An e-cientpowersaving mechanism for delay-guaranteed services in IEEE by rar99983

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									IEICE TRANS. COMMUN., VOL.Exx–??, NO.xx XXXX 200x
                                                                                                                          1


PAPER
An efficient power saving mechanism for
delay-guaranteed services in IEEE 802.16e∗
                                                Yunju PARK† , Nonmember and Gang Uk HWANG†a) , Member



SUMMARY As the IEEE 802.16e Wireless Metropolitan Ac-               services. Power Saving Class of type II is recommended
cess Network (WMAN) supports the mobility of a mobile station       for the connections of Unsolicited grant service (UGS)
(MS), increasing MS power efficiency has become an important
                                                                    and Real-Time Variable Rate (RT-VR) services. Power
issue. In this paper, we analyze the sleep-mode operation for an
efficient power saving mechanism for delay-guaranteed services in     Saving Class of type III is recommended for multicast
the IEEE 802.16e WMAN and observe the effects of the operating       connections and for management operations.
parameters related to this operation. For the analysis we use the        Regarding the analysis of the sleep-mode opera-
M/GI/1/K queueing system with multiple vacations, exhaus-           tion in IEEE 802.16e, Nga et al. [4] analyzed a numeri-
tive services and setup times. In the analysis, we consider the
power consumption during the wake-mode period as well as the
                                                                    cal model to determine the operating parameters in the
sleep-mode period. As a performance measure for the power con-      sleep-mode operation and proposed a delay guaranteed
sumption, we propose the power consumption per unit time per        energy saving algorithm to minimize energy consump-
effective arrival which considers the power consumption and the      tion with a given MAC (Medium Access Control) SDU
packet blocking probability simultaneously. In addition, since we
                                                                    (Service Data Unit) response delay. Xiao [5] considered
consider delay-guaranteed services, the average packet response
delay is also considered as a performance measure. Based on the     a sleep-mode scheme for the power saving mechanism
performance measures, we obtain the optimal sleep-mode opera-       and analyzed the effects of the operating parameters.
tion which minimizes the power consumption per unit time per        Seo et al. [6] used the M/GI/1/K queueing system with
effective arrival with a given delay requirement. Numerical stud-    multiple vacations and considered the dropping proba-
ies are also provided to investigate the system performance and
to show how to achieve our objective.
                                                                    bility of packets and the mean waiting times of packets
key words: sleep-mode operation, power saving, IEEE 802.16e,        as the performance measures. Jang et al. [7] simulated
power consumption, average packet response delay                    the sleep-mode operation of the Power Saving Class of
                                                                    type I and II, and found the optimal values of oper-
1.   Introduction                                                   ating parameters to satisfy different QoS requirements.
                                                                    Kim et al. [8] introduced an efficient power management
As wireless internet services are rapidly expanding, it             mechanism which takes into account the remaining en-
is an important task to provide high speed and high                 ergy. Kim et al. [9] modeled the sleep mode operation in
quality wireless services. To meet these demands, in                the IEEE 802.16e MAC and evaluated the effect of op-
the Wireless Metropolitan Access Network (WMAN),                    erating parameters on the performance of power man-
mobility and power management of a Mobile Station                   agement by considering the average interarrival time
(MS) become important issues, and to solve the issues               of MAC frames. They used simulation to evaluate the
the IEEE 802.16 standard [1, 2] has been extended to                performance. Xu et al. [10] discussed a novel adaptive
the IEEE 802.16e standard where the handover process                sleep-mode scheme which considered quick responses to
and the sleep mode operation are included [3].                      the packet arrival events. Dong et al. [11] modelled the
     In the IEEE 802.16e standard [3], the sleep-mode               sleep-mode operation where two-type-returning from
operation has three types of the Power Saving Classes.              sleep mode is considered. Most studies modelled the
Power Saving Class is a group of connections which                  sleep-mode period, but they focused on the analysis of
have common demand properties. Power Saving Class                   the sleep-mode period only and did not consider the
of type I is recommended for the connections of Best Ef-            wake-mode period and the queueing effect of the sys-
fort (BE) and Non-Real-Time Variable Rate (NRT-VR)                  tem. However, in our paper, we analyze the packet
    †                                                               response delay and the power consumption during the
      The authors are with the Department of Mathemati-
cal Sciences and Telecommunication Engineering Program              wake-mode period as well as the sleep-mode period.
in Korea Advanced Institute of Science and Technology                    The operation of IEEE 802.16e system is based on
(KAIST), Daejeon, Republic of Korea                                 frame units of 5 ms, and the sleep-mode operation in
    †
      This paper was presented in part at the 65th IEEE             IEEE 802.16e is performed between one base station
Vehicular Technology Conference VTC2007-spring, Dublin,             (BS) and one MS. In this paper, we focus on downlink
Ireland, April 22-25, 2007.                                         transmission from a BS to an MS, and assume that
  a) E-mail: guhwang@kaist.edu
    ∗
      This research was supported by the Korea Science and          there is a finite size buffer in the BS to develop a sys-
Engineering Foundation (KOSEF) grant funded by the Ko-              tem model close to the real one. The finite size buffer
rea government (MOST) (No. R01-2007-000-20053-0).                   in the BS accommodated packets addressed to the MS
                                                                                               IEICE TRANS. COMMUN., VOL.Exx–??, NO.xx XXXX 200x
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with the Power Saving Class of type I packets. The
detailed sleep-mode operation of this type will be given                                                                                                                                                      buffer
                                                             one frame
in section 2. In an IEEE 802.16e network, the resources     BS
                                                                                                                                                                                                                                   time
are shared by all MS in the network, and some amount
of resources are assigned to our MS at each frame. So,




