Difficult optimal cable cost analysis

[algar32]

Homework Statement

http://snag.gy/z4LzZ.jpg

I was going through some old course work from when I was an undergrad and was unable to complete this one. If you could just provide me with the proper approach/steps it would be much appreciated. I know how to solve all of the tensions, but the cost analysis approach is confusing me.

Thanks

 

[tiny-tim]

hi algar32!

… I know how to solve all of the tensions, but the cost analysis approach is confusing me.

you should get an expression for the length, and an expression for the tension

multiply them together, that gives you the cost (as a function of x y and z, the coordinates of B)

then minimise it, probably using calculus …

show us what you get ​

 

[algar32]

hi algar32!


you should get an expression for the length, and an expression for the tension

multiply them together, that gives you the cost (as a function of x y and z, the coordinates of B)

then minimise it, probably using calculus …

show us what you get ​

How do I get an expression for length and an expression for tension? That is the part I am struggling with. Also, if I make the tension expression won't it have 2 variables in it? How can I handle that if I am trying to rref it? Any further steps/work would be appreciated.Thanks.

 

[SteamKing]

Well, the expression for the length is going to be a function of the unknown x and z position of the point B. You may not know what else if you don't write some equations.

 

[algar32]

Well, the expression for the length is going to be a function of the unknown x and z position of the point B. You may not know what else if you don't write some equations.

Here is my work thus far

http://i954.photobucket.com/albums/ae30/Algar32/20130201_201135.jpg

http://i954.photobucket.com/albums/ae30/Algar32/20130201_201035.jpg

http://i954.photobucket.com/albums/ae30/Algar32/20130201_201154.jpg

 

[tiny-tim]

??

can you type it out, please?​

 

[voko]

You are making it more difficult than it has to be. You know that the forces of tension in the cables act along the cables, so $\vec{T}_{AB} = b \vec {AB} = b(x, 4, z)$, $\vec{T}_{AC} = c \vec {AC} = c(-2, 4, -2)$, $\vec{T}_{AD} = d \vec {AD} = d(-3, 4, 3)$, where $b$, $c$ and $d$ are some positive constants. Letting $\vec{T} = (0, -mg, 0)$, we must have $\vec{T}_{AD} + \vec{T}_{AC} + \vec{T}_{AB} + \vec{T} = 0$. Write this equation component-wise, and use Gauss's elimination to determine b in terms of x and z.

 

[algar32]

You are making it more difficult than it has to be. You know that the forces of tension in the cables act along the cables, so $\vec{T}_{AB} = b \vec {AB} = b(x, 4, z)$, $\vec{T}_{AC} = c \vec {AC} = c(-2, 4, -2)$, $\vec{T}_{AD} = d \vec {AD} = d(-3, 4, 3)$, where $b$, $c$ and $d$ are some positive constants. Letting $\vec{T} = (0, -mg, 0)$, we must have $\vec{T}_{AD} + \vec{T}_{AC} + \vec{T}_{AB} + \vec{T} = 0$. Write this equation component-wise, and use Gauss's elimination to determine b in terms of x and z.

Okay thanks... Is this what you mean?

b(x,4,z) + c(-2,4,2)+d(-3,4,3)+(0,-981,0)=0


I don't see how I can solve that for x and z? Could you show this process? Was I supposed to put in any number (let's say 1) for b, c and d?

By doing this I will have the most cost effective rope based on the equation givin in the problem (length*tension=cost)?

Thanks.

 

[voko]

You don't solve for x and z. You solve just for b as a function of x and z.

 

[algar32]

You say to use gaussian elimination, so I have just been setting up a few matrices and plugging them into the calculator and using rref (reduced row ef). I am still not getting a value for x and z. Would it be possible to show me what matrix I should be using? thanks.

 

[algar32]

You don't solve for x and z. You solve just for b as a function of x and z.

I am unsure how to do this. Thanks.

 

[voko]

b(x,4,z) + c(-2,4,2)+d(-3,4,3)+(0,-981,0)=0 is a system of three linear equations, where the unknowns are b, c and d (assume x and z are known). Use Gauss's elimination to find b, it will be a function of x and z.

 

[voko]

Just to be clear. This has to be done manually, calculators won't do it. I recommend rewriting the equation as c(-2,4,2) + d(-3,4,3) + b(x,4,z) = (0,981,0), so that the b-terms are in the right-most column of the matrix, that will simplify finding b, requiring just two Gaussian steps.

 

[algar32]

Just to be clear. This has to be done manually, calculators won't do it. I recommend rewriting the equation as c(-2,4,2) + d(-3,4,3) + b(x,4,z) = (0,981,0), so that the b-terms are in the right-most column of the matrix, that will simplify finding b, requiring just two Gaussian steps.

Thanks. Here is my work:

x 4 z | 0 b
-2 4 2 | -981 c
-3 4 3 | 0 d

(3c/2)-d

x 4 z | 0 b
0 2 0 | -1471.5 c
-3 4 3 | 0 d

b - 2c

x 0 z | 2943 b
0 2 0 | -1471.5 c
-3 4 3 | 0 d

(x + z) b = 2943

What would my final answer be to optimize the costs?

