Age of Information for Single Buffer Systems with Vacation Server
AAge of Information for Single Buffer Systems with Vacation Server
Jin Xu , I-Hong Hou , and Natarajan Gautam ∗ Industrial and Systems Engineering, Texas A&M University Electrical and Computer Engineering, Texas A&M University
Abstract
In this research, we consider age-related metrics for queueing systems with vacation server. Assumingthat there is a single buffer at the queue to receive packets, we consider three variations of this single buffersystem, namely
Conventional Buffer System (CBS),
Buffer Relaxation System (BRS), and
ConventionalBuffer System with Preemption in Service (CBS-P). We introduce a decomposition approach to derivethe closed-form expressions for expected Age of Information (AoI), expected Peak Age of Information(PAoI) as well as the variance of peak age for these systems. We then consider these three systemswith non-independent vacations, and use polling system as an example to show that the decompositionapproach can be applied to derive closed-form expressions of PAoI for general situation. We explorethe conditions under which one of these systems has advantage over the others, and we further performnumerical studies to validate our results and develop insights.
Age of Information (AoI) has drawn wide attention of researchers recently due to its applications in sensor net-works, communication networks and autonomous vehicle systems [23, 7]. Different from the long-establishedqueueing metrics such as delay or waiting time, AoI measures the time elapsed since the generation time ofa packet that is most recently delivered. AoI is deemed as a new but useful metric to describe the freshnessof data and timeliness of information [59, 19, 2]. An example of application of AoI can be found in smartmanufacturing where data sensed by sensors at machines are further processed by edge devices and proces-sors with limited processing capabilities. Processed information would be used for estimating the RemainingUseful Life (RUL) [45], detecting defects of manufactured products [5], or making real-time process controls ∗ Corresponding Author, Email:[email protected] a r X i v : . [ c s . PF ] A p r λ .Packets generated by the data source are sent to the server for further processing. There is a buffer at theserver which can hold at most one packet at a time. We assume that only the most recent arrived packet iskept in the buffer as it contains the freshest information about the data source. If a new packet is generatedand there is an old packet waiting in the buffer, the old packet will be discarded when the new packetenters the buffer. There is a single server in the system which processes (serves) packets from the bufferonce it becomes available. The processing time of each packet is i.i.d. The age at time t is thus defined as ∆( t ) = t − max { r { l } : C { l } ≤ t } , where r { l } is the generation time of the l th packet that is processed by theserver, and C { l } is the time when this packet has been processed by the server. Note that the packets thatare not processed by the server are not included in the age calculation. The time-average age is then definedas ¯∆ = lim T →∞ T (cid:82) T ∆( t ) dt . By assuming the system being ergodic, we have E [∆] = lim t →∞ E [∆( t )] = ¯∆ , and in this paper we use the term “AoI” to refer E [∆] . While AoI is a useful metric to measure data freshness,many researchers also analyzed a metric called Peak Age of Information (PAoI) for its tractability [7, 19, 20].We let the l th peak of ∆( t ) be A { l } , and define the expectation of this peak value, i.e., E [ A { l } ] , as PAoI.Specifically, in this paper we consider AoI in scenarios where the server/processor in the communicationnetwork takes “vacations” over time. We assume that the server takes a vacation once a packet has beenprocessed. If the server finds no packet waiting in the buffer upon its returning from a vacation, it thentakes another vacation. This specific model is motivated by the application in smart manufacturing, wherekeeping the energy-consuming server/processor idling when there is no data packet, is not efficient in termsof energy saving. To save system energy as well as guarantee other system performance, a strategy is lettingthe server go for a low-energy-cost sleeping period when there is no packet waiting in the buffer, and wakeup if a packet is observed in buffer when a sleeping period is over. The advantage of this strategy in energysaving has been discussed in [14, 54], however its performance in AoI related metrics have not been studied.Another motivation of our model comes from the underwater sensor networks in the petroleum industryor underwater environment monitoring, where acoustic transmissions are not energy efficient and would2esult in the batteries needing frequent replacements (see [38, 17]). In such a case, an efficient way fordata transmission is to store the sensed data in an underwater node, and use a rechargeable autonomousunderwater vehicle that is sent from surface to upload or collect data from the node in a periodic way (see[16, 52]). So we abstract the underwater vehicle arrival as a completion of a vacation, collection of data asservice and leaving back to the surface as the start of the next vacation. A third application of our modelcan be found in remote health monitoring, where the health data is acquired by a wearable device from apatient and transmitted to the healthcare provider over time (see [8, 36]). Our model corresponds to thecase where a doctor at the healthcare center checks the patient’s updated information from time to time,where the time for the doctor to analyze the data can be modeled as the service time and the time betweenthe doctor checks the updated data for a specific patient can be regarded as the vacation time.In addition, this vacation server model has a wide range of other applications in smart manufacturingsensor networks and computer-communication systems where the server has additional tasks aside fromprocessing the primary data source of interest [3, 55, 25]. Whenever the server schedules these “non-primary”tasks during the idling period of the primary data source, we can regard this server as “taking vacations”.Many queueing network systems such as the priority queue system [55] and the polling system [48], canalso be regarded as vacation server systems. Systems with server maintenance (see [11]) or server turningon/off (see [35]) can be modeled as vacation server systems as well. Other vacation server models have beendiscussed in the queueing literature such as [31, 30, 12, 10], however age related metrics in vacation servermodels have not been fully studied yet.In this paper we consider three following variations of the vacation server model with single buffer system: • Conventional Buffer System (CBS) (see [29, 50, 6, 40]): In this system, the buffer becomes empty onlywhen the server finishes serving the packet. New arrivals during processing will be rejected. • Buffer Relaxation System (BRS) (see [50, 6]): In this system, the buffer becomes available once theserver starts serving. The vacation will start once a service is done. • Conventional Buffer System with Preemption in Service (CBS-P) : In this system, new arrival duringprocessing will preempt the packet in service. The preempted packet will be discarded.Note that in BRS, there could be at most two packets in the system at the same time, with one in processingand one waiting in the buffer. We also notice that in BRS, the server will anyway take a vacation afterprocessing a packet, and the packet arriving during processing will be processed only when the vacation isover. This service discipline is also called gated in some literatures about vacation servers (see [48, 47, 13]).In these three systems above, a packet arriving during vacation will always preempt the packet waiting inthe buffer. 3he major contributions of this paper are summarized as follows: • We provide a decomposition approach which decomposes the peak age of single buffer system intoindependent components, so that the Laplace-Stieltjes Transform (LST) of peak age in CBS, BRS andCBS-P can be obtained when the vacation time is i.i.d. We then provide closed-form expressions forAoI, PAoI and variance of peak age for CBS, BRS and CBS-P. • We prove that when vacation time is i.i.d., BRS always has smaller PAoI than CBS, regardless ofvacation or service time distribution. We also provide the condition under which CBS-P always hassmaller PAoI than CBS. • Unlike PAoI, for AoI we show that when vacation time is i.i.d., BRS does not always have smaller AoIthan CBS, and CBS-P does not always have smaller AoI than CBS. • We extend our discussion to systems with non-i.i.d vacation times, and provide an approach to calculatePAoI for polling systems with Markovian polling schemes. We show that in polling systems, BRS nolonger has advantage over CBS in terms of small PAoI, and CBS-P has smaller PAoI than CBS whenthe service time is exponential.The rest of this paper is organized as follows: A summary of the literature is provided in Section 2. InSection 3 we consider the cases where server takes i.i.d vacations. In Section 4 we consider the case withnon-i.i.d vacations, and discuss the polling system as an example of non-i.i.d vacation model. We performnumerical studies and develop insights in Section 5, and provide concluding remarks and ideas of future workin Section 6.
