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E&EC 710 (2)
Wireless Communications Networks
Xuemin (Sherman) Shen
Office: EIT 4155
Phone: x 32691
Email: [email protected]
Lecture Notes: #3
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M/M/1 Queue
Poisson Arrivals
λµ
n-1 n n+1
tut ∆−∆− λ1
t∆λ
t∆µ
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Balance Condition
(λ+ µ)Pn = λPn−1 + µPn+1
⇒ Pn = ρnP0 = (1− ρ)ρn, n = 0, 1, · · ·Utilization factor
ρ = λ/µ
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Fluid Source Modeling of Packet Voice
cells/sec
N
VC cells/sec
NV
N-1 N
λ
αN
0 1
λN
αcells/secNV
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Activity factorλ
λ+ α
Balance equations:
i = 0, Nλπ0 = απ1
1 ≤ i ≤ N − 1,
πi[iα+ (N − i)λ] = [N − (i− 1)]λπi−1 + (i+ 1)απi+1
i = N, NλπN = απN−1
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πi = P (i) =
(N
i
)(λ
λ+ α
)i(α
λ+ α
)N−i
Define V/α to be “unit of information”
N
Cα Unit of information /sec
N
1
x
Balance equations:
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Fi(t+∆t, x)
=P (i sources on at t+∆t,X ≤ x)
+ P (i− 1 sources on at t+∆t,X ≤ x) · [N − (i− 1)]λ∆t
+ P (i+ 1 sources on at t+∆t,X ≤ x) · (i+ 1)α∆t
+ {1− [(N − i)λ+ iα]∆t} · Fi[t, x− (i− c)α∆t] +O(∆t)
⇒ d−→F (x)
dxD =
−→F (x)M
∗ J. N. Daigle, and J. D. Langford, “Models for Analysis
of Packet Voice Communications Systems”, IEEE JSAC,
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Vol. 4, Issue 6, pp. 847-853, 1986.
For Voice Traffic:
∆x = (i−C)α∆t ⇒ ∆l = ∆x ·V/α = (i−C)V∆t cells
αi units of information are “pumping” into the buffer
when there are i sources into the buffer.
For Video Traffic: ∆x = (iA− C)K∆t, with K the
number of pixels (picture elements) per second required
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to transmit a video signal
K = 30 frames × 250, 000 pixels
= 7.5× 106 pixels / sec
KAi bits/sec
∗ letting ∆t → 0, ∆x → 0, we get
m− (i− 1)αFi−1(t, x) + (i+ 1)βFi+1(t, x)
−[(m− i)α+ iβ]Fi(t, x)− (iA− C)K∂Fi(t, x)
∂x= 0
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Note:
lim∆t→0
Fi(t, x−∆x)− Fi(t, x)
∆t= − lim
∆x→0
Fi(t, x)− Fi(t, x−∆t)
∆x/(iA− C)K
Fluid Source Modeling of Voice
−→F (x) =
N∑i=0
aiΦiezix
zjΦjD = ΦjM
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−→F (x) = [F0(x), F1(x), · · · , FN(x)], x is the buffer size
Since 0 ≤ Fj(x) ≤ 1, for any value of x
−→F (x) =
∑i:ReZi≤0
aiΦiezix (1)
(or ai = 0 for those zi > 0)
One of the eigenvalues must be equal to 0. (x → ∞)
Let zi − 0 ⇒ Φ0m = 0 ⇒ π = Φ0 ⇒ a0 = 1 (2)
Since Fj(∞) is just the probability that j sources are in
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talk spurt. Fj(∞) = πj ⇒
−→F (∞) = π = a0Φ0 ⇒ a0 = 1
−→F (x) = π +
∑i:ReZ+i<0
aiΦiezix
Fj(0) = 0 for i > C (3)
Fj(0) = 0 = πj +
N∑j=J0
aiΦij, J0 ≤ j ≤ N
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Note: the number of ai equals N − ⌊C⌋
G(x) = 1− F (x) = 1−N∑j=0
Fj(x)
The probability F (x) which is the buffer occupancy is
less than or equal to x is
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F (x) = P (X ≤ x) = P
N∪j=0
j on , X ≤ x
=
N∑j=0
P (j on , X ≤ x)
=N∑j=0
Fj(x)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 14
For N multiplexed independent video sources
λ(t) =
N∑i=1
λi(t)
c(τ) =
N∑i=1
ci(τ)
σ2 = c(0) =
N∑i=1
σ2i
Objectives: use m minisources (on-off model) to
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represent N multiplexed sources.
off on
α
β bits/pixelA
c(τ) = mcj(τ)
c(τ) =N∑i=1
ci(τ) = Nc(τ) identical moments
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cj(τ) = A2 αβ
(α+ β)2e−(α+β)τ
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 17
Chapter 3 Traffic Routing
Routing is one of the main functions of the network
layer. Routing algorithms are designed to find the best
path for a particular transmission from the source to the
destination. The “best” means (minimum) low delay and
maximum throughput.
A Router is a box with input and output links. It has
access to the routing table which includes a set S of N
routers (nodes) and a set of link lengths between these
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routers dij, i, j ∈ S; dij = ∞ if there is no link from i to
j. Routing table is updated at regular time intervals. (30
seconds).
Routers contain software that enables them to determine
which of the several possible paths between the source
and the destination is the best for a particular
transmission.
The Bellman-Ford Algorithm
Initial Condition
At step h = 0, node i has the estimate d0(i) = ∞ for all
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i ∈ S with i = D and d0(D) = 0, where D is the
destination node.
At step h ≥ 1, node j updates its estimate of its distance
to D to some value dh(j) and advertises that value to its
neighbors.
