# Modeling and Analysis of Mobility Management in Mobile Communication Networks

Authors: Woon Min Baek; Ji Hyun Yoon; Chesoong Kim
Journal: The Scientific World Journal
 (2014)

---

## Abstract

Many strategies have been proposed to reduce the mobility management cost in mobile communication networks. This paper studies the zone-based registration methods that have been adopted by most mobile communication networks. We focus on two special zone-based registration methods, called two-zone registration (2Z) and two-zone registration with implicit registration by outgoing calls (2Zi). We provide a new mathematical model to analyze the exact performance of 2Z and 2Zi. We also present various numerical results, to compare the performance of 2Zi with those of 2Z and one-zone registration (1Z), and show that 2Zi is superior to 2Z as well as 1Z in most cases.


---

## Body

## 1. Introduction

The number of mobile subscribers has been increasing, and the more accelerated growth of smart phone subscribers is expected with 4G networks. In a mobile communication network, since a mobile (mobile phone) is continually moving due to its basic characteristic, mobility management of the mobile is essential, to provide communication services with high quality.One of the most important issues in mobility management is the location tracking. The location of a mobile must be maintained to connect an incoming call to the mobile if required. Location registration and paging are two basic functions to locate a mobile. Location registration is the series of processes to register a mobile’s new location information in the system database, and paging is the series of processes to page the mobile in current location, to find mobile’s exact cell and connect an incoming call, when an incoming call arrives. Since there is a tradeoff between location registration cost and paging cost, it is essential to analyze location registration cost and paging cost, in order to find the optimal location tracking method.Various location registration methods have been proposed for mobile communication networks [1–9]. However, the most important location registration method is zone-based registration [5–8], since it is adopted by most mobile communication networks.In this study, zone-based registration is considered. In zone-based registration, each mobile has azone_list, where the visited zone is stored. If a mobile moves to a new zone, which is not in its currentzone_list, the new zone is stored in thezone_list, and the mobile registers its new location information in the system database. A mobile may have more than one zone in zone-based registration [5], but most mobile communication networks adopt one-zone registration (1Z) because of ease of operation.Lin [6] suggested a precise mathematical model, to analyze the performance of the case where a mobile has two zones, and compared the performance of the two cases, in which the number of zones is one and two. In addition, Jang et al. [7] considered implicit registration effects by outgoing calls, to improve the performance of the case where a mobile has two zones. However, Jang et al. [7] assumed a special mobility model and provided just a rough approximation of the performance.Even Lin [6] provided a precise mathematical model for the performance of the case where a mobile has two zones, but his model is too complex to be applied to Jang et al.’s study [7], which considers implicit registration by outgoing calls under two registered zones, to improve the performance.In this study, we derive a new mathematical model to analyze the exact performance of two-zone registration (2Z) and two-zone registration with implicit registration by outgoing calls (2Zi). Section 2 briefly describes general zone-based registration methods. Section 3 describes the mathematical model that we derived, to analyze the performance of 2Z and 2Zi. Section 4 presents the computational results of the signaling cost on radio channels, using our model. Section 5 summarizes the results and suggests future research directions.


## 2. Zone-Based Registration and 2Z

In zone-based registration, whenever a mobile moves to a new zone that is not in its currentzone_list, this new zone is stored in itszone_list, and the mobile registers new location information in the system database. If a mobile can have only one registered zone (i.e., one-zone registration, 1Z), the mobile stores the newly entered zone in itszone_list every time it moves from one zone to another. Thus, the system knows the zone in which the mobile is located, and the paging process always meets with success for incoming calls. Figure 1 shows the movement of a mobile.Figure 1
Movement of a mobile.However, if the mobile can have two registered zones (i.e., two-zone registration, 2Z), the system sometimes does not know the exact zone in which the mobile is located. For example, two zones, A and B, are stored in thezone_list, and the mobile is currently in zone B. Let the left-hand zone denote the most recently registered zone in the followingzone_list andsystem_DB:   z
o
n
e
_
l
i
s
t
B
A
s
y
s
t
e
m
_
D
B
B
AWhen the mobile enters a new zone, C, then thezone_list andsystem_DB are changed as follows:z
o
n
e
_
l
i
s
t
C
B
s
y
s
t
e
m
_
D
B
C
BIf an incoming call arrives in this situation, then the system pages the mobile in zone C and this paging succeeds (this is referred to aszone hit).Consider another case. If the mobile reenters zone B, thenzone_list is changed as below, but location registration does not occur, because zone B is already stored in the currentzone_list:   z
o
n
e
_
l
i
s
t
B
C
s
y
s
t
e
m
_
D
B
C
BIn this case, the system does not know the correct zone (B) in which the mobile is located, and the paging process is somewhat complicated. If an incoming call arrives, the system pages the mobile in zone C, since the mobile is known to be in this zone. If there is no response to paging after a predetermined time, the system recognizes that the mobile is not in zone C, but in zone B, and next pages the mobile in zone B (this is referred to aszone miss; note that the second paging always succeeds). This is one of the disadvantages of 2Z. That is, when the mobile reenters a previously visited zone and an incoming call occurs, the system must make two-step paging because ofzone miss.However, two-step paging can be avoided, if two-zone registration with implicit registration by outgoing calls (2Zi) is employed. For example, if the mobile enters zones in the order A→B→C→B and the outgoing call occurs in the last zone, B, then the call setup messages of an outgoing call can provide the system with the exact zone, B, and the system can successfully page the mobile in zone B at this time. In other words, when the mobile makes the outgoing call, call setup messages can provide the correct location information of the mobile implicitly, without an additional location registration message. This is termed implicit registration [5, 7].Henceforth, for convenience, we refer to location registration by entering a new zone as regular location registration (RR) and the location registration effect by an outgoing call as implicit registration (IR).2Zi was considered and analyzed by Jang et al. [7], assuming the following 4-direction mobility model.(i)
A mobile moves in a straight line, until it reaches a turning point.
(ii)
When it reaches the turning point, it can choose one of 4 directions with equal probability.
(iii)
The distance between two consecutive turning points is exponentially distributed.When the above mobility model is assumed, it is impossible to express the exact equations for the performance measures such as registration cost and paging cost. Thus, for convenience, Jang et al. [7] assumed that, once a mobile enters a zone and makes one direction change, it is located at a random point in the zone. This wild assumption makes it possible to obtain some performance measures, but they are inherently rough approximations of the exact performance. Lin [6] provided a precise model for the performance of 2Z, but Lin’s model is too complex to be applied to Jang et al.’s study, which considers implicit registration by outgoing calls under two registered zones.


## 3. New Mathematical Model and Performance Analysis

In this section, we propose a new mathematical model, to analyze the exact performance of 2Z and 2Zi. The radio channel is the most important resource determining the network performance in mobile communication networks. Although, thanks to technological enhancements, the capacity of mobile communication systems has been greatly improved, radio channels still have their own capacity and technological limit. Thus, the signaling cost on radio channels determines the performance of the entire mobile communication system. The performance analysis of 2Z and 2Zi is conducted from this viewpoint.