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single server queueing system. The amount of resources      MS                  active           sleep     awake                     sleep     awake                         sleep   awake                             active


allocated to our MS determines the packet transmission                                            T0                   L              T1          L                           T2     L
                                                                                                                                                                                                                                   time


time. The amount of resources allocated to our MS is                                              v0                            v1                                      v2            S

not fixed in general. In addition, the mobility and the       wake mode
                                                                                       start time of
                                                                                                                                         sleep mode
                                                                                                                                                                                              start time of
                                                                                                                                                                                                                       wake mode


wireless channel condition may affect the packet ser-                                   sleep mode                                                                                             wake mode


vice process through packet transmission errors. Such                                                                                 vacation period                                setup
                                                                                                                                                                                      time
                                                                                                                                                                                                                busy period


effects can be approximately taken into consideration
                                                                                                Fig. 1                          The sleep-mode operation in IEEE802.16e
by using a suitable service time distribution. Here, the
service time is defined by the time period needed to
transmit the HOL (Head Of Line) packet in the buffer
successfully. Accordingly, to consider such effects, we       2.                                The Operation of Sleep-Mode in IEEE802.16e
assume that the service times of packets have a gen-
eral service time distribution. As the arrival process       We consider downlink transmission from a BS to an MS
to the buffer in BS, we adopt the Poisson process for         where the Power Saving Class of type I is used. In this
the convenience in the analysis [4–11]. In addition, due     case, the MS has two modes: sleep-mode and wake-
to the sleep-mode operation, our queueing system has         mode. During a sleep-mode period, the MS powers
vacations.                                                   down to reduce the battery consumption and there is
     Based on above assumptions, to analyze the sleep-       no packet transmission between the BS and the MS. So,
mode operation mathematically, the buffer in the BS           the buffer in the BS accommodates incoming packets
is modelled by the M/GI/1/K queueing system with             addressed to the MS until the sleep-mode period ends.
multiple vacations, exhaustive services and setup times      When the sleep-mode period ends, a wake-mode period
[12, 13]. The contributions of this paper are as fol-        starts immediately and there are packet transmissions
lows. First, we mathematically model the sleep-mode          between the BS and the MS during the wake-mode pe-
operation as exactly as possible and obtain the packet       riod. After the wake-mode period, a new sleep-mode
blocking probability, the power consumption per unit         period begins again and this alternating procedure will
time and the average packet response delay. Second, as       continue.
performance measures, we consider the average packet               To enter a new sleep-mode period after a wake-
response delay and the power consumption per unit            mode period, the MS sends a Sleep Request message
time per effective arrival, which is newly proposed to        (MOB SLP-REQ) to the BS and waits for a Sleep Re-
combine the average packet blocking probability and          sponse message (MOB SLP-RSP) from the BS. The
the power consumption per unit time simultaneously.          MOB SLP-REQ contains the relevant parameters re-
Third, we provide a detailed procedure to get the opti-      garding the sleep-mode period such as initial-sleep win-
mal sleep-mode operation which satisfies the delay re-        dow, final-sleep window base, listening window, final-
quirement and minimizes the power consumption per            sleep window exponent and start frame number for
unit time per effective arrival.                              the first sleep window. When the BS receives the
     The rest of this paper is organized as follows.         MOB SLP-REQ, if there is no downlink traffic for the
In section 2, the sleep-mode operation in the IEEE           MS, then the BS sends a positive MOB SLP-RSP mes-
802.16e standard is described. In section 3, we use          sage to the MS with the same parameters as in the
the M/GI/1/K queueing system with multiple vaca-             MOB SLP-REQ message that it has received. Other-
tions, exhaustive services and setup times to analyze        wise, the BS sends a negative MOB SLP-RSP message.
the sleep-mode operation. In section 4 and 5, we ana-        After receiving the MOB SLP-RSP, the MS can deter-
lyze the system behaviors during the vacation and busy       mine whether to begin a new sleep-mode period or not.
periods, respectively. In section 6, we obtain the packet    If it has received a negative MOB SLP-RSP (i.e., no
blocking probability, the power consumption per unit         approval message), it continues to be in wake-mode
time and the average packet response delay. Based on         and waits for another packet transmissions. If it has
our analysis, we propose a detailed procedure to get the     received a positive MOB SLP-RSP (i.e., approval mes-
optimal sleep-mode operation. In section 7, we give our      sage), it begins a new sleep-mode period at the frame
conclusions.                                                 specified as start frame number for the first sleep win-
                                                             dow. A sleep-mode period may consist of a single or
                                                             multiple sleep intervals as shown in Fig. 1. The length
                                                             of the first sleep interval is equal to the initial-sleep
PARK and HWANG: AN EFFICIENT POWER SAVING MECHANISM FOR DELAY-GUARANTEED SERVICES IN IEEE 802.16E
                                                                                                                          3