At step h+ 1, node i examines the message that it just
received from its neighbors at the previous step h. e.g., j
might be a neighbor of i, that is, d(i, j) < ∞, and j
informs i at step h that the shortest estimated distance
from j to D is equal to dh(j). Node i gets similar
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 20
messages from all its neighbors. Node i then knows that
if it chooses to send a message D by first sending it to
its neighbor j, then the distance that the message will
travel to reach D is estimated to be d(i, j) + dh(j).
Since node i can choose the neighbor to which it sends
the message destinated to D, node i estimates the
shortest distance to D as
dh+1(i) = min{d(i, j) + dh(j), j ∈ S}Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 21
2
3
4
5
1
1
4
8
2
241 2
Destination node
Shortest path problem arc lengths as indicated
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011 =D
112 =D
413 =D
∞=14D
∞=15D
Shortest path using at most 1 arc
041 =D
142 =D
243 =D 44
5 =D
844 =D
Final tree of shortest paths
031 =D
132 =D
233 =D
934 =D
435 =D
Shortest path using at most 3 arc
021 =D
122 =D 92
4 =D
223 =D
625 =D
Shortest path using at most 2 arc
Successive iterations of the Bellman-Ford method. In this example, the shortest
(≤ h) walks are paths because all are lengths are positive and therefore all cycles
have positive length. The shortest paths are found after N − 1 iterations, which is
equal to 4 in this example.Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 23
The Bellman-Ford Algorithm
The shortest path length from node i to the destination
node is the sum of the length of the arc to the node
following i on the shortest path length from that node to
the destination node.
Algorithm:
Dh1 = 0 for all h(i = 1)
Dh+1i = minj[dij +Dh
j ] for all i = 1
Starting form the initial conditions
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D0i = ∞, for all i = 1.
Note: node 1 is the destination node. Di is the distance
from node i to node 1. h is the number of iterations. dijis the distance from node i to node j.
Example:
Given a digraph G(N,A) with the following
specifications.
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N ={1, 2, 3, 4, 5, 6}A ={(2, 1) = 1, (2, 3) = 2, (3, 1) = 5,
(3, 2) = 1, (4, 2) = 6, (4, 5) = 2,
(5, 3) = 2, (5, 4) = 3, (6, 4) = 1, (6, 5) = 2}
(a) Draw and label the arcs of the digraph.
(b) Discuss how the Bellman-Ford algorithm can be sued
to determine the shortest path from any node to a
destination node on the digraph.
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(c) Find the shortest paths from each of the other nodes
to destination node 1 on this digraph using the
Bellman-Ford algorithm.
Solution:
(a) The labeled digraph is as follows:
4
5
2
3
1
2
6
2
1
5
2 26 13 1
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(b) The Bellman-Ford algorithm can be implemented on
the above diagram.
1. Label each node i with (di, j) where di (the first
element of the label) is the distance from node i to the
destination node through the neighboring node j.
2. Initialization: label node i with (d(0)i , 0) where
d(0)i =
{0,i = A
∞,i = A
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3. Shortest distance labeling of all nodes: for each
successive run h ≥ 0, determine j which minimizes
d(h+1)i = d(i, j) + d
(h)j , and update label at each node i
with (d(h+1)i , j)
4. Repeat step 3 until no change of any di occurs.
(c) The execution results of Bellman-Ford algorithm on
the above digraph is shown in the following table.
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h 1 2 3 4 5 6
0 (0, ·) (∞, ·) (∞, ·) (∞, ·) (∞, ·) (∞, ·)1 (0, ·) (1, 1) (5, 1) (∞, ·) (∞, ·) (∞, ·)2 (0, ·) (1, 1) (2, 2) (7, 2) (7, 3) (∞, ·)3 (0, ·) (1, 1) (2, 2) (7, 2) (4, 3) (9, 4)
4 (0, ·) (1, 1) (2, 2) (6, 5) (4, 3) (6, 5)
5 (0, ·) (1, 1) (2, 2) (6, 5) (4, 3) (6, 5)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 30
From table, we can get all the shortest paths:
2 → 1,
3 → 2 → 1,
4 → 5 → 3 → 2 → 1,
5 → 3 → 2 → 1,
6 → 5 → 3 → 2 → 1,
Advantage: the algorithm is very flexible with respect to
the choice of initial estimates Dj and the ordering of
communications and updates. The Bellman-Ford
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algorithm can be executed at each node i in parallel with
every other nodes.
Disadvantage: the routing uses only one path per
origin-destination pair, thereby potentially limiting the
throughput of the network; its capability to adapt to
changing traffic conditions is limited by its susceptibility
to oscillations, which is due to the abrupt traffic shifts
resulting when some of the shortest paths change due to
changes in link lengths.
Optimal routing can eliminate both of these
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disadvantages.
Stability issues in datagram networks (Internet)
suppose that there are two paths along which an origin
can send traffic to a destination. Let the traffic input
(pakcets/sec) at each node i = 1, 2, · · · , 7, 9, · · · , 15 be
one packet and let the traffic input of node 8 be ϵ > 0,
where ϵ is very small. Assume that the length of link
(i, j) is
dij = Fij
where Fij is the arrival rate at the link counting input
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 33
and relayed traffic. Suppose that all nodes compute their
shortest path to the destination every T seconds using as
link lengths the arrival rates Fij during the preceding T
seconds, and route all their traffic along that path for the
next T seconds. Assume that we start with nodes 1
through 7 routing clockwise and nodes 8 through 15
routing counter clockwise.
