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UNIVERSITY OF SOUTHERN CALIFORNIA, SPRING 2008
1
Lecture Notes 5
EE 549 — Queueing Theory
Instructor: Michael Neely
I. BACKLOG AND DELAY BOUNDS FOR LEAKY BUCKET
(
r, σ
)
INPUTS
Recall that
X
(
t
)
∼
(
r, σ
)
if
X
[
t
1
, t
2
]
≤
r
(
t
2

t
1
) +
σ
for all intervals
[
t
1
, t
2
]
. Consider a single server work
conserving queue, and assume:
•
X
(
t
)
∼
(
r, σ
)
input stream
•
single server queue
•
constant server rate
μ
•
initially empty system
(
U
(0) = 0)
Fact 1:
(Queue Bound) If
r
≤
μ
, then
U
(
t
)
≤
σ
for all time
t
.
Proof:
Fix a particular time
t
. We want to show that
U
(
t
)
≤
σ
. If
U
(
t
) = 0
, then we are done. Else,
U
(
t
)
>
0
¡
and so we are in a busy period that started at some time
t
b
, where
t
b
≤
t
. Thus:
U
(
t
)
=
X
[
t
b
, t
]

μ
·
(
t

t
b
)
≤
r
(
t

t
b
) +
σ

μ
·
(
t

t
b
)
≤
σ
Fig. 1.
An illustration of the unﬁnished work bound for leaky bucket inputs.
Fig. 1 illustrates the bound on unﬁnished work.
Corollary 1:
(Delay Bound) If service is FIFO then delay
≤
σ
μ
.
Proof:
Take any packet
i
. Let
t
i
be the time of the packet arrival, and let
U
(
t
i
)
be the unﬁnished work at this
time (which includes all bits that were in the queue just before the packet arrived, plus the total number of bits in
packet
i
). Then under FIFO, the delay of packet
i
is exactly
U
(
t
i
)
/μ
, which is less than or equal to
σ/μ
.
Fact 2:
(Summing Leaky Bucket Inputs) If
X
1
(
t
)
∼
(
r
1
, σ
1
)
and
X
2
(
t
)
∼
(
r
2
, σ
2
)
, then
X
1
(
t
) +
X
2
(
t
)
∼
(
r
1
+
r
2
, σ
1
+
σ
2
)
.
Proof:
X
[
t, t
+
T
]
=
X
1
[
t, t
+
T
] +
X
2
[
t, t
+
T
]
≤
(
r
1
T
+
σ
1
) + (
r
2
T
+
σ
2
)
=
(
r
1
+
r
2
)
T
+ (
σ
1
+
σ
2
)
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View Full DocumentUNIVERSITY OF SOUTHERN CALIFORNIA, SPRING 2008
2
Fact 3:
(InputOutput Parameter Invariance) Consider a leaky bucket input stream
X
(
t
)
with parameters
(
r, σ
)
entering a single server work conserving queue with constant processing rate
μ
bits/sec. Assume
r
≤
μ
, and let
Y
(
t
)
represent the departure process. Then
Y
(
t
)
is leaky bucket with parameters
(
r, σ
)
.
Proof:
We want to show that for any time
t
and any interval duration
T
≥
0
, we have:
Y
[
t, t
+
T
]
≤
rT
+
σ
To show this, ﬁx any time
t
and any
T
≥
0
, and consider the interval
[
t, t
+
T
]
. Note that the maximum amount
of bits that depart over this interval is no more than
U
(
t
)
plus the total new bits that enter during
(
t, t
+
T
]
:
Y
[
t, t
+
T
]
≤
U
(
t
) +
X
(
t, t
+
T
]
(1)
We now have two cases:
•
Case 1:
U
(
t
) = 0
. In this case, we have from (1):
Y
[
t, t
+
T
]
≤
0 +
X
(
t, t
+
T
]
≤
rT
+
σ
and we are done.
•
Case 2:
U
(
t
)
>
0
. In this case, there is a time
t
b
≤
t
(at which the current busy period started) such that:
U
(
t
) =
X
[
t
b
, t
]

μ
·
(
t

t
b
)
(2)
Using (1) and (2)
Y
[
t, t
+
T
]
≤
X
[
t
b
, t
]

μ
·
(
t

t
b
) +
X
(
t, t
+
T
]
=
X
[
t
b
, t
+
T
]

μ
·
(
t

t
b
)
≤
r
(
t
+
T

t
b
) +
σ

μ
·
(
t

t
b
)
=
rT
+
σ

(
μ

r
)
·
(
t

t
b
)
≤
rT
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 Spring '08
 Neely

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