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Wolfram Language Optimization

Solving 2 unknowns by using 12 equations

INTAN SUPRABA

INTAN SUPRABA, Hokkaido University

Posted 11 years ago

Dear All, Usually we can obtain n unknowns by using n equations. But I want to obtain the best fit 2 unknowns for 12 equations. I tried to use below command: FindRoot[{a b Sech[266 b] == 0.391, a b Sech[128.3 b] == 0.3169, a b Sech[158.5 b] == 0.3969, a b Sech[240.25 b] == 0.3117, a b Sech[156.8 b] == 0.6311, a b Sech[136.6 b] == 0.2162, a b Sech[363.95 b] == 0.2727, a b Sech[106.35 b] == 0.3987, a b Sech[54.3 b] == 0.3596, a b Sech[127.9 b] == 0.5211, a b Sech[179.7 b] == 0.27, a b Sech[105.3 b] == 0.889}, {{a, 110}, {b, 0.007}}] But it didnt work. Then I tried the below command: NSolve[{a b Sech[266 b] == 0.391, a b Sech[128.3 b] == 0.3169, a b Sech[158.5 b] == 0.3969, a b Sech[240.25 b] == 0.3117, a b Sech[156.8 b] == 0.6311, a b Sech[136.6 b] == 0.2162, a b Sech[363.95 b] == 0.2727, a b Sech[106.35 b] == 0.3987, a b Sech[54.3 b] == 0.3596, a b Sech[127.9 b] == 0.5211, a b Sech[179.7 b] == 0.27, a b Sech[105.3 b] == 0.889}, {a, b}, Reals] But it also didnt work. So is that possible to obtain a and b by using 12 equations? Thanks for your attention. Best Regards, Intan

Dear All,

Usually we can obtain n unknowns by using n equations. But I want to obtain the best fit 2 unknowns for 12 equations. I tried to use below command:

FindRoot[{a b Sech[266 b] == 0.391, a b Sech[128.3 b] == 0.3169, a b Sech[158.5 b] == 0.3969, a b Sech[240.25 b] == 0.3117,
a b Sech[156.8 b] == 0.6311, a b Sech[136.6 b] == 0.2162, a b Sech[363.95 b] == 0.2727, a b Sech[106.35 b] == 0.3987, 
a b Sech[54.3 b] == 0.3596, a b Sech[127.9 b] == 0.5211, a b Sech[179.7 b] == 0.27, a b Sech[105.3 b] == 0.889},
{{a, 110}, {b, 0.007}}]

But it didnt work. Then I tried the below command:

NSolve[{a b Sech[266 b] == 0.391, a b Sech[128.3 b] == 0.3169,  a b Sech[158.5 b] == 0.3969, a b Sech[240.25 b] == 0.3117, 
a b Sech[156.8 b] == 0.6311, a b Sech[136.6 b] == 0.2162, a b Sech[363.95 b] == 0.2727,  a b Sech[106.35 b] == 0.3987, 
a b Sech[54.3 b] == 0.3596, a b Sech[127.9 b] == 0.5211, a b Sech[179.7 b] == 0.27, a b Sech[105.3 b] == 0.889}, {a, b}, Reals]

But it also didnt work. So is that possible to obtain a and b by using 12 equations? Thanks for your attention. Best Regards, Intan

POSTED BY: INTAN SUPRABA

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Bill Simpson

Posted 11 years ago

Perhaps this? In[1]:= sol=NMinimize[{ (a b Sech[266 b] - 0.391)^2 + (a b Sech[128.3 b] - 0.3169)^2 + (a b Sech[158.5 b] - 0.3969)^2 + (a b Sech[240.25 b] - 0.3117)^2 + (a b Sech[156.8 b] - 0.6311)^2 + (a b Sech[136.6 b] - 0.2162)^2 + (a b Sech[363.95 b] - 0.2727)^2 + (a b Sech[106.35 b] - 0.3987)^2 + (a b Sech[54.3 b] - 0.3596)^2 + (a b Sech[127.9 b] - 0.5211)^2 + (a b Sech[179.7 b] - 0.27)^2 + (a b Sech[105.3 b] - 0.889)^2, a > 100}, {a, b}] During evaluation of In[1]:= NMinimize::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. >> Out[1]= {0.342799, {a -> 141.072, b -> 0.00350932}} (* See how much error there is *) In[2]:= {a b Sech[266 b] - 0.391, a b Sech[128.3 b] - 0.3169, a b Sech[158.5 b] - 0.3969, a b Sech[240.25 b] - 0.3117, a b Sech[156.8 b] - 0.6311, a b Sech[136.6 b] - 0.2162, a b Sech[363.95 b] - 0.2727, a b Sech[106.35 b] - 0.3987, a b Sech[54.3 b] - 0.3596, a b Sech[127.9 b] - 0.5211, a b Sech[179.7 b] - 0.27, a b Sech[105.3 b] - 0.889} /.sol[[2]] Out[2]= {-0.0538205, 0.131903, 0.0303526, 0.0478318, -0.202564, 0.226966, -0.01655, 0.0637825, 0.126613, -0.0720312, 0.140667, -0.425912}

Perhaps this?

