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Mathematical Impossibility?

Mathematical Impossibility?

Mathematical Impossibility?

Is this scenario really mathematically impossible or am I entering my information incorrectly into MathCad?

RE: Mathematical Impossibility?

divide eq 2 by eq 1, b=2, divide eq 4 by eq 3, b= 1.333

so yes you are asking a fairly tricky question. Try replacing find by minerr, it will then try to satisfy your contradictory equations as best it can.



Greg Locock

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RE: Mathematical Impossibility?

Thank you.

This gets me closer to what I am looking for but is there a way to get these answers to converge. As I change the guesses the answers jump around. I am not looking for a exact match to the original 5,10,15,20 pattern but some way to get them close or even averaged between the two based on the 2/1.3333 disagreement.

RE: Mathematical Impossibility?

The only possibility I see for this set of equations to make sense would perhaps be complex numbers... Try solving it that way, you might get a result (but then it's 8 equations with 8 unknowns). HTH

RE: Mathematical Impossibility?

If changing the guesses changes the answer your solve block doesn't converge or you're not restraining your system enough.

Instead of banging your head against a solve block in MathCAD on this one try using a piece of paper (Try graphing it - in log scale you have 4 lines that intersect at 4 separate points... the issue is non convergence), if you can't rationalize out a path to a solution chances are you need more information, more assumptions, or a different problem.

RE: Mathematical Impossibility?

I'm with Greg and IRstuff. The equations contradict themselves.... no solution exists.

If you want to do some kind of "fit", you have to define an objective function to decide what constitutes a best match. I doubt that would be productive from what little I know about the problem... certainly looks like a textbook problem aimed to get you to use your noggin rather than blindly plugging and chugging.  

(2B)+(2B)'  ?

RE: Mathematical Impossibility?


I am not looking for a exact match to the original 5,10,15,20 pattern but some way to get them close or even averaged between the two based on the 2/1.3333 disagreement.
Sorry if I misunderstood in my previous post.

You have a non-linear optimization problem. SRSS of residual errors is a traditional objective function.  It is to be expected that different starting points can lead to different solutions (which may be local minimum).  Solution is not necessarily easy.

(2B)+(2B)'  ?

RE: Mathematical Impossibility?

Attached is an analysis using mupad.

We used SRSS of residual errors as objective function.

Taking derivative of objective function with respect to each of the four independent variables (a,b,x,y) and solving for locations where they are all zero reveals  local minima, local maxima, and saddle points.

Two solutions are found:

2 classes of local minima:
Solution 1:[a = (3315*z - 205*221^(1/2)*z)/(442*z^2), b = (2210*z - 155*221^(1/2)*z)/(221*z^2), x = - (10*z)/11 - (221^(1/2)*z)/11, y = z]

Solution 2: [a = (3315*z + 205*221^(1/2)*z)/(442*z^2), b = (2210*z + 155*221^(1/2)*z)/(221*z^2), x = (221^(1/2)*z)/11 - (10*z)/11, y = z]

Both solutions are parameterized by lower case "z".  Experimentation revealed objective function does not depend on z, so we are free to choose any value and the solution is equally good wrt objective function.  We choose z=1, which gives:

solution 1: [a = 0.6051038615, b = -0.4264283069, x = -2.260551704, y = 1.0]
Objective function for soln1 is 746.6.

solution 2: [a = 14.39489614, b = 20.42642831, x = 0.4423698861, y = 1.0]
Objective function for soln2 is 3.348281317

Plotting reveals these are both local minima (*). Conclude solution 2 is sort of a global minimum (*).

I have perhaps misused the terms local and global minimum here since each solution refers not just to a single point but an infinite number of points generated by picking different values of z (all give the same objective function).

There weird behavior of having many different solutions with same objective function may be a reason this one was particularly difficult to solve using brute force trial and error.

Seems like my attachment is having a problem uploading. I will post and try again. If not will come back later with attachment.

(2B)+(2B)'  ?

RE: Mathematical Impossibility?

File is too big. 11.4 mb including some 2-d graphs. I'll have to delete some graphs.

(2B)+(2B)'  ?

RE: Mathematical Impossibility?

Quote (electricpete):

There weird behavior of having many different solutions with same objective function may be a reason this one was particularly difficult to solve using brute force trial and error.
Just to elaborate, numerical solutions will lead you to many solutions with objective functions near/equal to that minimum 3.348281317. But this minimum objective function exists at many different coordinates (a,b,x,y).  So numerical solution will not ocnverge to one value of (a,b,x,y) since there are many solutions equally good.


(2B)+(2B)'  ?

RE: Mathematical Impossibility?


Experimentation revealed objective function does not depend on z
Actually I should have known this without experimenting.  Mupad identified that z can lie anywhere in C (C is the set of complex numbers). So it is an arbitrary consant.... any choice of z is still a solution.


(2B)+(2B)'  ?

RE: Mathematical Impossibility?

Another aspect of the solution.

Within the set of solutions identified as "solution 2" (which attain the global min value of objective fnction =3.348), as we change the value of the arbitrary parameter z, the values of a, b, x, y all change, but the values of the products a*x, b*x, a*y, b*y do not.

Exp    Computed Value    Desired Value
=a*x    6.367868565    5
=b*x    9.036036764    10
=a*y    14.39489614    15
=b*y    20.42642831    20

This makes sense insce inspection of the algebraic solution shows that x and y are proportional to z and a and b are inversely proportional to z
x = Cx*z
y = Cy*z
a = Ca/z
b = Cb/z
ax = Cx*Ca ... does not depend on z. Same for ay, bx, by.

So, all numerical solutions should  converge to the same products ax, ay, bx, by listed above, but not to the same values a, b, x, y.

(2B)+(2B)'  ?

RE: Mathematical Impossibility?

Based on the last post, I think we we can develop an intuitive understanding for why there are infinite number of best-fit solutions, rather than a single one.

The performance specification (or objective function) depends only on products containing one item from the set (a,b) times one item  from the set (x,y).  That can be seen from the original attachment where we only impose specifications on ax, ay, bx, by.

So it should not come as any suprise that for any set (a0, b0, x0, y0) that meets the performance specification to a certain level, we can develop another set  (a0/z, b0/z, x0*z, y0*z) which meets the performance specification in exactly the same manner.

So, by inspecting the original problem to begin with, we should have been able to anticipate the behavior that mupad told us.... that we have not one best-fit solution but a parameterized set (a0/z, b0/z, x0*z, y0*z) for arbitrary parameter z.


(2B)+(2B)'  ?

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