The Age of Updates in a Simple Relay Network
TThe Age of Updates in a Simple Relay Network
Ali Maatouk * , Mohamad Assaad * , and Anthony Ephremides † * TCL Chair on 5G, Laboratoire des Signaux et Syst`emes, CentraleSup´elec, Gif-sur-Yvette, France † ECE Dept., University of Maryland, College Park, MD 20742
Abstract —In this paper, we examine a system where statusupdates are generated by a source and are forwarded in a First-Come-First-Served (FCFS) manner to the monitor. We considerthe case where the server has other tasks to fulfill, a simpleexample being relaying the packets of another stream. Due to theserver’s necessity to go on vacations, the age process of the streamof interest becomes complicated to evaluate. By leveraging specificqueuing theory tools, we provide a closed form of the averageage for both streams which enables us to optimize the generationrate of packets belonging to each stream to achieve the minimumpossible average age. The tools used can be further adopted toprovide insights on more general multi-hop scenarios. Numericalresults are provided to corroborate the theoretical findings andhighlight the interaction between the two streams.
I. I
NTRODUCTION T H e Age of Information (AoI) is a new concept that hasbeen introduced in [1] and is considered of broad inter-est in communication systems. More particularly, ubiquitousconnectivity and the cheap hardware cost have created newapplications where sensors are able to send status updates to acertain receiver. The status can range from being as simple asthe temperature of a room [2] to as complicated as a vehicle’sposition and velocity [3]. In these applications, a sourcegenerates time-stamped updates that are forwarded through anetwork to reach the intended monitor. These updates shouldtherefore be as timely as possible, i.e., the receiver shouldhave the freshest possible information about what is beingmonitored. Knowing that the transmission of the packetsmay include channel errors and back-off timers to mitigateinterference from neighboring sensors, the characterization ofthis metric is far from being straightforward.Although practical scenarios of interest can be quite com-plex, the investigation of this metric for even the simplest mod-els was found to be challenging. In [1], the average age wasformulated for the case of First-Come-First-Served (FCFS)disciplines: M/M/1, M/D/1 and D/M/1 where it was shown thatan optimal packets generation rate can be found to minimizethe average age. The case where multiple information sourcesshare the same queue was also evaluated in [4].In [5], it was shown that if the source had the ability tomanage packets, the average age of the stream can be furtherreduced. Moreover, a related metric called the average peakage was introduced in [5]. With this metric being generallymore tractable than the average age, the authors in [6] for-mulated the average peak age of information for multipleinformation sources under general service time distributions. With the AoI metric gaining more attention, a surge ofpapers have been recently published on the subject [7]. Theage of information for energy harvesting sources was intesivelyinvestigated in the literature (e.g. [8]). More recently, a shiftof interest to multi-hop scenarios can be witnessed. This canbe justified by the fact that the AoI is of high interest inwireless sensor networks where, in general, sensors can bemultiple hops away from the monitor. In [9], it was shownthat when the transmission times of packets in the networkare exponentially distributed across all nodes, the Last-Come-First-Served (LCFS) preemptive policy at the relaying nodesminimizes the average age of the stream. Although this workgives insights on the disciplines to be used, it does not provideways to explicitly calculate the average age of each stream.More recently, an explicit calculation of the average age wascalculated for a single information stream over a network ofpreemptive servers in [10]. In another work [11], due to thecomplexity of the average age calculations, an upper boundand lower bound were provided for the case where a serverhas two streams to serve: a FCFS stream and a LCFS withpreemption stream.In the present work, we overcome the complexity that wasfound in [11] by leveraging some key queuing theory tools.By doing so, a study of a simple relay scenario will befeasible as will be depicted in the sequel. More precisely,we are able to formulate a closed form of the average ageof both streams along with a discussion on the interactionbetween the two streams in terms of both stability and theirrespective age. In fact, it will be shown in the sequel thatthe minimization of the age of the stream of interest admits aunique optimal packets generation rate. The