Wikipedia:Reference desk/Archives/Mathematics/2007 January 16

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[edit] January 16

[edit] Inverse of Erlang C formula

The Erlang C formula describes probabilities of queueing in an Automatic call distributor environment. If you know the total traffic A in Erlang_units and the number of agents N taking calls, you can calculate the probability PW of a call having to queue (making some assumptions about random distribution of calls, etc.) using the formula:

P_W = {{\frac{A^N}{N!} \frac{N}{N - A}} \over \sum_{x=0}^{N-1} \frac{A^i}{i!} + \frac{A^N}{N!} \frac{N}{N - A}} \,

I'm looking to find the inverse of the formula, so that if I know A, and know the maximum accepatble value of PW, I can calculate the lowest value of N - can the formula be reversed to do this. So far I've been using trial and improvement, which is rather tedious - unfortunately the maths needed to reverse this is beyond me.

Many thanks, Davidprior 18:45, 16 January 2007 (UTC)

This isn't an answer, but I managed to simplify the expression to P_W^{-1}=\sum_{i=1}^N \frac{i(N)_i}{NA^i} (note the inversion and see falling factorial). That should make evaluation a lot less tedious; beyond that, all I can suggest at the moment is to do a binary search — if your PW is very close to 0 or 1 you should try starting on an interval like [A,4A] or [1,A] respectively. Meanwhile, if someone could check my math (I can of course provide the derivation if needed), we can simplify the expression in the article a lot. --Tardis 21:54, 16 January 2007 (UTC)
Thanks for this, I think I'll have a go at putting together a wee program (for the first time in many years) to solve this for the various inputs I need to think about. As an aside, you mention your working - I don't know enough to check it, but would be interested in seeing as a learning excercise - if you could pop it here, or on my userpage (in wikitext, or anything that OpenOffice.org Math can read) I'd very much appreciate it. Cheers, Davidprior 19:55, 20 January 2007 (UTC)
OK — posted here for you and for checking (by anyone interested):
P_W=\left(\frac{N!}{A^N}\frac{N-A}N\sum_{i=0}^{N-1}\frac{A^i}{i!}+1\right)^{-1}=\left(\left(1-\frac AN\right)\sum_{i=0}^{N-1}\frac{N!A^{i-N}}{i!}+1\right)^{-1} (divide numerator and denominator by numerator; write as new denominator to the -1; move multipliers into sum and distribute)
P_W^{-1}=\left(1-\frac AN\right)\sum_{i=1}^N\frac{N!}{A^i(N-i)!}+1=\left(1-\frac AN\right)\sum_{i=1}^N(N)_iA^{-i}+1 (move the -1, then do the hard bit: substitute i\rightarrow N-i in the sum, and start using the falling factorial)
\dots=\sum_{i=1}^N(N)_iA^{-i}-\sum_{i=1}^N(N-1)_{i-1}A^{-i+1}+1=\sum_{i=1}^N(N)_iA^{-i}-\sum_{i=0}^{N-1}(N-1)_iA^{-i}+1 (distribute again, noting that \frac{(N)_i}{(N-1)_{i-1}}=N; then shift indices by 1 to simplify summand)
\dots=\sum_{i=0}^N(N)_iA^{-i}-\sum_{i=0}^N(N-1)_iA^{-i} (match the indices: combine 1 into left sum to make its i = 0 term; the i = N term of the right sum is (N − 1)N = 0 anyway)
\dots=\sum_{i=0}^NA^{-i}\left((N)_i-(N-1)_i\right)=\sum_{i=0}^NA^{-i}(N)_i\left(1-\frac{N-i}N\right) (combine sums, then use \frac{(N)_i}{(N-1)_i}=\frac N{N-i} (compare to earlier falling factorial ratio formula))
The formula in my earlier post follows from this last line by simplifying the fraction. Hope this helps, --Tardis 17:22, 23 January 2007 (UTC)

[edit] Algebra

My question is Find the equation of a line through it's q points. I believe I have it but im pretty sure its wrong. The q points are (39, 157) and (52, 183) —The preceding unsigned comment was added by 80.148.21.51 (talk) 19:24, 16 January 2007 (UTC). OK i posted this question and yet no one can help me. . . Please I need this for a HW assignment due in the mornig. Help@@

First of all, this isn't really meant to do your homework for you, so I'll limit my answer to giving you pointers:
The general equation of the line is y=mx+c
m can be found by dividing the difference in the 2 y values by the difference in the 2 x values.
once you have m, the three known quantities (x,y&m - either value of x&y can be used, so long as they both refer to the same point) into the equation and solving to give c
The equation can be checked uing the other of the 2 known points on the line.
Cheers, Davidprior 19:51, 16 January 2007 (UTC)

Also, if you post your answer, we will tell you if it's right, and if you post your work, we will even point out any mistakes you may have made. StuRat 18:08, 17 January 2007 (UTC)

[edit] Asymptote article

"A curve may or may not touch or cross its asymptote. In fact, the curve may intersect the asymptote an infinite number of times." (Asymptote) What is that about?? Do I just not understand, or can somebody fix that? X [Mac Davis] (DESK|How's my driving?) 22:41, 16 January 2007 (UTC)

The picture (the one with the diagonal line and the swirl closing on it) seems to make it pretty clear to me, although I hadn't honestly thought of it before. It might help to consider normal functions instead of parametric ones: consider the sinc function with its asymptote at y = 0 in either direction, and the function \frac{\ln x}x (with the same asymptote on the right) for an example with a finite number of intersections (here, one). --Tardis 23:26, 16 January 2007 (UTC)
How can a line not be touched or crossed, but still intersected many times? X [Mac Davis] (DESK|How's my driving?) 00:22, 18 January 2007 (UTC)
Read your own quote again: "may or may not touch or cross"... --mglg(talk) 02:19, 18 January 2007 (UTC)
To my understanding, there are both vertical and horizontal asymptotes. I seem to recall that vertical ones aren't crossed, but the horizontal ones (usually x axis for trigonometric functions) are sometimes. hope this helps...--Root2 05:48, 18 January 2007 (UTC)
For the graph of a function, that's correct: if the function is going to approach a vertical asymptote, it doesn't have the option of crossing it and then turning back to close on it like other curves do. Of course, we should really say that the vertical line is an asymptote for part of the graph of the function. Consider the (admittedly contrived) function f(x)=\frac{1}{1+x^2+\sgn x}. It diverges as it approaches 0 from the left, approaches 1/2 from the right, and at 0 is 1. It thus has an isolated point on its "asymptote" which is really only an asymptote for the graph of f(x):x < 0. (It can be made continuous on the right there with the further contrivation f(x)=\frac{1}{x^2+H_1(x)}.) --Tardis 22:24, 18 January 2007 (UTC)
Consider the function 1/x, for positive x. As x approaches zero, the function approaches a vertical line, the y axis. As x approaches infinity, the function approaches a horizontal line, the x axis.
Now consider the function
f(x) = \frac{(2-x)(8-x)}{(4+x)^3} .
At x = 0 it evaluates to 1⁄4, and for all non-negative values of x the denominator is positive. But the numerator evaluates to zero when x equals 2 or 8, and to −2⁄125 at 31⁄2, between. So as we go from zero towards infinity, the function starts positive, crosses zero to negative values briefly, crosses zero again to positive values, and then asymptotically approaches zero from above. We consider the x axis an asymptote of the function, despite crossing it twice. And with a double root in the numerator we could touch the x axis without crossing it at a chosen value. --KSmrqT 14:16, 18 January 2007 (UTC)