window, denoted by T0 . After the first sleep interval,        the first sub-vacation interval. Similarly, the MS has
the MS wakes for a fixed time period, called the lis-          the (i + 1)-th sub-vacation interval if there is no packet
tening interval, of length listening window L, to check       during the i-th sub-vacation interval which consists of
the Traffic Indication message (MOB TRF-IND). The               a listening interval and the (i + 1)-th sleep interval.
message is broadcasted by the BS during the listening               Now, assume that the MS is at the end of the n-
interval. It indicates the presence of the buffered traffic      th sub-vacation interval and that there is at least one
addressed to the MS in the BS. If the MS has received a       packet arrival during the n-th sub-vacation interval. In
negative MOB TRF-IND, i.e., no buffered traffic for it,          this case, the MS receives a positive MOB TRF-IND
it goes back to the sleep-mode again to start the second      message from the BS during the listening interval fol-
sleep interval with the doubled length T1 (= 2T0 ). Oth-      lowing the n-th sub-vacation interval. So, the MS ends
erwise, the MS begins a new wake-mode period after            the sleep-mode period consisting of n + 1 sub-vacation
the listening interval. The MS keeps the above pro-           intervals, called a vacation period in our model, and
cedure during the sleep-mode period until it receives         starts a new wake-mode period, called a busy period in
a positive MOB TRF-IND. In the IEEE 802.16e stan-             our model, to receive packets from the BS. Note that
dard [3], the length Ti of the i-th sleep interval in a       there is always a listening interval between a vacation
sleep-mode period, if any, is computed as follows:            period and the following busy period, which is called
                                                              the setup time period in our model. For details, refer to
    T0 = initial-sleep window,                                Fig. 1.
    Ti = min{2Ti−1 , Tmax }, i ≥ 1,                                 From the above assumptions and explanation,
Tmax = final-sleep window base · 2final-sleep window exponent . our system can be modelled by the discrete time
                                                              M/GI/1/K queueing system with multiple vacations,
                                                              exhaustive services and setup times [12, 13]. Here, GI
3. Mathematical Modelling of The Sleep-Mode                   implies that service times are general and independent.
      Operation                                               Note that our system has the exhaustive service disci-
                                                              pline because the MS begins the vacation period only
In this section, we consider the buffer in the BS which        when there is no packet for the BS to transmit. If
accommodates packets addressed to the MS, and ana-            we assume that the time to transmit the control mes-
lyze the operation of the sleep-mode in IEEE 802.16e          sages such as MOB SLP-REQ, MOB SLP-RSP and
explained in section 2.                                       MOB TRF-IND is zero, the i-th sub-vacation interval,
      We assume that the packet arrival process follows       denoted by vi , is given by
a Poisson process with rate λ and the service times
of packets are independent and identically distributed                     T0 ,      i = 0,
                                                                    vi =
(i.i.d) with common distribution function F (x). We                        L + Ti , i = 1, 2, ....
also assume that the buffer in the BS for the MS is
of finite size K − 1, and single server for packet trans-      Recall that L is the length of a listening interval, T0 is
mission. Then, if there are K packets in the system           the initial sleep interval, and Ti is given by
including the packet being transmitted, newly arriving             Ti = min{2Ti−1 , Tmax },          i = 1, 2, ....
packets are blocked and discarded. For this reason, K
is called the system size from now on. The packets in         Since Ti ’s are all fixed, all sub-vacation intervals, vi , i =
the buffer are transmitted based on the FCFS (First-           0, 1, 2, ..., are of fixed lengths.
Come-First-Served) service discipline.                              Let V , S and B be the vacation period, setup
     Since the operation of IEEE 802.16e is based on          time period and busy period in our model, respectively.
frame times, a frame time is considered as a unit time        Then, we can define a service cycle C by C = V +S +B,
and we assume that the time axis is divided into unit         i.e., a service cycle consists of a vacation period, a setup
times in our analysis. To model our system, we first           time period and a busy period.
consider a sleep-mode period. The sleep-mode period                 In the subsequent sections, we analyze the system
starts with the first sleep interval of length T0 . This       behaviors during the vacation period and busy period
first sleep interval is called the 0-th sub-vacation in-       separately and obtain the performance measures such
terval in our model. If there is no packet during the         as the power consumption per unit time per effective
0-th sub-vacation interval, the MS receives a negative        arrival and the average packet response delay.
MOB TRF-IND message during the following listening
interval and starts immediately the second sleep inter-       4.   The Vacation Period Analysis
val after the listening interval. Since the switching to
the wake-mode for the MS after the second sleep inter-        In this section, we analyze the length of a vacation pe-
val depends on whether there is at least one packet ar-       riod and the number of sub-vacation intervals in steady
rival during the listening and second sleep intervals, the    state. To do this, we first obtain the probability mass
sum of the listening and second sleep intervals is called     function of a vacation period. By the definition of a
                                                                      IEICE TRANS. COMMUN., VOL.Exx–??, NO.xx XXXX 200x
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vacation period V given in section 3, the probability           steady state. Observing that there should be at least
mass function of V is given by                                  one packet at the end of a vacation period, we have the
                                                                probability mass function of NV as follows:
            S0    with prob. 1 − e−λS0 ,
      V =                                               (1) Theorem 1: The probability mass function of the
            Sn    with prob. e−λSn−1 (1 − e−λvn ), n ≥ 1,   number NV of backlogged packets at the end of a va-
                n                                           cation period is given by
where Sn = i=0 vi and S0 = v0 .
      For our analysis, we use M to denote the index of                       M
                                                                                 (λvn )i e−λSn
                                                                                                  (λvM )i −λvM
                                                                                                         e
the sub-vacation interval such that vj = L + Tmax for        Pr[NV = i] =                      + i! −λvM e−λSM ,
                                                                                       i!          1−e
all j ≥ M and vj < L + Tmax for all 0 ≤ j < M . That                         n=0
is, M = min{j : vj = L + Tmax }. Note that the value                                            (i = 1, 2, ..., K − 1)
of M depends on T0 and Tmax . For example, if T0 = 2                             K−1
and Tmax = 1024, then M = 9. Then, from equation            Pr[NV = K] = 1 −         Pr[NV = i].
(1) the expectation E[V ] of a vacation period is given                          i=1
by
                                                            The proof of Theorem 1 is given in Appendix A.3.
                 ∞
      E[V ] =         Sj Pr[V = Sj ]                            5.2   Distribution of the number of packets at the be-
                j=0                                                   ginning of a busy period
                        M
                                               vM
            = S0 +           e−λSj−1 vj +             e−λSM .   Let NS be the number of backlogged packets in the
                       j=1
                                            1 − e−λvM           system at the beginning of a busy period in steady
                                                                state. Since there is always a setup time period be-
The detail derivation is given in Appendix A.1.                 tween a vacation period and a busy period, NS satis-
     Next, let NI denote the maximum of sub-vacation            fies NS = NV + AS where AS denotes the number of
interval indexes during a vacation period in steady             packets newly arriving during the setup time period.
state. By a similar argument as above, we also obtain           Then, by Theorem 1 the distribution of NS is derived
the expectation E[NI ] as follows:                              as follows.
                 ∞
                                                                Theorem 2: The probability mass function of the
      E[NI ] =         jPr[V = Sj−1 ]
                                                                number NS of backlogged packets at the beginning of
                 j=1
                                                                a busy period is given by
                        M
                                         e−λvM
            = 1+             e−λSj +             e−λSM .                          j
                                                                                       (λS)j−i e−λS
                       j=0
                                       1 − e−λvM                 Pr[NS = j] =                       ×
                                                                                 i=1
                                                                                         (j − i)!
The detail derivation is given in Appendix A.2.                                       M                    (λvM )i e−λSM −λvM
                                                                                       (λvn )i e−λSn             i!     e
                                                                                                     +                          ,
5.    The Busy Period Analysis                                                     n=0
                                                                                             i!                1 − e−λvM
                                                                                                    (j = 1, 2, ..., K − 1)
In this section, we analyze the length of a busy period                                K−1
in steady state. Since the length of a busy period is           Pr[NS = K] = 1 −             Pr[NS = j].
closely related with the number of backlogged packets                                  j=1
at the beginning of the busy period, we start with the
analysis of the number of backlogged packets at the end         The proof of Theorem 2 is given in Appendix A.4.
of a vacation period.
                                                                5.3   The length of a busy period
5.1   Distribution of the number of packets at the end
      of a vacation period                                      In this subsection, we analyze the length of a busy pe-
                                                                riod based on the results obtained in subsections 5.1
Let NV be the number of backlogged packets at the end           and 5.2. Recall that packets in the buffer are transmit-
of a vacation period in steady state. By the definition,         ted based on the FCFS discipline. However, the length
the distribution of NV is obtained as follows:                  of a busy period does not depend on the order in which
                                                                packets in the buffer are transmitted [12, 14–16]. That
                         ∞
                                                                is, the length of a busy period of the system with the
      Pr[NV = i] =            Pr[NV = i|NI = n]Pr[NI = n].
                                                                FCFS discipline is identical to that of the system with
                        n=0
                                                                the LCFS (Last-Come-First-Served) discipline. So, for
Here, recall that NI denotes the maximum of sub-                convenience in the analysis, we consider a new system
vacation interval indexes during a vacation period in           which is identical to our system except that the LCFS
PARK and HWANG: AN EFFICIENT POWER SAVING MECHANISM FOR DELAY-GUARANTEED SERVICES IN IEEE 802.16E
                                                                                                                          5