The system with vacation server has been studied by many researchers due to its wide applications [31,30, 12, 10, 25, 48]. However, most of the previous papers focused on metrics such as average waiting time,queue length, throughput and rejection rate. The AoI related metrics of the system with vacation serverhas not been fully studied. In the last few years, the data freshness has drawn much attention due to theneed of timely information processing and sharing. AoI and PAoI as metrics that measure data freshness,have been studied mostly from a queueing perspective. Kaul et al [23] first introduced the idea of AoI,and provided the average AoI for M/M/1, M/D/1 and D/M/1 queues. Costa et al [7] provided the averageAoI and PAoI for M/M/1/1, M/M/1/2 and M/M/1/2* queues (the asterisk means keeping the most recentpacket in the buffer), and pointed out that retaining the most recent packet is more efficient than keeping all4ackets that the data source generates. Najm and Telatar [43] considered M/G/1/1 queue with preemption.Najm and Nasser [41] considered Last Come First Serve (LCFS) scheme with and without preemption in asingle queue system with single buffer and gamma service time. M/G/1/1 queue systems with hybrid ARQ(HARQ) protocols are discussed in Najm et al[44]. Soysal and Ulukus [46] considered G/G/1/1 type queuesand provided bounds of AoI for different arrival or service processes. Zou et al [60] discussed the waitingprocedure in M/G/1/1 and M/G/1/2* systems. Inoue et al [20] discussed the relationship between AoI andPAoI for single queue systems, and derived the LST of AoI and PAoI for different variations of single queuesystems. Some recent papers have considered the system with single server but multiple queues. Huangand Modiano [19] considered PAoI of multi-class M/G/1 and M/G/1/1 queues, and they assumed thereis one single buffer for all queues in the M/G/1/1 case. Kaul and Yates [24] also considered a model withpriority queues with and without waiting rooms for preempted packets. Moltafet et al [39] derived the closed-form of AoI for multi-class M/G/1 queues with First Come First Serve (FCFS) scheme. The system withpacket deadlines is considered in Kam et al[22]. Many research articles considered AoI/PAoI in slotted timesystems, such as [27, 21, 15, 18, 51]. However, as pointed out by Talak et al [51], PAoI/AoI for the discretetime systems may differ significantly from their continuous time counterpart. For continuous time systems,in many cases the average AoI or PAoI is difficult to obtain, and advanced modeling and mathematicalmethodologies are thus needed, such as the Stochastic Hybrid System (SHS) used in [58, 34, 24]. Amongall the AoI related literature, to the best of our knowledge, there are very few papers discussing systemswith vacation server. Maatouk et al [33] considered a special case where server vacations occur in a randommanner. Najm et al [42] considered a system with two streams with different priorities and discussed severalservice disciplines for the low priority stream. Xu and Gautam [55] discussed the M/G/1/2* and M/G/1priority queues and allowed each queue to have an individual buffer. Closed-form expressions for PAoI underdifferent service disciplines are derived in [55] by modeling the system as vacation server system. However,the general system with single buffer and vacation server has not been fully studied. It is still unclear whichvariation of the single buffer system, which we introduced in Section 1, has smallest AoI or PAoI. And itis unknown how vacation times will influence the AoI/PAoI performance of each system. In this paper, weextend our discussion of vacation server in [55], and introduce a simple but useful decomposition approachto derive closed-form expressions of AoI and PAoI for different variation of the single buffer system. Usingthe decomposition approach we can also obtain the variance of peak age for general queueing systems withsingle buffer. We further discuss the advantage of each system under certain conditions. We then extend ourdiscussion to polling system with a single buffer at each queue, and show that the decomposition approachcan be easily applied to multi-queue systems to derive closed-form expression of PAoI.5
Age of Information for Systems with Independent Vacations
In this section we consider the system in which the vacations that the server takes are i.i.d. If we regarda vacation as the period when server is sleeping for energy saving, then this setting is the same as the multi-sleep scheme that was discussed in [14]. Later in Section 4 we will discuss the case where vacationsare non-i.i.d. Throughout this section, we assume that each vacation V that the server takes is a randomvariable with Laplace–Stieltjes transform (LST) V ∗ ( s ) . The server will take a vacation once the service isover. When the server comes back from a vacation, it will process the packet if the buffer is non-empty;otherwise a new vacation is taken. In this section we consider three variations of the system by varying theassumption of buffer availability: CBS, BRS and CBS-P, which we defined in Section 1.Figure 1: Age of Information Decomposition For Non-preemptive Service Systems. Variables r { l } , S { l } and C { l } are the generation time, time to start service and completion time for l th packet that is served by the server. The second age peak A { } is decomposed into threecomponents A { } = G { } + I { } + H { } . The first component G { } is the waiting time of thefirst served packet. The second component I { } is the time between the server starts servingtwo packets. The third component H { } is the service time of the second served packet.In our previous work [55], we decomposed PAoI into four components where each component could bederived easily. However, such a decomposition cannot be used to derive AoI as the decomposed componentsare not mutually independent. Therefore, here we introduce a new decomposition method for computingAoI and PAoI in non-preemptive service systems, i.e., CBS and BRS. Since the decomposition approach forCBS-P differs from the decomposition approach for CBS and BRS, we leave our discussion for CBS-P inAppendix B. From Figure 1, we find that the peak age of CBS or BRS is always the time span from thecompletion time of the recently processed packet, to the generation time (arrival time) of the previously6rocessed packet. This time span can then be divided into three components: waiting time G (in queue)of the previous packet, inter-service-starting time I between the recent and previous processed packets, andservice time H of the recent packet. These three components are mutually independent. This is becausehow long the packet has waited in the buffer has no influence on its processing time or the vacation that theserver will take next, thus G is independent of I and H . Also because the service time does not depend onhow long the previous vacation lasted, then G , I and H are mutually independent. Thus the PAoI of thissystem can be given as E [ A ] = E [ G ] + E [ I ] + E [ H ] , (1)and the AoI can be given as E [∆] = E [( G + I + H ) ] − E [( G + H ) ]2 E [ I ] = E [ I ]2 E [ I ] + E [ G ] + E [ H ] . (2)Note that Equation (1) and (2) are for non-preemptive service systems, i.e., CBS and BRS. The discussionfor CBS-P is left in Appendix A. It is important to point out that Equations (1) and (2) still hold true evenwhen the vacations are non-i.i.d., as the independence of the three decomposed component does not rely onthe assumption about vacations.Assume the LST of G, I and H exist and are given by G ∗ ( s ) , I ∗ ( s ) and H ∗ ( s ) . Since those three compo-nents are mutually independent, we have the LST of A as A ∗ ( s ) = G ∗ ( s ) I ∗ ( s ) H ∗ ( s ) . (3)PAoI can be easily obtained by calculating the first moment of A ∗ ( s ) . The variance of peak age can beused as a metric to measure the age violations, and the variance of peak age can be given as V ar ( A ) = G ∗ (2) (0) + I ∗ (2) (0) + H ∗ (2) (0) − { G ∗ (1) (0) } − { I ∗ (1) (0) } − { H ∗ (1) (0) } (4) = V ar ( G ) + V ar ( I ) + V ar ( H ) . (5)It is shown in [55] that E [ G ] = 1 λ (1 − E [ e − λW ]) , (6)where W is the time period that the buffer is occupied. Note that if we consider a different system where7e only keep the first packet that arrives in the buffer and reject the others, W is also the waiting time ofthe packet that enters the buffer. If W ∗ ( s ) is the LST of W , then E [ G ] = λ (1 − W ∗ ( λ )) . Therefore, oncewe have W ∗ ( s ) and I ∗ ( s ) , we are able to obtain E [∆] and E [ A ] . In order to obtain the variance of A, onealso needs to know the LST of G . The LST of G can also be written as a function of W ∗ ( s ) , as shown inthe following lemma. Lemma 1. G ∗ ( s ) = λλ + s + sλ + s W ∗ ( λ + s ) for the system with single buffer.Proof. It is shown in [55] that P ( G ≤ x | m ( t ) = m, W = t ) = 1 − ( t − xt ) m if there are m ( t ) = m packetsarriving during time W . From the fact that E [ e − sG | m ( t ) = 0 , W = t ] = e − st we have E [ e − sG | W = t ] = (cid:90) tx =0 e − sx ∞ (cid:88) m =1 m ( t − x ) m − t m e − λt ( λt ) m m ! dx + e − st e − λt = λλ + s + sλ + s e − ( λ + s ) t . By unconditioning on W = t we can prove the lemma.In the remaining part of this section we introduce the way to derive I ∗ ( s ) and W ∗ ( s ) for CBS, BRS andCBS-P. In this subsection we mainly derive the E [∆] , E [ A ] and V ar ( A ) of the CBS where arrivals are rejected whenthe server is serving. Recall that in CBS, the buffer will not be available until the processing is done, and theserver will start a vacation once the buffer becomes empty. We provide the results for CBS in the followingtheorem. Theorem 2.
The AoI of CBS is given as E [∆ CBS ] = − H ∗ (2) (0)+2 H ∗ (1) (0) V ∗ (1)(0)1 − V ∗ ( λ ) + V ∗ (2)(0)1 − V ∗ ( λ ) +2 V ∗ (1)(0) V ∗ (1)( λ )(1 − V ∗ ( λ ))2 (cid:18) H ∗ (1) (0)+ V ∗ (1)(0)1 − V ∗ ( λ ) (cid:19) + λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) − H ∗ (1) (0) , the PAoI of CBS is given as E [ A CBS ] = λ + V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) − H ∗ (1) (0) , andthe variance of peak age of CBS is given by V ar ( A CBS ) = λ + V ∗ (2) (0) − V ∗ (2) ( λ )1 − V ∗ ( λ ) − (cid:16) V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) (cid:17) +2 H ∗ (2) (0) − (cid:0) H ∗ (1) (0) (cid:1) . Proof.
We first show that I ∗ ( s ) = H ∗ ( s ) V ∗ ( s ) − V ∗ ( s + λ )1 − V ∗ ( s + λ ) for CBS. Notice that the period I starts once theserver starts serving, and ends when the server comes back from a vacation and observes a packet waitingin the buffer. Therefore we have I ∗ ( s ) = E [ e − s ( H + B ) ] , where B is the time period during which theserver is continuously in vacation. Let B ∗ ( s ) be the LST of B . By conditioning on the length of the firstvacation V that the server takes after serving a packet, and also conditioning on the number of arrivals8including the rejected ones) during V , i.e., m ( V ) , we have E [ e − sB | V = v , m ( V ) ≥
1] = e − sv and E [ e − sB | V = v , m ( V ) = 0] = e − sv B ∗ ( s ) . Unconditioning on m ( v ) we have E [ e − sB | V = v ] = e − sv (1 − e − λv )+ e − sv B ∗ ( s ) e − λv . Then by unconditioning on V , we have B ∗ ( s ) = V ∗ ( s ) − V ∗ ( s + λ )+ B ∗ ( s ) V ∗ ( s + λ ) and B ∗ ( s ) = V ∗ ( s ) − V ∗ ( s + λ )1 − V ∗ ( s + λ ) . Therefore we have I ∗ ( s ) = H ∗ ( s ) V ∗ ( s ) − V ∗ ( s + λ )1 − V ∗ ( s + λ ) .We next derive the expression for E [ G ] . From Equation (6) we know that E [ G ] can be given using theformula of the LST of W , where W is the time period when the buffer is occupied. So we now derive theLST of W . Since when the buffer becomes occupied, the server must be on a vacation. From Campbell’sTheorem (P173, Theorem 5.14 in [28]) we have E [ e − sW | m ( t ) = m, V = t ] = (cid:90) t e − sx mx m − t m dx. Unconditioning on m ( t ) = m and using the fact that P ( m ( t ) = m | m ( t ) ≥
1) = ( λt ) m m ! e − λt − e − λt , we have E [ e − sW | V = t, m ( V ) ≥
1] = ∞ (cid:88) m =1 (cid:90) tx =0 e − sx mx m − t m e − λt − e − λt ( λt ) m m ! dx = (cid:90) tx =0 e − sx e − λt − e − λt ∞ (cid:88) m =1 ( λx ) m − ( m − λdx = e − λt − e − st ( s − λ )(1 − e − λt ) λ. Now we need to find P ( t < V ≤ t + dt | m ( V ) ≥ . From P ( V ≤ x | m ( V ) ≥
1) = P ( V ≤ x, m ( V ) ≥ P ( m ( V ) ≥
1) = (cid:82) x dV ( u )(1 − e − λu ) (cid:82) ∞ dV ( u )(1 − e − λu )= (cid:82) x dV ( u )(1 − e − λu )1 − V ∗ ( λ ) , we have P ( t < V ≤ t + dt | m ( V ) ≥
1) = dV ( t )(1 − e − λt )1 − V ∗ ( λ ) . Therefore E [ e − sW | m ( V ) ≥
1] = (cid:90) ∞ E [ e − sW | V = t, m ( V ) ≥ dV ( t )(1 − e − λt )1 − V ∗ ( λ )= (cid:90) ∞ e − λt − e − st ( s − λ )(1 − e − λt ) λ dV ( t )(1 − e − λt )1 − V ∗ ( λ )= V ∗ ( λ ) − V ∗ ( s )( s − λ )(1 − V ∗ ( λ )) λ. Since W is the period that the buffer is occupied, and the buffer is only occupied when m ( V ) ≥ , so that E [ e − sW | m ( V ) ≥
1] = E [ e − sW ] . We thus have W ∗ ( s ) = V ∗ ( λ ) − V ∗ ( s )( s − λ )(1 − V ∗ ( λ )) λ . Using L’Hospital rule at s = λ we9ave E [ e − λW ] = − λV ∗ (1) ( λ )1 − V ∗ ( λ ) . From the fact that E [ G ] = λ (1 − W ∗ ( λ )) , we have E [ G ] = λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) . Thenfrom Equation (1) and (2) we can obtain the closed-form expressions for AoI and PAoI using the expressionfor E [ G ] and I ∗ ( s ) . By Lemma 1 we have G ∗ ( s ) = λλ + s − V ∗ ( s + λ )1 − V ∗ ( λ ) . By taking the second derivative of G ∗ ( s ) we have G ∗ (2) (0) = 2 λ + 2 λ V ∗ (1) ( λ )1 − V ∗ ( λ ) − V ∗ (2) ( λ )1 − V ∗ ( λ ) . Using Equation (4), the variance A is given by V ar ( A CBS ) = 1 λ + V ∗ (2) (0) − V ∗ (2) ( λ )1 − V ∗ ( λ ) − (cid:18) V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) (cid:19) + 2 H ∗ (2) (0) − (cid:16) H ∗ (1) (0) (cid:17) . In the following corollary we provide the E [∆] , E [ A ] and V ar ( A ) for the case where service times andvacation times are both exponential. Corollary 3.
