Serkan Kirbas
No: 1203207
25.11.2002
Homework 2
1) First, let’s analyze and deeply understand the current system :
Figure 1 – Leaky Bucket Policer 1
So; in the first system according to above parameters;
Number of packets admitted by network can not be more than (r t + b) where
t is a time interval (to be consisted with this figure assume t is in seconds).
Average rate (long term average of packets) will be less than (b+rt)/t packets
per second(pps). And also burst size (number of packets sent consecutively
with no intervening idle) is less than b+rt packets. So; the parameters
b (tokens in the bucket) and r (tokens generating rate) determine the average
rate and burst size of the packet flow. In the solution in fact the only thing
we have to determine will be these two parameters of the other leaky bucket
policer.
As the second step, let’s move to the problem.
In problem, we are supposed to police average rate, burst size and the
peak rate(p) of a packet flow. Currently, we are able to police average rate
and burst size by Leaky Bucket Policer 1(Figure-1). So don’t disturb that
and put the output of Leaky Bucket Policer 1 to the input of Leaky Bucket
Policer 2. And than only thing that remains for Leaky Bucket Policer 2 to
check if determined peak rate is satisfied. See the below figure.
Figure 2 – Operation of Leaky Bucket Policer 1 and 2
In fact the peak rate is similar to average rate, the only difference is the t
variable(more smaller than r of average rate).(look PS) So we can use the
same formula.
Assume we have a peak rate constraint pmax , the maximum peak rate
supported. So Leaky Bucket Policer 2 must satisfy p< pmax at any time .
Then according to general formula ( rt+b) (as described at the beginning) we
can set token generation rate(r) of Leaky Bucket Policer 2 to pmax and bucket
size(b) of Leaky Bucket Policer 2 to 0 to police the peak rate p.
So if we put these into the equation we will get (pmax * t + 0) and the peak
rate will satify p < ((pmax * t + 0) / t) = pmax
Actually this is what we need. So;
Answer:
token generation rate(r) of Leaky Bucket Policer 2
bucket size(b) of Leaky Bucket Policer 2
= pmax
=0
PS: Peak rate :
While the average rate constraint limits the amount of traffic that can be sent into the
network over a relatively long period of time, a peak-rate constraint limits the maximum number of packets
that can be sent over a shorter period of time. The network may police a flow at an average rate of 6,000
packets per minute, while limiting the flow’s peak rate to 1,500 packets per second.
2)
First, calculate the amount of storage required:
In the question, it is given that for 1 second of a film, 4 Mb storage is
needed. So for a 100 min film we need
For one film
= 100*60 seconds * 4Mb
= 24000 Mb
( 1min=60seconds)
For 1500 film = 1500 * 24000 Mb = 36 000 000 Mb =
 So amount of storage needed is 36 000 000 Mb = 4.29*106 MB
Second, find the actual transfer bandwith:
Since only the %10 of all homes will be served independently only
1 000 000 *10 / 100 = 100 000 homes will need 4 Mbps bandwith
independently. So the other homes can only use the streams which
currently in system. In other words, they can only watch the films
which the lucky %10 is watching currently.
(For each second it delivers 400 000 Mb data)
As a result, the server should have a bandwith of
 (100 000)*(4 Mbps) = 400 000 Mbps
( 1 Mb = 106 bits)