discipline is used. Note that the method of considering         3, we can derive the expectation of a busy period B in
a system with the LCFS discipline is widely used for            our system as follows.
the busy period analysis, e.g., [14].
     Now we assume that there are j backlogged packets          Theorem 4: The expectation of a busy period in our
in the buffer at the beginning of a busy period. Then,           system is given by
due to the LCFS service discipline the busy period gen-
erated by the j backlogged packets can be divided into                         K−1 K−1
j sub-busy periods, each of which is generated by the                 E[B] =               E[Bi ]Pr [NS = j]
service of a backlogged packet as follows. When a busy                         j=1 i=K−j
period starts, our system immediately starts the ser-
                                                                                            K−1
vice of the j-th backlogged packet and continue its ser-
                                                                               + E[X] +           E[Bi ] Pr[NS = K],
vice for all subsequent packets that newly arrive at the
                                                                                            i=1
system until it can start the service of the (j − 1)-th
backlogged packet. That is, the service of the j-th
                                                                where X is the service time of a packet.
backlogged packet generates the first sub-busy period
which ends with the start of the service of the (j − 1)-
th backlogged packet. In addition, since there are only         The proof of Theorem 4 is given in Appendix A.5.
K − j empty waiting rooms in the buffer for this case,
the length of the first sub-busy period in our system
is identical to the length of a busy period of the ordi-        6.    Performance Analysis
nary M/GI/1/K − j + 1 queueing system. Similarly,
the i-th sub-busy period is generated by the service
                                                                  In this section, we propose two performance measures,
of the (j − i + 1)-th backlogged packet. Since there
                                                                  called the the power consumption per unit time per ef-
are K − j + i − 1 empty waiting rooms in the buffer
                                                                  fective arrival and the average packet response delay.
for this case, the length of the i-th sub-busy period is
                                                                  Based on these two performance measures, for given
identical to the length of a busy period of the ordinary
                                                                  arrival rate λ, the first sleep interval T0 , the system
M/GI/1/K − j + i queueing system.
                     ∗                                            size K and the mean service time of a packet E[X], we
     Let Bm and Bm (s) be the length of a busy period
                                                                  obtain the optimal Tmax value. Then, by using the re-
and its LST (Laplace-Stieltjes Transform) in the ordi-
                                                                  sults of Tmax for each T0 , we will determine the optimal
nary M/GI/1/m + 1 queueing system. Then, we have
                                                                  set of (T0 , Tmax ) for a given delay requirement to mini-
the following Theorem [16].
                                                                  mize the power consumption per unit time per effective
Theorem 3: The LST of the length of a busy period                 arrival.
for the ordinary M/GI/1/m + 1 queueing system with                     The parameters related with the power consump-
m waiting rooms is
                                                                  tion are as follows. Let ES , EW and EL be the power
                                  u0 (s)
Bm ∗ (s) =                                                      . consumption units per unit time in sleep-mode, wake-
              m−1          m−1             ∞         m−1
                                   ∗                       ∗      mode and a listening interval, respectively. In addition,
           1−     uk (s)         Bj (s) −     uk (s)     Bj (s)
              k=1        j=m−k+1          k=m        j=1
                                                                  since the MS consumes the power to switch the mode,
                                                                  let Eon−switch and Eof f −switch be the power consump-
The expectation E[Bm ] is                                         tion units for the switch-on action and the switch-off
                                                                action, respectively. The switch-on (switch-off, resp.)
                1                             
                              m−1
                                                                  action means that the MS changes its state from sleep
      E[Bm ] =       E[X] +       E[Bj ]Qm−j ,
               q0                                               (wake or listen, resp.) to listen (sleep, resp.).
                             j=1