### 3.1. Notations and Assumptions

The following notations are defined, to analyze the signaling cost on radio channels:N
1
(
Z
): number of location registrations between two incoming calls in the location registration method Z;
C
1
(
Z
): total location registration cost between two incoming calls in the location registration method Z;
N
2
(
Z
): number of cells for paging for an incoming call in the location registration method Z;
C
2
(
Z
): total paging cost per zone for an incoming call in the location registration method Z;
C
p: signaling cost for paging per cell on radio channels;
C
u: signaling cost for one location registration on radio channels;
θ: probability of returning to the registered zone;
n: number of cells per zone;
t
c: interval between two incoming calls (r. v.);
t
o
c: interval between two outgoing calls (r. v.);
t
m: sojourn time in a zone (r. v.);
λ
c: arrival rate of incoming calls;
λ
o
c: arrival rate of outgoing calls;
1
/
λ
m: mean of sojourn time in a zone;
ρ: call-to-mobility ratio (CMR), ρ
=
λ
c
/
λ
m;
f
m
*
(
s
): Laplace-Stieltjes Transform for
t
m (=
∫
t
=
0
∞
e
-
s
t
f
m
(
t
)
d
t
);
Λ: probability that an outgoing call occurs before an incoming call, Λ
=
Pr
⁡
[
t
o
c
≤
t
c
]
=
λ
o
c
/
(
λ
c
+
λ
o
c
);
p: probability that an outgoing call occurs while in the zone, p
=
Pr
⁡
[
t
o
c
≤
t
m
]
=
(
λ
m
/
λ
o
c
)
[
1
-
f
m
*
(
λ
o
c
)
];
q: probability that an outgoing call does not occur while in the zone, q
=
1
-
p.In addition, the following assumptions are necessary to analyze the signaling cost on radio channels:(i)
the incoming calls to a mobile form a Poisson process withλ
c;
(ii)
the outgoing calls from a mobile form a Poisson process withλ
o
c;
(iii)
the sojourn time in a zone follows a general distribution with a mean of1
/
λ
m;
(iv)
the first paging is applied to the most recently registered zone. If there is no response, the second paging is applied to the other zone.


### 3.2. Performance Analysis of 2Z and 2Zi

This section studies the performance analysis for 2Z and 2Zi. We first estimate registration cost and paging cost between two incoming calls, which constitute the total signaling cost.To find registration cost and paging cost between two incoming calls, let us introduce the probabilityα
(
K
) that the mobile moves across K zones between two incoming calls. We use Lin’s result on α
(
K
) [6], because this is closely related to our study:
(1)
α
(
K
)
=
{
1
-
1
ρ
[
1
-
f
m
*
(
λ
c
)
]
,
K
=
0
,
1
ρ
[
1
-
f
m
*
(
λ
c
)
]
2
[
f
m
*
(
λ
c
)
]
K
-
1
,
K
≥
1
.


#### 3.2.1. Registration Cost

The number of RRs between two incoming calls is [6]
(2)
N
1
(
2
Z
)
=
∑
k
=
0
∞
∑
i
=
0
K
i
(
k
i
)
θ
k
-
i
(
1
-
θ
)
i
α
(
k
)
=
(
1
-
θ
)
ρ
.The number of RRs between two incoming calls in 2Zi is the same as that in 2Z:
(3)
N
1
(
2
Z
i
)
=
(
1
-
θ
)
ρ
.Thus, the location registration cost between two incoming calls is(4)
C
1
(
2
Z
)
=
C
1
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
.


#### 3.2.2. Paging Cost

Next, consider the paging cost. The paging cost can be derived by multiplying the number of cells to page by the paging cost per cell. In the case of 2Z and 2Zi, the number of cells to page for an incoming call is the sum of the cases where the system has correct location information (first paging success orzone hit) and incorrect location information (first paging failure orzone miss):
(5)
N
2
(
2
Z
)
=
n
+
n
·
Pr
⁡
(
zone  miss  in
2
Z
)
=
n
+
n
[
1
-
Pr
⁡
(
zone  hit  in
2
Z
)
]
,
N
2
(
2
Z
i
)
=
n
+
n
[
1
-
Pr
⁡
(
zone  hit  in
2
Z
i
)
]
.To obtain the probability of the first paging success in (5), let us define the conditional probability S
Z
(
K
) that, given that the mobile moves across K zones between two incoming calls, the first page succeeds in the location registration methodZ. Then, the probabilities that the first page succeeds in 2Z and 2Zi are, respectively,
(6)
Pr
⁡
[
zone  hit  in
2
Z
]
=
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
,
Pr
⁡
[
zone  hit  in
2
Z
i
]
=
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
.To obtain the above probabilities, we need to derive the general expressions ofS
2
Z
(
K
)  and  S
2
Z
i
(
K
).For the sake of convenience, we first derive the probabilityS
2
Z
i
(
K
) that the first page succeeds in 2Zi, given that the mobile moves across K zones between two incoming calls.(1
) Derivation of the Conditional Probability
S
2
Z
i
(
K
)
. Let us consider the following sequential procedure for deriving the general expression of S
2
Z
i
(
K
)
.
(i)
ForK
=
0, no movement occurs between two incoming calls, so the first page always succeeds, and S
2
Z
i
(
0
)
=
1.
(ii)
ForK
=
1, the probability that the first page succeeds can be obtained in the following two cases.
(1)
Case 1. The mobile moves to a new zone, with probability (1
-
θ
). In this case, thezone_list is updated by RR; thus, the probability that the first page succeeds is the same as that when K
=
0.
(2)
Case 2. The mobile moves back to the zone from whence it came, with probabilityθ. In this case, the system has incorrect information as to which zone the mobile is located in, but thezone_list can be updated by IR, with probability Λ that an outgoing call occurs before an incoming call. Therefore,
(7)
S
2
Z
i
(
1
)
=
(
1
-
θ
)
S
2
Z
i
(
0
)
+
θ
Λ
.
(iii)
ForK
=
2, the probability that the first page succeeds can also be obtained in the following two cases.
(1)
Case 1. The first movement of the mobile is to a new zone, with probability (1
-
θ
). In this case, thezone_list is updated; thus, the probability that the first page succeeds is the same as that when K
=
1.
(2)
Case 2. The first movement of the mobile is back to the zone from whence it came, with probabilityθ. In this case, if thezone_list is updated by IR, with probabilityp that an outgoing call occurs while the mobile is in the zone, then the probability that the first page succeeds is the same as that when K
=
1. If an outgoing call does not occur while the mobile is in the zone, then the system has incorrect location information, and the probability that the first page succeeds is the same as that when K
=
0, because the system will have correct location information when the mobile either enters a new zone or moves back to the zone from whence it came, in the second movement. Therefore,
(8)
S
2
Z
i
(
2
)
=
(
1
-
θ
)
S
2
Z
i
(
1
)
+
θ
[
p
S
2
Z
i
(
1
)
+
(
1
-
p
)
S
2
Z
i
(
0
)
]
.
(iv)
The process can be generalized, whenK
=
k. In the first movement after an incoming call, the mobile moves to a new zone with probability (1
-
θ
) or moves back to the zone from whence it came, with probability θ.
(1)
Case 1. In the case where the mobile moves to a new zone, the probability that the first page succeeds is the same as that whenK
=
k
-
1
.
(2)
Case 2. In the case where the mobile moves back to the zone from whence it came, if an outgoing call occurs while the mobile is in the zone with probabilityp, then thezone_list is updated by IR, and the probability that the first page succeeds is the same as that when K
=
k
-
1. Otherwise, the system will have the correct location information after the mobile makes one more movement, by either entering a new zone or moving back to the zone from whence it came. Thus, the probability that the first page succeeds is the same as that when K
=
k
-
2. Therefore, we get a recurrence formula for S
2
Z
i
(
k
) as follows:
(9)
S
2
Z
i
(
k
)
=
(
1
-
θ
)
S
2
Z
i
(
k
-
1
)
+
θ
[
p
S
2
Z
i
(
k
-
1
)
+
(
1
-
p
)
S
2
Z
i
(
k
-
2
)
]
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
for
k
=
2,3
,
…
,
S
2
Z
i
(
0
)
=
1
,
S
2
Z
i
(
1
)
=
(
1
-
θ
)
+
θ
Λ
.Note thatΛ
=
0 and p
=
0 in the case of 2Z, because IR by an outgoing call is not employed. Therefore,
(10)
S
2
Z
(
k
)
=
(
1
-
θ
)
S
2
Z
(
k
-
1
)
+
θ
S
2
Z
(
k
-
2
)
,
hhhhhhhhhhhhhhhhhhhhhh
for
k
=
2,3
,
…
,
S
2
Z
(
0
)
=
1
,
S
2
Z
(
1
)
=
(
1
-
θ
)
.(2
) Paging Cost for an Incoming Call. Using (9) and (10), for 2Z and 2Zi, respectively, (5) can be written by
(11)
N
2
(
2
Z
)
=
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
,
N
2
(
2
Z
i
)
=
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
.Using the above results, we can compute the number of cells required when an incoming call occurs, and we can find the paging cost, by multiplying this number by the paging cost per cell. Finally, the total paging costs for an incoming call for 2Z and 2Zi are, respectively,
(12)
C
2
(
2
Z
)
=
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
]
C
p
,
C
2
(
2
Z
i
)
=
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
]
C
p
.