If the vacation time is exponentially distributed with parameter v , and service time is expo-nentially distributed with parameter µ, then we have E [∆ CBS ] = λ + v − λ + v + µvλ + µλ + µv + v + λ + µ , E [ A CBS ] = λ + v + v + λ + µ and V ar ( A CBS ) = λ + v ) + λ + v + µ . Proof.
When the vacation time is exponentially distributed, we have I ∗ ( s ) = µvλ ( µ + s )( v + s )( λ + s ) . So from E [ I ] = vλ + µλ + µvµvλ and E [ I ] = 2 ( vλ + µλ + µv ) µ v λ − λ + v + µµvλ , we have V ar ( I ) = µ + v + λ . Also we know E [ G ] = v + λ and E [ G ] = v + λ ) . So that we have the results from Equation (1) and (2).
In this subsection we derive the AoI and PAoI for the BRS, where the buffer becomes available as soon asthe service starts. Recall that the server will go on a vacation after serving one packet. The arrival duringprocessing a packet will be processed after the vacation is over. A potential benefit by applying this “gated”policy is that it prevents the server from serving the buffer continuously without taking vacations, when thearrival rate is large. As discussed in Section 1, many systems can be modeled as queueing systems withvacation server, and the vacation is also important for some systems such as the priority queue systems [55],in which “vacation” actually corresponds to “serving the non-primary queues”. Therefore, it is sometimecrucial for the server to take vacations. Also, as we will see in this subsection, BRS may have advantageover CBS in some cases.Next we derive the AoI and PAoI for BRS in the following theorem.10 heorem 4.
The AoI of BRS is given by E [∆ BRS ] = − I ∗ (2) (0)2 I ∗ (1) (0) + λ + V ∗ (1) ( λ ) H ∗ ( λ ) + V ∗ ( λ ) H ∗ (1) ( λ ) + V ∗ (1) ( λ )1 − V ∗ ( λ ) V ∗ ( λ ) H ∗ ( λ ) − H ∗ (1) (0) , and PAoI of BRS is given by E [ A BRS ] = − I ∗ (1) (0) + λ + V ∗ (1) ( λ ) H ∗ ( λ ) + V ∗ ( λ ) H ∗ (1) ( λ ) + V ∗ (1) ( λ )1 − V ∗ ( λ ) V ∗ ( λ ) H ∗ ( λ ) − H ∗ (1) (0) , where I ∗ ( s ) = H ∗ ( s ) V ∗ ( s ) + H ∗ ( λ + s ) V ∗ ( s + λ )( V ∗ ( s ) − − V ∗ ( s + λ ) .Proof. We first show that in BRS, I ∗ ( s ) = H ∗ ( s ) V ∗ ( s ) + H ∗ ( λ + s ) V ∗ ( s + λ )( V ∗ ( s ) − − V ∗ ( s + λ ) . Since each I startswith processing a packet with processing time H, if there is more than one arrival during the processing time H , then the server only takes one vacation after processing the current packet. If there is no arrival duringthis processing time, the server takes vacations until a packet is observed in buffer when a vacation is over.By conditioning on scenarios during H , we have E [ e − sI | H = h, m ( H ) ≥
1] = e − sh V ∗ ( s ) , and E [ e − sI | H = h, m ( H ) = 0] = e − sh B ∗ ( s ) . We thus have E [ e − sI | H = h ] = e − sh V ∗ ( s )(1 − e − λh )+ e − sh B ∗ ( s ) e − λh . Therefore E [ e − sI ] = H ∗ ( s ) V ∗ ( s ) − H ∗ ( λ + s ) V ∗ ( s ) + H ∗ ( λ + s ) B ∗ ( s ) , where B ∗ ( s ) = V ∗ ( s ) − V ∗ ( s + λ )1 − V ∗ ( s + λ ) . We next derive E [ G ] for BRS. From Equation (6) we know that E [ G ] can be written as a formula of theLST of W , which is the time period when the buffer is occupied. So in the following we first derive the LSTof W . If there is more than one arrival before the server returns from the first vacation, then E [ e − sW | m ( V + H ) ≥
1] = V ∗ ( λ ) H ∗ ( λ ) − V ∗ ( s ) H ∗ ( s )( s − λ )(1 − V ∗ ( λ ) H ∗ ( λ )) λ. If there is no arrival before the server returns from the first vacation, we have E [ e − sW | m ( V + H ) = 0] = V ∗ ( λ ) − V ∗ ( s )( s − λ )(1 − V ∗ ( λ )) λ. We thus have E [ e − sW ] = V ∗ ( λ ) H ∗ ( λ ) − V ∗ ( s ) H ∗ ( s )( s − λ )(1 − V ∗ ( λ ) H ∗ ( λ )) λ { − V ∗ ( λ ) H ∗ ( λ ) } + V ∗ ( λ ) − V ∗ ( s )( s − λ )(1 − V ∗ ( λ )) λV ∗ ( λ ) H ∗ ( λ ) . Using L’Hospital rule at s = λ , we have E [ e − λW ] = − λV ∗ (1) ( λ ) H ∗ ( λ ) − λV ∗ ( λ ) H ∗ (1) ( λ ) − V ∗ (1) ( λ )1 − V ∗ ( λ ) λV ∗ ( λ ) H ∗ ( λ ) . Therefore E [ G ] = − V ∗ (1) ( λ ) H ∗ ( λ ) − V ∗ ( λ ) H ∗ (1) ( λ ) − V ∗ (1) ( λ )1 − V ∗ ( λ ) λV ∗ ( λ ) H ∗ ( λ ) . Using Equation (1) and (2)we can then obtain the PAoI and AoI of BRS.We can also obtain the variance of peak age for BRS, although its closed-form expression is complex. To11btain the variance of peak age, we need the LST of G , I and H as we show in Equation (4). The LST of I has been given in Theorem 4, which is I ∗ ( s ) = H ∗ ( s ) V ∗ ( s ) + H ∗ ( λ + s ) V ∗ ( s + λ )( V ∗ ( s ) − − V ∗ ( s + λ ) . We also have G ∗ ( s ) = λλ + s (cid:26) V ∗ ( λ ) H ∗ ( λ )1 − V ∗ ( λ ) (1 − V ∗ ( λ + s )) − V ∗ ( λ + s ) H ∗ ( λ + s ) (cid:27) from the fact that G ∗ ( s ) = λλ + s + sλ + s W ∗ ( λ + s ) where W ∗ ( s ) is given in Theorem 4. We will show thenumerical results for the variance of peak age for BRS in Section 5. In the next corollary we show the resultsfor the system with exponential service time and exponential vacation time. Corollary 5.
For exponential vacation time with parameter v and exponential service time with parame-ter µ , we have E [∆ BRS ] = v + vµ + µ + µλv ( λ + µ ) + µλ λ + µ ) + µλ ( λ + µ )21 v + µ + λ − λ + µ + λ + v + λv ( λ + µ ) ( λ + v ) + µ and E [ A BRS ] = µ − µv + λµ ( λ + µ ) ( λ + v ) + v + µ + λ . Proof.
The results follow from Theorem 4 with V ∗ ( s ) = vv + s and H ∗ ( s ) = µµ + s .A question is whether BRS always has smaller AoI or PAoI than CBS. In the next theorem we show thatBRS always has smaller PAoI than CBS, for all arbitrary service and vacation distributions. Theorem 6.
PAoI in BRS is always smaller than PAoI in CBS, if the arrival process is Poisson, and servicetimes as well as vacation times are i.i.d.Proof.
From Theorem 2 we have E [ A CBS ] = − H ∗ (1) (0) + 1 λ + V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) , and from Theorem 4 we have E [ A BRS ] = − H ∗ (1) (0) − V ∗ (1) (0) + 1 λ + V ∗ (1) ( λ ) H ∗ ( λ ) + V ∗ ( λ ) H ∗ (1) ( λ )+ H ∗ ( λ ) V ∗ ( λ )1 − V ∗ ( λ ) ( V ∗ (1) ( λ ) − V ∗ (1) (0)) . We then have E [ A CBS ] − E [ A BRS ] = V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) (1 − H ∗ ( λ ) V ∗ ( λ )) + V ∗ (1) (0) − V ∗ (1) ( λ ) H ∗ ( λ ) − V ∗ ( λ ) H ∗ (1) ( λ )= (cid:2) V ∗ (1) ( λ ) − V ∗ (1) (0) V ∗ ( λ ) (cid:3) (1 − H ∗ ( λ )) + V ∗ ( λ ) H ∗ (1) ( λ )( V ∗ ( λ ) − − V ∗ ( λ ) . Notice that H ∗ (1) ( λ ) ≤ and ≤ V ∗ ( λ ) ≤ , we have V ∗ ( λ ) H ∗ (1) ( λ )( V ∗ ( λ ) − ≥ . Since ≤ H ∗ ( λ ) ≤ , to show that E [ A CBS ] − E [ A BRS ] ≥ , we only need to show V ∗ (1) ( λ ) − V ∗ (1) (0) V ∗ ( λ ) ≥ . V ∗ (1) ( λ ) − V ∗ (1) (0) V ∗ ( λ ) = − E [ V e − λV ] + E [ V ] E [ e − λV ] , we let X = V , Y = e − λV with CDF F X ( x ) ,F Y ( x ) and joint CDF F ( x, y ) . We now show that P ( X ≤ x, Y ≤ y ) ≤ P ( X ≤ x ) P ( Y ≤ y ) . Notice that F ( x, y ) = P ( X ≤ x, Y ≤ y ) = P ( V ≤ x, e − λV ≤ y ) = P ( − ln yλ ≤ V ≤ x )= P ( V ≤ x ) − P ( V ≤ − ln yλ ) ≤ P ( V ≤ x ) − P ( V ≤ − ln yλ ) P ( V ≤ x )= F X ( x ) F Y ( y ) . From [32] we know E [ XY ] − E [ X ] E [ Y ] = (cid:82) ∞−∞ (cid:82) ∞−∞ [ F ( x, y ) − F X ( x ) F Y ( y )] dxdy. Therefore V ∗ (1) ( λ ) − V ∗ (1) (0) V ∗ ( λ ) = E [ X ] E [ Y ] − E [ XY ] ≥ and E [ A CBS ] − E [ A BRS ] ≥ .However, BRS does not always have smaller AoI than CBS. A graph for comparison is provided in Figure2 where service time and vacation time are both exponential. As we see from Figure 2, when vacation timeis large (i.e., small v ), E [∆ BRS ] is smaller than E [∆ CBS ] . However when vacation time is small, the CBShas smaller AoI than BRS.