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Destination
161
2
3
4
5
6
7 8 9
10
11
12
13
1415
1
1
1
1
1
1
11
1
1
1
1
1
1
ε
Sixteen-node ring network of Example 2. Node 16 is the only destination.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 35
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 36
Oscillations in a ring network for link lengths dij equal to
the link arrival rates Fij. Each node sends one unit of
input traffic to the destination except for the middle
node 8, which sends ϵ > 0, where ϵ is very small. The
numbers next to the links are the link rates in each of
the two directions. As an example of the shortest path
calculations, at the first iteration the middle node 8
computes the length of the clockwise path as
28(= 0 + 1 + 2 + · · ·+ 7), and the length of the
counterclockwise path as
28 + 8ϵ(= ϵ+ 1 + ϵ+ 2 + ϵ+ · · ·+ 7 + ϵ), and switches
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its traffic to the shortest (clockwise) path at the second
routing. The corresponding numbers for 9 are 28 and
28 + 7ϵ, so node 9 also switches its traffic to the
clockwise path. All other nodes find that the path used
at the first routing is shortest, and therefore they do not
switch their traffic to the other path.
After three shortest path updates, the algorithm is locked
into an oscillatory pattern whereby all traffic swings from
the clockwise to the counter clockwise direction and
back at alternate updates.
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Actually, the type of instability illustrated above will
occur generically if the length dij of link (i, j) increases
continuously and monotonically with the link arrival rate
Fij, and dij = 0 where Fij = 0. It is possible to damp
the oscillations by adding a positive constant to the link
length so that dij = α > 0 when Fij = 0. The scalar α is
known as a bias factor.
α ↑⇔ insensitive to traffic congestion ↓
The oscillations are less severe in virtual circuit networks,
where every session is assigned a fixed communication
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path at the time it is first established. There the average
duration of a virtual circuit is often large relative to the
shortest path updating period. As a result, the network
reaction to a shortest path update is much more gradual
since old sessions continue to use their established
communication paths and only new sessions are assigned
to the most calculated shortest paths.
Optimal Routing
Optimal routing directs traffic exclusively along paths
which are shortest with respect to some link lengths that
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depend on the flows carried by the links.
Define the cost function∑(i,j)
Dij(Fij)
where Fij is the total flow (in packets/sec) carried by
link (i, j) and given by
Fij =∑
all pathsP containing (i,j)
xP
Here xP is the flow of path P . For every
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 41
origin-destination pair w, there are the constraints∑P∈Pw
xP = rw, xP ≥ 0 for all P ∈ Pw
where rw is the given traffic input of the OD pair w and
Pw is the set of directed path of w. In terms of the
unknown path flow vector x = {xP |P ∈ Pw, w ∈ W},the problem is written as
minimize∑(i,j)
Dij[∑
all pathsP containing (i,j)
xP ]
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 42
subject to
∑xP = rw, for all w ∈ W
xp ≥ 0, for all p ∈ Pw, w ∈ W
Assume that each Dij is a differentiable function of Fij
and is defined in an interval [0, Cij), where Cij is either a
positive number (typically representing the capacity of
the link) or else infinity. Let x be the vector of path
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 43
flows xp. Denote by D(x) the cost function
D(x) =∑(i,j)
Dij[∑
all pathsP containing (i,j)
xP ]
∂D(x)
∂xP=
∑all links(i,j) on path P
D′ij
∂D(x)∂xP
is the length of path P when the length of each
link (i, j) is taken to be the first derivative D′ij evaluated
at x. ∂D(x)∂xP
is called the first derivative length of path P .
Let x∗ = {x∗P} be an optimal path flow vector. Then if
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 44
x∗P > 0 for some path P of an OD pair w, we must be
able to shift a small amount δ > 0 from path P to any
other path P ′ of the same OD pair without improving the
cost; otherwise, the optimality of x∗ would be violated.
To first order, the change in cost from this shift is
δ∂D(x∗)
∂xP ′− δ
∂D(x∗)
∂xP
and since this change must be non-negative, we obtain
x∗P > 0 ⇒ ∂D(x∗)
∂xP ′≥ ∂D(x∗)
∂xP
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for all P ′ ∈ Pw.
This is a necessary condition for optimality of x∗. When
Dij exist and are positive in [0, Cij), the optimality is
also sufficient.
Ex. Consider the two-link network shown in the figure,
where nodes 1 and 2 are the only origin and destination,
respectively. The given input r is to be divided into the
two path flows x1 and x2 so as to minimize a cost
function based on the M/M/1 approximation
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 46
High capacity C1
Low capacity C2
x1
x2
1 2 DestinationOrigin
r
D(x) = D1(x1) +D2(x2)
where for i = 1, 2,
Di(xi) =xi
Ci − xi
and Ci is the capacity of link i, r < C1 + C2.
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At the optimum, the shortest path condition must be
satisfied the constraints
x∗1 + x∗
2 = r, x∗1 ≥ 0, x∗
2 ≥ 0
Assume that C1 ≥ C2, then x∗1 ≥ x∗
2.
1) x∗1 = r and x∗
2 = 0, then
dD1(r)
dx1≤ dD2(0)
dx2
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i.e.,C1
(C1 − r)2 ≤ 1C2
or r ≤ C1 −√
C1C2
(the derivative of xc−x is c
(c−x)2)
2) x∗1 > 0 and x∗
2 > 0, then the lengths of paths 1 and 2
are equaldD1(x
∗1)
dx1=
dD2(x∗2)
dx2
i.e.,C1
(C1 − x∗1)
2=
C2
(C2 − x∗2)
2
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Since x∗1 + x∗
2 = r, we have
x∗1 =
√C1[r − (C2 −
√C1C2)]√
C1 +√C2
x∗2 =
√C2[r − (C1 −
√C1C2)]√
C1 +√C2
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x1* x2*x1*
x2*0 C1+C2 r211 CCC −
Optimal path flows for the routing example. When the traffic input is low, only the
high-capability link is used. A the input increases beyond the threshold C1−√C1C2,
some traffic is routed on the low-capability link.
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The optimal solution is shown for r in the range
[0, C1 + C2) of possible inputs.
For r ≤ C1 −√C1C2, the faster link is used exclusively.