In[1]:= sol=NMinimize[{
 (a b Sech[266 b] - 0.391)^2 + (a b Sech[128.3 b] - 0.3169)^2 + (a b Sech[158.5 b] - 0.3969)^2 +
 (a b Sech[240.25 b] - 0.3117)^2 + (a b Sech[156.8 b] - 0.6311)^2 + (a b Sech[136.6 b] - 0.2162)^2 +
 (a b Sech[363.95 b] - 0.2727)^2 + (a b Sech[106.35 b] - 0.3987)^2 + (a b Sech[54.3 b] - 0.3596)^2 +
 (a b Sech[127.9 b] - 0.5211)^2 + (a b Sech[179.7 b] - 0.27)^2 + (a b Sech[105.3 b] - 0.889)^2, a > 100}, {a, b}]

During evaluation of In[1]:= NMinimize::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. >>

Out[1]= {0.342799, {a -> 141.072, b -> 0.00350932}}

(* See how much error there is *)

In[2]:= {a b Sech[266 b] - 0.391, a b Sech[128.3 b] - 0.3169, a b Sech[158.5 b] - 0.3969,
 a b Sech[240.25 b] - 0.3117, a b Sech[156.8 b] - 0.6311, a b Sech[136.6 b] - 0.2162, 
  a b Sech[363.95 b] - 0.2727, a b Sech[106.35 b] - 0.3987, a b Sech[54.3 b] - 0.3596,
 a b Sech[127.9 b] - 0.5211, a b Sech[179.7 b] - 0.27, a b Sech[105.3 b] - 0.889} /.sol[[2]]

Out[2]= {-0.0538205, 0.131903, 0.0303526,  0.0478318, -0.202564, 0.226966,
 -0.01655, 0.0637825, 0.126613, -0.0720312, 0.140667, -0.425912}

POSTED BY: Bill Simpson

INTAN SUPRABA

INTAN SUPRABA, Hokkaido University

Posted 11 years ago

Dear Bill, Thanks a lot for your answer, it really helps! I tried it out and yes, I got the same results as yours. I just know about the function of this NMinimize command, so may I ask you further questions as follows: a. The function of this NMinimize command is to find the global minimum of problems with/without constraints. In this case, the constraint is a > 100. If I added one more constraint for the b, I obtained the values of a and b are similar to the threshold itself. In[1] :=solution = NMinimize[{(a b Sech[266 b] - 0.391)^2 + (a b Sech[128.3 b] - 0.3169)^2 + (a b Sech[158.5 b] - 0.3969)^2 + (a b Sech[240.25 b] - 0.3117)^2 + (a b Sech[156.8 b] - 0.6311)^2 + (a b Sech[136.6 b] - 0.2162)^2 + (a b Sech[363.95 b] - 0.2727)^2 +(a b Sech[106.35 b] - 0.3987)^2 + (a b Sech[54.3 b] - 0.3596)^2 + (a b Sech[127.9 b] -0.5211)^2 + (a b Sech[179.7 b] - 0.27)^2 + (a b Sech[105.3 b] - 0.889)^2, {a > 100, b > 0.006}}, {a, b}] Out[1] ={0.387513, {a -> 100.576, b -> 0.006}} Similarly if I changed the constraints into a > 110 and b > 0.006, then the obtained a and b become 110 and 0.006: In[2] :=solution =NMinimize[{(a b Sech[266 b] - 0.391)^2 + (a b Sech[128.3 b] - 0.3169)^2 + (a b Sech[158.5 b] - 0.3969)^2 + (a b Sech[240.25 b] - 0.3117)^2 + (a b Sech[156.8 b] - 0.6311)^2 + (a b Sech[136.6 b] - 0.2162)^2 + (a b Sech[363.95 b] - 0.2727)^2 + (a b Sech[106.35 b] - 0.3987)^2 + (a b Sech[54.3 b] - 0.3596)^2 + (a b Sech[127.9 b] - 0.5211)^2 + (a b Sech[179.7 b] - 0.27)^2 + (a b Sech[105.3 b] - 0.889)^2, {a > 110, b > 0.006}}, {a, b}] Out[2] ={0.405625, {a -> 110., b -> 0.006}} So does it mean that actually using 2 constraints for solving 2 unknows is not acceptable? b. About the error/ discrepancies between the real solution and the estimated solution, is that possible to add on the boundary condition for the error such as defined the error < 0.001? c. About the command itself, the equations were rearranged by moving the rhs to the lhs for each equations then sum up all of those equations. Is it the fixed rule for finding global solution for several equations? Then how about the meaning of square/quadratic of each equation? Is there any reference to further understand about this command? I only know the reference from this link: http://reference.wolfram.com/mathematica/ref/NMinimize.html?q=NMinimize&lang=en Thanks in advance for your reply. Best Regards, Intan

POSTED BY: INTAN SUPRABA

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 11 years ago

When searching for "best approximation" solutions to overdetermnied systems of equations, a fairly common tactic is to minimize a sum of squares. This is not the only approach but it is perhaps the one most commonly used. As for the results in the constrained examples, it just means NMinimize is finding its best values on the constraint boundary.