interest in thosetools used in the formulation come from the fact that theycan be used to evaluate even more complicated multi-hopscenarios. Moreover, we study the case where the secondstream (i.e. the relayed packets) is not age-sensitive. Thiscomes from the fact that some applications are more restrictedby throughput than age. In this case, we aim to find the bestway to minimize its effect on the age sensitive stream. It isworth mentioning that our system model is general and that isit not bound to the relay scenario taken into account. It can beused for any type of scenarios where the server is allowed tobe unavailable for a certain amount time. Some non-exclusiveexamples include the node transitioning to SLEEP mode ordoing necessary offline processing.The paper is organized as follows: Section II describes thesystem model. Section III presents the theoretical results on a r X i v : . [ c s . I T ] M a y he average age of the streams. Section IV provides numericalresults that corroborate the theoretical findings while SectionV concludes the paper.II. S YSTEM M ODEL
Consider an information source sending status updates to amonitor. In this scenario, the instantaneous age of informationat the receiver at time instant t is defined as: ∆( t ) = t − U ( t ) (1)where U ( t ) is the time stamp of the last successfully receivedpacket by the monitor. Suppose that packet j is generated attime instant t j and is received by the monitor at time t j , theevolution of the instantaneous age in this case can be depictedin Fig. 1 where A j and T j denote the inter-arrival and sojourntime of packet j respectively.Fig. 1: The time evolution of the instantaneous age ∆( t ) The main interest lays in the computation, and eventualminimization, of the average age of information in the aimof keeping the information as fresh as possible at the monitor.The average age is nothing but the saw-tooth area highlightedin Fig.1 and is defined as: ∆ = lim τ →∞ τ Z τ ∆( t ) dt (2)The computation of this metric is however challenging, evenin the simplest scenarios [1]. With the metric being relativelynew, giving insights on the computation and minimization ofthe average age in particular realistic scenarios is of paramountimportance.The scenario of interest that we investigate in the followingconsists of an information source generating packets, denotedas stream , and buffering them in a single queue (FCFSdiscipline) where the packet j inter-arrival time A j is supposedto be exponentially distributed with mean λ . In other words, λ can be seen as generation rate of packets belonging tostream . The physical service time of the packet G j issupposed to be exponential as well with mean µ . However,we consider the case where the server is not always availableand can take eventual vacations. When a vacation takes place,the transmission of the packet of stream is interrupted (witheventual resume) until the vacation finishes. The decision togo on vacation is dictated by an appropriate timer S that issupposed to be exponentially distributed with mean s . The vacation duration W is also supposed to be exponentiallydistributed with mean w . What makes this model of broadinterest is the fact that, in realistic scenarios, a device may havedifferent functionality than just being occupied by forwardingits own generated packets. Some non-exclusive examples ofvacations are: • The device may transition into
SLEEP mode for acertain amount of time after which it wakes up. The mo-tivation behind this transition is to prolong the battery’slife • The device may proceed to do necessary offline process-ing • The device may relay packets of different streams otherthan its ownIn the sequel, for clearer presentation, we focus on thecase where the vacation is motivated by relaying packets ofdifferent streams although the analysis is still valid for any ofthe aforementioned cases. Sensor is considered as the sensorof interest that is generating age-sensitive data and forwardingthem to a monitor that is only one hop away. However, at thesame time, it keeps a separate queue for packets generated bysensor in the aim of forwarding them to the monitor as seenin both Fig. 2 and 3.Fig. 2: A simple relay networkWe consider in the following work two different applicationsfor the simple relay network in question: • The relayed stream is not age-sensitive and we aresimply restricted by a certain target average throughput • The relayed stream is age-sensitiveWe first start by providing the general mathematical formulaused to calculate the age of the stream of interest (stream ).When the system is in steady state, both the inter-arrival andthe sojourn times of the packets of stream are stochasticallyidentical, i.e., A j st = A j − = A and T j st = T j − = T . Wetherefore