In this subsection we consider the system when preemption is allowed in service, i.e., CBS-P. Note that whenallowing preemption in service, both CBS and BRS will reduce to CBS-P. Different from the non-preemptiveservice case, in CBS-P, the age peak cannot be decomposed as shown in Equation (1) simply because the13acket that result in age peak may not have waiting time G (as it may be a preemptive packet). A detaileddecomposition approach for CBS-P is given in Appendix A. The AoI and PAoI in CBS-P is given in thefollowing theorem. Theorem 7.
The PAoI for CBS-P is given by E [ A CBS − P ] = − H ∗ ( λ ) − λH ∗ (1) ( λ )+ H ∗ ( λ ) λH ∗ ( λ ) + H ∗ ( λ ) V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) , and the AoI for this system is given by E [∆ CBS − P ] = V ∗ (2) (0)1 − V ∗ ( λ ) + 2 V ∗ (1) (0) V ∗ (1) ( λ )(1 − V ∗ ( λ )) − V ∗ (1) (0)1 − V ∗ ( λ ) 1 − H ∗ ( λ ) λH ∗ ( λ ) + λH ∗ ( λ ) (cid:104) λ − H ∗ ( λ ) λ + H ∗ (1) ( λ ) (cid:105) − V ∗ (1) (0)1 − V ∗ ( λ ) + − H ∗ ( λ ) λH ∗ ( λ ) ) − H ∗ (1) ( λ ) H ∗ ( λ ) + H ∗ ( λ ) (cid:18) λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) (cid:19) . Proof.
Detailed proof is shown in Appendix A.
Corollary 8.
For exponential vacation time with parameter v and exponential service time with parameter µ , we have E [ A CBS − P ] = λ + µ + v + λ + µ + p ( λ + µ )( λ + v ) and E [∆ CBS − P ] = v + λv λ + λ + v + λλvµ + µ λ + µ + v + µ + λ + µ ( µ + λ )( v + λ ) . Theorem 9.
If the service time is exponentially distributed, then the system CBS-P has both AoI and PAoIsmaller than CBS.Proof.
Detailed proof is shown in Appendix B.Notice that Theorem 9 holds true for exponential service times only. A question is whether the inequalitiesstill holds between CBS and CBS-P if the service is not exponential. In the next theorem, we provide asufficient condition under which CBS-P will always have smaller PAoI than CBS.
Theorem 10.
If the service time H satisfies E [ H ] ≥ − H ∗ ( s ) sH ∗ ( s ) for all s > , then CBS-P always has smallerPAoI than CBS.Proof. Detailed proof is shown in Appendix C.Theorem 10 provides a useful sufficient condition for checking whether CBS-P has smaller PAoI thanCBS, and this sufficient condition does not rely on the vacation time distribution. Necessary condition canbe obtained by directly comparing the closed-form expression of PAoI for CBS-P and CBS, however it isquite involved. We can also provide sufficient and necessary conditions for CBS-P to have smaller AoI orvariance of peak age than CBS or BRS, by simply comparing their closed-form expressions. However thoseconditions are specific and complicated. In Section 5 we will compare E [ A ] , . E [∆] and V ar ( A ) numerically.We now provide some examples on how Theorem 10 can be applied. When the service time is exponentialwith parameter µ , we have − H ∗ ( s ) sH ∗ ( s ) = − µµ + s s µµ + s = µ = E [ H ] . Then by Theorem 10 we can conclude that14 a) H ∼ Gamma ( , (b) H ∼ Gamma (2 , Figure 3: PAoI in CBS vs PAoI in CBS-P. Service time is Gamma distributed. Vacation timeis exponentially distributed.CBS-P has smaller PAoI than CBS, which is the same as our conclusion in Theorem 9. We next give anexample where the service time is Gamma distributed with parameters α and β . Since the LST of Gammadistribution is given by H ∗ ( s ) = (1 + βs ) − α , we have − H ∗ ( s ) sH ∗ ( s ) = (1+ βs ) α − s . By Bernoulli’s inequality we havethat (1 + βs ) α ≥ αβs when α ≥ , and (1 + βs ) α < αβs when α < . From the fact that E [ H ] = αβ ,we have − H ∗ ( s ) sH ∗ ( s ) ≥ E [ H ] when α ≥ and − H ∗ ( s ) sH ∗ ( s ) ≤ E [ H ] when α < . A numerical study of this exampleis given in Figure 3, where service time is Gamma distributed and vacation time is exponential. In Figure3(a), we find that when α = , CBS-P does not always have smaller PAoI than CBS. However, in Figure3(b) where α = 2 , the service distribution satisfies E [ H ] ≥ − H ∗ ( s ) sH ∗ ( s ) , we find that CBS-P has smaller PAoIthan CBS, for all the positive values of λ and v . We realize that when the server takes no vacations or takes vacation infinitely fast, then CBS is equal tothe M/G/1/1 non-preemptive system, BRS is equal to the M/G/1/2* system (the asterisk means that onlythe most recent packet is kept in the buffer as defined in [7, 60]), and CBS-P becomes M/G/1/1/preemptivesystem. Different variations of these systems has been discussed in [43, 19, 60, 7, 22, 20], however the varianceof peak age in these single buffer systems has not been studied. We here provide the variance of peak agefor the systems without server vacations, as an extension of our discussion about vacation server systems.With the decomposition approach that we introduced earlier, we are able to provide the variance of peakage for M/G/1/1, M/G/1/2* and M/G/1/1/preemptive systems, as shown in the following theorem.
Theorem 11.
For M/G/1/1 system we have E [ A M/G/ / ] = λ − H ∗ (1) (0) , E [∆ M/G/ / ] = λ − λ H ∗ (1) (0)+ H ∗ (2) (0) λ − H ∗ (1) (0) − µ = 1 . H ∗ (1) (0) and V ar ( A M/G/ / ) = λ + 2 H ∗ (2) (0) − { H ∗ (1) (0) } . For M/G/1/1/preemptive system we have E [ A M/G/ / /preemptive ] = − H ∗ (1) ( λ ) H ∗ ( λ ) + λH ∗ ( λ ) , E [∆ M/G/ / /preemptive ] = λH ∗ ( λ ) , and V ar ( A M/G/ / /preemptive ) = H ∗ (2) ( λ ) H ∗ ( λ ) − { H ∗ (1) ( λ ) } H ∗ ( λ ) + λ H ∗ ( λ ) + H ∗ (1) ( λ ) λH ∗ ( λ ) . For M/G/1/2* system we have E [ A M/G/ / ∗ ] = − H ∗ (1) (0) + λ + H ∗ (1) ( λ ) , E [∆ M/G/ / ∗ ] = H ∗ (2) (0)+ λ H ∗ ( λ ) − λ H ∗ (1) ( λ ) − H ∗ (1) (0)+ H ∗ ( λ ) λ + λ − λ H ∗ ( λ ) + H ∗ (1) ( λ ) − H ∗ (1) (0) , and V ar ( A M/G/ / ∗ ) = 2 H ∗ (2) (0) − H ∗ (1) (0) + H ∗ ( λ )[ H ∗ (1) (0)+ H ∗ (1) ( λ )] λ + H ∗ ( λ )(1 − H ∗ ( λ )) λ + λ − H ∗ (2) ( λ ) − λ H ∗ (1) ( λ ) − H ∗ (1) ( λ ) . Proof.
The detailed proof is shown in Appendix D.
Corollary 12.
