Queuing Analysis of the Leaky
Bucket
• A “leaky bucket” is a mechanism for managing
buffers to smooth the downstream flow.
• What is described here is what is sometimes called a
“token bucket”.
• A queue holds a stock of “permits” which arrive at a
rate r (one every 1/r seconds) up to W permits may
be held.
• A packet cannot leave the queue if there is no permit
stored.
• The idea is that the scheme limits downstream flow
but can deal with bursts of traffic.
Modelling the Leaky Bucket
• Let us assume that the arrival process is a
Poisson process with a rate 
• Consider how many packets arrive in 1/r
seconds. The prob ak that k packets arrive is:
Queue of
permits (arrive
at 1/r seconds)
Queue of
packets (Poisson)
e   / r ( / r ) k
ak 
k!
Exit queue for packets
with permits
Exit of
buffer
A Markov Model
• Model this as a Markov Chain which changes state
every 1/r seconds.
• States 0iW represent no packets waiting and i-W
permits available. States W+i (where i > 1)
represent 0 permits and i packets waiting.
 a0  a1 i  j  0
• Transition probabilities:

pij  ai  j 1
j  i 1
0
otherwise

a2
0
a0+a1
a0
a2
1
a1
a0
2
a1
...
a2
W
a1
a0
W+1
a1
...
Solving the Markov Model
• By solving the balance equations we get:
 0  a0 1  (a0  a1 ) 0
i 1
 i   ai  j 1 j
i 1
j 0
 1  (1  a0  a1 ) 0 / a0
 1  a2 0  a1 1  a0 2
 0  (1  a0  a1 )(1  a1 )

 2  
 a2 
a0 
a0

Similarly, we can get expressions for 3 in terms
of 2 ,1 and 0. And so on...
Solving the Markov Model (2)
• Normally we would solve this using the remaining

balance equation:
i 1
• This is difficult analytically in this case.
i 0
• Instead we note that permits are generated every
step except when we are in state 0 and no packets
arrive (W permits none used).
• This means permits are generated at a rate (1-0a0)r
• This must be equal to  since each packet gets a
permit (assume none dropped while waiting).

r 
0 
ra0
And Finally
• The average delay for a packet to get a permit
is given by:
No of iterations
Time taken for each
iteration of chain
Amount of time
spent in given state
taken to get out
of queue from
state j
1 
T   j ( j  W )
r j W 1
For those states with queue
• Of course this is not a closed form expression.
To complete this analysis, look at Bertsekas
P515