#### 3.2.3. Total Signaling Cost

The total signaling cost on radio channels is derived by combining the registration cost and the paging cost as follows:(13)
C
(
2
Z
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
]
C
p
,
C
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
]
C
p
.


#### 3.2.4. Propositions for Explicit Expressions of Costs

Proposition 1.
The general solution of (9) is
(14)
S
2
Z
i
(
n
)
=
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
n
]
1
+
θ
q
.Proof.
For convenience, let us omit subscripts. Rearranging the above equation, we can get(15)
S
(
n
)
-
S
(
n
-
1
)
=
-
θ
q
[
S
(
n
-
1
)
-
S
(
n
-
2
)
]
,
hhhhhhhhhhh
for
n
=
2,3
,
…
.
It can be seen that differences of the progression form a geometric progression with equal ratio (-
θ
q
). Then, the general term of the progression S
(
n
)  can be easily obtained as follows:
(16)
S
(
n
)
=
S
(
0
)
+
∑
i
=
1
n
(
-
θ
q
)
i
=
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
n
]
1
+
θ
q
=
{
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
θ
q
)
n
]
1
+
θ
q
if
n
is  even  number
1
+
-
θ
(
1
-
Λ
)
[
1
+
(
θ
q
)
n
]
1
+
θ
q
if
n
is  odd  number
.Proposition 2.
The general solution of (10) is  S
2
Z
R
(
n
)
=
[
1
-
(
-
θ
)
n
+
1
]
/
(
1
+
θ
).Proof.
For convenience, let us omit subscripts. In the case of 2Z, since IR by an outgoing call is not employed,Λ
=
0 and p
=
0. Inserting these values into (10), we have
(17)
S
(
n
)
=
[
1
-
(
-
θ
)
n
+
1
]
1
+
θ
=
{
1
+
θ
n
+
1
1
+
θ
if
n
is  even  number
1
-
θ
n
+
1
1
+
θ
if
n
is  odd  number
.Proposition 3.
Equation (17) gives the same P
[
z
o
n
e
h
i
t
i
n
2
Z
] as in Lin’s study [6].Proof.
From (18), (19), and (20) of Lin’s study [6],
(18)
P
[
z
o
n
e
h
i
t
i
n
2
Z
]
=
ω
1
+
ω
2
+
ω
3
=
α
(
0
)
+
∑
K
=
1
∞
ω
2
(
K
)
α
(
K
)
+
∑
i
=
1
∞
θ
2
i
α
(
2
i
)
=
α
(
0
)
+
∑
K
=
1
∞
1
-
θ
2
⌊
(
K
-
1
)
/
2
⌋
+
2
1
+
θ
α
(
K
)
+
∑
i
=
1
∞
θ
2
i
α
(
2
i
)
.
If we express the previous equation as∑
i
=
1
∞
z
(
K
)
α
(
K
),   it is easy to show that z
(
K
)  is as follows:
(19)
z
(
K
)
=
{
1
,
if
K
=
0
,
1
-
θ
K
1
+
θ
+
θ
K
=
1
+
θ
K
+
1
1
+
θ
,
if
K
is  even  number,
1
-
θ
K
+
1
1
+
θ
+
0
=
1
-
θ
K
+
1
1
+
θ
,
if
K
is  odd  number
.
Sincez
(
K
)
=
S
(
K
)  for all  K
≥
0, the proof is complete.Note that the above proposition implies that our model includes Lin’s model on 2Z [6].As shown in the appendix, the probability that the first paging succeeds in 2Zi, given that the mobile moves acrossk zones between two incoming calls, is composed of three probabilities for three exclusive cases, ω
1
(
k
),  ω
2
(
k
),   and  ω
3
(
k
)
,  and their sum, ω
(
k
), is
(20)
ω
(
k
)
=
{
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
+
1
q
k
+
1
]
1
-
θ
2
q
2
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
-
1
q
k
-
1
]
1
-
θ
2
q
2
+
Λ
,
if
K
is  odd  number
(
k
≥
3
)
,
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
Λ
,
if
K
is  even  number
(
k
≥
2
)
,
(
1
-
θ
)
(
1
-
Λ
)
+
Λ
=
1
-
θ
+
θ
Λ
,
if
k
=
1
,
1
,
if
k
=
0
.Proposition 4.
Equation (14),  S
(
k
)
=
1
+
(
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
k
]
/
(
1
+
θ
q
)
), is the same as (20).Proof.
(i) Whenk
=
0 and k
=
1, it is trivial.
(ii) Whenk is even number (
k
≥
2
),
(21)
ω
(
k
)
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
Λ
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
+
θ
2
q
2
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
]
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
×
[
1
+
θ
2
q
2
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
]
+
Λ
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
q
+
θ
2
q
2
-
θ
3
q
3
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
-
θ
k
-
1
q
k
-
1
]
1
+
θ
q
1
+
θ
q
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
+
θ
q
.
(iii) Whenk is odd number (
k
≥
3
),
(22)
ω
(
k
)
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
+
1
q
k
+
1
]
1
-
θ
2
q
2
+
p
(
1
-
Λ
)
θ
2
q
×
[
1
-
θ
k
-
1
q
k
-
1
]
1
-
θ
2
q
2
+
Λ
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
q
+
θ
2
q
2
-
θ
3
q
3
+
θ
4
q
4
+
⋯
-
θ
k
-
2
q
k
-
2
+
θ
k
-
1
q
k
-
1
]
1
+
θ
q
1
+
θ
q
=
1
-
θ
(
1
-
Λ
)
[
1
+
θ
k
q
k
]
1
+
θ
q
.Proposition 5.
The explicit expressions ofC
(
2
Z
) and C
(
2
Z
i
) are
(23)
C
(
2
Z
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
θ
[
1
-
f
m
*
(
λ
c
)
]
ρ
[
1
+
θ
f
m
*
(
λ
c
)
]
)
]
C
p
,
(24)
C
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
θ
[
1
-
f
m
*
(
λ
c
)
]
(
1
-
Λ
)
(
1
+
θ
q
)
ρ
(
1
+
θ
q
)
[
1
+
θ
q
f
m
*
(
λ
c
)
]
)
]
C
p
.Proof.
The result follows from(25)
∑
k
=
0
∞
S
2
Z
i
(
k
)
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
∑
k
=
1
∞
{
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
k
]
1
+
θ
q
}
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
1
+
θ
(
q
+
Λ
-
1
)
1
+
θ
q
×
∑
k
=
1
∞
α
(
k
)
-
θ
(
Λ
-
1
)
1
+
θ
q
∑
k
=
1
∞
(
-
θ
q
)
k
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
[
1
+
θ
(
q
+
Λ
-
1
)
]
[
1
-
f
m
*
(
λ
c
)
]
2
ρ
(
1
+
θ
q
)
×
[
1
1
-
f
m
*
(
λ
c
)
]
-
θ
(
Λ
-
1
)
[
1
-
f
m
*
(
λ
c
)
]
2
ρ
(
1
+
θ
q
)
×
[
-
θ
q
1
+
θ
q
f
m
*
(
λ
c
)
]
=
1
-
θ
[
1
-
f
m
*
(
λ
c
)
]
(
1
-
Λ
)
(
1
+
θ
q
)
ρ
(
1
+
θ
q
)
[
1
+
θ
q
f
m
*
(
λ
c
)
]
,
(26)
∑
k
=
0
∞
S
2
Z
(
k
)
α
(
k
)
=
1
-
θ
[
1
-
f
m
*
(
λ
c
)
]
ρ
[
1
+
θ
f
m
*
(
λ
c
)
]
.


