Need help solving system of equations to min. variable y

[ikoukas]

I am trying to minimize the value of a variable y = f(x[2], x[3], ... x[N]). All symbols below are known constants except for x[2], x[3], ... , x[N]. Vector x is a variable vector of N-1 items, starting from x[2] and ending to x[N]. All x are positive real numbers and so are the constants.

. . .y = g[1]*(x[2] + Q[1]) / CIN + g[2]*(x[3] + Q[2]) / x[2] + ... + g[N-1]*(x[N] + Q[N-1]) / x[N-1] + g[N]*COUT / x[N].

I have used the method of partial derivatives to minimize the y quantity. After doing so, I have ended up with these equations:

(Note: By x^2 I mean the square of x, or, equivalently: x*x.)

. . .(1) x[N-1] = x[N]^2*g[N-1] / (COUT*g[N])

. . .(2) for n E [3, N-1]

. . . . .x[n-1] = x[n]^2*g[n-1] / ((x[n+1] + Q[n])*g[n])

. . .(3) x[2]^2 = g[2]*CIN*(x[3]+ Q[2]) / g[1]

How can I calculate the values of x where i E [2, N] out of these equations?

If I did know the correct value for x[N] I could calculate successively the values of x[N-1] through equation (1), and then I could calculate all values of x through equation (2).

If I did that though, there would be an extra equation (3) which should remain satisfied for the calculated values of x[3] and x[2] which makes me think that there is a way to solve this system of equations and that there are enough equations for doing so.

Please help!!!

 

[Deleted member 4993]

From (1) and (2) we could deduce:

((x[n+1] + Q[n])*g[n]) = (COUT*g[N])

x[n+1] = COUT - Q[n] ...assuming g[n]<>0

from this you can find x(4)

Then using (2) and (3) you can find two equations and two unknowns (x(2) & x(3))

 

[ikoukas]

RE:

You can't combine these equations like that because the (2) is valid for n E [3, N-1], not n-1 E[3, N-1] so it goes like x[N-2] = x[N-1]^2*g[N-2] / ((x[N] + Q[N-1])*g[N-1]) and you can't combine this one with (1) like that unfortunately. Be careful with upcase N and lowcase n and also with valid ranges for each equation. I think it takes an unknown to me trick to solve this system of equations. Thank you for your help though, I appreciate it.