Difficult optimal cable cost analysis

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algar32 said:
the magnitude is just the sqrt( x^2+y^2+z^2)

in our case that would be sqrt(x^2+16+z^2)

You are telling me that is the tension?
If that is the case where does the equation of b come into play?Thanks.

Tension magnitude = sqrt(x^2+16+z^2) times b

b was a scalar.

I may be wrong. Let someone confirm.
 
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is it Tab= b(tension)

=(2963/5x+z+12) (x^2+16+z^2)
 
rollingstein said:
sqrt(x^2+16+z^2) times b

b was a scalar.

Okay just saw this now. I think I got the same in my last post so... length x tension
= (x^2+16+z^2) x (2963/5x+z+12) (x^2+16+z^2)

Then minimize this equation?
 
algar32 said:
Okay just saw this now. I think I got the same in my last post so...


length x tension
= (x^2+16+z^2) x (2963/5x+z+12) (x^2+16+z^2)

No.

sqrt(x^2+16+z^2) x (2963/5x+z+12) sqrt(x^2+16+z^2)
 
rollingstein said:
No.

sqrt(x^2+16+z^2) x (2963/5x+z+12) sqrt(x^2+16+z^2)

eh my bad. I meant to put those in there... I need sleep :(. Anyway thanks for the help. I understand where this eq. comes from now. When voko did his minimization the 2963 is never included. Could you provide some insight as to where this goes? Thanks.
 
algar32 said:
eh my bad. I meant to put those in there... I need sleep :(. Anyway thanks for the help. I understand where this eq. comes from now. When voko did his minimization the 2963 is never included. Could you provide some insight as to where this goes? Thanks.


min(k*f(x,y)) = k * min(f(x,y))
 
rollingstein said:
min(k*f(x,y)) = k * min(f(x,y))

Okay so the k in your equation is just the 2963 constant being brought out because it is no longer needed for the calculation of the cost function.

It completely unrelated to the value of k in voko's work correct?

voko said:
Indeed b > 0 is a physical constraint that must be satisfied. This constrains the domain of the cost function. And it must be minimized in that domain.

For a general multivariate function, global minimization is a very difficult problem. In this case, however, the problem can be analyzed with elementary means. Let's take some value k and see if the cost function can be less than that: i.e.,

(x^2 + z^2 +16)/(5*x+z+12) < k

(x^2 + z^2 +16) < k(5*x+z+12)

(x^2 + z^2 +16) - k(5*x+z+12) < 0

x^2 - 5kx + (2.5k)^2 - (2.5k)^2 + z^2 - kz + (0.5k)^2 - (0.5k)^2 + 16 - 12k < 0

(x - 2.5k)^2 - (2.5k)^2 + (z - 0.5k)^2 - (0.5k)^2 + 16 - 12k < 0

(x - 2.5k)^2 + (z - 0.5k)^2 - 6.5k^2 - 12k + 16 < 0

It is obvious that the minimum of the left-hand side expression is -6.5k^2 - 12k + 16, so -6.5k^2 - 12k + 16 < 0. What does this inequality mean? It means that for any k satisfying it, there are lesser values of the cost function, i.e., this value is not minimal.

There are two ranges of k satisfying this inequality: ## k > 4 \frac {\sqrt{35} - 3 } {13} ## and ## k < -4 \frac {\sqrt{35} + 3 } {13} ##. The second range is invalid, because we require that the cost-function be positive, so we have only one range, and that means that ## \displaystyle k = 4 \frac {\sqrt{35} - 3 } {13} ## is the greatest value, for which no lesser value exists, i.e., it is the global minimum. And it is exactly the value you obtained for the local minimum, so the local minimum is also global.
 
Also I followed vokos steps for global minimization. Wolfram says there is no global minima. Why is this?

Also now that I have the global minima, what is the best way to find x an zfrom this? Thanks.
 
Last edited:
Regarding the global minimum, you have to specify that 5x+z+12 > 0.

To get x and z, go back to (x - 2.5k)^2 + (z - 0.5k)^2 - 6.5k^2 - 12k + 16 < 0. The minimum is attained when (x - 2.5k) = 0 and (z - 0.5k)^2 = 0.