When r exceeds the threshold value C1 −√C1C2 for
which the first derivative lengths of the two links are
equalized, the slower link 2 is also utilized. As r
increases, the flows on both links increase while the
equality of the first derivative lengths is maintained.
This behavior is typical of optimal routing when the cost
function is based on the M/M/1 approximate for low
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input traffic, each OD pair tends to use only path for
routing, and as traffic input increases, additional paths
are used to avoid overloading the fastest path. More
generally, this type of behavior is associated with link
cost functions Dij with the property that the derivative
D′ij(0) at zero flow depends only on the link capacity and
decreases as the link capacity increases.
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Call Admission Control
∗ How much traffic can a network handle if a prescribed
quality of service for each traffic class is to be maintained
while the network utilization, i.e., traffic throughout
(hence revenue), is to meet some minimum goal?
∗ Given a call arriving, requiring a virtual connection
with specified (QoS) such as bandwidth, loss probability,
delay, etc., should it be admitted?
(Providing the desired QoS for each connection, while at
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the same time allowing efficient use of the network)
On
Off
On
Off Time
...... Rp
α-1β-1
a.Traffic model
b.Markov model
α
Rp
Basic on-off traffic source
β
Assuming that there are k traffic classes, each with its
own traffic descriptors, to be multiplexed onto one
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network access link, of capacity CL cells/sec.
class 1
class 2
class k
Scheduler
network access link
CL cells/sec
Network access node with access scheduler
The network throughput and traffic class quality of
service are closely related to the way in which each buffer
is served by the access link.
Note that there must be a maximum number of
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connections or calls of each class for the given strategy
that can be handled by the access link. These maximum
numbers represent an admissible region for this system.
The admissible region depends on the scheduling
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strategy, the access capacity CL, the number of classes,
and the QoS requirements for each.
Given the admissible region, in principle a call is admitted
if the system, with the call present, still operates within
the admissible region. A call is blocked if its acceptance
would take the system outside of the admissible region.
Example: Given N homogeneous sources. The
probability a source s in the “on” state is P = αα+β, PRP
is the average rate of transmission.
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1
NCL
Statistical multiplexing of homogeneous sources .
The utilization is ρ = NPRP/CL
Let ρ = 1 ⇒ the maximum N can be obtained.
For k class sources:
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k∑i=1
niPiRPi = CL
(tradeoffs are possible among the maximum number of
calls allowed per user class.)
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For homogeneous case:
Peak bandwidth assignment is equivalent to P ⇒ 1, or
NRP = CL, no statistical multiplexing gain.
Average bandwidth assignment (NPRP = CL) provides
maximum multiplexing gain, but it may be unacceptable
in terms of cell loss.
∗ For a given QoS (cell loss) and CL. Find N .
Let m = PN as the mean number of source “on”. The
mean bit rate inputted by these m connections is mRP .
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With the cell loss constrain, CL ≥ mRP .
Let CL = (m+ kδ)RP
where k ≥ 0 is a constant to be determined, which varies
with the quality of service specified δ is the standard
deviation. δ = NP (1− P ).
Find k is equivalent to find m (or N).
Procedures for finding k.
Let C = CL/RP = m+ kδ = NP + k√
NP (1− P )
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Jo = ⌈C⌉, Ju = Jo − 1 = ⌊C⌋
ρ =m
C< 1
If the buffer in the system is ignored by assuming that all
the cells arriving beyond capacity C in the overload stats
will be lost (Conservative estimate of loss probability),
then in the overload state i, cell loss rate is (i− c)RP .
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 63
The combined cell loss rate over all the overload states is
N∑i=Jo
(i− C)RPπi
where πi is the probability of the system in state i
πi =
(N
i
)P i(1− P )N−i
P =α
α+ β
The average cell flow is given by mRP .
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 64
Therefore, the loss probability
PL =
∑Ni=Jo
(i− C)RPπi
mRP=
N∑i=Jo
(i− C)πi
m
Another measure of loss probability is simply the
probability of the system being in overload states,
ε =N∑
i=Jo
πi
If N >> 1 and P << 1, the binomial distribution
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 65
solution for πi can be approximated by Gaussian
distribution with the same mean value m = NP and
variance σ2 = NP (1− P )
We obtain
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 66
PL =
N∑i=Jo
πi(i− c)
m
=1
m
∫ ∞
Jo
e−(x−m)2
2σ2(x−c)
√2πσ2
dx
ε =
∫ ∞
Jo
e−(x−m)2
2σ2
√2πσ2
dx
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 67
∗ PL and ε are equivalent.
Proof: By replacing Jo by C and rewriting x− C as
(x−m)− (C −m), we have
ε =
∫ ∞
C
e−(x−m)2
2σ2
√2πσ2
[(x−m
m)− (
C −m
m)]dx
=
∫ ∞
C−m√2σ
√2σye−y2
√πm
dy − C −m
m· ε
=σ
m
e−(C−m)2/2σ2
√2π
− C −m
m· ε
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 68
On the other hand,
ε ≈ 1√π
e−(C−m√
2σ)2
2(C−m√2σ)− 1√
π
e−(C−m√
2σ)2
4(C−m√2σ)3
PL ≈ σ
m
e−(C−m)2/2σ2
√2π
− (C −m)ε
m
=σ3e−(C−m)2/2σ2
√2πm(C −m)2
=(1− P )
C −mε < ε,
(C −m)√2σ
> 3
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 69
Take ln on both sides of PL equation
ln√2πPL = ln
(σ3
m(C −m)2
)− (C −m)2
2σ2
C ≈ m+ σ√
− ln(2π)− 2 lnPL
For PL = 10−5 ,
C = m+ 4.6σ = NP + 4.6√
NP (1− P )
Note: σ2 = m(1− P ) = NP (1− P )
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 70
Generally speaking, the ε approach is more conservative
than the PL approach resulting in a larger capacity
requirement if used.
