Chap
.
22
Unconstrained Optimization
.
Linear Programming
Sec
.
22
.
1
Basic Concepts
.
Unconstrained Optimization
Problem Set 22
.
1
.
Page 939
3
.
Cauchy
’
s method of steepest descent
. The given function is
f
u
x
U
u
3
x
1
2
U
2
x
2
2
u
12
x
1
U
16
x
2
.
(A)
The given starting value is
x
0
u
±
1
1
²
T
. Proceed as in Example 1, beginning with the general
formulas and using the starting value later. To simplify notations, denote the components of the gradient
of
f
by
f
1
and
f
2
. The gradient of
f
is
U
f
u
x
U
u
±
f
1
f
2
²
T
u
±
6
x
1
u
12
4
x
2
U
16
²
T
.
In terms of components,
f
1
u
6
x
1
u
12,
f
2
u
4
x
2
U
16.
(B)
Furthermore,
z
u
t
U
u
±
z
1
z
2
²
T
u
x
u
t
U
f
u
x
U
u
±
x
1
u
tf
1
x
2
u
tf
2
²
T
.
In terms of components,
z
1
u
t
U
u
x
1
u
tf
1
,
z
2
u
t
U
u
x
2
u
tf
2
.
(C)
Now obtain
g
u
t
U
u
f
u
z
u
t
UU
from
f
u
x
U
in (A) by replacing
x
1
with
z
1
and
x
2
with
z
2
. This gives
g
u
t
U
u
3
z
1
2
U
2
z
2
2
u
12
z
1
U
16
z
2
.
Calculate the derivative of
g
u
t
U
with respect to
t
, obtaining
g
±
u
t
U
u
6
z
1
z
1
±
U
4
z
2
z
2
±
u
12
z
1
±
U
16
z
2
±
.
From (C) you see that
z
1
±
u
u
f
1
and
z
2
±
u
u
f
2
. Substitute this and
z
1
and
z
2
from (C) into
g
±
u
t
U
, obtaining
g
±
u
t
U
u
6
u
x
1
u
tf
1
U u
u
f
1
U
U
4
u
x
2
u
tf
2
U u
u
f
2
U
U
12
f
1
u
16
f
2
.
Order the terms as follows. Collect the terms containing
t
and denote their sum by
D
(suggesting
“denominator” in what follows). This gives
tD
u
t
u
6
f
1
2
U
4
f
2
2
U
.
(D)
Denote the sum of the other terms by
N
(suggesting “numerator”), obtaining
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Optimization, Graphs
Part F
N
u
u
6
x
1
f
1
u
4
x
2
f
2
U
12
f
1
u
16
f
2
.
(E)
With these notations you have
g
±
u
t
U
u
tD
U
N
. Solving
g
±
u
t
U
u
0 for
t
gives
t
u
u
N
D
.
Step 1
. For the given
x
u
x
0
u
±
1
1
²
T
you have
x
1
u
1,
x
2
u
1 and from (B)
f
1
u
u
6,
f
2
u
20,
tD
u
1816
t
,
N
u
u
436,
so that
t
u
t
0
u
u
N
/
D
u
0.240088.
From this and (B) and (C) you obtain the next approximation
x
1
of the desired solution in the form
x
1
u
z
u
t
0
U
u
±
1
u
t
0
u
u
6
U
1
u
t
0
U
20
²
T
u
±
1
U
6
t
0
1
u
20
t
0
²
T
u
±
2.44053
u
3.80176
²
T
.
This completes the first step.
Step 2
. Instead of
x
0
now use
x
1
, in terms of components,
x
1
u
2.44053,
x
2
u
u
3.80176,
and do the computations in Step 2 with these data. And so on.
The results for the first six steps are as follows.
n
x
f
0
1.00000
1.00000
9.0000
1
2.44053
u
3.80176
u
43.3392
2
1.98753
u
3.93766
u
43.9918
3
2.00549
u
3.99753
u
44.0000
4
1.99985
u
3.99922
u
43.9999
5
2.00006
u
3.99994
u
44.0000
6
2.00006
u
3.99994
u
44.0000
You can see that the convergence is fast, and you also notice the effect of roundoff in the last few digits.
Completing squares you find that
f
can be written
f
u
x
U
u
3
u
x
1
u
2
U
2
U
2
u
x
2
U
4
U
2
u
44.
This explains the numeric results. It also shows that the level curves
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 Spring '12
 M.lee
 Linear Programming, Optimization, Row

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