                 ∞      k
where uk (s) = 0 (λx) e−(λ+s)x dF (x), F (x) is the dis-
                     k!                                         6.1   The power consumption per unit time per effective
tribution function of the service time, qk = uk (0) and               arrival
            j
Qj = 1 − k=0 qk . (Note that when the lower limit of a
product (a summation, resp.) is greater that the upper
limit, the product (the summation, resp.) is taken to           In this subsection, we obtain the power consumption
be 1 (0, resp.).)                                               per unit time per effective arrival in our system. To
                                                                do this, we first derive the power consumption per unit
    Then, from our observation above the length of a
                                                                time of an MS. Then, we get the packet blocking prob-
busy period B(j) generated by j backlogged packets in
                                                                ability. Let pV , pS and pB be the amounts of total
the buffer is given by
                                                                power consumption during a vacation period, a setup
           d
    B(j) = BK−j + BK−j+1 + · · · + BK−1 .               (2)     time period and a busy period, respectively. Then, by
                                                                the definitions the expectations E[pV ], E[pS ] and E[pB ]
Therefore, using equation (2), Theorem 2 and Theorem            are given as follows:
                                                                           IEICE TRANS. COMMUN., VOL.Exx–??, NO.xx XXXX 200x
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                          M
                                                                       packets leave the system after service completion. Let
      E[pV ] = ES S0 +         e−λSj−1 (ES Tj + EL L)                    d
                                                                       πj be the steady state probability that j packets are
                         j=1                                           left in the system immediately after service completion
                 ES TM + EL L −λSM                                     (0 ≤ j ≤ K − 1). Let Ln be the number of packets
               +               e   ,
                  1 − e−λvM                                            left behind in the system immediately after the n-th
                                                                                                               d
      E[pS ] = EL E[S] = EL L,                                         Markov point (n = 1, 2, ...). Then the πj is represented
      E[pB ] = EW E[B].                                                as follows.
                                                                                                                     d
                                                                       Theorem 5: The steady state probability πj that j
Note that E[pV ] can be derived from E[V ] in section 4.
                                                                       packets are left in the system immediately after a ser-
Then, the power consumption per unit time of the MS,
                                                                       vice completion is given by
which is denoted by P C and defined by
                                                                            d
           E[total power consumption in a cycle]                           πj = lim Pr[Ln = j],       0≤j ≤K −1
                                                                                n→∞
      PC =                                       ,                              
                 E[total length of a cycle]                                            1
                                                                                      K−1
                                                                                             , j=0
is given by                                                                   =       j=0 πj
                                                                                 d
                                                                                   π0 πj ,     1 ≤ j ≤ K − 1,
       E[pV ]+ E[pS ]+ E[pB ]+ E[NI ](Eon−switch + Eof f −switch )
PC =                                                           (3) .
                                 E[C]                                  where
Let PB and ρ be the packet blocking probability and                      π0 = 1,
the probability that the server is busy at an arbitrary                                         j                  j+1
                                                                                1
time in steady state, respectively. Since our system has               πj+1 =           πj −         πi aj−i+1 −         aj−k+1 Pr[NS = k]     ,
                                                                                a0
a single server and is of finite size K, it satisfies that                                       i=1                 k=1
                                                                                                                   for 0 ≤ j ≤ K − 2,
      ρ = λ(1 − PB )E[X]                                    (4)                     ∞
                                                                                        (λx)k −λx
                                                                            ∆
                                                                         ak =                e    dF (x), k = 0, 1, ...,
where E[X] denotes the mean service time of a packet.                           0         k!
On the other hand, by its definition ρ can be also ob-                  and Pr[NS = k] are given in Theorem 2.
tained as follows:
                                                                       The proof of Theorem 5 is given in Appendix A.6.
           E[B]                                                             Let Qk be the probability that there are k packets
      ρ=        .                                           (5)
           E[C]                                                        in the system including the packet being transmitted
Then, by combining equations (4) and (5), we finally                    at an arbitrary time (k = 0, 1, ..., K). To derive Qk by
                                                                              d           a
obtain PB as follows:                                                  using πj , we let πj be the probability that an arriving
                                                                       packet finds j packets in the system, (j = 0, 1, ..., K).
                   E[B]                                                From the PASTA (Poisson Arrivals See Time Average)
      PB = 1 −             .                                (6)
                 λE[X]E[C]                                             property,
To consider two performance measures PB and P C si-                         a
                                                                           πj = Q j ,         0 ≤ j ≤ K.                                     (7)
multaneously, we propose a new combined performance
measure P Ce , called the power consumption per unit                   Since the state changes only by unit steps, by Burke’s
time per effective arrival and defined by                                theorem [12],
                                                                            a             d
                  PC                                                       πj = (1 − QK )πj ,              0 ≤ j ≤ K − 1.                    (8)
      P Ce =              .
               λ(1 − PB )                                              Therefore, combining equations (7) and (8), we obtain
Note that P Ce is the average actual power consumption                                    d
                                                                           Qj = (1 − QK )πj ,              0 ≤ j ≤ K − 1,                    (9)
to transmit a packet.
                                                                       Note that QK is the blocking probability. So, by equa-
6.2   The average packet response delay                                tion (6) QK is, in fact, given as
                                                                                             E[B]
In this subsection, we propose the average packet re-                      QK = 1 −                  .
                                                                                           λE[X]E[C]
sponse delay defined by the sum of queueing delay
in the buffer and transmission delay from the BS to                     Then using equation (9) and Theorem 5 we can obtain
the MS. To do this, first, we obtain the distribution                   {Qj | 0 ≤ j ≤ K}. Let Le and D be the number of pack-
of the number of backlogged packets in the system                      ets in the system at an arbitrary time and the response
(called the queue length) immediately after a service                  delay of an arbitrary packet in the system, respectively.
completion by applying the embedded Markov chain                       Then the expectation E[Le ] is given by
method. Second, we derive the queue length distribu-                                      K
tion at an arbitrary time. We consider a set of embed-                     E[Le ] =            kQk                                      (10)
ded Markov points which are those points in time when                                    k=0
PARK and HWANG: AN EFFICIENT POWER SAVING MECHANISM FOR DELAY-GUARANTEED SERVICES IN IEEE 802.16E
                                                                                                                                                                                             7


Then, from Little’s formula, the average packet re-                                                                                                E[X]=2, K=10, T0=2

sponse delay E[D] is derived as follows:
                                                                                                                  35
                                                                                                                                                                          I =64(Num)
                                                                                                                                                                           A
                                                                                                                                                                          IA=64(Sim)
                 1                                                                                                30
      E[D] =            E[Le ].                      (11)                                                                                                                 I =32(Num)




                                                               Average Packet Response Delay ( E[D] )
                                                                                                                                                                           A
             λ(1 − QK )                                                                                                                                                   IA=32(Sim)
                                                                                                                  25                                                      I =16(Num)
                                                                                                                                                                           A
                                                                                                                                                                          I =16(Sim)
6.3   The Procedure to get the optimal sleep-mode op-                                                                                                                      A
                                                                                                                                                                          IA=8(Num)
                                                                                                                  20
      eration                                                                                                                                                             I =8(Sim)
                                                                                                                                                                           A

                                                                                                                  15
For simulation studies, we develop a MATLAB code
to simulate the system. The simulation condition is as                                                            10
follows. Since the sleep-mode operation is performed
between one BS and one MS and we consider downlink                                                                 5

transmission, one BS and one MS are considered in our
simulation. We assume an ideal wireless channel model                                                              0

and generate 3×105 frames for each simulation. We also
                                                                                                                           4 8 16    32     64             128                         256
                                                                                                                                                          T
                                                                                                                                                           max
assume that the time to transmit the control messages
                                                                                                                                                    E[X]=2, K=10, T0=2
is zero. Here, the control messages are MOB SLP-                                                                  900

REQ, MOB SLP-RSP, and MOB TRF-IND. Since the                                                                                                                              I =64(Num)
                                                                                                                                                                           A
                                                                                                                                                                          IA=64(Sim)
purpose of the standardized sleep-mode operation in             Power consumption per effective arrival ( PCe )
                                                                                                                  800
                                                                                                                                                                          I =32(Num)
IEEE 802.16e is to save the power consumption for low-                                                            700
                                                                                                                                                                           A
                                                                                                                                                                          IA=32(Sim)
rate traffic environment, we use the expected interar-                                                                                                                      I =16(Num)
                                                                                                                                                                           A
rival time IA (= 1/λ) = 8, 16, 32, 64 frames. For all ex-                                                         600
                                                                                                                                                                          I =16(Sim)
                                                                                                                                                                           A

amples in this subsection, otherwise mentioned, we as-                                                            500
                                                                                                                                                                          IA=8(Num)
                                                                                                                                                                          I =8(Sim)
sume that the service time of a packet has the geometric                                                                                                                   A


distribution with mean E[X] = 2, and the system size                                                              400