lose the packet’s index j in the sequel. In this case,the average age of stream is evaluated as [1]: ∆ = λ ( E ( AT ) + 1 /λ ) (3)The main difficulty in finding a closed form of the averageage of a certain stream lays in the term E ( AT ) . In fact, thewo variables are not independent (a large inter-arrival timeallows the system to be emptied and T to be smaller) andthe evaluation of this term, even in the simplest scenarios, canbe challenging. In the following, we provide a closed formexpression of the average of the stream of interest. Using theformulated expression, insights on the effect of the vacation ishighlighted along with finding the optimal packets generationrate λ ∗ to minimize the average age of stream is provided. Todo so, we first model the system using a D Markov chain. Wethen proceed to leverage the notion of probability generatingfunctions and Little’s distributional law. After appropriateanalysis, the closed form is formulated along with differentdiscussions on practical scenarios considerations.Fig. 3: The perspective of Sensor 1III. A GE C ALCULATION
Let us consider the 2D continuous time stochastic process { (cid:0) N ( t ) , M ( t ) (cid:1) : t ≥ } where N ( t ) ∈ N and M ( t ) ∈ { V, B } denote the number of packets of stream inside the systemand the server status at time instant t respectively. M ( t ) canonly take two values: 1) V when the server is on vacation and2) B when the server is not on vacation and is serving or isready to serve the stream of interest. By taking into accountthe model’s assumptions, { (cid:0) N ( t ) , M ( t ) (cid:1) : t ≥ } becomesa Markovian process and the evolution of the system can berepresented by a 2D continuous time Markov chain as seenin Fig. 4 where the states are all the possible combinations of ( N, M ) . Fig. 4: Markov ChainLet π B ( i ) be the stationary distribution for the state where theserver is not on vacation while i packets of stream are inthe system. On the counterpart, let π V ( i ) be the case wherethe server is on vacation with i packets of stream in thesystem. Based on the preceding, we can define the ProbabilityGenerating Function ( p.g.f ) of the random variable N as: P ( x ) = E ( x N ) = ∞ P i =0 π ( i ) x i = ∞ P i =0 π V ( i ) x i + ∞ P i =0 π B ( i ) x i = P V ( x ) + P B ( x ) . Theorem 1.
In the aforementioned system, the p.g.f of thenumber of packets of stream in the system is given by: P ( x ) = µ π B (0)( λ + w + s − λ x ) λ x − ( λ w + λ + λ s + λ µ ) x + λ µ + wµ (4) where : π B (0) = µ − λ − λ sw µ (1 + sw ) (5) Proof:
The proof can be found in Appendix A.
Theorem 2.
The sojourn time T of packets belonging tostream in the system has the following distribution: f T ( t ) = C exp( α t ) + C exp( α t ) (6) where α and α are the roots of the second degree equation: z + z ( w + s + µ − λ ) + wµ − λ w − λ s = 0 (7) Moreover, C and C are given by: C = ( µ − λ − λ sw )( w + s + α )(1 + sw )( α − α ) (8) C = ( µ − λ − λ sw )( w + s + α )(1 + sw )( α − α ) (9) Proof:
The proof can be found in Appendix B.
A. Application 1: Relay of non age sensitive data
Suppose that the packets that are being relayed are not agesensitive and our aim is to have a desired average throughputfor stream . The average throughput experienced by therelayed stream is equal to the portion of time the server spendson vacation . Corollary 1.
The average amount of time the server spendson vacation is: π V = swsw + 1 (10) Proof:
The proof can be found in Appendix C.We can see from the results of this corollary that the averagethroughput of stream depends on the ratio of sw and notindividually on them. These results are of paramount interest:In fact, investigating the average age of stream , for a fixed sw ,answers the important question; which has the least effect onthe age of stream , short and numerous vacations, or long rarevacations? If it turns out to be the first, then packets of stream are better off being served in fragments by assigning shortmean vacation times. If it is the latter, then packets of stream are better served as a batch (i.e assigning long mean vacation The system is stable if and only if π B (0) > which translates into having µ > λ + λ sw It is assumed that the queue corresponding to stream is always back-logged, i.e., whenever the server goes on vacation, it finds a packet to relay.If not, dummy packets are supposed to be sent which is a natural assumptionto evaluate the average throughput performance [12][13][14, p.8] ime which would result in numerous packets being servedin a single vacation). In both cases, the average throughputof stream is the same as the frequency of the vacations isdifferent. To be able to answer this question, we provide aclosed form of the average age of stream in the following. Proposition 1.