## 3.1. Notations and Assumptions

The following notations are defined, to analyze the signaling cost on radio channels:N
1
(
Z
): number of location registrations between two incoming calls in the location registration method Z;
C
1
(
Z
): total location registration cost between two incoming calls in the location registration method Z;
N
2
(
Z
): number of cells for paging for an incoming call in the location registration method Z;
C
2
(
Z
): total paging cost per zone for an incoming call in the location registration method Z;
C
p: signaling cost for paging per cell on radio channels;
C
u: signaling cost for one location registration on radio channels;
θ: probability of returning to the registered zone;
n: number of cells per zone;
t
c: interval between two incoming calls (r. v.);
t
o
c: interval between two outgoing calls (r. v.);
t
m: sojourn time in a zone (r. v.);
λ
c: arrival rate of incoming calls;
λ
o
c: arrival rate of outgoing calls;
1
/
λ
m: mean of sojourn time in a zone;
ρ: call-to-mobility ratio (CMR), ρ
=
λ
c
/
λ
m;
f
m
*
(
s
): Laplace-Stieltjes Transform for
t
m (=
∫
t
=
0
∞
e
-
s
t
f
m
(
t
)
d
t
);
Λ: probability that an outgoing call occurs before an incoming call, Λ
=
Pr
⁡
[
t
o
c
≤
t
c
]
=
λ
o
c
/
(
λ
c
+
λ
o
c
);
p: probability that an outgoing call occurs while in the zone, p
=
Pr
⁡
[
t
o
c
≤
t
m
]
=
(
λ
m
/
λ
o
c
)
[
1
-
f
m
*
(
λ
o
c
)
];
q: probability that an outgoing call does not occur while in the zone, q
=
1
-
p.In addition, the following assumptions are necessary to analyze the signaling cost on radio channels:(i)
the incoming calls to a mobile form a Poisson process withλ
c;
(ii)
the outgoing calls from a mobile form a Poisson process withλ
o
c;
(iii)
the sojourn time in a zone follows a general distribution with a mean of1
/
λ
m;
(iv)
the first paging is applied to the most recently registered zone. If there is no response, the second paging is applied to the other zone.


## 3.2. Performance Analysis of 2Z and 2Zi

This section studies the performance analysis for 2Z and 2Zi. We first estimate registration cost and paging cost between two incoming calls, which constitute the total signaling cost.To find registration cost and paging cost between two incoming calls, let us introduce the probabilityα
(
K
) that the mobile moves across K zones between two incoming calls. We use Lin’s result on α
(
K
) [6], because this is closely related to our study:
(1)
α
(
K
)
=
{
1
-
1
ρ
[
1
-
f
m
*
(
λ
c
)
]
,
K
=
0
,
1
ρ
[
1
-
f
m
*
(
λ
c
)
]
2
[
f
m
*
(
λ
c
)
]
K
-
1
,
K
≥
1
.


### 3.2.1. Registration Cost

The number of RRs between two incoming calls is [6]
(2)
N
1
(
2
Z
)
=
∑
k
=
0
∞
∑
i
=
0
K
i
(
k
i
)
θ
k
-
i
(
1
-
θ
)
i
α
(
k
)
=
(
1
-
θ
)
ρ
.The number of RRs between two incoming calls in 2Zi is the same as that in 2Z:
(3)
N
1
(
2
Z
i
)
=
(
1
-
θ
)
ρ
.Thus, the location registration cost between two incoming calls is(4)
C
1
(
2
Z
)
=
C
1
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
.


### 3.2.2. Paging Cost

Next, consider the paging cost. The paging cost can be derived by multiplying the number of cells to page by the paging cost per cell. In the case of 2Z and 2Zi, the number of cells to page for an incoming call is the sum of the cases where the system has correct location information (first paging success orzone hit) and incorrect location information (first paging failure orzone miss):
(5)
N
2
(
2
Z
)
=
n
+
n
·
Pr
⁡
(
zone  miss  in
2
Z
)
=
n
+
n
[
1
-
Pr
⁡
(
zone  hit  in
2
Z
)
]
,
N
2
(
2
Z
i
)
=
n
+
n
[
1
-
Pr
⁡
(
zone  hit  in
2
Z
i
)
]
.To obtain the probability of the first paging success in (5), let us define the conditional probability S
Z
(
K
) that, given that the mobile moves across K zones between two incoming calls, the first page succeeds in the location registration methodZ. Then, the probabilities that the first page succeeds in 2Z and 2Zi are, respectively,
(6)
Pr
⁡
[
zone  hit  in
2
Z
]
=
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
,
Pr
⁡
[
zone  hit  in
2
Z
i
]
=
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
.To obtain the above probabilities, we need to derive the general expressions ofS
2
Z
(
K
)  and  S
2
Z
i
(
K
).For the sake of convenience, we first derive the probabilityS
2
Z
i
(
K
) that the first page succeeds in 2Zi, given that the mobile moves across K zones between two incoming calls.(1
) Derivation of the Conditional Probability
S
2
Z
i
(
K
)
. Let us consider the following sequential procedure for deriving the general expression of S
2
Z
i
(
K
)
.
(i)
ForK
=
0, no movement occurs between two incoming calls, so the first page always succeeds, and S
2
Z
i
(
0
)
=
1.
(ii)
ForK
=
1, the probability that the first page succeeds can be obtained in the following two cases.
(1)
Case 1. The mobile moves to a new zone, with probability (1
-
θ
). In this case, thezone_list is updated by RR; thus, the probability that the first page succeeds is the same as that when K
=
0.
(2)
Case 2. The mobile moves back to the zone from whence it came, with probabilityθ. In this case, the system has incorrect information as to which zone the mobile is located in, but thezone_list can be updated by IR, with probability Λ that an outgoing call occurs before an incoming call. Therefore,
(7)
S
2
Z
i
(
1
)
=
(
1
-
θ
)
S
2
Z
i
(
0
)
+
θ
Λ
.
(iii)
ForK
=
2, the probability that the first page succeeds can also be obtained in the following two cases.
(1)
Case 1. The first movement of the mobile is to a new zone, with probability (1
-
θ
). In this case, thezone_list is updated; thus, the probability that the first page succeeds is the same as that when K
=
1.
(2)
Case 2. The first movement of the mobile is back to the zone from whence it came, with probabilityθ. In this case, if thezone_list is updated by IR, with probabilityp that an outgoing call occurs while the mobile is in the zone, then the probability that the first page succeeds is the same as that when K
=
1. If an outgoing call does not occur while the mobile is in the zone, then the system has incorrect location information, and the probability that the first page succeeds is the same as that when K
=
0, because the system will have correct location information when the mobile either enters a new zone or moves back to the zone from whence it came, in the second movement. Therefore,
(8)
S
2
Z
i
(
2
)
=
(
1
-
θ
)
S
2
Z
i
(
1
)
+
θ
[
p
S
2
Z
i
(
1
)
+
(
1
-
p
)
S
2
Z
i
(
0
)
]
.
(iv)
The process can be generalized, whenK
=
k. In the first movement after an incoming call, the mobile moves to a new zone with probability (1
-
θ
) or moves back to the zone from whence it came, with probability θ.
(1)
Case 1. In the case where the mobile moves to a new zone, the probability that the first page succeeds is the same as that whenK
=
k
-
1
.
(2)
Case 2. In the case where the mobile moves back to the zone from whence it came, if an outgoing call occurs while the mobile is in the zone with probabilityp, then thezone_list is updated by IR, and the probability that the first page succeeds is the same as that when K
=
k
-
1. Otherwise, the system will have the correct location information after the mobile makes one more movement, by either entering a new zone or moving back to the zone from whence it came. Thus, the probability that the first page succeeds is the same as that when K
=
k
-
2. Therefore, we get a recurrence formula for S
2
Z
i
(
k
) as follows:
(9)
S
2
Z
i
(
k
)
=
(
1
-
θ
)
S
2
Z
i
(
k
-
1
)
+
θ
[
p
S
2
Z
i
(
k
-
1
)
+
(
1
-
p
)
S
2
Z
i
(
k
-
2
)
]
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
for
k
=
2,3
,
…
,
S
2
Z
i
(
0
)
=
1
,
S
2
Z
i
(
1
)
=
(
1
-
θ
)
+
θ
Λ
.Note thatΛ
=
0 and p
=
0 in the case of 2Z, because IR by an outgoing call is not employed. Therefore,
(10)
S
2
Z
(
k
)
=
(
1
-
θ
)
S
2
Z
(
k
-
1
)
+
θ
S
2
Z
(
k
-
2
)
,
hhhhhhhhhhhhhhhhhhhhhh
for
k
=
2,3
,
…
,
S
2
Z
(
0
)
=
1
,
S
2
Z
(
1
)
=
(
1
-
θ
)
.(2
) Paging Cost for an Incoming Call. Using (9) and (10), for 2Z and 2Zi, respectively, (5) can be written by
(11)
N
2
(
2
Z
)
=
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
,
N
2
(
2
Z
i
)
=
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
.Using the above results, we can compute the number of cells required when an incoming call occurs, and we can find the paging cost, by multiplying this number by the paging cost per cell. Finally, the total paging costs for an incoming call for 2Z and 2Zi are, respectively,
(12)
C
2
(
2
Z
)
=
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
]
C
p
,
C
2
(
2
Z
i
)
=
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
]
C
p
.


