       Re: A question on interval arithmetic

• To: mathgroup at smc.vnet.net
• Subject: [mg43760] Re: A question on interval arithmetic
• From: "Steve Luttrell" <luttrell at _removemefirst_westmal.demon.co.uk>
• Date: Fri, 3 Oct 2003 02:28:46 -0400 (EDT)
• References: <blcq5l\$p5f\$1@smc.vnet.net> <blgjf4\$i1d\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Groan! I didn't read the question properly. My suggestion computed the
expected value (mean) of the parallel resistance. I see now that all you
wanted was the "range of resistance of the combination", like everyone else
noticed first time around.

Apologies for the confusion.

--
Steve Luttrell
West Malvern, UK

"Steve Luttrell" <luttrell at _removemefirst_westmal.demon.co.uk> wrote in
message news:blgjf4\$i1d\$1 at smc.vnet.net...
> You need to average  (r1*r2)/(r1+r2) over the required interval.
>
> Here is how I did it evaluating over the 2-dimensional tolerance region
> around r1=r10 and r2=r20. I assume each resistor independently has a
> tolerance +/-a which is uniformly distributed over an interval of length
2a,
> which gives a probability density 1/(4a^2) over the 2-dimensional
tolerance
> region for the pair of resistors. I had to evaluate indefinite integrals
and
> then substitute in limits because definite integration seemed to lock
> Mathematica up (I have version 5 for Windows).
>
> Integrate[((r1*r2)/(r1 + r2))*(1/(4*a^2)), r1]
> Simplify[(% /. {r1 -> r10 + a}) - (% /. {r1 -> r10 - a})]
> Integrate[%, r2]
> Simplify[(% /. {r2 -> r20 + a}) - (% /. {r2 -> r20 - a})]
>
> --
> Steve Luttrell
> West Malvern, UK
>
> "Oliver Friedrich" <oliver.friedrich at tzm.de> wrote in message
> news:blcq5l\$p5f\$1 at smc.vnet.net...
> > Hallo,
> >
> > 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?
> >
> > Oliver Friedrich
> >
>
>

```

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