The average age of the stream of interest isgiven by: ∆ ( λ , µ , s, w ) = λ C α ( α − λ ) + λ C α ( α − λ ) + 1 λ + µ s ( µ − λ − λ sw ) + ( s + w )( λ + w )( µ (1 + sw ))( wµ )( λ + w )( µ (1 + sw )) (11) Proof:
The proof can be found in Appendix D.The next step revolves around finding, for a fixed setting ( µ , s, w ) , the optimal packets generation rate λ ∗ to minimizethe average age of stream . Insights on this optimization isprovided in the following proposition. Proposition 2.
For a fixed service rate µ > , mean vacationtime s > and mean vacation duration w > , the followingoptimization problem admits a unique minimizer λ ∗ :minimize λ ∆ ( λ ) subject to < λ < µ sw (12) Proof:
The proof can be found in Appendix E.By using the results of Proposition 2, we can answer thequestion previously stated as for the nature of vacations thathave the least effect on the average age of stream . Aswe have shown in the proof in Appendix E, the optimalsolution of the problem (12) in the interval ]0 , µ sw [ can beachieved by finding λ ∗ such that d ∆ ( λ ) dλ | λ = λ ∗ = 0 (pleaserefer to Appendix E for more details). After formulation of λ ∗ , we can compare the minimal achievable average age ∆ ( λ ∗ , µ , s , w ) and ∆ ( λ , µ , s , w ) such that s > s and w > w while preserving s w = s w to find theanswer to our question. It turns out, as will be seen in thenumerical results section, that it is better to have numerousshort vacations rather than long rare vacations to reduce theeffect of the vacations on the average age of stream . B. Application 2: Relay of age sensitive packets
As previously mentioned in the introduction, the averageage of information in multi-hop networks was investigated in[9] and it was shown that, for exponential packets transmis-sion time, relayed packets should follow LCFS policy withpreemption to minimize their age. In this case, the vacationcan be thought to be an arrival of a packet of stream thatpreempts any packet being transmitted, either from stream or stream . In this case, the relayed packets see the systemas an M/M/1/1 queue and their average age is therefore [4]: ∆ = 1 µ + 1 λ (13) where λ and µ are the arrival and service rate of stream respectively. This type of scenarios was investigated in [11]where a lower and upper bound on the average age of stream was provided using the detour flow graph tool. Unlikethe work in [11], we develop a closed form of the averageage of stream and give insights on choosing the optimumpackets generation rate λ to minimize the age of stream .This highlights the importance of the tools presented in theAppendices A and B as they are able to reduce the difficultyof the average age calculations for certain complex scenarios. Proposition 3.
The average age of stream is given byreplacing s = λ and w = µ in eq. (11) ∆ ( λ , µ , λ , µ ) = λ C α ( α − λ ) + λ C α ( α − λ ) + 1 λ + µ λ ( µ − λ − λ λ µ ) + ( µ + λ )( λ + µ )( µ (1 + λ µ ))( µ µ )( λ + µ )( µ (1 + λ µ )) (14) Proof:
The analysis is similar to that of Appendix D. Inthis scenario, the vacation starts when a packet from stream arrives to the system and preempts the service of packetsof stream . The vacation finishes once the packet of stream that is being served finishes. The same Markov chain ofFig. 4 holds, by replacing s = λ and w = µ . However,the difference between the two scenarios is the fact that newpackets of stream can preempt the previous stream packetin service. In other words, the vacation duration timer canbe reset due to a new arrival from stream . Due to thememoryless property of the exponential vacation timer, thisreset of the vacation timer does not affect stream and thesame sojourn and virtual service time for packets of stream as in Application 1 is to anticipated and the same proof ofAppendix D holds.IV. N UMERICAL R ESULTS
This section provides numerical results to give insights onthe theoretical conclusions previously stated.
