Jtdylktuy

(Use a Calculator) Use geometric programming to solve the following program:

Minimine: $g(x) = frac{1000}{xy} + 2x + 2y + xy$, $forall x,y > 0$

I am a little confused on how to begin this program, can anyone give me any tips? Thanks!

This is exercise 26 from Chapter 2 , Peressini.
Solution is solve Dual geometric Programing: Maximize $$V(delta)=(frac{1000}{delta_1})^{delta_1}(frac{2}{delta_2})^{delta_2}(frac{2}{delta_3})^{delta_3}(frac{1}{delta_4})^{delta_4}$$
Where $delta = (delta_1 ,delta_2 ,delta_3 ,delta_4)$
with the conditions normality and ortogonality
$$ left{ begin{array}{2}
delta_1 +delta_2+delta_3+delta_4 =1 \[2ex]
-delta_1 +delta_2+0delta_3+delta_4 =0 \[3ex]
-delta_1 +0delta_2+delta_3+delta_4 =0end{array} right.
$$
And positivity condition $delta_1>0,delta_2>0,delta_3>0,delta_4>0$
Where the solution set from the system is $$K={(delta_1 ,delta_2 ,delta_3 ,delta_4) / delta_2=delta_3=1-2delta_1 , delta_4 = 3delta_1-1, frac{1}{3}<delta_1<frac{1}{2} }$$
Now Maximize $$f(s)=(frac{1000}{s})^{s}(frac{2}{1-2s})^{1-2s}(frac{2}{1-2s})^{1-2s}(frac{1}{3s-1})^{3s-1}$$ on the interval $1/3<s<1/2$.Taking logs,we maximize
$$ u(s) =ln f(s)=-sln frac{s}{1000}-2(1-2s)ln frac{1-2s}{2}-(3s-1)ln (3s-1)$$
Using Newton Method in $s_0 = 0.4$
$$s^{k+1}=s^k-frac{u'(s^k)}{u''(s^k)}approx s^k-frac{frac{u(s^k)-u(s^k-h)}{h}}{frac{u(s^k+h)-2u(s^k)+u(s^k-h)}{h^2}}$$
with $h = 0.001$ and 100 iterations, $s^*=0.4394 Rightarrow u(s^*)=4.4405$
Then $$f(s^*) =e^{u(s^*)}=84.8197 $$
This is the minimun of $g(mathbf{x})$

Without a calculator, I would just use the fact that at the minimum you must have $frac {partial g(x,y)}{partial x}=frac {partial g(x,y)}{partial y}=0$ That will give you two equations in two unknowns. The symmetry indicates that $x=y$ at the minimum (though you should check that).

My graphing calculator, Alpha, finds the minimum to be near $x=y=5.18284$ As it does not offer an exact result, I think you are just supposed to search around to find the minimum.

The problem can also be solved numerically using cvxpy's geometric programming capabilities, like so:

#!/usr/bin/env python3



import cvxpy as cp



x = cp.Variable(pos=True)

y = cp.Variable(pos=True)



constraints = 

objective   = 1000/x/y+2*x+2*y+x*y



prob = cp.Problem(cp.Minimize(objective), constraints)



objval = prob.solve(gp=True)



print("Objective value = {0}".format(objval))

print("x value         = {0}".format(x.value))

print("y value         = {0}".format(y.value))

As output, this gives:

Objective value = 84.82073466127639

x value         = 5.182852109768804

y value         = 5.18285210938629