Let
ε =
∫ ∞
C
e−(x−m)2/2σ2
√2πσ2
dx
If (C −m) > 3√2σ
ε =σe−(C−m)2/2σ2
√2π(C −m)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 71
Taking natural logs,
ln(√2πε) = ln(
σ
C −m)− (C −m)2
2σ2
⇒ C = m+ σ√
− ln(2π)− 2 ln ε
(neglecting the first right-hand side term)
K =√
− ln(2π)− 2 ln ε
e.g. ε = 10−5 ⇒ K = 4.6
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 72
C = m+ 4.6σ
or
C = NP + 4.6√
NP (1− P )
C = 0.02N + 0.64√N, (P = 0.02)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 73
In general
C = NP +K√
NP (1− P )
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 74
⇒ N =C
P− 1
P[√
4α(C + α)− 2α]
where α ≈ K2(1− P )/4
An admission control using this relation would thus
admit a new call if fewer than N calls were connected.
In summary, if peak bit rate admission control is used,
the number of calls allowed into the system would be
CL/RP = C (ϵ = 0)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 75
The multiplexing gain Gϵ is thus defined as
Gϵ = N/C = NP/CP
The maximum multiplexing G(C = m = NP ) is
G = 1/P
when P ↓ (P << 1), G ↑
when P ↑, G ↓Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 76
Gϵ can also be written as
Gϵ =1
P− 1
P
√
K2(1− P )
C
[1 +
K2(1− P )
4C
]− K2(1− P )
2C
Note: a greater gain is possible as either the capacity C
increases or K decreases.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 77
The above figure is a plot of Gϵ on a log-log scale to
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 78
show that over most of the range of P they are very
nearly linear. (GϵP = constant)
Since GϵP = NP/C = ρ < 1
GϵP = ρ ≈ 0.64 (C = 100)
GϵP = ρ ≈ 0.27 (C = 10)
To include the effect of the access buffer and resultant
cell loss probability on admission control, as well as the
buffer delay, the fluid -flow analysis approach is used.
Assume there are m on-off minisources statistically
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 79
multiplexed at an access buffer.
m is replaced N
KA bits/sec is replaced RP
(K = 7.5× 106 pixels/sec)
KC bits/sec is replaced CL
The survivor function G(x) for the M- minisource model
(the probability the buffer occupancy exceeds x bits) can
be approximated by
G(x) ≈ ANρNe−βrx/RP
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 80
with the parameter r given by
r = (1− ρ)(1 + α/β)/(1− CL/NRP)
and
ρ = NPRP/CL < 1
Letting G(x) be an approximation to PL and letting all
the parameters in the equation be specified, except CL,
one can get CL required to obtain a specified quality of
service PL.
The larger the value of x, the smaller the loss probability
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 81
or, with PL fixed, the smaller the capacity required.
Using the exponent e−βrx/RP only
PL ≈ e−βrx/RP
⇒ βrx/RP = − lnPL
⇒ CL
RPN=
1− k
2+
√(1− k
2
)2
+ kP
with the parameter k defined as
k =βx
RP(1− P ) ln(1/PL)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 82
Finally, we can find N .
In the case of multiplexed heterogeneous,
∗C = m+ σ
√− ln(2π)− 2 ln ε
only involves two moments: the mean bit rate and
standard deviation.
The basic assumption is that the central limit theorem
applies, i.e., the mean bit rates and variances of the bit
rate of the multiplexed source is the summation of the
means and variances of each individual source.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 83
∗ Using fluid approximation
CL
N=
RP
2− (α+ β)x
2 ln(1/PL)+
√(RP
2− (α+ β)x
2 ln(1/PL)
)2
+αRPx
ln(1/PL)
Consider N heterogeneous on-off sources multiplexed
together, for the i-th source, 1 ≤ i ≤ N , we have RiP ,
1/βi and 1/αi, and
CL =N∑i=1
RiP
2− (αi + βi)x
2 ln(1/PL)+
√(Ri
P
2− (αi + βi)x
2 ln(1/PL)
)2
+αiRi
Px
ln(1/PL)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 84
(each source has the same QoS parameters PL and x)
Example
N bursty on-off sources are to be multiplexed together at
an ATM access port. Each source has
exponentially-distributed on- and off-times, with average
values of 1 sec and 10 sec, respectively. When “on”, a
source transmits at its peak rate of 5 Mbps. The
outgoing link capacity of the multiplexer is 100 Mbps.
a) Find the number of sources that may be
accommodated if (1) peak rate allocation is used; (2)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 85
average rate allocation is used. What is the probability
of loss with peak-rate allocation?
b) Find the number of sources that may be multiplexed if
the probability of loss is PL = 10−6. PL is
approximated by the average time the multiplexer is in
the overload region.
c) Repeat b. if approximate fluid-flow anlaysis is used,
with the probability of loss defined as
PL = P [buffer occupancy > x], x chosen such that the
maximum buffer delay is 100 msec. Compare all four
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 86
values of N in a., b.,c.
d) Repeat a., b.,c., if the average off-time is reduced to 5
sec. Compare with the previous case of 10 sec.
Solution: Rp = 5 Mbps, CL = 100 Mbps, N bursty
on-off sources, “on” = 1 sec, “off” 10 sec,
P = αα+β = 1
1+10 =111
a)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 87
1) For peak rate allocation
N = CL/RP =100
5= 20
2) For average rate allocation
N =CL
RP · P=
100
5 · ( 110+1)
= 220
The probability of loss with peak rate allocation PL = 0.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 88
b) PL = 10−6, PL = ε
C = CL/RP = NP +K√
NP (1− P )
K =√− ln(2π)− 2 ln(PL) = 5.1
⇒ 100
5= N × 1
11+
√N × 1
11× 10
11× 5.1
⇒ N = 77
c) PL = P [buffer occupancy > x].
