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Mathe IIILecture 9Mathe IIILecture 9
2
max
s.t. j
x
g x,r = 0, j = 1, ...,m
f x,r
1 n 1 2 kx = x , .....x , r = r ,r , .....r
the solution is
and the maximum is
,
*
* *
x r
f r = f x r ,r
Envelope Theorem
3
Envelope Theorem
*
j
*
j
f x r ,rf r=
r r
j
*f x r ,r
r
First, consider the case where there are no costraints:
maxx
f x,r
the solution is
and the maximum is
,
*
* *
x r
f r = f x r ,r
4
Envelope Theorem (with no onstraints)
h
h j j
*
j h
xf f=x r r
f rr
maxx
f x,r h
f = 0x
h
fx
j
fr
0
j
*f x r ,r
r
*
j=
f rr
* *f r = f x r ,r
5
*
j
f r
r
Envelope Theorem
max
s.t. j
x
g x,r = 0, j = 1, ...,m
f x,r
-m
i ii=1
x,λ,r = f x,r λ g x,rLAt the maximum point:
* *
i*i
h h
m
i=1
f x r ,r g x r ,r
x xλ
*nh
h=1 h j j
xf f=
x r r
(with constraints)
6
*
j
f r
r
Envelope Theorem
*nh
h=1 h j j
xf f=
x r r
(with constraints)
7
**
j j
f x r ,rf r=
r r
Envelope Theorem
*nh
h=1 h j j
xf f=
x r r
(with constraints)
* i
ih h
=
m
i=1
gf
x xλ
*h
h j
i
j
n
h
m
ii=1=1
g x f=
rx rλ
*
i hi
h j j
m n
i=1 h=1
g x f= λ
x r r
*ig x r ,r = 0
*ni h i
h=1 h j j
g x g+ = 0
x r r
*n
i h i
h=1 h j j
g x g
x r r
* ii
j j
m
i=1
g f= - λ
r r
Lagrange:
8
Envelope Theorem (with constraints)
*
* ii
j j j
m
i=1
f r g f= - λ
r r r
-m
i ii=1
x,λ,r = f x,r λ g x,rL
*
j
x , λ,r=
r
L
* **
j j
x , λ ,rf r=
r r
L
9
Non Linear Programming
max s.t.
1 1
m m
x
g x c
g x c
f x ......
1 nx = x , ....x
A simple case:
max s.t. x,y
g x, y cf x, y
10
max s.t. x,y
g x, y cf x, y
x
y
.constf x, y =+ ++ g x, y c
g x, y = c
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max s.t. x,y
g x, y cf x, y
x
y
.constf x, y =+ ++ g x, y c
g x, y = c
The max is
a small change in c
has no effect on the max.
i.e. the constraint is binding,
the shadow price of
is c
not
λ 0
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max s.t. x,y
g x, y cf x, y
x
y
g x, y c
g x, y = c
If the situation is :
the max. is: the constraint is binding.
Increasing , relaxes the constraint c
the shadow price of is positive. λ c
and the max.increases
13
Non Linear Programming
max s.t.
1 1
m m
x
g x c
g x c
f x ......
a general treatment
Let be a solution of the problem.* x
Let for
and for
*i
*i
i
i
i = 1,2, ....,k
i = k, ......,m
g x = c
g x < c
14
Non Linear Programming a general treatment
-
k
i i ii=1
x,λ = f x λ g x -cL
The solution is a stationary point of the Lagrangian:
*
*h
h
f x c
cλ c
clearly an increase in ci increases the maximum. Hence: 0
h i
= 0, = 0, h = 1, ...,n i = 1, ...kx λ
L L
k
15
Non Linear Programming a general treatment
if is a solution of the problem*x
j j = 1,3 .λ 0,. ..,m
is a stationary point (w.r.t. ) of the Lagrangian :
-
*i
m
i i ii=1
2 x. x
x = f x λ g x -c L
*j j1. j = 1, .g x mc , ..,
if then *j j jg x < c λ = j = 1, .4. 0, ..,m
m
16
Non Linear Programming an example
max . .s t
2 2 2
2 2 2x
z xy4z - x - y - z
x + y + z 3
- - 2 2 2 2 2 2x, y, z = 4z - x - y - z λ z - xy x + y + z - 3L
-2x + λy - 2μx = 0x
L
-2y + λx - 2μy = 0y
L
4 - 2z - λ - 2μz = 0z
L
λ, μ 0
z < xy 0 λ =2 2 2x + y + z < 3 0 μ =
17
Non Linear Programming an example
-2x + λy - 2μx = 0x
L
-2y + λx - 2μy = 0y
L
4 - 2z - λ - 2μz = 0z
L
if both constraints are2 2 2
inactiv
z < xy, x + y + z <
e :
3 λ = μ = 0
This candidate does not satisfy the constraints
x = y = 0, z = 2
18
Non Linear Programming an example
-2x + λy - 2μx = 0x
L
-2y + λx - 2μy = 0y
L
4 - 2z - λ - 2μz = 0z
L
if only the first constraint is : 2 2 2
ina
λ = 0, z < xy, x + y +
tive
z
c
= 3
z < xy = 0 z = - 3
x = 0
y = 0
z = 3
2z = > 0
1+ μ no solution
19
Non Linear Programming an example
-2x + λy - 2μx = 0x
L
-2y + λx - 2μy = 0y
L
4 - 2z - λ - 2μz = 0z
L
if only the second constraint is :
μ = 0,
ina
ctive
xy = z
z = 1 λ 0 λ = 2 x = y
if or : 2 2 x = y0 x =x y
2 2x + y = 22 2 2x + y + z = 3 no solution
20
Non Linear Programming an example
-2x + λy - 2μx = 0x
L
-2y + λx - 2μy = 0y
L
4 - 2z - λ - 2μz = 0z
L
if only the second constraint is :
μ = 0,
ina
ctive
xy = z
indeed 2 2 2 λ = 4, x + y + z = 0 < 3
if x = 0 y = 0 z = 0
and is a candidate(0,0,0)
(0,0,0)
21
Non Linear Programming an example
-2x + λy - 2μx = 0x
L
-2y + λx - 2μy = 0y
L
4 - 2z - λ - 2μz = 0z
L
if both constraints are : 2 2 2
act
xy = z, x + y +
ve
z
i
= 3
y - x 2 + λ + 2μ = 0
(0,0,0)
x = y 2 z = x
2 2 4x + x + x = 3 2 z = x = -1 2 = 1,-3
z = 1, x = y = 1 μ = 0, λ = 2 are candidates(1,1,1), (-1,-1,1)
(1,1,1)
(-1,-1,1)
22
Non Linear Programming an example
(0,0,0) (1,1,1) (-1,-1,1)
max . .s t
2 2 2
2 2 2x
z xy4z - x - y - z
x + y + z 3
2 2 2f(x, y, z) = 4z - x - y - z
f(0,0,0) = 0, f 1,1,1 = 1, f -1,-1,1 = 1
(-1,-1,1)(1,1,1) are maxima
