From: Charilaos Skiadas <cskiadas_at_gmail.com>

Date: Fri, 09 May 2008 06:18:37 -0400

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Fri 09 May 2008 - 10:22:25 GMT

Date: Fri, 09 May 2008 06:18:37 -0400

On May 9, 2008, at 5:39 AM, Dieter Menne wrote:

> Dr. Ottorino-Luca Pantani <ottorino-luca.pantani <at> unifi.it>

*> writes:
**>>
**>> Imagine that for a particular cuvette (I have 112 different
**>> cuvettes !!)
**>> you have to mix the following volumes of solution A, B, and C
**>> respectively.
**>>
**>> c(1803.02, 193.51, 3.47)
**>>
**>> Each solution is to be taken with 3 different pipettes (5000, 250
**>> and 10
**>> µL Volume max) and each of those delivers volumes in steps of 50
**>> µL, 5
**>> µL or 1µL, respectively
*

>> Since the above values would eventually become

*>>
**>> c(1800, 195, 3)
**>>
**>> it is then necessary to recalculate all the final concentrations
**>> of A, B and C, because the volumes are changed.
**>
**> A first guess would be
**>
**> a = c(1803.02, 193.51, 3.47)
**> round(a / 5)*5
**>
**> which gives
**>
**> 1805 195 5
*

If I understand the OP's question properly, the first value is to be a multiple of 50, the second a multiple of 5, and the third a multiple of 1. This can be done with this slight variation on the above theme:

a <- c(1803.02, 193.51, 3.47)

b <- c(50,5,1)

round(a/b) *b

> This is not exactly what you want, but it shows that the problem is

*> a bit
**> ill-defined. In the example you gave, why do you want 1800, and not
**> 1805, which
**> is possible with the pipettes? I assume that you laboratory
**> experience is
**> working in the background, telling you to stop pipetmanning when
**> you are close
**> to the result in some "percentage" feeling.
**>
**> Dieter
*

Haris Skiadas

Department of Mathematics and Computer Science
Hanover College

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