K is 10. Note that the choice of E[X] = 2 and K = 10                                                              300

is an example. The system size K is not a critical pa-
rameter in our model due to the fact that the packet                                                              200


blocking probability is very low in low-rate traffic envi-                                                          100

ronment. We can choose any other values of E[X] and
K in our analysis. We follow the guidelines of IEEE                                                                    0
                                                                                                                            4 8 16   32      64               128                      256
                                                                                                                                                          T
802.16e standard for the initial sleep window T0 , the                                                                                                        max


final sleep window Tmax , and the listening interval L.                                                                     Fig. 2         E[D] and P Ce for different IA and Tmax
The length L of a listening interval is fixed. We assume
that the length of L is equal to 1. Since there is no gen-
eral information of the actual parameters for the power      by simulation, the results obtained by our analysis and
consumption units, we use ES : EL : EW = 1 : 10 : 10         the expected interarrival time (= 1/λ) of the packets
and Eon−switch : Eof f −switch = 30 : 20, as given           (in frame), respectively. As seen from the figure, our
in [5, 10, 17–19].                                           analytic results are well matched with the simulation
     In this subsection, we first propose a procedure to      results, which partially verifies the validity of our anal-
get the optimal value of Tmax for the delay-guaranteed       ysis. From the figure, we also see that the value of Tmax
services. That is, for given mean service time E[X],         affects E[D] and P Ce for each fixed value of IA , and
system size K and the initial sleep interval T0 we show      the effect of Tmax on E[D] and P Ce becomes more sig-
how to get the optimal value of Tmax based on two per-       nificant as IA increases. The reason for this is that, as
formance measures – the power consumption per unit           IA decreases, it occurs more frequently that the sleep-
time per effective arrival in subsection 6.1 and the av-      mode period is terminated before the length of a va-
erage packet response delay in subsection 6.2. Then,         cation period reaches the maximum possible value (re-
by investigating the results, we show how to determine       lated with Tmax ).
the optimal values of T0 and Tmax for a given delay re-           Another observation from Fig. 2 is that, in most
quirement to minimize the power consumption per unit         cases except IA = 64 frame, when Tmax is greater than
time per effective arrival.                                   128, Tmax does not affect E[D] and P Ce any more.
     In Fig. 2, we change the value of Tmax and the          This result is in accordance with previously known re-
packet arrival rate λ, and plot the resulting value of       sults in [11,21]. That is, when the value IA is small (the
E[D] and P Ce for given mean service time E[X], sys-         high-rate traffic environment in this case), the param-
tem size K and the initial sleep interval T0 . In the fig-    eter Tmax does not affect the system performance sig-
ure, ‘Sim’, ‘Num’ and ‘IA ’ denotes the results obtained     nificantly. However, since the sleep-mode operation is
                                                                                                                 IEICE TRANS. COMMUN., VOL.Exx–??, NO.xx XXXX 200x
8


Table 1     Optimal value of Tmax given delay requirement                                                                                     E[X]=2, K=10
(T0 =2)                                                                                                         40

    PP                IA
                                                                                                                                                                T =4, I =64
                                                                                                                                                                 0    A
          PP                                                                                                                                                    T0=4, IA=32
    delay   PP             64      32     16     8                                                              35
                                                                                                                                                                T =4, I =16
    requirement PP




                                                             Average Packet Response Delay ( E[D] )
                                                                                                                                                                 0    A
                 P                                                                                              30
                                                                                                                                                                T0=4, IA=8
                                                                                                                                                                T =16, I =64
          10 frames        16      16     16    1024                                                                                                             0        A
                                                                                                                                                                T =16, I =32
                                                                                                                25                                               0        A
                                                                                                                                                                T0=16, IA=16
          15 frames        16      32    1024   1024
                                                                                                                                                                T =16, I =8
                                                                                                                20                                               0        A

          20 frames        32     1024   1024   1024
                                                                                                                15
          25 frames        64     1024   1024   1024
                                                                                                                10
          30 frames        128    1024   1024   1024
                                                                                                                 5



                                                                                                                 0
                                                                                                                         8 16   32   64           128                          256
considered to save the power consumption for low-rate                                                                                            T
                                                                                                                                                   max

traffic environment (under which the effect of Tmax is                                                                                           E[X]=2, K=10
significant), finding a suitable value of Tmax is impor-                                                          600
                                                                                                                                                                T =4, I =64
                                                                                                                                                                 0    A
tant for low-rate traffic environment.                                                                                                                            T0=4, IA=32

                                                              Power consumption per effective arrival ( PCe )
      To get the optimal value of Tmax , first note that                                                         500                                             T =4, I =16
                                                                                                                                                                 0    A

the average packet response delay E[D] increases and                                                                                                            T0=4, IA=8
                                                                                                                                                                T =16, I =64
the power consumption per unit time per effective ar-                                                            400
                                                                                                                                                                 0        A
                                                                                                                                                                T =16, I =32
rival P Ce decreases as we change the value of Tmax .                                                                                                            0        A
                                                                                                                                                                T0=16, IA=16
This implies that we can not improve E[D] and P Ce                                                              300
                                                                                                                                                                T =16, I =8
                                                                                                                                                                 0        A
simultaneously. Hence, we assume that a certain level
of delay requirement is given. Then we consider the
following twofold procedure. In the first step, for given                                                        200


mean service time E[X], system size K, initial sleep in-
terval T0 and interarrival time IA , we select the values                                                       100

of Tmax with which the value of E[D] is lower than the
delay requirement. In the second step, among the se-                                                                 0
                                                                                                                         8 16   32   64              128                       256
lected values of Tmax , we then choose the optimal value                                                                                          T
                                                                                                                                                     max
of Tmax which minimizes the value of P Ce . In all pro-
                                                                                                                                     Fig. 3   The effect of T0
cedures in this subsection, we assume the value of Tmax
is of the form 2M T0 for simplicity.
      Using Fig. 2 and the twofold procedure, we can
choose the optimal value of Tmax as follows when the        cordingly the number of packets stored in the buffer
other parameters are given as E[X] = 2, K = 10 and          during the sleep intervals increases. This results in
T0 = 2. For instance, assume that the delay require-        the increase in E[D]. On the contrary, as T0 increases,
ment is 15 frames and IA = 32 frames. Then, from the        the MS has longer sleep-mode periods and switches its
delay requirement we can first select Tmax = 4, 8, 16, 32    mode less frequently. Noting that the power consump-
because they result in the average packet response de-      tion units Eon−switch and Eof f −switch are greater than
lay lower than the delay requirement. Among these           the power consumption units ES , EW and EL in sleep-
selected values of Tmax , we check the corresponding val-   mode, wake-mode and listening intervals, the less the
ues of P Ce from Fig. 2 and finally choose Tmax = 32         MS switches its mode, the less it consumes its power.
as the optimal value which minimizes P Ce . For other       Thus, P Ce decreases. The optimal values of Tmax for
values of delay requirement and interarrival times, we      various values of T0 are summarized in Table 2 and
can perform the same procedure to obtain the optimal        3. In Table 3, the notation * means that the optimal
value of Tmax and the results are summarized in Table       value of Tmax does not exist. This can happen since
1. Note that the maximum length of Tmax is fixed and         the value of T0 is relatively large compared with the
equal to 1024 [3], so we do not consider values greater     delay requirement. That is, if T0 is large, then the
than 1024 for Tmax to obtain Table 1.                       length of sleep-mode period is large. So, the average
      In Fig. 3, we change the value of T0 and plot         packet response delay is large and accordingly the de-
the resulting value of E[D] and P Ce . From the fig-         lay requirement is violated for any value of Tmax in this
ure, we see that, as T0 increases, E[D] increases and       case.
P Ce decreases. The reason for this is as follows. As            Finally, using the above procedure for each given
T0 increases, the sleep intervals become longer and ac-     T0 and other parameters, we can determine the opti-
PARK and HWANG: AN EFFICIENT POWER SAVING MECHANISM FOR DELAY-GUARANTEED SERVICES IN IEEE 802.16E
                                                                                                                          9