The variance of peak age in M/M/1/1 is
V ar ( A M/M/ / ) = λ + µ , the variance of peakage in M/M/1/1/preemptive is V ar ( A M/M/ / /preemtive ) = λ + µ ) + λ + µ , and the variance of peak agein M/M/1/2* is V ar ( A M/M/ / ∗ ) = λ + µ − λ +4 λµ +3 µ ( λ + µ ) . From Corollary 12 we find that the variance of PAoI in M/M/1/2* system is smaller than PAoI inM/M/1/1 system. Also we find that the variance of peak age in M/M/1/1/preemptive system is smallerthan it in M/M/1/1 system. Now we compare the variance of peak age for M/M/1/1/preemptive systemwith M/M/1/2* system. First we have
V ar ( A M/M/ / /preemptive ) − V ar ( A M/M/ / ∗ ) = − λ − λ µ − λ µ + 2 λµ + 3 µ ( λ + µ ) µ . We find that only when λ is large can we have V ar ( A M/M/ / /preemptive ) − V ar ( A M/M/ / ∗ ) ≤ . Ademonstrative graph of the case where µ = 1 is shown in Figure 4. We can get from numerical study and Fig-ure 4 that when λ ≤ . , M/M/1/2* system has smaller variance of peak age than M/M/1/1/preemptivesystem. 16 Peak Age of Information for Systems with Dependent Vacations
In this section we extend our discussion in Section 3 to a more general case by allowing the vacations to benon-i.i.d. In this case, obtaining LST or the second moment of I is difficult, and the closed-form expressionfor AoI may become intractable. However, the PAoI is still solvable in this case. In this section, we willdiscuss the approach for deriving the exact solution for PAoI.From our discussion in Section 3 we have E [ G ] = λ (1 − W ∗ ( λ )) , where W is the time period when thebuffer is occupied. Also from Equation (1) and our discussion in Appendix E, we have E [ A ] = − λ W ∗ ( λ ) + λ + E [ W ] + 2 E [ H ] for CBS, − λ W ∗ ( λ ) + λ + E [ W ] + E [ H ] for BRS, and − H ∗ (1) ( λ ) H ∗ ( λ ) + H ∗ ( λ ) λ (1 − W ∗ ( λ )) + E [ W ] + λH ∗ ( λ ) for CBS-P.When W ∗ ( s ) is available, then the closed-form of PAoI can be obtained. In the remaining part of this section,we will focus our discussion on the polling system as it is a system where the server takes non-i.i.d vacations(see [26]). We will next show the approach to calculate exact PAoI for polling system by obtaining W ∗ ( s ) .A polling system is a queueing system that contains a single server and k classes of packets. Each packetclass would have its own queue, so there are k queues in the system. The server serves packets by switchingbetween queues, and a switchover time is incurred when the server switches from one queue to another. Ademonstrative graph of polling systems is provided in Figure 5. Polling systems have a wide application incommunication networks and other networks (see [53, 3, 56]), while the PAoI in polling systems has not beenfully studied. Specifically, if there are multiple data nodes in the underwater sensor network example whichwe discussed in Section 1 (also see [52, 16]), we can then model the underwater system as a polling system,where each data node can be modeled as a queue/buffer and the autonomous vehicle can be regarded as theserver that collects/processes data from each node in a periodic manner.In this paper we are interested in single buffer systems, so we assume that each queue has a single bufferthat can hold only one packet at a time. Similar to our discussion in Section 3, we assume that only the mostrecently arrived packet is kept in the buffer, and we consider three variations of the polling system by makingdifferent assumptions about the buffer and service preemption. To distinguish from the names in Section 4,we call these three polling systems Conventional Buffer Polling System (CBPS) , Buffer Relaxation PollingSystem (BRPS) , and
Conventional Buffer Polling System with Preemption in Service (CBPS-P) respectively . In CBPS, the buffer is not available until the current packet completes its service. When the server is busy17rocessing, newly arrived packets in this queue will be rejected. So that each queue can only have at mostone packet at any time. In BRPS, the buffer becomes available once the service has started, however thenew arrival during the service time will be served in the next polling instant. In CBPS-P, the new arrivalwill preempt the packet in service, and the preempted packet will be discarded. The server will switch tonext queue when the service of a packet is complete. In all these three systems, the server will start anotherswitching process if the it observes an empty queue. We assume that the arrival process of packets in eachqueue i follows a Poisson process with rate λ i , and the service time H i for packets at each queue is i.i.d.with mean h i and LST H ∗ i ( s ) . The switchover time from queue i to queue j has mean u ij , CDF U ij ( x ) andLST U ∗ ij ( s ) . In the remaining part of this section we use the subscript i to denote the parameter for queue i in the polling system.There are multiple widely used routing schemes that determine which queue to switch to next for theserver. Routing schemes include cyclic [40, 50, 49, 9], Markovian polling [6, 4] and random polling [29]. Formost of those polling systems with single buffer in each queue, W ∗ ( s ) can be derived (see [50, 6, 29]). In thispaper we mainly discuss the Markovian polling scheme, since random polling and cyclic polling schemes areboth special cases of the Markovian polling scheme, as we will see later. In the Markovian polling scheme,after serving queue i , the probability of serving queue j next is given by p ij . Considering all the possiblestates for the current queue and next queue, the switching process can be characterized by a discrete Markovchain with transition matrix P = [ p ij ] . In this paper we assume that P is irreducible positive recurrent. Forthe cyclic polling scheme, the transition matrix is given by for i, j ∈ { , , ..., k } ,p ij = if j = i + 1 , otherwise . Two special polling schemes were discussed in [6]. One is called load-oriented-policy (LOP), which isdefined by the transition matrix with p ij = λ j (cid:80) kl =1 λ l for all i and j . The other polling scheme is called symmetric random polling, in which p ij = k for all i and j . We will show the performance of these schemesnumerically in Section 5.The service process for each individual queue in polling systems can be modeled as a single server withmultiple vacations: when the server polls the queue, it serves the packet if the queue is not empty, and takesa vacation (switches out and serves other queues) once the service completes; if the queue is empty whenpolled, the server takes another vacation. It is important to note that as pointed out by Kofman in [26],even when cyclic polling scheme is applied, the vacations that the server takes in a polling system are noti.i.d. Suppose W i is the time period that the buffer i is occupied, with LST W ∗ i ( s ) . Then our methods for18igure 5: A k -queue Polling System with Cyclic Polling Schemederiving W ∗ ( s ) for i.i.d vacations in Section 3 cannot be applied here for deriving W ∗ i ( s ) in general pollingsystems.We now summarize how W ∗ i ( s ) is obtained by Chung et al[6] and use it to derive the PAoI for queue i (i.e., E [ A i ] ). The main idea in [6] of deriving W ∗ i ( s ) is to solve the following linear system: F i ( z , ..., z k ) = k (cid:88) j =1 π j π i p ji ˜ U ∗ ij (cid:110) (1 − ˜ H ∗ j ) F j ( z , ..., z k ) z j =0 + ˜ H ∗ j F j ( z , ..., z k ) z j =1 (cid:111) for i = 1 , ..., k, (7)where F i ( z , ..., z k ) is a probability generating function with F i (1 , ...,
1) = 1 , ( π , ..., π k ) is the stationarydistribution of the transition matrix P , ˜ U ∗ ij = U ∗ ij ( (cid:80) kl =1 λ l (1 − z l )) , and ˜ H ∗ j = H ∗ j ( (cid:80) kl =1 ,l (cid:54) = j λ l (1 − z l )) for CBPS, H ∗ j ( (cid:80) kl =1 λ l (1 − z l )) for BRPS, and H ∗ j ( (cid:80) kl =1 ,l (cid:54) = j λ l (1 − z l )+ λ j ) (cid:80) kl =1 ,l (cid:54) = j λl (1 − zl ) (cid:80) kl =1 ,l (cid:54) = j λl (1 − zl )+ λj + λj (cid:80) kl =1 ,l (cid:54) = j λl (1 − zl )+ λj H ∗ j ( (cid:80) kl =1 ,l (cid:54) = j λ l (1 − z l )+ λ j ) for CBPS-P. (8)In [6] only ˜ H ∗ j in CBPS and BRPS are discussed. In both CBPS and BRPS, the server would switchout from queue j after serving a packet from queue j . Here we also discuss the case of CBPS-P. Noticethat in CBPS-P, the server switches out from queue j only when one packet has been completely served. Ifwe consider the time period when the server is continuously serving in CBPS-P as the service time for “one19acket”, then we can also regard CBPS-P as CBPS. The only difference is that in CBPS, each service periodis H j for queue j . While in CBPS-P, the service period is L j with LST L ∗ j ( s ) = H ∗ j ( s + λ j ) ss + λ j + λ j s + λ j H ∗ j ( s + λ j ) . (9)A detailed derivation of Equation (9) can be found in Appendix A. Then, the formula of ˜ H ∗ j for CBPS-P inEquation (8) is obtained by simply using Equation (9) and the formula ˜ H ∗ j for CBPS.To solve the system (7) analytically is quite involved as shown in [6], however the expected value of W i can be obtained by solving the system (7) with z j = 0 or for j = 1 , ..., k , where only k (2 k − linearequations need to be solved. The expected time W i is then given as E [ W i ] = γ i λ i α i − λ i , where α i = 1 − F i (1 , ..., i , ..., (the notation F i (1 , ..., i , ..., means that z i = 0 and z l (cid:54) = i = 1 in F i ( z , ..., z k ) ) and γ i = λ i π i (cid:80) kj =1 π j ( α j h j + (cid:80) kl =1 p jl u jl ) − λ i α i h i for CBPS, λ i π i (cid:80) kj =1 π j ( α j h j + (cid:80) kl =1 p jl u jl ) for BRPS, and λ i π i (cid:80) kj =1 π j ( α j − H ∗ j ( λ j ) λH ∗ j ( λ j ) + (cid:80) kl =1 p jl u jl ) − λ i α i − H ∗ i ( λ i ) λ i H ∗ ( λ i ) for CBPS-P. (10)To obtain E [ G i ] , we need to get W ∗ i ( λ i ) . From [6, 50] we have W ∗ i ( s ) = 1 α i λ i s − λ i (cid:26) − α i − f i (1 − sλ i ) (cid:27) , where f i ( z ) = F i (1 , ..., i z, ..., . Using L’Hospital rule we have W ∗ i ( λ i ) = f (1) i (0) α i = 1 α i ∂F i (1 , ..., z, ..., ∂z | z =0 , in which the derivative of F i (1 , ..., z, ..., is needed. Therefore we need to compute the partial derivative ofEquation (7) with respect to z l for l = 1 , ..., k , which is to solve the following linear system: ∂F i ( z , ..., z k ) ∂z l = ∂∂z l k (cid:88) j =1 π j π i p ji ˜ U ∗ ij (cid:16) (1 − ˜ H ∗ j ) F j ( z , ..., z k ) z j =0 + ˜ H ∗ j F j ( z , ..., z k ) z j =1 (cid:17) for i = 1 , ..., k and l = 1 , ..., k. (11)20ote here we only need to solve system (11) for z j = 0 or for j = 1 , ..., k to obtain W ∗ i ( λ i ) , so that k k number of equations need to be solved. After solving system (7) and (11), the closed-form of PAoI canbe obtained from the following equations: E [ A i ] = − λ i W ∗ i ( λ i ) + λ i + E [ W i ] + 2 E [ H i ] for CBPS, − λ i W ∗ i ( λ i ) + λ i + E [ W i ] + E [ H i ] for BRPS, and − H ∗ (1) i ( λ i ) H ∗ i ( λ i ) + H ∗ i ( λ i ) λ i (1 − W ∗ i ( λ i )) + E [ W i ] + λ i H ∗ i ( λ i ) for CBPS-P.Similar to our discussion in Section 3, in the next theorem we show that for polling system with exponentialservice time, CBPS-P always has smaller PAoI than CBPS. Theorem 13.
If the service time for each queue is exponentially distributed, then CBPS-P will always havesmaller PAoI than CBPS.Proof.
A detailed proof is shown in Appendix F.However, when the service time is not exponential, CBPS-P will not always have smaller PAoI thanCBPS. We will show more computational results in Section 5.
In this section, we first perform a set of numerical experiments for systems with i.i.d. vacations that weintroduced in Section 3, and then provide the numerical results to verify the exact solution of PAoI forpolling systems. We then describe the results for polling system under different Markovian polling schemes.
