Re: A question on interval arithmetic

*To*: mathgroup at smc.vnet.net*Subject*: [mg43735] Re: A question on interval arithmetic*From*: "Steve Luttrell" <luttrell at _removemefirst_westmal.demon.co.uk>*Date*: Thu, 2 Oct 2003 02:51:50 -0400 (EDT)*References*: <blcq5l$p5f$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

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 >

**Follow-Ups**:**Re: Re: A question on interval arithmetic***From:*Daniel Lichtblau <danl@wolfram.com>