Difficult optimal cable cost analysis

[rollingstein]

2c+3d+zb=0

This one's wrong I think.

-2c+3d+zb=0

What's rref BTW?

 

[voko]

Nevertheless, how about this: The global minimum has to occur either at one of the local minima or at the boundary of the domain (including ±∞).

It is not correct to call limits at infinity global extrema, because the function never attains such values. It is not correct to call ±∞ a boundary, either.

In this case, analytical differentiation reveals only two optima.

These are not optima, these are critical points till you analyze them further.

The only other limit is x,z --> +∞ but that offers no better cost function (it blows up actually).

You are forgetting the boundary of the domain: 5x + z + 12 = 0.

At this point am I not assured that I have a global minimum? What makes your strategy necessary or is that only an alternate way of saying what I just said?

You have an open domain. You can show that the function is continuous and differentiable there. That still does not guarantee that a global minimum exists. You have to prove in one way or another that there is a global minimum. My approach is in a way simplest because it does not require any advanced technique - not even calculus - to prove that.

 

[algar32]

If I am just solving for b I don't understand why I can't use rref in the calculator? Can you confirm that the following is the correct system of equations is correct?

-2c -3d +xb=0
4c+4d+4b=981
2c+3d+zb=0

ussing rref: I am getting b to be equal to 2943/(x-5)*(z-2.4) . Is this what I am looking for (b in terms of x and z)?
Thanks.

Here is my matrix with the correct values:
-2c -3d +xb=0
4c+4d+4b=981
-2c+3d+zb=0
when I use rref I get : b = 2943/(5x+z+12)
Is this correct?


rref is reduced row echelon form

 

[voko]

What actually happens if we do let x,z lie where 5x + z + 12 =0.

The cables AC and AD lie in a certain plane. This plane is not vertical, so there is component of gravity normal to this plane, which must be balanced by cable AC; naturally, AC must not be in that plane, otherwise it won't be able to balance that force. The equation above is the equation of the line where that plane intersects with the plane y = 0, which is why AB must not satisfy it. If you do attach AC anywhere on that line, the system of the cables will not be able to maintain its angle to the vertical.

Is that a statically indeterminate structure?

It will also be statically indeterminate, but, more importantly, it won't be able to have the required configuration.

 

[voko]

when I use rref I get : b = 2943/(5x+z+12)
Is this correct?

Correct.

 

[algar32]

Correct.

So from this information how would I determine a value of x and z that optimizes costs? Thanks.

 

[voko]

Review the discussion from #18 on.

 

[rollingstein]

It is not correct to call limits at infinity global extrema, because the function never attains such values. It is not correct to call ±∞ a boundary, either.



These are not optima, these are critical points till you analyze them further.



You are forgetting the boundary of the domain: 5x + z + 12 = 0.



You have an open domain. You can show that the function is continuous and differentiable there. That still does not guarantee that a global minimum exists. You have to prove in one way or another that there is a global minimum. My approach is in a way simplest because it does not require any advanced technique - not even calculus - to prove that.

In hindsight, I hope it didn't sound like I was criticizing your approach, sorry! I'm only trying to understand all strategies! Thanks again for the exposition. :)

 

[algar32]

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.

Where does this part come from?


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

I see that you have squared each of the coordinates for b on top and divided by the denominator of what we solved for b. Why was that done? Could you explain this part? Thanks.

 

[voko]

In hindsight, I hope it didn't sound like I was criticizing your approach, sorry!

You don't have to be sorry, there was nothing wrong with your questions.

 

[voko]

Where does this part come from?(x^2 + z^2 +16)/(5*x+z+12) < k

I see that you have squared each of the coordinates for b on top and divided by the denominator of what we solved for b.

Recall how the cost function was defined. It was tension multiplied by length. What are these, now that b has been determined?

 

[rollingstein]

It is not correct to call limits at infinity global extrema, because the function never attains such values. It is not correct to call ±∞ a boundary, either.

True. OTOH it may still be needed to check those limits? Say I have a function with one local minima (m1) and one local maxima (M1). Yet it might slope down from M2 so that the limit at ∞ was a value lower than m1? In that case you get a range of values lower than m1 attained much before ∞ but which won't be captured by the second partial derivative approach.

These are not optima, these are critical points till you analyze them further.

Agreed. Sloppy notation. My bad. I should have written: "In this case, analytical differentiation reveals at most two optima."

You have an open domain. You can show that the function is continuous and differentiable there. That still does not guarantee that a global minimum exists. You have to prove in one way or another that there is a global minimum.