A. Application 1
In this scenario, we have a fixed average service rate ofstream and is set to µ = 1 . Moreover, we have a certainaverage throughput requirement for the relayed stream toensure its queuing stability. This requirement is supposed tobe equal to in the sequel, i.e., the portion of time where theserver should be relaying packets from stream is equal to . One question arises: knowing that the throughput dependson the ratio of sw rather than individually on them, how couldwe calibrate these two parameters in a way to have the leastimpact on the average age of stream ?For this purpose, we simulate different scenarios of s and w such that we keep achieving the same required averagethroughput ( sw = 1 ). As seen in Fig. 5, for the same ratio sw = 1 , it is better to opt for high values of s and w . In otherwords, as previously mentioned, it is better to go often onvacation but for short amount of time. Consequently, packetsf stream are better served as fragments by making w be ashigh as practically possible while providing a vacation rate s high enough to keep the same ratio of sw = 1 . Moreover, thereis a certain optimal rate λ ∗ that achieves the lowest possibleaverage age of stream for a fixed setting. Discussions onthis matter will be provided in the next subsection. A ve r a g e A g e o f S t r ea m s=0.5,w=0.5s=1,w=1s=5,w=5 Fig. 5: Illustration of the effect of s and w on the average ageof stream B. Application 2
In this scenario, we have a fixed average service rate of µ = 1 and fixed vacation timer w = µ = 4 which is dictatedby the size of stream packets. In this case ∆ = + λ ,therefore once can clearly that increasing s = λ reducesthe average age of stream . Our goal becomes to find thepackets generation rate that minimizes the age of stream for a certain fixed relayed packets rate λ . This can bethought as an optimal packets generation rate λ ∗ that mitigatethe effect of the relay of stream as much as possible.In fact, one can see that as the arrival rate λ of stream grows larger, the optimum λ ∗ grows smaller and the achievedminimal average age is larger. This is due to the serverbeing unavailable for a higher duration of time and therefore,delays are anticipated by waiting for vacations to finish whichresults in a necessity to reduce the packets generation rate λ to mitigate this delay and therefore reduce the average age.V. C ONCLUSION
In this paper, we have investigated the age of informationin the case where the server is not always available and isallowed to go on vacation. This setting is of broad interest asit incorporates several practical scenarios, such as the relayof packets from different streams. By leveraging several keyqueuing theory concepts, we were able to provide a closedform expression of the average age of the stream of interest.Insights on the theoretical results were provided to furtherunderstand the interaction between the age of the stream ofinterest and the vacations considered. The authors believe thatthe analysis provided in the paper can be used to explore moregeneral scenarios where more than one stream are relayed by A ve r a g e A g e o f S t r ea m =4 =6 =8 Fig. 6: Illustration of the optimal packets generation ratethe node of interest and therefore be a basis for multi-hopsystems analysis. R
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PPENDIX AP ROOF OF T HEOREM M = V : ( λ + w ) π V ( i ) = λ { i = 0 } π V ( i −