 

[voko]

Even assuming that the system is written down properly, I do not see how your steps would result in anything. I think you should review systems of linear equations.

Re writing down the system, observe that the equation c(-2,4,2) + d(-3,4,3) + b(x,4,z) = (0,981,0) is a vector equation; the entities in brackets are vectors, while the coefficients in front of them are numbers. A vector equation means that the same equation holds for each of its components. For example, taking the first component (X-direction), we obtain -2c - 3d + xb = 0. This is a linear equation for c, d, and b (x is assumed known). Writing the equations for the two other components, we obtain a system of three linear equations.

 

[rollingstein]

One interesting outcome is x and z are always related by

x=-z

Unless having a tensionless chain is ok.

 

[voko]

I just noticed that in #8 and later the equation is not correct. It should be c(-2,4,-2) + d(-3,4,3) + b(x,4,z) = (0,981,0).

 

[rollingstein]

I just noticed that in #8 and later the equation is not correct. It should be c(-2,4,-2) + d(-3,4,3) + b(x,4,z) = (0,981,0).

Yup. That was key. Now the weird x=-z constraint disappears.

We get:

Length^2 = x^2 + z^2 +16

Therefore we minimise the product:

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

(x,z)=(2.243,0.449)

Not sure if this is correct.
Of course, I cheated using Wolfram. Analytically is another challenge. I'd love to see that minimization done by hand.

 

[rollingstein]

Not even sure how one could do those minimizations by hand. It'd require solving simultanously two quadratics in x and z.

 

[voko]

That result is correct. Of course, the report requested must prove that this really is the point resulting in a minimal cost.

 

[rollingstein]

Not even sure how one could do those minimizations by hand. It'd require solving simultanously two quadratics in x and z.

On deeper thought doing it by hand isn't hard.

We can show x=5z pretty easily and that leads to a quadratic in z. Not so hard.

Nice problem!

 

[rollingstein]

That result is correct. Of course, the report requested must prove that this really is the point resulting in a minimal cost.

What do you mean? Proving the sufficiency? The Hessian test?

 

[voko]

The Hessian test would only prove that a local minimum exists. It says nothing whether a global minimum exists, which is what really is required. In fact, the cost function you found has no global minimum, because it can be trivially shown that it goes to negative infinity in certain directions. So the question remains open.

 

[pongo38]

The calculation methods already discussed are error prone. Why not fix z, and then try different values of x and plot x v. cost. If you can do it in 3 dimensions, plot x v. z v. cost. Once you have say six points, you will probably home in on a solution that is sufficiently precise.

 

[rollingstein]

The Hessian test would only prove that a local minimum exists. It says nothing whether a global minimum exists, which is what really is required. In fact, the cost function you found has no global minimum, because it can be trivially shown that it goes to negative infinity in certain directions. So the question remains open.

You are right. The local minimum of (x^2 + z^2 +16)/(5*x+z+12) at (x,z)=(2.243,0.449) is equal to 0.897.

Question is whether we can do better.

I think your - Infinity limits are not valid; we ought to add a b>0 constraint. You can only go to -ive values of my cost function by violating the b>0 constraint. Chains can only support tension not compression. Unless making the chain redundant is acceptable. I'm not sure. At that point might as well get rid of said chain?

What do you think?

OTOH, is there a point where the cost function is greater than zero yet less than 0.897? I don't know.

 

[voko]

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.

 

[rollingstein]

The calculation methods already discussed are error prone. Why not fix z, and then try different values of x and plot x v. cost. If you can do it in 3 dimensions, plot x v. z v. cost. Once you have say six points, you will probably home in on a solution that is sufficiently precise.

I don't see how your approach is better; isn't that just a brute force grid search?

 

[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.

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.

I'm still trying to understand your approach.

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 ±∞). In this case, analytical differentiation reveals only two optima. One of them being the minima I reported; other one is a maxima (and anyways outside the b>0 region).

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

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?

Does this make sense? Or am I missing possibilities?

 

[rollingstein]

Another aside that puzzles me:

Note that 5x + z + 12 =0 is forbidden by the solution (even prior to any optimization).

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

Say, x=0 z=-12

Is that a statically indeterminate structure? Or...?

 

[algar32]

Even assuming that the system is written down properly, I do not see how your steps would result in anything. I think you should review systems of linear equations.

Re writing down the system, observe that the equation c(-2,4,2) + d(-3,4,3) + b(x,4,z) = (0,981,0) is a vector equation; the entities in brackets are vectors, while the coefficients in front of them are numbers. A vector equation means that the same equation holds for each of its components. For example, taking the first component (X-direction), we obtain -2c - 3d + xb = 0. This is a linear equation for c, d, and b (x is assumed known). Writing the equations for the two other components, we obtain a system of three linear equations.

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.

 

[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!