NON LINEAR PROGRAMMING
Prof. Stephen Graves
Consider the example we used to introduce Lagrange multipliers:
MIN
s.t.
3 SD
H Q
f(Q , Q , Q ) = ∑ i i + i i
1
2 3
2
i=1 Q i
3 TD
g(Q , Q , Q ) = ∑ i i = K
1
2 3
i=1 Q i
Some definitions:
Gradient of f:
⎛∂f
∂f
∂f ⎞
∇f = ⎜
,
,
⎟⎟
⎜ ∂Q
⎝ 1 ∂Q2 ∂Q3 ⎠
⎛ SD
S3D3 H3 ⎞
H1
S2D2 H 2
1
1
⎜
⎟
= −
+
, −
+
, −
+
2
2
2
⎜
⎟
2
2
2
Q1
Q2
Q3
⎝
⎠
Gradient of g:
⎛ ∂g
∂g
∂g ⎞
,
,
∇g = ⎜
⎟⎟
⎜ ∂Q
⎝ 1 ∂Q 2 ∂Q 3 ⎠
⎛ TD
T3 D 3 ⎞
T2 D 2
1
1
⎜
⎟
= −
, −
, −
2
2
2
⎜
Q1
Q2
Q3 ⎟
⎝
⎠
Directional derivatives:
Let x = (x1, x2, x3 ) denote a direction. The directional derivative for f and g are given
by:
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Summer 2003
3
df
= ∑
dx
i=1
3 ⎛ SD
⎛∂f ⎞
Hi ⎞⎟
i
i
⎜
+
x
⎜⎜
⎟⎟ xi = ∑ −
2 ⎟ i
Qi2
⎝ ∂Qi ⎠
i=1 ⎜⎝
⎠
3
dg
= ∑
dx
i=1
3 ⎛ TD ⎞
⎛∂g⎞
⎜⎜
⎟⎟ xi = ∑ ⎜ − i 2i ⎟ x i
Qi ⎟
⎝ ∂Qi ⎠
i=1 ⎜⎝
⎠
Note: the directional derivative is the "dot product" of the gradient and the direction
vector.
For a given point, say (Q1, Q2, Q3 ), what is the direction of steepest ascent for the
objective function f? That is, what direction provides the largest value for the directional
derivative? The direction of steepest ascent will be the solution to the following
optimization problem:
df
MAX
=
dx
s. t.
⎛ SD
Hi ⎞⎟
i
i
⎜
∑ − 2 + 2 ⎟ xi
Qi
i=1 ⎜⎝
⎠
3
x12 + x 22 + x 32 = 1
By using a Lagrange multiplier to solve this problem, you can show that the direction of
steepest ascent is given by
x * = (x1, x 2 , x 3 ) = ∇f/ ∇f
That is, the "best" direction is the (normalized) gradient; for our example, the direction of
steepest ascent (not normalized) is
⎛ SD
S D
H ⎞
H
S D
H
x* = ⎜ − 1 1 + 1 , − 2 2 + 2 , − 3 3 + 3 ⎟
⎜
2
2
2 ⎟
Q12
Q22
Q32
⎝
⎠
Note that since the actual problem is a minimization, we actually want the direction of
steepest descent, which would just be the negative of the gradient.
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Now suppose that the given point (Q1, Q2, Q3 ) is feasible; that is, it satisfies the
constraint:
g(Q1, Q2, Q3 ) = K
What is the best feasible direction?
The best feasible direction (for ascent) will be the solution to the following optimization
problem:
⎛ SD
Hi ⎞⎟
i
i
⎜
∑ − 2 + 2 ⎟ xi
Q
i=1 ⎜⎝
i
⎠
3
df
MAX
=
dx
s. t..
x12 + x 22 + x 32 = 1
dg
=
dx
⎛ TD ⎞
∑ ⎜⎜ − i 2i ⎟⎟ xi = 0
Qi
i=1 ⎝
⎠
3
By using two Lagrange multipliers to solve the problem, you can find that the best
feasible direction (called the reduced gradient) is given by:
x* =
( x1* , x*2 , x*3 ) = ∇f - ⎛⎜⎝ ∇∇fg •• ∇∇gg ⎞⎟⎠ ∇g
where ∇f • ∇g =
3
⎛∂f ∂g⎞
,
and
∇
g
•
∇
g
=
⎜
⎟
∑ ⎜
∑
⎟
i=1 ⎝ ∂Qi ∂Qi ⎠
i=1
3
⎛∂g ∂g⎞
⎜⎜
⎟⎟
∂
Q
∂
Q
i⎠
⎝ i
The reduced gradient can then be used as a search direction to improve the current
solution. To see that the reduced gradient is a feasible direction, we note that
⎛ ∇f • ∇ g ⎞
x* • ∇g = ∇f • ∇g - ⎜
⎟ ∇g • ∇g = 0
⎝ ∇g • ∇g ⎠
Besides determining the reduced gradient, an algorithm would also need to determine the
"step size:" namely how far to move along the reduced gradient. An algorithm stops
when the reduced gradient equals (approximately) the zero vector.
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