Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2003
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2003

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: A question on interval arithmetic

  • To: mathgroup at smc.vnet.net
  • Subject: [mg43747] Re: A question on interval arithmetic
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Thu, 2 Oct 2003 02:52:35 -0400 (EDT)
  • Organization: The University of Western Australia
  • Sender: owner-wri-mathgroup at wolfram.com

In article <blcq5l$p5f$1 at smc.vnet.net>,
 Oliver Friedrich <oliver.friedrich at tzm.de> wrote:

> the resistance of 2 resistors in parallel is r1*r2/(r1+r2). Now I want to
> introduce tolerances in the resistors and ask for the range of resistance
> of the combination. One may think that e.g
>
> (r1*r2)/(r1+r2)/.{r1->Interval[{10,20}],r2->Interval[{20,40}]}
>
> would lead to the correct result, but there's a trap. If I replace the
> expressions by the intervals, Mathematica evaluates the new expression
> assuming that all four intervals are independant from each other. And that's not
> correct. Taken either the minimum or the maximum from a certain interval ,
> Mathematica should stick to that, because it is nonsense to take Min[r1] and
> Max[r1] within the same expression, r1 can have only one value at a time.
>
> How can I avoid this problem?

Further to my previous posting, here are some additional comments.

Define the resistance of 2 resistors in parallel:

  R[r1_,r2_] = 1/(1/r1+1/r2)

Note that Mathematica does not simplify this result to r1*r2/(r1+r2) and
that r1 and r2 only appear _once_ in the result. Hence

  R[Interval[{10,20}],Interval[{20,40}]]

works as you would like.

You can also compute the extremal values directly:

  NMinimize[{R[r1, r2], 10 < r1 < 20 && 20 < r2 < 40}, {r1, r2}]

  NMaximize[{R[r1, r2], 10 < r1 < 20 && 20 < r2 < 40}, {r1, r2}]

Alternatively, you could use a statistical approach (other
correspondents could easily improve on the following primitive effort)
assuming the resistor values are normally distributed:

  << "Statistics`

 r1 := Random[NormalDistribution[15,5]]
 r2 := Random[NormalDistribution[30,10]]

 Table[R[r1,r2], {10000}];

 {Mean[%], StandardDeviation[%]}

But calculus of errors is probably the best approach.

Cheers,
Paul

--
Paul Abbott                                   Phone: +61 8 9380 2734
School of Physics, M013                         Fax: +61 8 9380 1014
The University of Western Australia      (CRICOS Provider No 00126G)
35 Stirling Highway
Crawley WA 6009                      mailto:paul at physics.uwa.edu.au
AUSTRALIA            


  • Prev by Date: Re: A question on interval arithmetic
  • Next by Date: Re: Computational aspects of Galois theory?
  • Previous by thread: Re: A question on interval arithmetic
  • Next by thread: Re: A question on interval arithmetic