### 3.2.3. Total Signaling Cost

The total signaling cost on radio channels is derived by combining the registration cost and the paging cost as follows:(13)
C
(
2
Z
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
]
C
p
,
C
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
]
C
p
.


### 3.2.4. Propositions for Explicit Expressions of Costs

Proposition 1.
The general solution of (9) is
(14)
S
2
Z
i
(
n
)
=
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
n
]
1
+
θ
q
.Proof.
For convenience, let us omit subscripts. Rearranging the above equation, we can get(15)
S
(
n
)
-
S
(
n
-
1
)
=
-
θ
q
[
S
(
n
-
1
)
-
S
(
n
-
2
)
]
,
hhhhhhhhhhh
for
n
=
2,3
,
…
.
It can be seen that differences of the progression form a geometric progression with equal ratio (-
θ
q
). Then, the general term of the progression S
(
n
)  can be easily obtained as follows:
(16)
S
(
n
)
=
S
(
0
)
+
∑
i
=
1
n
(
-
θ
q
)
i
=
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
n
]
1
+
θ
q
=
{
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
θ
q
)
n
]
1
+
θ
q
if
n
is  even  number
1
+
-
θ
(
1
-
Λ
)
[
1
+
(
θ
q
)
n
]
1
+
θ
q
if
n
is  odd  number
.Proposition 2.
The general solution of (10) is  S
2
Z
R
(
n
)
=
[
1
-
(
-
θ
)
n
+
1
]
/
(
1
+
θ
).Proof.
For convenience, let us omit subscripts. In the case of 2Z, since IR by an outgoing call is not employed,Λ
=
0 and p
=
0. Inserting these values into (10), we have
(17)
S
(
n
)
=
[
1
-
(
-
θ
)
n
+
1
]
1
+
θ
=
{
1
+
θ
n
+
1
1
+
θ
if
n
is  even  number
1
-
θ
n
+
1
1
+
θ
if
n
is  odd  number
.Proposition 3.
Equation (17) gives the same P
[
z
o
n
e
h
i
t
i
n
2
Z
] as in Lin’s study [6].Proof.
From (18), (19), and (20) of Lin’s study [6],
(18)
P
[
z
o
n
e
h
i
t
i
n
2
Z
]
=
ω
1
+
ω
2
+
ω
3
=
α
(
0
)
+
∑
K
=
1
∞
ω
2
(
K
)
α
(
K
)
+
∑
i
=
1
∞
θ
2
i
α
(
2
i
)
=
α
(
0
)
+
∑
K
=
1
∞
1
-
θ
2
⌊
(
K
-
1
)
/
2
⌋
+
2
1
+
θ
α
(
K
)
+
∑
i
=
1
∞
θ
2
i
α
(
2
i
)
.
If we express the previous equation as∑
i
=
1
∞
z
(
K
)
α
(
K
),   it is easy to show that z
(
K
)  is as follows:
(19)
z
(
K
)
=
{
1
,
if
K
=
0
,
1
-
θ
K
1
+
θ
+
θ
K
=
1
+
θ
K
+
1
1
+
θ
,
if
K
is  even  number,
1
-
θ
K
+
1
1
+
θ
+
0
=
1
-
θ
K
+
1
1
+
θ
,
if
K
is  odd  number
.
Sincez
(
K
)
=
S
(
K
)  for all  K
≥
0, the proof is complete.Note that the above proposition implies that our model includes Lin’s model on 2Z [6].As shown in the appendix, the probability that the first paging succeeds in 2Zi, given that the mobile moves acrossk zones between two incoming calls, is composed of three probabilities for three exclusive cases, ω
1
(
k
),  ω
2
(
k
),   and  ω
3
(
k
)
,  and their sum, ω
(
k
), is
(20)
ω
(
k
)
=
{
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
+
1
q
k
+
1
]
1
-
θ
2
q
2
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
-
1
q
k
-
1
]
1
-
θ
2
q
2
+
Λ
,
if
K
is  odd  number
(
k
≥
3
)
,
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
Λ
,
if
K
is  even  number
(
k
≥
2
)
,
(
1
-
θ
)
(
1
-
Λ
)
+
Λ
=
1
-
θ
+
θ
Λ
,
if
k
=
1
,
1
,
if
k
=
0
.Proposition 4.
Equation (14),  S
(
k
)
=
1
+
(
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
k
]
/
(
1
+
θ
q
)
), is the same as (20).Proof.
(i) Whenk
=
0 and k
=
1, it is trivial.
(ii) Whenk is even number (
k
≥
2
),
(21)
ω
(
k
)
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
Λ
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
+
θ
2
q
2
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
]
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
×
[
1
+
θ
2
q
2
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
]
+
Λ
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
q
+
θ
2
q
2
-
θ
3
q
3
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
-
θ
k
-
1
q
k
-
1
]
1
+
θ
q
1
+
θ
q
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
+
θ
q
.
(iii) Whenk is odd number (
k
≥
3
),
(22)
ω
(
k
)
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
+
1
q
k
+
1
]
1
-
θ
2
q
2
+
p
(
1
-
Λ
)
θ
2
q
×
[
1
-
θ
k
-
1
q
k
-
1
]
1
-
θ
2
q
2
+
Λ
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
q
+
θ
2
q
2
-
θ
3
q
3
+
θ
4
q
4
+
⋯
-
θ
k
-
2
q
k
-
2
+
θ
k
-
1
q
k
-
1
]
1
+
θ
q
1
+
θ
q
=
1
-
θ
(
1
-
Λ
)
[
1
+
θ
k
q
k
]
1
+
θ
q
.Proposition 5.
The explicit expressions ofC
(
2
Z
) and C
(
2
Z
i
) are
(23)
C
(
2
Z
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
θ
[
1
-
f
m
*
(
λ
c
)
]
ρ
[
1
+
θ
f
m
*
(
λ
c
)
]
)
]
C
p
,
(24)
C
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
θ
[
1
-
f
m
*
(
λ
c
)
]
(
1
-
Λ
)
(
1
+
θ
q
)
ρ
(
1
+
θ
q
)
[
1
+
θ
q
f
m
*
(
λ
c
)
]
)
]
C
p
.Proof.
The result follows from(25)
∑
k
=
0
∞
S
2
Z
i
(
k
)
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
∑
k
=
1
∞
{
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
k
]
1
+
θ
q
}
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
1
+
θ
(
q
+
Λ
-
1
)
1
+
θ
q
×
∑
k
=
1
∞
α
(
k
)
-
θ
(
Λ
-
1
)
1
+
θ
q
∑
k
=
1
∞
(
-
θ
q
)
k
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
[
1
+
θ
(
q
+
Λ
-
1
)
]
[
1
-
f
m
*
(
λ
c
)
]
2
ρ
(
1
+
θ
q
)
×
[
1
1
-
f
m
*
(
λ
c
)
]
-
θ
(
Λ
-
1
)
[
1
-
f
m
*
(
λ
c
)
]
2
ρ
(
1
+
θ
q
)
×
[
-
θ
q
1
+
θ
q
f
m
*
(
λ
c
)
]
=
1
-
θ
[
1
-
f
m
*
(
λ
c
)
]
(
1
-
Λ
)
(
1
+
θ
q
)
ρ
(
1
+
θ
q
)
[
1
+
θ
q
f
m
*
(
λ
c
)
]
,
(26)
∑
k
=
0
∞
S
2
Z
(
k
)
α
(
k
)
=
1
-
θ
[
1
-
f
m
*
(
λ
c
)
]
ρ
[
1
+
θ
f
m
*
(
λ
c
)
]
.


