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
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