I didn't get that part. Say, I prove it is continous. and differentiable; then I obtain and test a finite number of optima that the partial derivative condition yields. Then I test any boundaries.

At this point, haven't I exhausted any possibility that there's any global minimum outside of the set I enumerated? Is there anything else that'd be needed?

 

[algar32]

Recall how the cost function was defined. It was tension multiplied by length. What are these, now that b has been determined?

The length should be sqrt(Sqrt(z^2+x^2)^2 +y^2)^2
I see that someone squared it to get length squared (x^2+z^2+16)... I don't know why this was done.

The tension on b we know to be 2943/(5x+z+12).

So in my mind the length*tension would just be:

(sqrt(Sqrt(z^2+x^2)^2 +y^2)^2)*2943/(5x+z+12)

 

[rollingstein]

The length should be sqrt(Sqrt(z^2+x^2)^2 +y^2)^2
I see that someone squared it to get length squared (x^2+z^2+16)... I don't know why this was done.

I did it to make life easier. r^2 should attain a minimum at the same spot r does for r>0.

Don't know how rigorous that isl.

 

[algar32]

I did it to make life easier. r^2 should attain a minimum at the same spot r does for r>0.

Don't know how rigorous that isl.

Thanks. That is what I figured, but would you have to square the tensions as well if you were to do that?

also where does the 2943 go when you multiply the length and tension. It appears to just disappear...

 

[voko]

At this point, haven't I exhausted any possibility that there's any global minimum outside of the set I enumerated? Is there anything else that'd be needed?

Having all that done, you have indeed exhausted all the possibilities.

 

[voko]

The length should be sqrt(Sqrt(z^2+x^2)^2 +y^2)^2
I see that someone squared it to get length squared (x^2+z^2+16)... I don't know why this was done.

The tension on b we know to be 2943/(5x+z+12).

$b$ is not the tension. Recall how it is related to tension.

 

[algar32]

$b$ is not the tension. Recall how it is related to tension.

b is a function of x and z. I don't recall it's relation to tension.

Sorry this is taking me so long to get :(. Thanks for your help.

 

[voko]

b is a function of x and z. I don't recall it's relation to tension.

Review #7.

 

[rollingstein]

I did it to make life easier. r^2 should attain a minimum at the same spot r does for r>0.

Don't know how rigorous that isl.

In hindsight I was wrong. That is not the reason for " someone squared it to get length squared ".

I was just lucky! Two wrongs made a right!

 

[algar32]

Review #7.

It just says b is some constant. If b is some constant that is multiplied by the tension, I still need the actual tension of the cable to multiply by the length.

 

[algar32]

In hindsight I was wrong. That is not the reason for " someone squared it to get length squared ".

I was just lucky! Two wrongs made a right!

Okay thanks :)

 

[rollingstein]

Okay thanks :)

It got squared because the other identical square root came from the tension term.

I mistakenly assumed all other than b was a unit vector. It was not.

 

[algar32]

It got squared because the other identical square root came from the tension term.

I mistakenly assumed all other than b was a unit vector. It was not.

so now I have an equation for

Length x tension

(sqrt(Sqrt(z^2+x^2)^2 +y^2)^2)*tension

I am told b is not equal to tension, but is related and I was never quite sure of the relation. B is just some made up positive constant. So how can I find tension from what I have done thus far? If I can find tension all I will have to do left is do the minimization of Lenght x tension. Thanks.

 

[rollingstein]

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.

On further thought, if b>0 why do we not need to also ensure that c>0 and d>0?

They all being chains violation of these constraints would cause one of the other chains to go slack too?

Or are those constraints automatically met?

Shouldn't we be adding and testing boundaries like x+z=0 and x-z=0? Maybe I was just lucky with that too!

 

[rollingstein]

I am told b is not equal to tension, but is related and I was never quite sure of the relation. B is just some made up positive constant. So how can I find tension from what I have done thus far?

Think harder.

 

[algar32]

On further thought, if b>0 why do we not need to also ensure that c>0 and d>0?

They all being chains violation of these constraints would cause one of the other chains to go slack too?

Or are those constraints automatically met?

Shouldn't we be adding constraints like x+z=0 and x-z=0?

Is it only for b simply because that is the only cable we care to optimize costs for in this problem?

 

[algar32]

Think harder.

Not sure this helped :(. Could you point me in the right direction?

 

[rollingstein]

Not sure this helped :(. Could you point me in the right direction?

Formula for Magnitude of a vector? Tension's a vector.

 

[algar32]

Formula for Magnitude of a vector? Tension's a vector.

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.