## 3.2.1. Registration Cost

The number of RRs between two incoming calls is [6]
(2)
N
1
(
2
Z
)
=
∑
k
=
0
∞
∑
i
=
0
K
i
(
k
i
)
θ
k
-
i
(
1
-
θ
)
i
α
(
k
)
=
(
1
-
θ
)
ρ
.The number of RRs between two incoming calls in 2Zi is the same as that in 2Z:
(3)
N
1
(
2
Z
i
)
=
(
1
-
θ
)
ρ
.Thus, the location registration cost between two incoming calls is(4)
C
1
(
2
Z
)
=
C
1
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
.


## 3.2.2. Paging Cost

Next, consider the paging cost. The paging cost can be derived by multiplying the number of cells to page by the paging cost per cell. In the case of 2Z and 2Zi, the number of cells to page for an incoming call is the sum of the cases where the system has correct location information (first paging success orzone hit) and incorrect location information (first paging failure orzone miss):
(5)
N
2
(
2
Z
)
=
n
+
n
·
Pr
⁡
(
zone  miss  in
2
Z
)
=
n
+
n
[
1
-
Pr
⁡
(
zone  hit  in
2
Z
)
]
,
N
2
(
2
Z
i
)
=
n
+
n
[
1
-
Pr
⁡
(
zone  hit  in
2
Z
i
)
]
.To obtain the probability of the first paging success in (5), let us define the conditional probability S
Z
(
K
) that, given that the mobile moves across K zones between two incoming calls, the first page succeeds in the location registration methodZ. Then, the probabilities that the first page succeeds in 2Z and 2Zi are, respectively,
(6)
Pr
⁡
[
zone  hit  in
2
Z
]
=
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
,
Pr
⁡
[
zone  hit  in
2
Z
i
]
=
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
.To obtain the above probabilities, we need to derive the general expressions ofS
2
Z
(
K
)  and  S
2
Z
i
(
K
).For the sake of convenience, we first derive the probabilityS
2
Z
i
(
K
) that the first page succeeds in 2Zi, given that the mobile moves across K zones between two incoming calls.(1
) Derivation of the Conditional Probability
S
2
Z
i
(
K
)
. Let us consider the following sequential procedure for deriving the general expression of S
2
Z
i
(
K
)
.
(i)
ForK
=
0, no movement occurs between two incoming calls, so the first page always succeeds, and S
2
Z
i
(
0
)
=
1.
(ii)
ForK
=
1, the probability that the first page succeeds can be obtained in the following two cases.
(1)
Case 1. The mobile moves to a new zone, with probability (1
-
θ
). In this case, thezone_list is updated by RR; thus, the probability that the first page succeeds is the same as that when K
=
0.
(2)
Case 2. The mobile moves back to the zone from whence it came, with probabilityθ. In this case, the system has incorrect information as to which zone the mobile is located in, but thezone_list can be updated by IR, with probability Λ that an outgoing call occurs before an incoming call. Therefore,
(7)
S
2
Z
i
(
1
)
=
(
1
-
θ
)
S
2
Z
i
(
0
)
+
θ
Λ
.
(iii)
ForK
=
2, the probability that the first page succeeds can also be obtained in the following two cases.
(1)
Case 1. The first movement of the mobile is to a new zone, with probability (1
-
θ
). In this case, thezone_list is updated; thus, the probability that the first page succeeds is the same as that when K
=
1.
(2)
Case 2. The first movement of the mobile is back to the zone from whence it came, with probabilityθ. In this case, if thezone_list is updated by IR, with probabilityp that an outgoing call occurs while the mobile is in the zone, then the probability that the first page succeeds is the same as that when K
=
1. If an outgoing call does not occur while the mobile is in the zone, then the system has incorrect location information, and the probability that the first page succeeds is the same as that when K
=
0, because the system will have correct location information when the mobile either enters a new zone or moves back to the zone from whence it came, in the second movement. Therefore,
(8)
S
2
Z
i
(
2
)
=
(
1
-
θ
)
S
2
Z
i
(
1
)
+
θ
[
p
S
2
Z
i
(
1
)
+
(
1
-
p
)
S
2
Z
i
(
0
)
]
.
(iv)
The process can be generalized, whenK
=
k. In the first movement after an incoming call, the mobile moves to a new zone with probability (1
-
θ
) or moves back to the zone from whence it came, with probability θ.
(1)
Case 1. In the case where the mobile moves to a new zone, the probability that the first page succeeds is the same as that whenK
=
k
-
1
.
(2)
Case 2. In the case where the mobile moves back to the zone from whence it came, if an outgoing call occurs while the mobile is in the zone with probabilityp, then thezone_list is updated by IR, and the probability that the first page succeeds is the same as that when K
=
k
-
1. Otherwise, the system will have the correct location information after the mobile makes one more movement, by either entering a new zone or moving back to the zone from whence it came. Thus, the probability that the first page succeeds is the same as that when K
=
k
-
2. Therefore, we get a recurrence formula for S
2
Z
i
(
k
) as follows:
(9)
S
2
Z
i
(
k
)
=
(
1
-
θ
)
S
2
Z
i
(
k
-
1
)
+
θ
[
p
S
2
Z
i
(
k
-
1
)
+
(
1
-
p
)
S
2
Z
i
(
k
-
2
)
]
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
for
k
=
2,3
,
…
,
S
2
Z
i
(
0
)
=
1
,
S
2
Z
i
(
1
)
=
(
1
-
θ
)
+
θ
Λ
.Note thatΛ
=
0 and p
=
0 in the case of 2Z, because IR by an outgoing call is not employed. Therefore,
(10)
S
2
Z
(
k
)
=
(
1
-
θ
)
S
2
Z
(
k
-
1
)
+
θ
S
2
Z
(
k
-
2
)
,
hhhhhhhhhhhhhhhhhhhhhh
for
k
=
2,3
,
…
,
S
2
Z
(
0
)
=
1
,
S
2
Z
(
1
)
=
(
1
-
θ
)
.(2
) Paging Cost for an Incoming Call. Using (9) and (10), for 2Z and 2Zi, respectively, (5) can be written by
(11)
N
2
(
2
Z
)
=
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
,
N
2
(
2
Z
i
)
=
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
.Using the above results, we can compute the number of cells required when an incoming call occurs, and we can find the paging cost, by multiplying this number by the paging cost per cell. Finally, the total paging costs for an incoming call for 2Z and 2Zi are, respectively,
(12)
C
2
(
2
Z
)
=
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
]
C
p
,
C
2
(
2
Z
i
)
=
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
]
C
p
.


