Post on 30-Dec-2015
description
Lecture III
A Slightly More Complicated ProblemStarting with a hyperbolic tangent production
function
We define the profit function as
1 2 3 1 2
3 1 2 2 2
3 3 1 2 1 3 2 3
15, , 1 tanh 0.50 0.2624 0.3125
20.3500 0.0025 0.0028
0.0025 0.00035 0.0028 0.000032
f x x x x x
x x x x x
x x x x x x x x
1 2 3 1 2 3 1 2 3max , , , , 0.375 0.450 0.475x x x f x x x x x x
As a starting point, we assume
Starting with some feasible step s1
3kB I
21 1 1 1 1( ) ( ) ( )x x xxx s x x s
If we let 1 1( ) 0x x s
21 1 1
121 1 1
( ) ( ) 0
( ) ( )
x xx
xx x
x x s
s x x
In the hyperbolic tangent formulation
1 2 3 1 2 3
1 2 3 1 2 3
1 2 3 1 2 3
150.375 0.2625 0.005 0.00035 0.00028 , ,
215
0.450 0.3125 0.00035 0.00560 0.000032 , ,215
0.475 0.350 0.00028 0.000032 0.00500 , ,2
x
x x x h x x x
x x x x h x x x
x x x h x x x
1 2 3 1 2 3 1 2
2 2 3 3 1 2 1 3 2 3
, , sech 0.50 0.2624 0.3125 0.3500 0.0025
0.0028 0.0025 0.00035 0.0028 0.000032
h x x x x x x x x
x x x x x x x x x x
Let us start by considering
1 1 1 1x
11.25872
1.49480
1.71050x x
Given that B1 is the identity matrix
Thus, in this case we have
1 3 1
1.25872
1.49480
1.71050xs I x
2 1 1
1 1.25872 2.25872
1 1.49480 2.49480
1 1.71050 2.71050
x x s
Next, to update the Hessian matrix following the BFGS algorithm we have
1 1
1
1
( ) ( )
0.16973
( ) 0.20692
0.20127
1.25872
1.49480
1.71050
0.16973 1.25872 1.42845
0.20692 1.49480 1.70172
0.20127 1.71050 1.91177
k x k k x k
x
x
y x s x
x s
x
y
Thus, following the BFGS update for the Hessian matrix
1
1
1 1' '
' '
1.03315 0.04038 0.03954
0.04038 1.04915 0.04830
0.03954 0.04830 1.04636
k k k k k k k kk k k k k
B B B s s B y ys B s y s
B
Updating the solution, next we have the gradient at the new solution as
1 1 2
0.16973
( ) ( ) 0.20692
0.20127x xx s x
Updating the point of approximation
3 2 2
2.25872 0.150297 2.10842
2.49480 0.183235 2.31136
2.71050 0.178212 2.53229
x x s
Updating the gradient
2 2 3
0.10095
( ) ( ) 0.12524
0.10956x xx s x
2 2 2 2
0.06878
( ) ( ) 0.08168
0.09172x xy x s x
New Hessian
2
0.8567 0.1783 0.1532
0.1783 0.7783 0.1915
0.1532 0.1915 0.8407
B
R-Codefr <- function(b){-15/2*(1+tanh(-0.50+0.2625*b[1]+0.3125*b[2]+0.3500*b[3]- 0.0025*b[1]*b[1]-0.0028*b[2]*b[2]-0.0025*b[3]*b[3]+ 0.00035*b[1]*b[2]+0.00028*b[1]*b[3]+0.000032*b[2]*b[3])) +0.375*b[1]+0.450*b[2]+0.475*b[3]}
dfr <- function(b){ dd <- cosh(0.50-0.2625*b[1]-0.3125*b[2]-
0.3500*b[3]+0.0025*b[1]*b[1]+0.0028*b[2]*b[2]+ 0.0025*b[3]*b[3]-0.00035*b[1]*b[2]-0.00028*b[1]*b[3]-0.000032*b[2]*b[3]) as.matrix(cbind(0.375+(15/2)*(-0.2625+0.00500*b[1]-0.00035*b[2]-
0.00028*b[3])/(dd*dd),
0.450+(15/2)*(-0.3125-0.00035*b[1]+0.00560*b[2]-0.000032*b[3])/(dd*dd),
0.475+(15/2)*(-0.3500-0.00028*b[1]-0.000032*b[2]+0.0050*b[3])/(dd*dd)))}
b0 <- cbind(1.0,1.0,1.0)
res.fr <- optim(b0,fr,dfr,method="BFGS")
print(res.fr)
Results$par [,1] [,2] [,3][1,] 1.062955 0.4756912 4.537824
$value[1] -11.46800
$countsfunction gradient 28 23
$convergence[1] 0
$messageNULL
Starting with a simple 2*3 example, assume that we have the matrix equation
Note that by row operations the A matrix can be transformed to
1
2
3
1 1 1 5
0 1 1 6
x
Ax b x
x
1 1 1 1 1 1 0 2
0 1 0 1 1 0 1 1
This expression implies the following homogeneous relationships:
Setting x3=1 yields
1 3
2 3
2 0
0
x x
x x
1
2
2
1
x
x
Or in vector form:
Next we confirm that z is the nullspace.
2
1
1
z
We do this by confirming that both vectors of the matrix are orthogonal to the nullspace:
2
1 1 1 1 2 1 1 0
1
2
0 1 1 1 0 1 1 0
1
Implications of the nullspace:Start by defining a feasible solution:
5 1 1 1 1 1 5 1 0 2 11
6 0 1 0 1 1 6 0 1 1 6
111 1 1 11 6 0 5
60 1 1 6 0 6
0
Ax
Next, we would like to generate another feasible point based on this solution and the nullspace matrix:
Letting yields p = 5
1
*2
3
11 2
6 1
0 1
x
x x Zp x p
x
1
*2
3
11 2 11 10 1
6 1 5 6 5 1
0 1 0 5 5
x
x x Zp x
x
Checking the original solution:
11 1 1 1 1 5 5
10 1 1 0 1 5 6
5