We begin our discussion by comparing the AoI, PAoI and variance of peak age for CBS, BRS, and CBS-P,as shown in Figure 6. In each subfigure of Figure 6 we plot simulation and exact results. As we observe fromeach subfigure of Figure 6, the simulation result matches the exact result for each system.In Figure 6(a) and Figure 6(d) we compare the AoI for these three systems under different service andvacation time assumptions. As we see from Figure 6(a), CBS-P has advantage over the other two systems interms of smaller AoI, when service time is exponentially distributed. However, in Figure 6(d) when servicetime is deterministic, CBS-P does not have smaller AoI than the other two systems when the arrival rate islarge. This is because in CBS-P, the server would start a new packet when an arrival preempts the service.21he server will continuously serve only until an inter-arrival time is smaller than the constant service time.If arrival rate is large (which means the expected inter-arrival time is small), then the probability of theinter-arrival time being smaller than the constant service time is small. Thus the AoI of CBS-P becomeslarge when arrival rate is large, for deterministic service time cases. In Section 5.2 we will see that thisobservation is mainly due to the constant service time and not the change of vacation distributions, since inthe cases when the server does not take vacations, we observe similar phenomenons.In Figure 6(b) and Figure 6(e) we compare the PAoI of these three systems. We find that CBS always haslarger PAoI than BRS for both exponential and deterministic service cases, as we proved in Theorem 6. Alsowe find that when service time is deterministic, the PAoI of CBS-P will increase drastically as arrival rateincreases. From Figure 6(a) and Figure 6(b) we also find that CBS-P has smaller AoI and PAoI than CBSfor the exponential service cases, as we proved in Theorem 9. In Figure 6(c) and Figure 6(f) we comparethe variance of peak age for these three systems. We find that when service time is exponential, the CBSas relatively larger variance than the other two systems, and when service time is deterministic, CBS-P willhave large variance as arrival rate becomes large. From all the subfigures in Figure 6, we find that for bothCBS and BRS, increasing the arrival rate would reduce the AoI, PAoI and variance of peak age. (a) AoI Comparison, H ∼ exp (1) ,V ∼ exp ( ) (b) PAoI Comparison, H ∼ exp (1) ,V ∼ exp ( ) (c) Variance of Peak AgeComparison, H = exp (1) , V ∼ exp ( ) (d) AoI Comparison, H =1 , V ∼ gamma (2 , (e) PAoI Comparison, H = 1 , V ∼ gamma (2 , (f) Variance of Peak AgeComparison, H = 1 , V ∼ gamma (2 , Figure 6: Vacation Server Systems with E [ H ] = 1 and E [ V ] = 2 a) AoI Comparison with H ∼ exp (1) (b) PAoI Comparison with H ∼ exp (1) (c) Variance of Peak AgeComparison with H ∼ exp (1) (d) AoI Comparison with H = 1 (e) PAoI Comparison with H = 1 (f) Variance of Peak AgeComparison with H = 1 Figure 7: Single Queue System with E [ H ] = 1 We next compare the AoI, PAoI and variance of peak age for M/G/1/1, M/G/1/2* and M/G/1/1/preemptivesystems, under exponential and deterministic service cases. As we discussed in Section 3.4, when the vacationtime becomes zero, CBS would become M/G/1/1 system, BRS would be M/G/1/2* system, and CBS-P nowbecomes M/G/1/1/preemptive system. The comparison results are shown in Figure 7. In each subfigure ofFigure 7, we plot the exact result and simulation result for each system, and we find that the simulationresult matches the exact result that we provided in Section 3.4 for each system. Similar to the vacationserver cases, we find that when there is no vacation for the server, AoI, PAoI and variance of peak age willstill decrease in M/G/1/1 and M/G/1/2* as arrival rate increases. For M/G/1/1/preemptive system, theAoI, PAoI and variance of peak age will increase dramatically when arrival rate becomes large, when theservice time is deterministic. We also find that when service time is exponential, the variance of peak agein M/M/1/2* system is smaller than the variance in M/M/1/1 system, and the variance of peak age inM/M/1/1/preemptive system is smaller than the variance in M/M/1/1 system, as we showed in Section 3.4.When the service time is deterministic, M/D/1/2* system has lower variance of peak age than the other twosystems. 23 .3 Polling Systems
We now perform numerical studies for different polling systems. In Figure 8 we compare the exact solutionsof PAoI that we provided in Section 4 with the simulation results for the polling system with k = 3 and cyclicpolling scheme. We find that the exact results match the simulation results from Figure 8. Interestingly,we find that increasing the traffic load will not always reduce the PAoI for CBPS, BRPS and CBPS-P. Aswe observed from Figure 8(c), when the traffic load increases, the PAoI of queue 3 in all three systems willincrease. This is mainly because the numerical test of Figure 8 is based on the cyclic polling scheme, andincreasing the traffic load for all queues will reduce the chance that the server observes an empty buffer whenthe queue is polled. For queue 3, the vacation time actually increases since the other queues are more likelyto be served during the server’s vacation. Although we know that increasing the traffic load will reduce thewaiting time of the packet that is eventually processed (i.e., the server is more likely to find a fresh packetwhen vacation is over), the vacation time for queue 3 will also be affected as the server becomes more likely toserve queue 1 and queue 2 during vacation. Therefore, the increase in vacation time for queue 3 overshadowsthe reduction in G , so that we see in Figure 8 that the PAoI is increasing for queue 3 as total traffic loadincreases. (a) PAoI of Queue 1 (b) PAoI of Queue 2 (c) PAoI of Queue 3 Figure 8: PAoI of Polling Systems with Cyclic Scheme, λ = (0 . , . , . ∗ T otal Load , H i = H ∼ exp (1) , U i = U = 0 . The numerical study for a polling system with k = 8 and cyclic scheme is provided in Table 1. We choosethe same system parameters as the numerical study in [50] by letting two queues be heavily loaded (eachtakes 45% of the total load). As we proved in Theorem 6, BRS always has smaller PAoI than CBS whenthe server’s vacations are i.i.d. However, we observe in polling system that PAoI of BRPS is not alwayssmaller than PAoI of CBPS. This is because in polling systems the vacations that the server takes are noti.i.d., as pointed out in [26]. Moreover, the vacation that the server takes also depends on the service timeof the previous packet. If the service time of a packet is long, the other queues will become more likely toreceive packets during this service time, resulting in a longer vacation time in the next cycle. The complexity24f the vacation in polling system thus prevents Theorem 6 from holding true. However, from Table 1 wecan see that it still holds true that PAoI in CBPS is larger than PAoI in CBPS-P when the service time isexponential, as we proved in Theorem 13. Queue CBPS BRPS CBPS-PPAoI Simu PAoI Simu PAoI Simu1 5.4396 5.4235 5.0996 5.1078 5.0688 5.05672 74.2941 75.7875 73.6306 73.9982 74.2684 74.10013 74.2984 74.6491 73.6372 74.9442 74.2726 72.96714 5.4386 5.4292 5.0985 5.1076 5.0677 5.08045 74.2897 73.3433 73.6236 75.2181 74.2639 74.62256 74.2938 73.2033 73.6300 74.3852 74.2680 75.74377 74.2980 75.8521 73.6366 74.2756 74.2723 75.82498 74.3024 75.7529 73.6433 73.2163 74.2766 73.6263 (a) Total load = 0.85
Queue CBPS BRPS CBPS-PPAoI Simu PAoI Simu PAoI Simu1 8.7298 8.7368 8.8934 8.8892 7.7298 7.73602 10.9433 10.9366 10.9663 10.9606 10.0502 10.08333 10.9513 10.9366 10.9697 10.9589 10.0584 10.06994 8.7296 8.7433 8.8935 8.8942 7.72963 7.73575 10.9352 10.9290 10.9630 10.9432 10.0419 10.04196 10.9426 10.9026 10.9662 10.9835 10.0494 10.08177 10.9509 10.9874 10.9698 10.9990 10.0578 10.07998 10.9601 10.9653 10.9735 10.9509 10.0672 10.0768 (b) Total load = 30
Table 1: Exact PAoI for the system with k = 8 and cyclic scheme. Queue 1 and 4 are heavilyloaded: each with 45% total load. H i = H ∼ exp (1) , U i = U = . Now we consider the PAoI of the polling system under different polling schemes that are described inSection 4. We keep the same set of parameters for service and switching time for Table 2 and 3, and providethe computational results for cyclic, LOP and symmetric random polling schemes with different total trafficloads. From both Table 2 and 3, we find that cyclic scheme and symmetric random scheme perform similarlywhen total traffic load is low. When traffic load is high, symmetric scheme provides lower PAoI for thosequeues with high arrival rates than cyclic scheme, and provides higher PAoI for other queues than cyclicscheme. From both Table 2 and 3 we find that LOP has lower PAoI than the other two polling schemesfor queues with high arrival rates, especially when total traffic load is high. However LOP causes very largePAoI for queues with low arrival rates. This is because the server under LOP would serve queues with higharrival rates more frequently. Notice that in Theorem 13, we do not specify the polling scheme for CBPS orCBPS-P. So when service time is exponential, CBPS-P will always have smaller PAoI than CBPS, regardlessof the polling scheme. This can also be observed from Table 2 and 3.Queue CBPS BRPS CBPS-PCyclic LOP Symmetric Cyclic LOP Symmetric Cyclic LOP Symmetric1 7.0216 6.9340 7.1243 6.4694 6.3262 6.5428 6.7901 6.7137 6.88402 123.1109 125.6743 123.2638 122.2918 126.2261 122.6218 123.0980 125.5646 123.25043 123.1121 125.6743 123.2638 122.2935 126.2261 122.6218 123.0992 125.5646 123.25044 7.0212 6.9340 7.1243 6.4690 6.3262 6.5428 6.7897 6.7137 6.88405 123.1097 125.6743 123.2638 122.2900 126.2261 122.6218 123.0969 125.5646 123.25046 123.1108 125.6743 123.2638 122.2917 126.2261 122.6218 123.0980 125.5646 123.25047 123.1120 125.6743 123.2638 122.2933 126.2261 122.6218 123.0991 125.5646 123.25048 123.1131 125.6743 123.2638 122.2951 126.2261 122.6218 123.1003 125.5646 123.2504Table 2: Exact PAoI for the system with k = 8 and different polling schemes. Queue 1 and 4 areheavily loaded: each with 45% total load. Total load = 0.5. H i = H ∼ exp (1) , U i = U = . k = 8 and different polling schemes. Queue 1 and 4are heavily loaded: each with 45% total load. Total load = 20. H i = H ∼ exp (1) , U i = U = . (a) CBPS (b) BRPS (c) CBPS-P Figure 9: Average PAoI Across Queues, λ = (0 . , . , . ∗ T otal Load , H i = H ∼ exp (1) , U i = U = 0 . Next we consider the average PAoI across queues (i.e., (cid:80) ki =1 E [ A i ] ) under those three different Markovianpolling schemes, as shown in Figure 9. The average PAoI across queues was also considered in [1, 55]. InFigure 9 we find that cyclic scheme achieves the lowest average PAoI under different traffic loads for bothCBPS, BRPS and CBPS-P. LOP has the highest average PAoI among these three polling schemes. Thisis because under LOP, the server would serve the queues with high arrival rates more likely, and queueswith low arrival rates would be polled infrequently. Since PAoI is more sensitive to the arrival rate changewhen arrival rate is small (which we can observe from Figure 6 and 7), the PAoI reduction in queues withhigh arrival rates would be easily overshadowed by the PAoI increase caused by queues with low arrivalrates, when LOP is applied. This observation implies that if one wants to reduce the average PAoI for theentire system, a good strategy is to avoid polling certain queues too frequently. Therefore, policies with evenpolling frequency for queues such as symmetric or cyclic scheme are recommended for small average PAoI.26 Concluding Remarks
In this paper we considered AoI related metrics such as AoI, PAoI as well as variance of peak age in queueingsystems with server vacations. We discussed cases with both i.i.d vacations and non-i.i.d vacations, and fornon-i.i.d vacation systems we considered polling system specifically. We provided a general decompositionapproach that decomposes the system age into independent components, which can be used to derive AoI,PAoI as well as the variance of peak age for i.i.d vacation systems, and PAoI for non-i.i.d vacation systems.In these systems with vacation servers, we discussed three system variations, i.e., CBS, BRS and CBS-P,which differs in assumption about buffer availability and service preemptions. We proved that when vacationtime is i.i.d, PAoI in BRS is always smaller than PAoI in CBS. However, when vacation time is nom-i.i.d.,this result is no longer true. We derived the conditions under which CBS-P system has smaller PAoI thanCBS. We further provided numerical study to justify our findings, and discuss the advantage of each systemin terms of AoI, PAoI or variance of peak age. In our future work, we will consider the closed-form of AoI forsystem with non-i.i.d vacations such as polling systems. We will also consider the optimal switching schemeand scheduling scheme for polling systems in the future.