## 3.2.3. Total Signaling Cost

The total signaling cost on radio channels is derived by combining the registration cost and the paging cost as follows:(13)
C
(
2
Z
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
(
K
)
α
(
K
)
)
]
C
p
,
C
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
1
-
∑
K
=
0
∞
S
2
Z
i
(
K
)
α
(
K
)
)
]
C
p
.


## 3.2.4. Propositions for Explicit Expressions of Costs

Proposition 1.
The general solution of (9) is
(14)
S
2
Z
i
(
n
)
=
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
n
]
1
+
θ
q
.Proof.
For convenience, let us omit subscripts. Rearranging the above equation, we can get(15)
S
(
n
)
-
S
(
n
-
1
)
=
-
θ
q
[
S
(
n
-
1
)
-
S
(
n
-
2
)
]
,
hhhhhhhhhhh
for
n
=
2,3
,
…
.
It can be seen that differences of the progression form a geometric progression with equal ratio (-
θ
q
). Then, the general term of the progression S
(
n
)  can be easily obtained as follows:
(16)
S
(
n
)
=
S
(
0
)
+
∑
i
=
1
n
(
-
θ
q
)
i
=
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
n
]
1
+
θ
q
=
{
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
θ
q
)
n
]
1
+
θ
q
if
n
is  even  number
1
+
-
θ
(
1
-
Λ
)
[
1
+
(
θ
q
)
n
]
1
+
θ
q
if
n
is  odd  number
.Proposition 2.
The general solution of (10) is  S
2
Z
R
(
n
)
=
[
1
-
(
-
θ
)
n
+
1
]
/
(
1
+
θ
).Proof.
For convenience, let us omit subscripts. In the case of 2Z, since IR by an outgoing call is not employed,Λ
=
0 and p
=
0. Inserting these values into (10), we have
(17)
S
(
n
)
=
[
1
-
(
-
θ
)
n
+
1
]
1
+
θ
=
{
1
+
θ
n
+
1
1
+
θ
if
n
is  even  number
1
-
θ
n
+
1
1
+
θ
if
n
is  odd  number
.Proposition 3.
Equation (17) gives the same P
[
z
o
n
e
h
i
t
i
n
2
Z
] as in Lin’s study [6].Proof.
From (18), (19), and (20) of Lin’s study [6],
(18)
P
[
z
o
n
e
h
i
t
i
n
2
Z
]
=
ω
1
+
ω
2
+
ω
3
=
α
(
0
)
+
∑
K
=
1
∞
ω
2
(
K
)
α
(
K
)
+
∑
i
=
1
∞
θ
2
i
α
(
2
i
)
=
α
(
0
)
+
∑
K
=
1
∞
1
-
θ
2
⌊
(
K
-
1
)
/
2
⌋
+
2
1
+
θ
α
(
K
)
+
∑
i
=
1
∞
θ
2
i
α
(
2
i
)
.
If we express the previous equation as∑
i
=
1
∞
z
(
K
)
α
(
K
),   it is easy to show that z
(
K
)  is as follows:
(19)
z
(
K
)
=
{
1
,
if
K
=
0
,
1
-
θ
K
1
+
θ
+
θ
K
=
1
+
θ
K
+
1
1
+
θ
,
if
K
is  even  number,
1
-
θ
K
+
1
1
+
θ
+
0
=
1
-
θ
K
+
1
1
+
θ
,
if
K
is  odd  number
.
Sincez
(
K
)
=
S
(
K
)  for all  K
≥
0, the proof is complete.Note that the above proposition implies that our model includes Lin’s model on 2Z [6].As shown in the appendix, the probability that the first paging succeeds in 2Zi, given that the mobile moves acrossk zones between two incoming calls, is composed of three probabilities for three exclusive cases, ω
1
(
k
),  ω
2
(
k
),   and  ω
3
(
k
)
,  and their sum, ω
(
k
), is
(20)
ω
(
k
)
=
{
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
+
1
q
k
+
1
]
1
-
θ
2
q
2
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
-
1
q
k
-
1
]
1
-
θ
2
q
2
+
Λ
,
if
K
is  odd  number
(
k
≥
3
)
,
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
Λ
,
if
K
is  even  number
(
k
≥
2
)
,
(
1
-
θ
)
(
1
-
Λ
)
+
Λ
=
1
-
θ
+
θ
Λ
,
if
k
=
1
,
1
,
if
k
=
0
.Proposition 4.
Equation (14),  S
(
k
)
=
1
+
(
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
k
]
/
(
1
+
θ
q
)
), is the same as (20).Proof.
(i) Whenk
=
0 and k
=
1, it is trivial.
(ii) Whenk is even number (
k
≥
2
),
(21)
ω
(
k
)
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
[
1
-
θ
k
q
k
]
1
-
θ
2
q
2
+
Λ
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
+
θ
2
q
2
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
]
+
θ
k
q
k
(
1
-
Λ
)
+
p
(
1
-
Λ
)
θ
2
q
×
[
1
+
θ
2
q
2
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
]
+
Λ
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
q
+
θ
2
q
2
-
θ
3
q
3
+
θ
4
q
4
+
⋯
+
θ
k
-
2
q
k
-
2
-
θ
k
-
1
q
k
-
1
]
1
+
θ
q
1
+
θ
q
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
k
q
k
]
1
+
θ
q
.
(iii) Whenk is odd number (
k
≥
3
),
(22)
ω
(
k
)
=
(
1
-
θ
)
(
1
-
Λ
)
[
1
-
θ
k
+
1
q
k
+
1
]
1
-
θ
2
q
2
+
p
(
1
-
Λ
)
θ
2
q
×
[
1
-
θ
k
-
1
q
k
-
1
]
1
-
θ
2
q
2
+
Λ
=
1
-
θ
(
1
-
Λ
)
[
1
-
θ
q
+
θ
2
q
2
-
θ
3
q
3
+
θ
4
q
4
+
⋯
-
θ
k
-
2
q
k
-
2
+
θ
k
-
1
q
k
-
1
]
1
+
θ
q
1
+
θ
q
=
1
-
θ
(
1
-
Λ
)
[
1
+
θ
k
q
k
]
1
+
θ
q
.Proposition 5.
The explicit expressions ofC
(
2
Z
) and C
(
2
Z
i
) are
(23)
C
(
2
Z
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
θ
[
1
-
f
m
*
(
λ
c
)
]
ρ
[
1
+
θ
f
m
*
(
λ
c
)
]
)
]
C
p
,
(24)
C
(
2
Z
i
)
=
(
1
-
θ
)
C
u
ρ
+
[
n
+
n
(
θ
[
1
-
f
m
*
(
λ
c
)
]
(
1
-
Λ
)
(
1
+
θ
q
)
ρ
(
1
+
θ
q
)
[
1
+
θ
q
f
m
*
(
λ
c
)
]
)
]
C
p
.Proof.
The result follows from(25)
∑
k
=
0
∞
S
2
Z
i
(
k
)
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
∑
k
=
1
∞
{
1
+
-
θ
(
1
-
Λ
)
[
1
-
(
-
θ
q
)
k
]
1
+
θ
q
}
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
1
+
θ
(
q
+
Λ
-
1
)
1
+
θ
q
×
∑
k
=
1
∞
α
(
k
)
-
θ
(
Λ
-
1
)
1
+
θ
q
∑
k
=
1
∞
(
-
θ
q
)
k
α
(
k
)
=
1
-
(
1
-
f
m
*
(
λ
c
)
)
ρ
+
[
1
+
θ
(
q
+
Λ
-
1
)
]
[
1
-
f
m
*
(
λ
c
)
]
2
ρ
(
1
+
θ
q
)
×
[
1
1
-
f
m
*
(
λ
c
)
]
-
θ
(
Λ
-
1
)
[
1
-
f
m
*
(
λ
c
)
]
2
ρ
(
1
+
θ
q
)
×
[
-
θ
q
1
+
θ
q
f
m
*
(
λ
c
)
]
=
1
-
θ
[
1
-
f
m
*
(
λ
c
)
]
(
1
-
Λ
)
(
1
+
θ
q
)
ρ
(
1
+
θ
q
)
[
1
+
θ
q
f
m
*
(
λ
c
)
]
,
(26)
∑
k
=
0
∞
S
2
Z
(
k
)
α
(
k
)
=
1
-
θ
[
1
-
f
m
*
(
λ
c
)
]
ρ
[
1
+
θ
f
m
*
(
λ
c
)
]
.


