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Conjugate Gradient and Linear Equality Constraints

Conjugate Gradient and Linear Equality Constraints . Lecture III. A Slightly More Complicated Problem. Starting with a hyperbolic tangent production function We define the profit function as. As a starting point, we assume Starting with some feasible step s 1. If we let.

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Conjugate Gradient and Linear Equality Constraints

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  1. Conjugate Gradient and Linear Equality Constraints Lecture III

  2. A Slightly More Complicated Problem • Starting with a hyperbolic tangent production function • We define the profit function as

  3. As a starting point, we assume • Starting with some feasible step s1

  4. If we let

  5. In the hyperbolic tangent formulation

  6. Let us start by considering

  7. Given that B1 is the identity matrix • Thus, in this case we have

  8. Next, to update the Hessian matrix following the BFGS algorithm we have

  9. Thus, following the BFGS update for the Hessian matrix

  10. Updating the solution, next we have the gradient at the new solution as

  11. Updating the point of approximation

  12. Updating the gradient

  13. New Hessian

  14. R-Code fr <- 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)

  15. Results $par [,1] [,2] [,3] [1,] 1.062955 0.4756912 4.537824 $value [1] -11.46800 $counts function gradient 28 23 $convergence [1] 0 $message NULL

  16. 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

  17. This expression implies the following homogeneous relationships: • Setting x3=1 yields

  18. Or in vector form: • Next we confirm that z is the nullspace.

  19. We do this by confirming that both vectors of the matrix are orthogonal to the nullspace:

  20. Implications of the nullspace: • Start by defining a feasible solution:

  21. Next, we would like to generate another feasible point based on this solution and the nullspace matrix: • Letting yields p = 5

  22. Checking the original solution:

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