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 89
The maximum buffer delay is 100 msec, i.e.,
x
CL= 100 msec ⇒ x = 100 msec ×100 Mbps = 10 Mbits
Since
CL
RPN=
1− k
2+
√(1− k
2
)2
+ kP
k =βx
[RP(1− P ) ln(1/PL)]= 1.59
⇒ N = 107
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 90
This result lies between the peak assignment and average
assignment. It is also larger than the conservative
assignment in b.
d) For average off-time to be 5 sec.
P =α
α+ β=
1
1 + 5=
1
6
For the peak assignment, N is the same as in a, i.e.,N = 20.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 91
For the average assignment, N = CL/(RP · P ),
N =100
5 · 16= 120
For conservative assignment, PL = ε = 10−6,
100
5= N × 1
6+
√N × 1
6× 5
6× 5.1
⇒ N = 44
For the fluid flow analysis,
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 92
CL
RPN=
1− k
2+
√(1− k
2)2 + kP
k =βx
[RP(1− P ) ln(1/PL)]
⇒ N = 44
“Distributed Call Admission Control in Mobile/Wireless
Networks”, by M. Naghshineh and M. Schwrtz. IEEE
JSAC, Vol. 14, pp. 711-717, 1996.
Objective: The purpose of distributed call admission
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 93
control is to limit the call handoff dropping probability in
loss systems or the cell overload probability in lossless
systems.
(Handoff dropping or cell overload are consequences of
congestion in wireless networks.)
The call admission controller must take two factors into
consideration when admitting a new call: i) By admitting
the new call, the desired QoS of existing calls in the
system must be maintained and ii) the system must
provide the new call its desired QoS.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 94
Handoff dropping probability is the probability that after
a handoff a call is dropped due to unavailability of
wireless capacity.
Call blocking probability is the probability that a new call
is blocked based on the call admission algorithm.
∗ Approach: the call admission decision is made in a
distributed manner in which each base station makes an
admission decision by exchanging cells periodically
(without the involvement of the network call processor).
In particular, we estimate the state of the cell in which a
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 95
new call admission request is made in addition to the
state of its adjacent cells T units of time after the call
admission.
Model:
Cl
l
Cn
n
Cr
r handoffs
Call Arrivals
1-D cell array
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 96
Cn is the cell where a call admission request is made. Cr
and Cl are adjacent cells of Cn. n, r and l represent the
number of calls in cells Cn, Cr and Cl, respectively.
Let λ denote the new call arrival rate to any radio cell;
µ denotes the call departure rate, and
h denotes the call handoff rate.
Assume each radio cell can support up to N calls.
For any test mobile in radio cell Cn, it is assumed that
the test mobile remains in the same cell with probability
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 97
ρs, and that it hands-off to Cr (or Cl) with probability
Pm/2 during time T . (the probability that a call
hands-off more than once during T is negligible).
The handoff dropping probability of a radio cell
P0 = PQoS =∞∑
i=N+1
Pi (1)
where Pi represents the probability of having i calls in a
radio cell.
A new call is admitted to cell Cn at time t0 if and only if
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 98
the following two admission conditions are satisfied:
1) At time t0 + T , the overload probability of cell Cn
affected by handoffs from Cr or Cl to Cn, and including
handoffs from cell Cn to any other cell must be smaller
than PQoS.
2) At time t0 + T , the overload probability of cell Cr
affected by handoffs from cell Cn or the cell to the right
side of Cr to cell Cr and including handoffs from cell Cr
to any other cell, in addition to new calls admitted to cell
Cr during T must be smaller than PQoS.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 99
Let k be the number of calls in a cell at time t0, theprobability that i calls (out of k calls at time t0) are inthe same cell at time t0 + T
B(i, k, ρs) = P (i) =
(k
i
)ρis(1− ρs)
k−i, 0 ≤ i ≤ k (2)
the probability that j calls (out of k calls at time t0) are
in the cell to the right handside of the test cell at time
t0 + T
P (j) =
(k
i
)(Pm/2)
j(1− Pm/2)k−j, (3)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 100
Assume by admitting a call to cell Cn at time t0 there will
be n, r, and l calls in cells Cn, Cr, and Cl, respectively.
For the first admission condition to be satisfied, the
probability distribution of the number of calls in Cn at
time t0 + T , denoted by Pnt0+T)(k) can be found using a
convolution sum of three binomial distribution
B(in, n, ρs), B(il, l, Pm/2), B(ir, r, Pm/2), where
n ≥ in ≥ 0, l ≥ il ≥ 0, r ≥ ir ≥ 0, and in + il + ir = k.
The number of calls in cell Cn at time t0 + T can have a
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 101
Gaussian distribution given by
Pnt0+T(k) ≈ G(nρs + (l + r)Pm/2, (4)√
(nρs(1− ρs) + (l + r)Pm(1− Pm/2)/2))
The overload probability P0 can be calculated by
P0 =l+n+r∑N+1
Pnt0+T(k) (5)
≃ Q
(N − (nρs + (l + r)Pm/2)√
(nρs(1− ρs) + (l + r)Pm(1− Pm/2)/2)
)Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 102
For a given PQoS, there exists a value “a” where
PQoS = Q(a), and
N − nPs − (l + r)Pm/2 (6)
−a√(nPs(1− Ps) + (l + r)Pm(1− Pm/2)/2) = 0
Solving for n, an admission threshold n1, which satisfiesthe first admission condition
n1 =1
2Ps[a2(1− Ps) + 2N − Pm(r + l)− (7)
−a√a2(1− Ps)2 + 4N(1− Ps)− P 2
m(r + l) + 2PmPs(l + r)]
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 103
The second admission condition states that the overload
probability of the radio cell Cr must be less than (or
equal to ) PQoS. Using the similar approach
Prt0+T≃ G(rPs + (E(n) + n)Pm/2 + λT, (8)√(rPs(1− Ps) + (E(n) + n)Pm(1− Pm/2)/2 + λT )
P0 ≃ Q
(N − rPs − (E(n) + n)Pm/2− λT√
(rPs(1− Ps) + (E(n) + n)Pm(1− Pm/2)/2 + λT )
)(9)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 104
Using PQoS = Q(a), the number of calls that can be
admitted to cell Cn such that the second admission
condition for Cr is satisfied is given by
n2 =1
2Pm[a2(2− Pm) + 4N − 4λT − 2E(n)Pm − 4rPs − (10)
a√a2(2− Pm)2 + 16N + 8Pm(λT + rPs −N − 16rP 2
s )
Similarly (for Cl)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 105
n3 =1
2Pm[a2(2− Pm) + 4N − 4λT − 2E(n)Pm − 4lPs−
a√a2(2− Pm)2 + 16N + 8Pm(λT + lPs −N − 16lP 2
s )
The final admission threshold which satisfies all
admission conditions is given by
n = min(n1, n2, n3) (11)
∗ Here E(n) is the average number of calls per cell. It is
assumed the cell to the right-hand side of Cr has E(n)
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 106
calls at time t0.