## 4. Numerical Results

In this section, the performances of 1Z, 2Z, and 2Zi are investigated through various numerical results for the signaling cost on radio channels. The signaling cost of 1Z can be obtained by substituting θ
=
0  in (24). The performance of 2Z is analyzed, using both our proposed model and Lin’s model [6], and it can be seen that the results of both models are the same, in every case, as shown in Proposition 3. The performance of 2Zi is analyzed, using our proposed model, and is compared with those of 1Z and 2Z.We obtain the numerical results, assuming the following environments [2, 6, 7]:
(27)
C
p
=
1
,
C
u
=
4
,
θ
=
0.5
,
n
=
8
,
λ
c
=
1
,
λ
o
c
=
4
,
λ
m
=
4
.In our examples, the sojourn time in a zone (t
m) is assumed to follow an exponential distribution, for convenience. However, since the foregoing equations were derived under the assumption that t
m has a general distribution, any distribution can be assumed.Figure2 shows the signaling cost with respect to CMR (
=
λ
c
/
λ
m
). It shows the signaling cost when λ
c
=
1, with different levels of λ
m from 0.5 to 8.0. The same results are shown in Table 1. As shown in Figure 2 and Table 1, the signaling cost of 2Z is lower than that of 1Z, in most cases, and the signaling cost of 2Zi is lower than that of 2Z. Table 1 shows that the signaling cost of 2Zi is 22% lower than that of 1Z and 10.86% lower than that of 2Z, when CMR
=
1
/
2. In fact, the signaling cost of 2Zi is lower than those of the other two methods, 2Z and 1Z, in most cases.Table 1
Signaling cost with respect to CMR.
CMR (=λ
c
/
λ
m)
0.125
0.25
0.5
1
1.5
2
C
(
1
Z
)
40.00
24.00
16.00
12.00
10.68
10.00
C
(
2
Z
)
26.46
18.29
14.00
11.60
10.68
10.14
C
(
2
Z
i
)
24.74
16.54
12.48
10.38
9.65
9.26
Reduction ratio (%)=100
×
[
1
-
C
(
2
Z
i
)
/
C
(
1
Z
)
]
38.15
31.08
22.00
13.49
9.63
7.38
Reduction ratio (%)=100
×
[
1
-
C
(
2
Z
i
)
/
C
(
2
Z
)
]
6.50
9.54
10.86
10.51
9.60
8.69Figure 2
Signaling cost with respect to CMR.Table1 also shows that, as CMR increases (λ
m decreases), the signaling cost of 2Zi always remains lower than that of 1Z, but the reduction ratio of the signaling cost decreases. Conversely, the signaling cost of 2Zi is always lower than that of 2Z, for all CMR values, but the largest reduction of the signaling cost occurs when CMR
=
1
/
2. Another notable feature of Table 1 is that the signaling cost of 2Zi is lower than that of 1Z, whereas the signaling cost of 2Z is greater than that of 1Z, when CMR
=
2. When CMR is very large (i.e., λ
m is very small), there are very few location registrations, and 2Z, which has an increasing paging cost, may have a disadvantage, compared to 1Z. Although it is not shown in Table 1, we can infer that 2Zi also may have a disadvantage, compared to 1Z, when CMR is very large.The signaling cost of 2Zi and 2Z is lower than that of 1Z, because 2Zi and 2Z have lower location registration cost than 1Z. To show this feature clearly, we present the location registration cost with respect to λ
m, when λ
c
=
1, in Figure 3. As shown in Figure 3, the increase of the location registration cost is exactly proportional to the increase of λ
m. In addition, the location registration cost is directly related to θ, which is the probability of returning to the previous zone. The location registration cost of 2Z and 2Zi is 25% lower than that of 1Z, when θ is 0.25, and 50% lower, when θ is 0.5.Figure 3
Location registration cost with respect toλ
m.The location registration cost of 2Z and 2Zi is lower, but the paging cost is greater, than that of 1Z. To show this feature clearly, we present the paging cost with respect to λ
m, when λ
c
=
1, in Figure 4.Figure 4
Paging cost with respect toλ
m.To show this feature clearly, we present the paging cost with respect toλ
m, when λ
c
=
1, in Figure 4 and Table 2. As shown in Figure 4 and Table 2, the paging cost of 1Z remains constant, while that of 2Z and 2Zi increases, as λ
m increases. One of the notable results of this study is that, when λ
m
=
8, the paging cost of 2Z, which is 10.46 (30.77% greater than 8.00 of 1Z), can be reduced to 8.74 (9.25% greater than 8.00 of 1Z) if 2Zi is adopted, which causes the total signaling cost of 2Zi to be lower than that of 2Z, to an extent corresponding to this reduction, as shown in Table 2.Table 2
Paging cost with respect toλ
m.
λ
m
0.5
1
2
4
8
C
2
(
1
Z
)
8.00
8.00
8.00
8.00
8.00
C
2
(
2
Z
)
9.14
9.60
10.00
10.29
10.46
C
2
(
2
Z
i
)
8.26
8.38
8.48
8.54
8.74
Reduction ratio (%)=100
×
[
1
-
C
2
(
2
Z
i
)
/
C
2
(
1
Z
)
]
9.64
12.70
15.20
16.97
16.44
Reduction ratio (%)=100
×
[
C
2
(
2
Z
)
/
C
2
(
1
Z
)
-
1
]
14.29
20.00
25.00
28.57
30.77
Reduction ratio (%)=100
×
[
C
2
(
2
Z
i
)
/
C
2
(
1
Z
)
-
1
]
3.27
4.76
6.00
6.76
9.27Figure5 shows the signaling cost of each method with respect to θ, the probability of returning to the previous zone. As shown in Figure 5, the signaling cost of 2Z and 2Zi decreases, as θ increases. In particular, the signaling cost of 2Zi decreases more than that of 2Z. Even though it is clear that the signaling cost of 2Z and 2Zi decreases, as θ increases, it seems to be unreasonable to assume that θ is larger than 0.5, in a real-world mobile communication environment.Figure 5
Signaling cost with respect toθ.Finally, Figure6 shows the signaling cost with respect to n, the number of cells in a zone. In this case, since the location registration cost remains constant, the overall amount of the signaling cost will increase, as the number of cells in a zone increases, due to the increase of the paging cost. As shown in Figure 6, the signaling costs of 1Z, 2Z, and 2Zi all increase, as the number of cells in a zone increases, but the increased ratios of 2Zi and 1Z are lower than that of 2Z. That is, if the other conditions are the same, 2Z is more superior to 1Z, and 2Z is more superior to 2Zi, respectively, as the paging cost increases.Figure 6
Signaling cost with respect to the number of cells in a zone (λ
o
c
=
λ
m
=
5).


## 5. Conclusion

Many efficient mobility management methods have been suggested, to minimize the signaling cost on radio channels. This study considered the zone-based registration methods that are widely used in the majority of mobile communication networks.We provided a new mathematical model to analyze the performance of the zone-based registration methods, 2Z and 2Zi, by considering implicit registration effects of outgoing calls from a mobile, which were not considered properly in the previous studies. It should be noted that our mathematical model is simple, compared to the previous studies, but provides the exact performance of 2Zi for the first time. Also, our model can easily be applied to 2Z and 1Z and provides the same results as Lin’s previous study on 2Z and 1Z. From various numerical results by using our model, we showed that 2Zi is superior to 2Z as well as 1Z, in most cases.Our results are helpful in considering which registration scheme should be adopted. For further study, we will consider the case where a mobile can have multiple zones, to get the performance of every type of zone-based registration.



---

*Source: 250981-2014-05-05.xml*