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Topic review (newest first)

eddie
2006-01-14 08:08:07

ok, well i dont know if anyone will be able to help with this but i hope so.

it in my class notes for finding the steady state probability distribution of an M/M/m/n queue (erlang loss)

with arrival rate lamda and service time mu, for the second part of the proof, when m < k < n

P(k) = [lamda/(mu*m)] * P(k-1)

substituting a = lamda/mu

P(k) = (a/m) * P(k-1)

since we know k >= m that gives

P(k) = (a/m)^(k-m) * P(m)

and P(m) = [ a^m / m! ] *P (0)

which in my notes gives

P(k) = P(0) * (a^k)/[m^m * m!]

so there is something wrong somewhere, either in the proof or the formula at the end that we were trying to derive.

yonski
2006-01-14 06:40:02

Yes i get that too, the equation's wrong.

mathsyperson
2006-01-14 06:27:00

That's what I get it to be too, eddie.
Unless we've both made the same mistake, which is unlikely, the equation is wrong.

eddie
2006-01-14 05:40:43

apparently:

[(a/m)^(k-m)] * [(a^m)/m!] = (a^k)/[(m^m)*m!]


its part of a proof i need to learn for an exam on monday. i dont get it. surely it should be (a^k)/[(m^k-m)*m!] ?

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