Table 2      Optimal value of Tmax given delay requirement   Table 4    Optimal values of T0 and Tmax (IA = 64 and delay
(T0 =4)                                                      requirement=15 frames)
     PP           IA
         P
     delay PP               64     32     16     8                              (T0 , Tmax )     P Ce
                P
     requirement PPP
                                                                                  (2, 16)      394.3040
           10 frames        8      16     16    1024
                                                                                  (3, 24)      321.8611
           15 frames        16     32    1024   1024
                                                                                  (4, 16)      354.7521
           20 frames        32    1024   1024   1024
                                                                                  (5, 20)      312.7799
           25 frames        64    1024   1024   1024
                                                                                  (6, 24)      283.7988
           30 frames        64    1024   1024   1024
                                                                                  (7, 14)      349.4597

Table 3      Optimal value of Tmax given delay requirement                        (8, 16)      320.6569
(T0 =16)
     PP                                                                           (9, 18)      297.8579
           PP          IA
     delay   PP             64     32     16     8                                (10, 20)     279.3499
     requirement PP
                  P
                                                                                  (11, 22)     264.0157
           10 frames        *       *      *     *
                                                                                  (12, 24)     251.0954
           15 frames        *       *     32    1024
                                                                                  (13, *)          -
           20 frames        32     32    1024   1024
                                                                                  (14, *)          -
           25 frames        32    1024   1024   1024
                                                                                  (15, *)          -
           30 frames        64    1024   1024   1024
                                                                                  (16, *)          -


mal values of T0 and Tmax that minimize the power
consumption per unit time per effective arrival P Ce .        that satisfies a given delay requirement and minimizes
For instance, assume that the delay requirement is 15        the power consumption per unit time per effective ar-
frames and the interarrival time IA is 64 frames. In         rival.
this case, we compute the optimal value of Tmax and
the corresponding value of P Ce for each value of T0 , and   References
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                                                                                                                                              11




A.2. Derivation of E[NI ]                                                      A.4. Proof of Theorem 2

                                                                               For 1 ≤ j ≤ K − 1,
                                                                                                 K−2

           ∞                                                                   Pr[NS = j] =            Pr[NS = j|NV = i]Pr[NV = i]
                                                                                                 i=1
E[NI ] =         jPr[V = Sj−1 ]
                                                                                                  j
           j=1
                                                                                             =         Pr[AS = j − i]Pr[NV = i].
                                   ∞
                   −λS0                       −λSj−2        −λvj−1                               i=1
      = (1 − e              )+           je            (1 − e        )                            j
                                   j=2                                                                 (λS)j−i e−λS
                                                                                             =                      ×
                                    ∞                                                            i=1
                                                                                                         (j − i)!
      = (1 − e−λS0 ) +                   (je−λSj−2 − je−λSj−1 )                                       M                      i
                                                                                                                          (λvM ) −λvM
                                                                                                          (λvn )i e−λSn          e
                                   j=2                                                                                  +    i!
                                                                                                                                      e−λSM        .
                 ∞                                                                                n=0
                                                                                                                i!         1 − e−λvM
      = 1+             e−λSj
                                                                               Similarly as in the proof of Theorem 1, we get Pr[NS =
                 j=0
                                                                               K].
                 M                       ∞
                           −λSj
      = 1+             e          +               e−λSj
                 j=0                  j=M +1
                                                                               A.5. Proof of Theorem 4
                 M
      = 1+             e−λSj
                 j=0
                                                                                         K
           +e−λSM e−λvM + (e−λvM )2 + · · ·                                    E[B] =          E[B|NS = j]Pr[NS = j]
                 M
                                      e−λvM                                              j=1
      = 1+             e−λSj      +           e−λSM                                      K−1
                 j=0
                                    1 − e−λvM
                                                                                     =         E[B|NS = j]Pr[NS = j]
                                                                                         j=1
                                                                                         +E[B|NS = K]Pr[NS = K]
                                                                                         K−1
A.3. Proof of Theorem 1
                                                                                     =         E[B(j)]Pr[NS = j]
                                                                                         j=1
For 1 ≤ i ≤ K − 1,
                                                                                         + (E[X] + E[B(K − 1)]) Pr[NS = K]
                  ∞                                                                      K−1 K−1
Pr[NV = i] =           Pr[NV = i|NI = n]Pr[NI = n]                                   =                    E[Bi ]Pr[NS = j]
                 n=0
                                                                                         j=1 i=K−j
            = Pr[NV = i|NI = 0]Pr[NI = 0]                                                                  K−1
                       M
                                                                                         + E[X] +                E[Bi ] Pr[NS = K]
                 +          Pr[NV = i|NI = n]Pr[NI = n]
                                                                                                           i=1
                     n=1
                       ∞
                 +                Pr[NV = i|NI = n]Pr[NI = n]
                     n=M +1                                                    A.6. Proof of Theorem 5
                  (λv0 )i −λv0
                         e
            =       i!
                                      (1 − e−λv0 )                                    d
                                                                               Let πj be the steady state probability that j packets are
                   1 − e−λv0
                       M    (λvn )i −λvn                                       left in the system immediately after service completion
                                   e
                 +            i!
                                               e−λSn−1 (1 − e−λvn )            (0 ≤ j ≤ K − 1). Let Ln be the number of packets
                             1 − e−λvn
                     n=1                                                       left behind in the system immediately after the n-th
                     (λvM )i −λvM                                                                                       d
                            e
                                           ∞                                   Markov point (n = 1, 2, ...). Then the πj is represented
                 +      i!
                                                  e−λ(SM +kvM ) (1 − e−λvM )   as follows:
                      1 − e−λvM
                                          k=0
                                                                                     d
                  M
                       (λvn ) e   i −λSn          (λvM )i −λvM
                                                         e                          πj = lim Pr[Ln = j],                 0 ≤ j ≤ K − 1.
            =                                 +      i!
                                                                 e−λSM .                   n→∞
                 n=0
                             i!                    1 − e−λvM
                                                                               Let pij and ak be the one step transition probability
                                                                               in the Markov chain and the probability that k packets
                                               K−1
We have Pr[NV = K] = 1 −                       i=1    Pr[NV = i].              arrive during a service time, respectively. Then pij and
                                                                                           IEICE TRANS. COMMUN., VOL.Exx–??, NO.xx XXXX 200x
12