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A Proof for Theorem 7
Theorem 7 : The PAoI for CBS-P is given by E [ A CBS − P ] = − H ∗ ( λ ) − λH ∗ (1) ( λ )+ H ∗ ( λ ) λH ∗ ( λ ) + H ∗ ( λ ) V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) , and the AoI for this system is given by E [∆ CBS − P ] = V ∗ (2) (0)1 − V ∗ ( λ ) + 2 V ∗ (1) (0) V ∗ (1) ( λ )(1 − V ∗ ( λ )) − V ∗ (1) (0)1 − V ∗ ( λ ) 1 − H ∗ ( λ ) λH ∗ ( λ ) + λH ∗ ( λ ) (cid:104) λ − H ∗ ( λ ) λ + H ∗ (1) ( λ ) (cid:105) − V ∗ (1) (0)1 − V ∗ ( λ ) + − H ∗ ( λ ) λH ∗ ( λ ) ) − H ∗ (1) ( λ ) H ∗ ( λ ) + H ∗ ( λ ) (cid:18) λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) (cid:19) . Proof.
Different from the case of non-preemptive service, in the case where service is preempted by newarrivals, we decompose the age peak into three pieces E [ A ] = E [ D ] + E [ B ] + E [ L ] , (12)where D is the delay of a packet that is eventually processed by the server, B is the time period when theserver is on vacation (the same as we defined in Theorem 2), and L is the time when the server is busyin serving. A demonstrative graph is given by Figure 10. Note that these three components are mutuallyindependent. Therefore the AoI of this system can be given as E [∆] = E [ L ] + 2 E [ L ] E [ B ] + E [ B ]2( E [ L ] + E [ B ]) + E [ D ] . (13)We now derive the LST of D , denoted as D ∗ ( s ) . We first notice that if the service time of a packet H is smaller than the inter-arrival time T , then the packet is served without being preempted. Therefore, allthe packets that are eventually processed must have the service time smaller than the inter-arrival time.If the packet that we serve arrives during the last vacation, then its delay D is its waiting time G plusits service time. If it arrives during service (it preempts the previous packet in service), then the delay is32igure 10: Age of Information Decomposition for Preemptive Service Systems. The second agepeak is decomposed into three components: A { } = D { } + B { } + L { } , where D { } is the delayof packet 1, B { } is period when the server is on vacation, and L { } is the time period whenthe server is serving. Notice that in this example, packet 2 is preempted by packet 3 at time r { } , and packet 3 is not preempted by any packet.its service time only. Thus we have E [ e − sD | H < T ] = G ∗ ( s ) ˆ H ( s ) and E [ e − sD | H ≥ T ] = ˆ H ( s ) , where ˆ H ( s ) = E [ e − sH | H < T ] . Since the inter-arrival time is exponential, by letting F ( x ) be the CDF of service time H , we have ˆ H ( s ) = (cid:82) ∞ e − su dF ( u ) (cid:82) ∞ u λe − λx dx P ( H
Lemma 14.
It holds true for any LST function V ∗ ( s ) that V ∗ (1) ( λ )1 − V ∗ ( λ ) ≥ − λ for any positive λ .Proof. A non-rigorous but intuitive way of proving this is that by the fact that E [ G ] = λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) ≥ in CBS, then V ∗ (1) ( λ )1 − V ∗ ( λ ) ≥ − λ must hold. We now prove this inequality in another way without using theproperty of E [ G ] . 34ince V ∗ (1) ( λ )1 − V ∗ ( λ ) = − E [ V e − λV ]1 − E [ e − λV ]= − E [ λV − λ V + λ V − λ V + λ V − ... ] λ E [ λV − λ V + λ V − λ V + ... ] , we only need to show that λV − λ V + λ V − λ V + λ V − ... ≤ λV − λ V + λ V − λ V + λ V − ... for any V ≥ and λ ≥ . By letting β ( x ) = (1 −
12! ) x − ( 12! −
13! ) x + ( 13! −
14! ) x − ..., we now only need to show that β ( x ) ≥ for any x ≥ . Notice that β ( x ) = 12 x − x + 34! x − x + ... = x ( x − x + 13! x − x + ... ) + ( − x + 13! x − x + 15! x + ... )= x (1 − e − x ) + ( − e − x + 1 − x )= 1 − e − x − xe − x . Since ∂β ( x ) ∂x = e − x − e − x + xe − x ≥ and x ≥ , we have β ( x ) ≥ . Therefore we have V ∗ (1) ( λ )1 − V ∗ ( λ ) ≥ − λ . Theorem 9 : If the service time is exponentially distributed, then the system CBS-P has both AoI andPAoI smaller than CBS.
Proof.
We assume that the service time is exponentially distributed with parameter µ. We first show theconclusion holds true for AoI. When the service time is exponentially distributed, we have E [∆ CBS ] = − H ∗ (2) (0) + 2 H ∗ (1) (0) V ∗ (1) (0)1 − V ∗ ( λ ) + V ∗ (2) (0)1 − V ∗ ( λ ) + 2 V ∗ (1) (0) V ∗ (1) ( λ )(1 − V ∗ ( λ )) (cid:16) H ∗ (1) (0) + V ∗ (1) (0)1 − V ∗ ( λ ) (cid:17) + 1 λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) − H ∗ (1) (0)= µ − µ V ∗ (1) (0)1 − V ∗ ( λ ) + V ∗ (2) (0)1 − V ∗ ( λ ) + 2 V ∗ (1) (0) V ∗ (1) ( λ )(1 − V ∗ ( λ )) (cid:16) µ − V ∗ (1) (0)1 − V ∗ ( λ ) (cid:17) + 1 λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) + 1 µ E [∆ CBS − P ] = V ∗ (2) (0)1 − V ∗ ( λ ) + 2 V ∗ (1) (0) V ∗ (1) ( λ )(1 − V ∗ ( λ )) − V ∗ (1) (0)1 − V ∗ ( λ ) 1 − H ∗ ( λ ) λH ∗ ( λ ) + λH ∗ ( λ ) (cid:104) λ − H ∗ ( λ ) λ + H ∗ (1) ( λ ) (cid:105) − V ∗ (1) (0)1 − V ∗ ( λ ) + − H ∗ ( λ ) λH ∗ ( λ ) ) − H ∗ (1) ( λ ) H ∗ ( λ ) + H ∗ ( λ ) (cid:18) λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) (cid:19) = V ∗ (2) (0)1 − V ∗ ( λ ) + 2 V ∗ (1) (0) V ∗ (1) ( λ )(1 − V ∗ ( λ )) − V ∗ (1) (0)1 − V ∗ ( λ ) 1 µ + µ − V ∗ (1) (0)1 − V ∗ ( λ ) + µ ) + µ ( µ + λ ) µµ + λ + µµ + λ (cid:18) λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) (cid:19) . Therefore, by using Lemma 14, we have E [∆ CBS ] − E [∆ CBS − P ] = 1 µ + λµ + λ V ∗ (1) ( λ )1 − V ∗ ( λ ) ≥ µ − λµ + λ λ = λµ ( µ + λ ) ≥ . Now we show the result holds true for PAoI. Since we have E [ A CBS ] = 1 λ + V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) + 2 µ and E [ A CBS − P ] = 1 − H ∗ ( λ ) − λH ∗ (1) ( λ ) + H ∗ ( λ ) λH ∗ ( λ ) + H ∗ ( λ ) V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ )= 1 − µµ + λ + λµ ( µ + λ ) + ( µµ + λ ) λµµ + λ + µµ + λ V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) , then E [ A CBS ] − E [ A CBS − P ] = 1 λ + 1 µ − λ + λµ + λ V ∗ (1) ( λ )1 − V ∗ ( λ ) ≥ µ − µ + λ ≥ . Proof for Theorem 10
Theorem 10:
If the service time H satisfies E [ H ] ≥ − H ( s ) λH ( s ) for all s > , then CBS-P always has smallerPAoI than CBS. Proof.
