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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

Dept. Electrical and Computer Engineering , University of Waterloo, X. Shen First Previous Next Last 1

M/M/1 Queue

Poisson Arrivals

λµ

n-1 n n+1

tut ∆−∆− λ1

t∆λ

t∆µ


Balance Condition

(λ+ µ)Pn = λPn−1 + µPn+1

⇒ Pn = ρnP0 = (1− ρ)ρn, n = 0, 1, · · ·Utilization factor

ρ = λ/µ


Fluid Source Modeling of Packet Voice

cells/sec

N

VC cells/sec

NV

N-1 N

λ

αN

0 1

λN

αcells/secNV


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


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:


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,


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


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


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


−→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


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


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


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)


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


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


cj(τ) = A2 αβ

(α+ β)2e−(α+β)τ


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


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


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


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


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


021 =D

122 =D 92

4 =D

223 =D

625 =D


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


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.


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.


(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


(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


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.


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)


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


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


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


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.


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.


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


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.


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


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


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


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[∑


xP ]


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


flows xp. Denote by D(x) the cost function

D(x) =∑(i,j)

Dij[∑


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


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


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


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.


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


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


Since x∗1 + x∗

2 = r, we have

x∗1 =

√C1[r − (C2 −

√C1C2)]√

C1 +√C2

x∗2 =

√C2[r − (C1 −

√C1C2)]√

C1 +√C2


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.


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


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.


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


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


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


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


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.


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:


k∑i=1

niPiRPi = CL

(tradeoffs are possible among the maximum number of

calls allowed per user class.)


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 .


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 )


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 .


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 .


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


solution for πi can be approximated by Gaussian

distribution with the same mean value m = NP and

variance σ2 = NP (1− P )

We obtain


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


∗ 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


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


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 )


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)


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


C = m+ 4.6σ

or

C = NP + 4.6√

NP (1− P )

C = 0.02N + 0.64√N, (P = 0.02)


In general

C = NP +K√

NP (1− P )


⇒ 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)


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.


The above figure is a plot of Gϵ on a log-log scale to


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


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


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


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)


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.


∗ 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)


(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)


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


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)


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.


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].


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


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.


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,


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


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.


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


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


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


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


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.


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)


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


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)]


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)


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)


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)


calls at time t0.

New calls admitted to any cell are Poisson

(λT )inewe−λT/inew


These figures show n as a function of r and l, given Ps,

Pm, T , λ and E(n) and PQoS.


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 .


C1

C0

C2

C3

C4

C5

C6

C7

C8

C9

C10

C11

C12

C13

C14

C15

C16

C17

C18


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


Note: For the trunk reservation system, out of N

channels available in each cell, g channels are reserved


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:


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.


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.


• Introduction

• Mobility information model

• Adaptive fuzzy inference system

• Numerical results

• Conclusions


Introduction

Two main objectives associated with resource

management in communication networks are:


• 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


• 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


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.


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


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


• 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


The Adaptive Fuzzy Inference System

4Fuzzifier: converts the numerical values of the measured

data into linguistic values of the fuzzy set


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


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


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


• 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


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.


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:


Figure 7: The structure of the RLS predictor


Numerical Results

Parameters:

• D0 = 100 m

• D = 1500 m

• n = 2, 4, 6

• γi = 1


• 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


• p0 = 1− yMT/√3D

• p1 + p2 = 1− p0, p1/p2 = d2/d1

Performance of the fuzzy inference system

one-dimensional space:


Figure 8: Relation between p0 and p0 with δ = 2 dB


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:



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



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;


• 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.



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


Figure 9: The movement trajectory of a mobile user simulated


Figure 10: The comparison of the true probability and the

corresponding one-step predicted probability for the mobile user


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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