1) + sπ B ( i ) ∀ i ∈ N (15)where { i = 0 } = 1 if and only if i = 0 . We proceed bymultiplying eq. (15) by x i and summing over all the possiblevalues of i ∈ N : ( λ + w ) + ∞ X i =0 x i π V ( i ) = + ∞ X i =0 λ { i = 0 } x i π V ( i − s + ∞ X i =0 x i π B ( i ) (16)By manipulating the power series on the right handside(change of variables, appropriate multiplication/division by thevariable x ), and by making use of the definition of the p.g.f,we can conclude the following: P V ( x ) = sλ + w − λ x P B ( x ) (17)The next step consists of formulating the balance equationsfor the states where the server is not on vacation: ( λ + s + µ { i = 0 } ) π B ( i ) = µ π B ( i + 1)+ λ { i = 0 } π B ( i −
1) + wπ V ( i ) (18)As it has been previously done, we multiply by x i and sumover all the possible values of i ∈ N : ( λ + s + µ ) + ∞ X i =0 x i π B ( i ) − µ π B (0) = µ ∞ X i =0 x i π B ( i + 1)+ λ ∞ X i =0 x i { i = 0 } π B ( i −
1) + w + ∞ X i =0 x i π V ( i ) (19)After some manipulations, we get the following: ( λ + s + µ − λ x − µ x ) P B ( x ) = µ π B (0) − µ x π B (0)+ wP V ( x ) By using the results of eq.(17) and the fact that P ( x ) = P V ( x ) + P B ( x ) , the p.g.f of the number of packets of stream in the system, referred to as N , is: P ( x ) = (cid:0) xµ π B (0) − µ π B (0) (cid:1)(cid:0) λ + w − λ x + s (cid:1) ( λ + w − λ x )( λ x + sx + µ x − λ x − µ ) − wsx (20)What remains is to find the constant π B (0) . To do so, weeliminate the common factor ( x − from the numerator anddenominator by taking into account that: ( λ + w − λ x )( λ x + sx + µ x − λ x − µ ) − wsx = ( x − λ x − ( λ w + λ + λ s + µ λ ) x + ( λ µ + wµ )) (21)By doing so, the p.g.f can be rewritten as follows: P ( x ) = µ π B (0)( λ + w + s − λ x ) λ x − ( λ w + λ + λ s + λ µ ) x + λ µ + wµ (22) Knowing that P (1) = ∞ P i =0 π V ( i ) + ∞ P i =0 π B ( i ) = 1 , we canconclude by replacing x by the value of in eq. (22) that: π B (0) = µ − λ − λ sw µ (1 + sw ) (23)which concludes our proof.A PPENDIX BP ROOF OF T HEOREM are Poisson2) All packets that enter the system of stream , stay in thesystem until they are served (in other words, there is noblocking, balking or reneging)3) Packets of stream enter the system and leave one at atime in the order of their arrivals4) The arrival of packets of stream after time t and thetime in the system of any packet of stream arrivingbefore t are independent (i.e. new arriving packets donot influence the time of the system of previous arrivingpackets)These conditions are sufficient for the distributional form ofLittle Law to hold [15]. By defining the Laplace transform ofthe sojourn time of packets in the system as P T ( z ) , we cantherefore conclude from the distributional form of Little’s lawthat: P ( x ) = P T ( λ − λ x ) . By doing the appropriate changeof variable: x = λ − zλ and after some algebraic manipulations,we get the following: P T ( z ) = µ π B (0)( w + s + z ) z + z ( w + s + µ − λ ) + wµ − λ w − λ s (24)One can see that the roots of the denominator verify: ( α + α < if w + s + µ > λ α α > if µ > λ (1 + sw ) (25)In other words, the roots of the denominators are negative ifand only if these two conditions are satisfied. We recall thatthe negativity of the poles implies the stability of the queuingsystem. One can easily see that the second condition is morerestrictive (if condition is verified then condition is verifiedas well). Therefore, for the system to be stable, it is sufficientand necessary to have µ > λ (1 + sw ) which was previouslyconcluded in Theorem 1. Furthermore, the Laplace transformof sojourn time can be decomposed as follows: P T ( z ) = C z − α + C z − α (26)here α , α are the roots of the second polynomial in thedenominator and C and C are given by: C = ( µ − λ − λ sw )( w + s + α )(1 + sw )( α − α ) (27) C = ( µ − λ − λ sw )( w + s + α )(1 + sw )( α − α ) (28)By employing the inverse Laplace transform, the proof can beconcluded. A PPENDIX CP ROOF OF C OROLLARY
Definition 1.