                                                                                                        d
ak are represented as follows:                                                      with K unknowns {πj | 0 ≤ j ≤ K − 1}. An efficient
                                                                                                              d
      ∆                                                                             algorithm for computing {πj | 0 ≤ j ≤ K − 1} can be
pi,j = Pr[Ln+1 = j|Ln = i]                                                          given in terms of
               ∞
      ∆            (λx)k −λx                                                                 d
ak =                    e    dF (x)                   k = 0, 1, ....                   ∆    πj
           0         k!                                                             πj =               0 ≤ j ≤ K − 1.                            (20)
                                                                                             d
                                                                                            π0
Then the one step transition probability pij is derived
as follows:                                                                         This πj is called an upper Hessenberg matrix [12,22,23].
               j+1                                                                  It is easy to see from equation (18) that {πj | 0 ≤ j ≤
     P0,j =           aj−k+1 Pr[NS = k],                   0 ≤ j ≤ K − 2,           K − 1} can be recursively computed as follows. For
               k=1                                                                  0≤j ≤K −2
                                                                             (12)
               K          ∞
                                                                                    π0 = 1
                                                                                           d
P0,K−1 =                          al Pr[NS = k],           j = K − 1,        (13)        πj
               k=1 l=K−k                                                            πj = d
                                                                                         π0
                                                                                             d        j+1                           j+1    d
                                                                                            π0        k=1   aj−k+1 Pr[NS = k] +     i=1   πi aj−i+1
     Pi,j = aj−i+1 ,                  1 ≤ i ≤ K − 1, i − 1 ≤ j ≤ K − 2,                =
                                                                                                                       π0d

                                                                             (14)           j+1                           j+1
                  ∞                                                                    =          aj−k+1 Pr[NS = k] +           πi aj−i+1
Pi,K−1 =                  al ,        1 ≤ i ≤ K − 1, j = K − 1,              (15)           k=1                           i=1
              l=K−i
                                                                                            j+1                            j
     Pij = 0,                         otherwise.                                       =          aj−k+1 Pr[NS = k] +           πi aj−i+1 + πj+1 a0 .
                                                                                            k=1                           i=1
Note here that NS is the number of backlogged packets
at the beginning of a busy period. Also, the balance
                                                                                    Thus, for 0 ≤ j ≤ K − 2,
equations for the steady state probabilities are given
                                                                                                                                       
by                                                                                           1 
                                                                                                      j            j+1                  
                                                                                    πj+1   =      π −   π aj−i+1 −     aj−k+1 Pr[NS = k] . (21)
                K−1                                                                          a0  j i=1 i          k=1
                                                                                                                                        
       d                   d
      πj   =              πi pi,j ,           0 ≤ j ≤ K − 1,
                  i=0                                                               From equations (17) and (20), we have
                     j+1
                
                                                                                                 1
                
                                 d
                                 πi pi,j ,        0 ≤ j ≤ K − 2,                     d
                                                                                    π0 =                    .                                    (22)
                                                                                                K−1
                                                                                                       πj
                      i=0                                                                        j=0
           =          K−1                                                    (16)
                
                
                
                                 d
                                 πi pi,j ,        j = K − 1,                        Hence, using equation (20), (21) and (22), we can get
                                                                                     d
                        i=0                                                         {πj | 0 ≤ j ≤ K − 1}.
K−1
       d
      πj = 1.                                                                (17)
j=0

Then, for 0 ≤ j ≤ K − 2, by substituting (12) and (14)
into (16), we have
              j+1                                        j+1
 d    d                                                         d
πj = π0               aj−k+1 Pr[NS = k] +                      πi aj−i+1 .   (18)
              k=1                                        i=1

Similarly, for j = K − 1, by substituting (13) and (15)
into (16), we have
                      K          ∞
 d      d
πK−1 = π0                              al Pr[NS = k]
                   k=1 l=K−k
                   K−1               ∞
                            d
              +            πi                al                              (19)
                   i=1            l=K−i

Note that equation (19) is redundant, and that equa-
tions (17) and (18) provide K independent equations
PARK and HWANG: AN EFFICIENT POWER SAVING MECHANISM FOR DELAY-GUARANTEED SERVICES IN IEEE 802.16E
                                                                                                    13


                        Yunju Park          received the B.S. de-
                        gree in Mathematics from Kyungpook
                        National University in 2002 and the M.S.
                        degree in Mathematics from Korea Ad-
                        vanced Institute of Science and Technol-
                        ogy (KAIST), Daejeon, Korea, in 2004.
                        She is currently working toward the Ph.D.
                        degree in the Department of Mathemati-
                        cal Sciences at KAIST. Her research in-
                        terests include IEEE 802.16e and the
                        power saving and performance modelling
of wireless networks.



                       Gang Uk Hwang             received the B.Sc.,
                       M.Sc., and Ph. D. degrees in Mathemat-
                       ics (Applied Probability) from Korea Ad-
                       vanced Institute of Science and Technol-
                       ogy (KAIST), Daejeon, Korea, in 1991,
                       1993 and 1997, respectively. From Febru-
                       ary 1997 to March 2000, he was with
                       Electronics and Telecommunications Re-
                       search Institute (ETRI), Daejeon, Korea.
                       From March 2000 to February 2002, he
                       was a visiting scholar at the Department
of Computer Sciences and Electrical Engineering in University
of Missouri - Kansas City. Since March 2002, he has been with
the Department of Mathematical Sciences and Telecommunica-
tion Engineering Program at KAIST, where he is an Associate
Professor. His research interests include teletraffic theory, perfor-
mance evaluation of communication systems, quality of service
provisioning for wired/wireless networks and cross-layer design
for wireless networks.

								
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