New calls admitted to any cell are Poisson
(λT )inewe−λT/inew
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Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 108
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 109
These figures show n as a function of r and l, given Ps,
Pm, T , λ and E(n) and PQoS.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 110
Extension to a 2-D model
Assume a mobile hand off to any one of its six
neighboring cells with equal probability Pm/6 during the
time period T and that it remains in the same cell with
the probability Ps. Using the approach similar to the 1-D
case, all admission thresholds (na0, · · · , na
6) can be
calculated, based on the state of cell C0 and all its
immediate neighbors, the final call admission threshold is
given by n = min6i=0 nai .
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 111
C1
C0
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
C13
C14
C15
C16
C17
C18
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 112
Simulation: The performance results are based on the
simulation of a system consisting of 10 cells arranged on
a circle.
Pm = 0.1813, Ps = 0.7866
N is the number of calls that each cell can support
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 113
Note: For the trunk reservation system, out of N
channels available in each cell, g channels are reserved
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 114
for handoff calls. That is, at the time of call admission, a
call is admitted to a cell if at least g channels are
available in that cell.
Future research topics:
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 115
1) how to estimate Pm and Ps ?
2) how to take the effect of different traffic classes into
account?
3) how to take the effect of different channel assignment
schemes, and nonuniform traffic conditions into account?
4) the average overload time should also be considered.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 116
Mobility Profile Prediction for ResourceManagement – An Adaptive Fuzzy
Inference Approach
Xuemin Shen, Jon W. Mark, Jun Ye, Centre for Wireless
Communications, Department of Electrical and
Computer Engineering, University of Waterloo, Waterloo,
Ontario, Canada
ACM Wireless Networks, Vol. 6, pp. 363-374, 2002.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 117
• Introduction
• Mobility information model
• Adaptive fuzzy inference system
• Numerical results
• Conclusions
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 118
Introduction
Two main objectives associated with resource
management in communication networks are:
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 119
• to maximize the throughput rate
• to ensure the QoS requirements of all services
supported
Resource managementCongestion controlResource allocation
Call admission control
Medium access control
routing
Resource Management Functions
Challenges in resource management of wireless networks
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 120
• user mobility
• limited radio spectrum
• radio channel impairments
Approaches to dealing with handoffs
• treat handoff calls as new calls
• reserve resources for handoff calls
• make use of statistical multiplexing in resource
reservation
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 121
Mobility Profile
Definition: The mobility profile is characterized by a set
of probabilities that govern the mobile user’s movement
within a cellular network.
• Knowledge of the user mobility profile will allow the
design of effective and efficient resource management
functions.
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 122
Mobility Information Model
• Pilot signals from base stations (BSs) in the forward
link
• Pilot signal measurement at mobile terminals
• Mobility profile prediction in a mobile switching center
(MSC) based on the measurement
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 123
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 124
3Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 125
Based on a combination of path loss and lognormal
shadowing, the received pilot signal amplitudes are
ai(t) = γi[di(t)/D0]−n10ξi(t)/10, i = 0, 1, · · · , 6 (1)
• γi: a constant proportional to amplitude of the pilot
signal (γi = γ for i = 0, 1, · · · , 6, if the transmitted
pilot signals have the same power)
• di(t): the distance between the mobile user and BSi at
time t
• n: the path loss exponent
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 126
• D0: a close-in reference distance
• ξi(t): a normal random variable at any t with zero
mean and variance σ2 (ξi(t) and ξj(t) are independent
for i = j)
One possible approach for two BSs (BS0 and BS1)
p[d1(t) > d0(t)] = Φ
(10 log10 [a0(t)/a1(t)]√
2σ
)Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 127
where
Φ(u) =
∫ u
−∞exp(−x2/2)dx/
√2π
Drawbacks:
• difficult to extend to multiple (more than two) BSs
• does not take into account the measurement accuracy
Proposed approach: adaptive fuzzy inference system
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 128
The Adaptive Fuzzy Inference System
4Fuzzifier: converts the numerical values of the measured
data into linguistic values of the fuzzy set
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 129
Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 130
Fuzzy rule base: consists of a collection of fuzzy if -then
rules in the following form:
If a0 is A0k and a1 is A1k and · · · a6 is A6k, then p0 is
P0k and p1 is P1k and · · · and p6 is P6k, k = 1, 2, · · · ,K.
The fuzzy rule base can be created from training data
sequence (measured input-output pairs)
(a(1)0 , a
(1)1 , · · · , a(1)6 ; p
(1)0 , p
(1)1 , · · · , p(1)6 ),
(a(2)0 , a
(2)1 , · · · , a(2)6 ; p
(2)0 , p
(2)1 , · · · , p(2)6 ),
· · · · · · · · · · · ·Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 131
The training data can be generated in computer
simulation based on propagation model, cell structure,
and mobile user movement patterns.
