ok, for the same question, i made a little progress but i'm not sure if i did it right...

i found the expected number of arrivals(a/3 *30) then the expected no. of blocked calls(a/3 *30*B(n,a)) for each of the 2 half hours, add them together to get the expected no. of blocked calls in entire 1 hour(=27.055...), then divides this value by the expected no. of arrivals in the same period(5/3*30+15/3*30 = 200), i get the blocking probability of a randomly chosen arrival in this period(=0.135...).

mm... is this the right way to do it?

the blocking prob of 10 erlangs 15 circuits is 0.0364...
if i got the above right, then why is the blocking prob of a randomly chosen call in the 1 hour period(0.135...) different from B(15,10)? since the total arrivals of the 2 half hours(a=5&15) = arrivals of the 1 hour(a=10)=200.

Any help would be very much appriciated.
Thanks

pui

tough question.. Beyond me. The homework moderator asked for some extra Effort.

Using the Google search ..There's an erlang calculator you may use to check your work.

http://www.voip-calculator.com/calculator/ervp

Your numbers are correct, Pui: The probability of a blocked call is indeed 13.52% in the first case and only 3.65% when the load is evenly spread, although the expected number of calls is the same in either case.

This should not surprise you: When there's a quiet and a busy period, most of the calls occur during the busy period when the performance of the system is rather poor. The "average" performance is thus poorer than when the load is uniformly distributed... This is, in fact, the whole point of this question.

Hope this helps.