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A
aayyappan1-Visitor
1-Visitor
March 6, 2016
Question

Minimization

  • March 6, 2016
  • 1 reply
  • 3229 views

I have created a worksheet for minimization with 9 variational parameter. The running time of the calculation is seems to more than 16 hours. Is it  possible to reduce the  calculation time. I  have attached the worksheet.

1 reply

W
Werner_E25-Diamond I
25-Diamond I
March 6, 2016

Your sheet throws errors and does not run at all.

Your definition of L is incomplete, so L is undefined and in the following calculations L is interpreted as unit "liter". Thats the reason Mathcad thinks that the upper limit i should be dimension volume.

As you write that the sheet worked for you it looks like you sent a wrong version of your  sheet.

Concerning the speed I guess the reason is not the number of variables but the multiple integrals. Every one of them is calculated by an iteration process and this is done for every single change the solve block makes with your variables. As you have nested/multiple integrals, the number of iteration steps is multiplied, not just added which makes for a huge number of iteration steps for just one single evaluation.

So if you are able to reduce/replace those integrals (maybe by a symbolic solution) would help to speed up calculation.

A
aayyappan1-VisitorAuthor
1-Visitor
March 8, 2016

Sir,

Thank you for your kind reply. I have attached the final worksheet.

As you told, I tried for symbolic evaluation. It also having lot of time, still i didn't get answer.

For example, the following integral not evaluate symbolically.  Kindly help me please........

Capture.PNG

W
Werner_E25-Diamond I
25-Diamond I
March 9, 2016

I can see no way to speed up calculation unless you manage somehow to get rid of or at least reduce those multiple integrals.

For example, the following integral not evaluate symbolically.  Kindly help me please........

This is because of the absolute value you use. Mathcad is even able to find the indefinite integral

which would be all you need anyway, but it fails when you add limits. Nevertheless you could insert those limits in the symbolic result of the indefinite integral yourself.

But things are even simpler because you don't need those absolute values - you are squaring so the sign does not matter:

WE

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