ] )  x k
2
x k 1 
xk
(1  h  q[ x k 1 ] )
2
 xk
lim
x k 1  0
k  
FOR ALL
x0
AND ALL
h0
Why Sensitivities?
 USEFUL
IN OPTIMIZATION BASED DESIGN
 SENSITIVITIES HAVE MANY OTHER USES
 PRIORITIZE DESIGN & CONTROL VARIABLES
 EVALUATE DESIGNS & CONTROL LAWS
 NON- OPTIMIZATION BASED DESIGN
 FAST SOLVERS
 ANALYZE UNCERTAINTIES
 PREDICT “FAILURE” (FLOW SEPARATION, ETC.)
 SOME OBSERVATIONS
 DO NUMERICS CAREFULLY
 “ORDER’’ MATTERS
MAY REQUIRE COMPLEX
MATHEMATICAL THEORIES
------DIFFERENTIATION OF
SET-VALUED FUNCTIONS
END OF SHORT COURSE
BUT…
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