Reply To: Reverse Erlang

Hi,

For Erlang B :

E(N,Y) = [ (YexpN / N!) ] / [ (Sigma(j=0 to N) Yexpj / j!) ]

with N in Erlangs, Y=nb of circuits.

Recursive solution : E(N,Y) = [ (Y/N) * E(N-1,Y) ] / [ 1 + Y/N * E(N-1, Y) ]

Hope this can help.

Marc.