We first have E [ A CBS ] − E [ A CBS − P ] = 1 λ + V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ ) + 2 E [ H ] − − H ∗ ( λ ) − λH ∗ (1) ( λ ) + H ∗ ( λ ) λH ∗ ( λ ) − H ∗ ( λ ) V ∗ (1) ( λ ) − V ∗ (1) (0)1 − V ∗ ( λ )= (1 − H ∗ ( λ )) (cid:18) λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) (cid:19) + 2 E [ H ] − − H ∗ ( λ ) − λH ∗ (1) ( λ ) λH ∗ ( λ ) . From Lemma 14 we know that λ + V ∗ (1) ( λ )1 − V ∗ ( λ ) ≥ and − H ∗ ( λ ) ≥ − λH ∗ (1) ( λ ) , then we have E [ A CBS ] − E [ A CBS − P ] ≥ E [ H ] − − H ∗ ( λ ) H ∗ ( λ ) ≥ . D Proof for Theorem 11.
Theorem 11:
For M/G/1/1 we have E [ A M/G/ / ] = λ − H ∗ (1) (0) , E [∆ M/G/ / ] = λ − λ H ∗ (1) (0)+ H ∗ (2) (0) λ − H ∗ (1) (0) − H ∗ (1) (0) and V ar ( A M/G/ / ) = λ +2 H ∗ (2) (0) − { H ∗ (1) (0) } . For M/G/1/1/preemptive we have E [ A M/G/ / /preemptive ] = − H ∗ (1) ( λ ) H ∗ ( λ ) + λH ∗ ( λ ) , E [∆ M/G/ / /preemptive ] = λH ∗ ( λ ) , and V ar ( A M/G/ / /preemptive ) = H ∗ (2) ( λ ) H ∗ ( λ ) − { H ∗ (1) ( λ ) } H ∗ ( λ ) + λ H ∗ ( λ ) + H ∗ (1) ( λ ) λH ∗ ( λ ) . For M/G/1/2* we have E [ A M/G/ / ∗ ] = − H ∗ (1) (0) + λ + H ∗ (1) ( λ ) , E [∆ M/G/ / ∗ ] = H ∗ (2) (0)+ λ H ∗ ( λ ) − λ H ∗ (1) ( λ ) − H ∗ (1) (0)+ H ∗ ( λ ) λ + λ − λ H ∗ ( λ ) + H ∗ (1) ( λ ) − H ∗ (1) (0) , and V ar ( A M/G/ / ∗ ) = 2 H ∗ (2) (0) − H ∗ (1) (0) + H ∗ ( λ )[ H ∗ (1) (0)+ H ∗ (1) ( λ )] λ + H ∗ ( λ )(1 − H ∗ ( λ )) λ + λ − H ∗ (2) ( λ ) − λ H ∗ (1) ( λ ) − H ∗ (1) ( λ ) . Proof.
We first show the variance of peak age in M/G/1/1. Realizing that in M/G/1/1 system, once apacket arrives, the server will start processing it immediately. Thus there is no waiting time for all packets.Then the LST of peak age in M/G/1/1 can be given as A ∗ ( s ) = I ∗ ( s ) H ∗ ( s ) . The inter-service time I can befurther decomposed into the idling time T (exponentially distributed) and service time H , i.e., I = T + H .We thus have A ∗ ( s ) = T ∗ ( s ) H ∗ ( s ) . By some simple algebra we have E [ A M/G/ / ] = λ − H ∗ (1) (0) , E [∆ M/G/ / ] = λ − λ H ∗ (1) (0)+ H ∗ (2) (0) λ − H ∗ (1) (0) − H ∗ (1) (0) and V ar ( A M/G/ / ) = λ + 2 H ∗ (2) (0) − { H ∗ (1) (0) } . Similarly, for M/G/1/1/preemptive system, there is no waiting time for packets. Thus by the argument inAppendix A, we have D ∗ ( s ) = H ∗ ( s + λ ) H ∗ ( λ ) . Then the LST of peak age can be given as A ∗ ( s ) = D ∗ ( s ) T ∗ ( s ) L ∗ ( s ) , L ∗ ( s ) is given by Equation (15). We then have E [ A M/G/ / /preemptive ] = − H ∗ (1) ( λ ) H ∗ ( λ ) + λH ∗ ( λ ) and E [∆ M/G/ / /preemptive ] = λH ∗ ( λ ) . And the variance of peak age of M/G/1/1/preemptive system is given by
V ar ( A M/G/ / /preemptive ) = H ∗ (2) ( λ ) H ∗ ( λ ) − { H ∗ (1) ( λ ) } H ∗ ( λ ) + λ H ∗ ( λ ) + H ∗ (1) ( λ ) λH ∗ ( λ ) . For M/G/1/2* system, the inter-service time is H if there is an arrival during processing time. If thereis no arrival during processing time, the next service starts when the next arrival occurs. By memorylessproperty of Poisson arrivals, we have I = T in this case. Therefore I = max { H, T } . To calculate the LST of I, by letting F ( h ) be the CDF of H , we have I ∗ ( s ) = E [ e − sI ] = (cid:90) ∞ h =0 (cid:90) ∞ t = h λe − λt e − st dF ( h ) dt + (cid:90) ∞ h =0 (cid:90) ht =0 e − sh λe − λt dF ( h ) dt = λλ + s H ∗ ( s + λ ) + H ∗ ( s ) − H ∗ ( s + λ )= H ∗ ( s ) − sλ + s H ∗ ( s + λ ) . We can then have I ∗ (1) (0) = H ∗ (1) (0) − H ∗ ( λ ) λ , and I ∗ (2) (0) = H ∗ (2) (0) + 2 λ H ∗ ( λ ) − λ H ∗ (1) ( λ ) . The waiting time only occurs when there is an arrival during processing time H , so that W = max { H − T, } . The LST of W is thus be given as W ∗ ( s ) = (cid:90) ∞ h =0 (cid:90) ht =0 e − s ( h − t ) dF ( h ) λe − λt dt + (cid:90) ∞ h =0 dF ( h ) (cid:90) ∞ t = h λe − λt dt = λλ − s ( H ∗ ( s ) − H ∗ ( λ )) + H ∗ ( λ )= λλ − s H ∗ ( s ) − sλ − s H ∗ ( λ ) . From Lemma 1 we have G ∗ ( s ) = λλ + s + sλ + s W ∗ ( λ + s )= λλ + s − λλ + s H ∗ ( λ + s ) + H ∗ ( λ ) .
38y taking the first and second derivative of G ∗ ( s ) , we have G ∗ (1) (0) = − λ + 1 λ H ∗ ( λ ) − H ∗ (1) ( λ ) and G ∗ (2) (0) = 2 λ − λ H ∗ ( λ ) + 2 λ H ∗ (1) ( λ ) − H ∗ (2) ( λ ) . By Equation (1) and (2), we have E [ A M/G/ / ∗ ] = − H ∗ (1) (0) + 1 λ + H ∗ (1) ( λ ) and E [∆ M/G/ / ∗ ] = H ∗ (2) (0) + λ H ∗ ( λ ) − λ H ∗ (1) ( λ ) − H ∗ (1) (0) + H ∗ ( λ ) λ + 1 λ − λ H ∗ ( λ ) + H ∗ (1) ( λ ) − H ∗ (1) (0) . Using Equation (4), we have the variance of peak age
V ar ( A M/G/ / ∗ ) = 2 λ − λ H ∗ ( λ ) + 2 λ H ∗ (1) ( λ ) − H ∗ (2) ( λ ) − (cid:20) − λ + 1 λ H ∗ ( λ ) − H ∗ (1) ( λ ) (cid:21) + H ∗ (2) (0) + 2 λ H ∗ ( λ ) − λ H ∗ (1) ( λ ) − (cid:20) H ∗ (1) (0) − H ∗ ( λ ) λ (cid:21) + H ∗ (2) (0) − H ∗ (1) (0) = 2 H ∗ (2) (0) − H ∗ (1) (0) + 2 H ∗ ( λ )[ H ∗ (1) (0) + H ∗ (1) ( λ )] λ + 2 H ∗ ( λ )(1 − H ∗ ( λ )) λ + 1 λ − H ∗ (2) ( λ ) − λ H ∗ (1) ( λ ) − H ∗ (1) ( λ ) . E Exact Solution for PAoI in CBS-P with Dependent Vacations
Notice that in CBS-P, the server’s vacation time B can be divided into B = T + W , where T is the inter-arrival time of packets which is exponentially distributed, and W is the time when the buffer is occupied(the same as their definition in Section 4). Because of the memoryless property of exponential distribution,we have E [ B ] = λ + E [ W ] . 39y Equation (6), (12), (14), and (16), the PAoI for CBS-P can be written as E [ A ] = E [ D ] + E [ B ] + E [ L ]= − H ∗ (1) ( λ ) H ∗ ( λ ) + H ∗ ( λ ) 1 λ (1 − W ∗ ( λ )) + 1 λ + E [ W ] + 1 − H ∗ ( λ ) λH ∗ ( λ )= − H ∗ (1) ( λ ) H ∗ ( λ ) + H ∗ ( λ ) 1 λ (1 − W ∗ ( λ )) + E [ W ] + 1 λH ∗ ( λ ) . F Proof for Theorem 13
Theorem 13:
If the service time for each queue is exponentially distributed, then CBPS-P will always havesmaller PAoI than CBPS.
Proof.
When the service time is exponentially distributed, from Equation (9), we have L ∗ j ( s ) = H ∗ j ( s + λ j ) ss + λ j + λ j s + λ j H ∗ j ( s + λ j ) = µ j s + λ j + µ j ss + λ j + λ j s + λ j µ j s + λ j + µ j = µ j s + µ j . So that the expressions for ˜ H ∗ j in Equation (8) are identical for CBPS and CBPS-P. Both systems will havethe same F j ( z , ..., z k ) for all j after solving for Equation (7). Similarly, since − H ∗ j ( λ j ) λ j H ∗ ( λ j ) = µ j , both CBPSand CBPS-P will have the same expression for γ j in Equation (10) for all j . Therefore, CBPS and CBPS-Phave the same expressions for W ∗ j ( λ j ) and E [ W j ] for all queue j .We then have E [ A CBP Sj ] − E [ A CBP S − Pj ]= − λ j W ∗ j ( λ j ) + 2 λ j + E [ W j ] + 2 E [ H j ] − (cid:34) − H ∗ (1) j ( λ j ) H ∗ j ( λ j ) + H ∗ j ( λ j ) 1 λ j (1 − W ∗ j ( λ j )) + 1 λ j + E [ W j ] + 1 − H ∗ j ( λ j ) λ j H ∗ j ( λ j ) (cid:35) = (cid:0) − H ∗ j ( λ j ) (cid:1) λ j (cid:0) − W ∗ j ( λ j ) (cid:1) + 1 µ j − µ j + λ j ≥ ..