Suppose that the state-space of a Markov chainis divided into K disjoint subsets of states P = { S , . . . , S k } .A continuous-time Markov chain is said to be lumpable withrespect to the partition P if and only if, for any subset S i and S j in P , for any states c, c ∈ S i , we have: X m ∈ S j q ( c, m ) = X m ∈ S j q ( c , m ) (29) where q ( c, m ) is the transition rate between states c and m . Our proposed chain is characterized by being lumpable withrespect to the following partition: P = { S , S } where S isthe collection of states where the server is on vacation while S contains the states where the server is not on vacation.In fact, as seen in Fig. 7, the total transition rate from anystate in the upper part of the chain S to the lower one S is constant and is equal to w . In the other direction, this totaltransition is equal to s . This means that we can reduce ourstate space to two states: V and B to capture the state of theserver. By doing so, we can easily conclude the probability tobe in a vacation (and therefore the amount of time the serveris on vacation on average) by simply solving for π V in thefollowing equations which concludes our proof: ( wπ V = sπ B π V + π B = 1 (30)Fig. 7: Lumpable Markov chain A PPENDIX DP ROOF OF P ROPOSITION j denoted as T j can be decomposed as follows: T j = B j + Y j (31)where B j is the waiting time of the packet j in the queue and Y j being the virtual service time of each packet. The virtualservice time of packet j is defined as the time elapsed betweenthe time of delivery of packet j and the maximum betweenthe time of delivery of the previous packet j − or the arrivaltime of packet j : Y j = D j − max( D j − , t j ) (32)The virtual service time may be different than the actualphysical service time due to the possibility of the serviceof the packet to be interrupted by the vacation. In this case,the calculation of E ( AT ) can be decomposed into calculating E ( AB ) and E ( AT ) . E ( AB ) : The waiting time of a packet can be expressed as B = ( T − A ) + where ( . ) + = max( ., [1]. By the total lawof expectation, we have the following: E ( AB ) = E ( E ( AB | A = a )) = Z + ∞ Z + ∞ a a ( t − a ) f T ( t ) λ exp ( − λ a ) dtda (33)By using the results of Theorem 2, we get after some calcu-lations: E ( AB ) = λ C α ( α − λ ) + λ C α ( α − λ ) (34) E ( AY ) : To be able to characterize this term, it is importantto discuss the virtual service time of each packet. For thispurpose, we define the following event:
Υ = { Arriving packet of stream finds the systemin state π V (0) } When the complement event Υ takes place, one of the follow-ing happen: • The arriving packet finds the server ready to serve it andthe physical service of the packet starts • It finds buffered packets of stream ahead of itIn the first case, the physical service time starts immediately.In the second case, the packet waits for the packets ahead tobe served (which is counted in its waiting time B ). When thepacket ahead of it finishes being served, the physical servicetime of the packet of interest immediately starts. On the otherhand, if the event Υ takes place, the arriving packet findsitself at the head of the buffer but the server on vacation andit has to first wait for the vacation to finish before its physicalservice time starts. We can see from this discussion that the It is worth mentioning that we suppose that the system is in steady state,i.e., the sojourn times and virtual service times of the packets are stochasticallyidentical T j st = T j − = T , B j st = B j − = B and Y j st = Y j − = Y erm E ( AY ) is tricky to calculate since the arrival time clearlyaffect the service time of the packet. However, these randomvariables A and Y are conditionally independent given Υ or Υ . In fact, the value of A does not affect Y when we knowthe state of the system the packet arrives to. Based on that,and by leveraging the total law of expectation, we can writethe following: E ( AY ) = E ( AY | Υ) P (Υ) + E ( AY | Υ) P (Υ)= E ( A | Υ) E ( Y | Υ) P (Υ) + E ( A | Υ) E ( Y | Υ) P (Υ) (35)In order to deal with these terms, we first recall the PASTAproperty: an arrival from a Poisson process observes thesystem as if it was arriving at a random instant of time. Basedon that property and by taking into account that the arrival timeis independent of the state of the system, we can conclude: E ( AY ) = 1 λ E ( Y | Υ)(1 − π V (0)) + 1 λ E ( Y | Υ) π V (0) (36)What remains is to calculate the two terms: E ( Y | Υ) and E ( Y | Υ) . To proceed with that, we elaborate more on whathappens when a packet arrives to the system. Suppose thatthe event Υ takes place and the packet is now at head of thebuffer: due to the memoryless property of the vacation starttimer, two clocks start ticking: an exponential service time G of mean /µ and an exponential vacation start time S ofmean /s . The packet physical service time finishes beforethe vacation starts with a probability p = P ( G < S ) = µ µ + s .With a probability − p = sµ + s , the vacation interruptsthe physical service time of the packet for a duration W exponentially distributed with mean /w . After the vacationfinishes, the physical service of the packet is resumed. Dueto the memoryless property, when the service of the packetis resumed, it is as if a new an exponential service time G of mean /µ was taken. And therefore, the same probability − p holds to go for another vacation before the service ofthe packet finishes. One can see from this that the virtualservice time of the packet can include a random number ofexponentially distributed vacation times. We therefore define L as the number of vacations that take place during thepacket service time. Based on our analysis, L is geometricallydistributed with parameter p : P ( L = i ) = (1 − p ) i p (37)Knowing that each vacation adds an expected time of /w tothe virtual service time of the packet, and by using the totallaw of expectation, we have the following: E ( Y | Υ) = + ∞ X i =0 E ( Y | Υ , L = i ) P ( L = i ) = + ∞ X i =0 p (1 − p ) i ( 1 µ + i w )= 1 µ + pw (1 − p ) p = 1 µ + swµ = s + wwµ (38)As for E ( Y | Υ) , we know that the only difference betweenthe two virtual service time is that the packet has to waitfirst for the vacation to finish when the event Υ takes place. In other words, E ( Y | Υ) = E ( Y | Υ) + w = s + w + µ wµ . Bytaking into account that π V (0) = s ( µ − λ − λ sw )( λ + w )( µ (1+ sw )) , we havethe following: E ( AY ) = ( s + w + µ ) s ( µ − λ − λ sw )( λ wµ )( λ + w )( µ (1 + sw ))+ ( s + w ) λ wµ (1 − s ( µ − λ − λ sw )( λ + w )( µ (1 + sw )) ) (39)Combining the two results and by applying eq. (3), we end upwith the closed form of average age.A PPENDIX EP ROOF OF P ROPOSITION ∆ ( λ ) is decomposed into three parts:1) λ which is a convex function for any < λ < µ sw µ s ( µ − λ − λ sw )+( s + w )( λ + w )( µ (1+ sw ))( wµ )( λ + w )( µ (1+ sw )) which can berewritten as aλ + bcλ + d where: a = ( s + w ) µ w − µ s − µ s w b = µ s + ( s + w ) µ c = µ ( s + w ) d = wµ ( s + w ) (40)By deriving twice with respect to λ , one can verify that asufficient and necessary condition for convexity for λ > is: bc − ad > . By a simple calculation, this condition can befound to be always verified for any < λ < µ sw .3) The first two terms are also two convex functions. Theexact calculations are omitted for the sake of space due to thecomplexity of the objective function, but by deriving twicewith respect to λ , the convexity can be verified.The objective function is therefore a sum of several con-vex functions and is therefore convex itself. Knowing that ∆ −→ + ∞ when λ −→ and when λ −→ µ sw ,and by taking into account the convexity of the objectivefunction, we can assert that there exist a unique stationarypoint < λ ∗ < µ sw such that d ∆ ( λ ) dλ | λ = λ ∗ = 0 . Thisobservation renders the Lagrangian approach to the problemunnecessary as this stationary point is the uniqueunique