< previous page page_389 next page >

Page 389
6.6.2 Designing a M/M/1 Buffer Given a Mean Queue Size
In chapter 4, we discussed two bounds on a queue (Q) exceeding a fixed buffer size (BF).
For open queueing systems with a single server, we can introduce a third bound based on the M/M/1 model. Since the binomial request rate and the constant service rate have smaller variances than M/M/1, this bound should be safe for many memory and I/O design situations (i.e., so long as c2 < 1). Here:
d87111c01013bcda00bb8640fdff6754.gif
Prob (0 items in server) = (1 - r).
Now, intuitively, we compute the probability of exactly one item in the queue by computing the probability of an item arriving given an empty queue; i.e.,
d87111c01013bcda00bb8640fdff6754.gif
Prob (exactly 1 item in queue or server) = r(1 - r).
Again (intuitively) extending this, we get:
d87111c01013bcda00bb8640fdff6754.gif
Prob (exactly k items in queue or server) = (1 - r)rk.
Now we can compute the probability that there are k or fewer items in the queue or server:
0389-01.gif
This can be rewritten as:
d87111c01013bcda00bb8640fdff6754.gif
(1 - r)(1 + r + + rk),
or, simplifying,
d87111c01013bcda00bb8640fdff6754.gif
1 - rk+1,
so that the probability that there are more than k items in the queue or servers is:
d87111c01013bcda00bb8640fdff6754.gif
Prob (> k items in queue awaiting service) = rk+1.
Finally, since a buffer of size BF represents BF + 1 = k items in the queue (BF) or in the server:
d87111c01013bcda00bb8640fdff6754.gif
Prob (buffer of size BF overflows) = rBF+2.
or
d87111c01013bcda00bb8640fdff6754.gif
Prob (Q > BF items awaiting service) = rBF+2.
Our Chebyshev bound (section 4.4.4) used the concept of ''buffer full or overflowed" as the point of system slowdown. This corresponds to:
d87111c01013bcda00bb8640fdff6754.gif
Prob (Q ³ BF) = rBF+1.

 
< previous page page_389 next page >