The contributions of each input-output training data to
rule k is represented by a degree calculated using product
operations
Qk = µi
6∏i=0
µIik(ai)
6∏i=0
µOik(pi) (2)
• µIik(ai) : the degree of ai in the input region Iik of rule
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k
• µOik(pi) : the degree of pi in the output region Oik of
rule k
• µk: the degree of the data vector (a0, a1, · · · , a6) whichreflects the expert’s belief of the importance of the rule
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Table 1: The degree µ(a0, a1) assigned to input data
(a0, a1) which represents the usefulness of the data in
one-dimensional space.a0/a1 ES VS S SM ML L VL EL
ES 0.1 0.2 0.3 0.4 0.5 0.8 0.9 1.0VS 0.2 0.1 0.2 0.3 0.4 0.6 0.8 0.9S 0.3 0.2 0.1 0.2 0.3 0.4 0.6 0.8SM 0.4 0.3 0.2 0.1 0.2 0.3 0.4 0.6ML 0.5 0.4 0.3 0.2 0.1 0.2 0.3 0.4L 0.8 0.6 0.4 0.3 0.2 0.1 0.2 0.3VL 0.9 0.8 0.6 0.4 0.3 0.2 0.1 0.2EL 1.0 0.9 0.8 0.5 0.4 0.3 0.2 0.1
• The overall degree of rule k is obtained by
accumulating the degrees
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• There may be some conflicting rules, i.e., rules which
have the same if part but a different then part;
• For the conflicting rules, the rule that has the largest
degree is chosen.
Table 2: The fuzzy rule base for p0 (σ = 2dB) in one-
dimensional space.a0/a1 ES VS S SM ML L VL EL
ES SM S VS ZEVS SM SM S VS ZES SM SM SM S VS ESSM ML SM ML SM S S VSML ML ML ML SM SML L L L MLVL VL VL VLEL OE OE OE OE
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Fuzzy inference engine: maps fuzzy sets of fuzzifier to
fuzzy sets of defuzzifier based on the if -then fuzzy rule
base
Given Fact:
a0 is A0 and a1 is A1 and · · · and a6 is A6
Consequence:
p0 is P0 and p1 is P1 and · · · and p6 is P6
where Ai and Pi are linguistic terms for ai and pirespectively.
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Defuzzifier: converts the fuzzy sets into crisp values that
represent the mobility information.
The output of the defuzzifier is
pi =
∑Kk=1[Qk
∏6j=0 µIjk(aj)pik]∑K
k=1[Qk
∏6j=0 µIjk(aj)]
(3)
where Qk is the normalized degree of rule k and pik is
the center value of the output region of rule k.
Recursive least square (RLS) predictor:
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Figure 7: The structure of the RLS predictor
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Numerical Results
Parameters:
• D0 = 100 m
• D = 1500 m
• n = 2, 4, 6
• γi = 1
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• σ = 1, 2, · · · , 6 dB
• 50,000 mobile users uniformly distributed in the
shadow area for rule base generation and for system
performance testing respectively.
Equations for training the fuzzy rule base:
• d1 = xMT/ cos 30◦
• d2 = yMT + xMT tan 30◦ − d1
• p3 = p4 = p5 = p6 = 0
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• p0 = 1− yMT/√3D
• p1 + p2 = 1− p0, p1/p2 = d2/d1
Performance of the fuzzy inference system
one-dimensional space:
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Figure 8: Relation between p0 and p0 with δ = 2 dB
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Table 3: The mean and standard deviation of the
estimation error p0 − p0(n = 4).σ (dB) mean standard deviation
1 -3.55e-4 3.14e-22 4.10e-4 4.08e-23 -3.39e-3 4.80e-24 1.13e-3 6.25e-25 -1.75e-3 7.32e-26 -5.29e-3 8.32e-2
two-dimensional space:
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Table 4: The mean and standard deviation of the
estimation error p0 − p0(n = 4).σ (dB) mean standard deviation
1 3.35e-3 5.95e-22 1.39e-3 6.36e-23 -1.92e-3 7.66e-24 1.99e-3 9.04e-25 -4.76e-3 1.29e-16 2.51e-3 1.44e-1
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Table 5: The mean and standard deviation of the
estimation error p0 − p0(σ = 2dB).σ (dB) mean standard deviation
2 8.20e-4 6.94e-24 4.10e-4 4.08e-26 3.00e-4 3.92e-2
Performance of the RLS predictor:
500 mobile users are simulated with movement pattern
characterized by
• The initial location of each mobile user is uniformly
distributed in regions R1 of BS0 and R4 of BS1;
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• Each mobile user has a constant velocity uniformly
distributed in [36, 108] kilometers per hour;
• The initial direction of movement is uniformly
distributed in [0, 2π] and the movement direction is
then changed several times each being uniformly
distributed in [0, 2π] and independent of previous
direction (s);
• The time interval for updating the mobility information
is 1 second.
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Table 6: The mean and standard deviation of the
prediction error (pn+N,0 − pn+N,0, N = 1, 2, 3) of the
adaptive fuzzy inference system given n = 4 and σ2
dB.prediction mean standard deviationone-step 9.76e-3 9.90e-2two-step 1.08e-2 9.94e-2three-step 1.21e-2 1.01e-1
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Figure 9: The movement trajectory of a mobile user simulated
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Figure 10: The comparison of the true probability and the
corresponding one-step predicted probability for the mobile user
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Conclusions
The adaptive fuzzy inference system has the advantage
of simplicity, usefulness, low cost, and flexibility.
The system can be used to assist
• user mobility management (e.g., traffic routing)
• network resource management (e.g., admission control, congestion
control)
• handoff process in wireless/wired networks.
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