Resource Allocation for On-Demand Data Delivery to High-Speed Trains via Trackside Infostations

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1 Resource Allocaton for On-Demand Data Delvery to Hgh-Speed Trans va Tracksde Infostatons Hao Lang and Wehua Zhuang Department of Electrcal and Computer Engneerng, Unversty of Waterloo, Waterloo, Ontaro, Canada, N2L 3G1 Emal: Abstract In ths paper, we nvestgate the on-demand data delvery to hgh-speed trans va tracksde nfostatons. The optmal resource allocaton problem s formulated by consderng the trajectory of a tran, qualty of servce (QoS) requrements, and network resources. The orgnal problem s transformed nto a sngle-machne preemptve schedulng problem based on a tmecapacty mappng. As the servce demands are not known a pror, an onlne resource allocaton algorthm s proposed based on the Smth rato and exponental capacty. The performance of the proposed algorthm s evaluated based on a real hgh-speed tran schedule. Compared wth the exstng approaches, our proposed algorthm can acheve the best performance n terms of the total reward of delvered servces over the trp of a tran. I. INTRODUCTION Recently, the hgh-speed ral s fast growng all over the world [1]. Besdes the sgnfcantly shortened journey tmes, passenger comforts can be largely mproved by hgh-speed Internet servces [2]. The cellular networks (e.g., 3G) deployed near the ral lnes can provde seamless coverage. However, the data transmsson rate s lmted for trans movng at an extremely hgh speed because of the Doppler effect [3]. Wth hundreds of passengers onboard and an ever-ncreasng Internet servce demand, hgh congeston n the cellular networks s nevtable. One straghtforward soluton s to resort to the exstng low earth orbt (LEO) satelltes, but the servce cost s stll prohbtve for many passengers. An alternatve or complementary soluton s proposed n [3] [5], where low-cost tracksde nfostatons (or repeaters) are deployed n close vcnty to the ral lnes to offload the cellular traffc. Powerful antennas are nstalled around each tran to communcate wth the nfostatons. The antennas are further connected to a vehcle staton whch can be accessed by the passenger devces based on wreless local area network (WLAN) technologes. A medum access control (MAC) layer frame structure s proposed, whch can support a tran speed up to 360 km/h [3]. In order to acheve servce delvery based on tracksde nfostatons, an on-demand servce model can be appled [6]. The cellular networks are consdered to support control channels for servce requests and acknowledgements, whle the data delvery s acheved va tracksde nfostatons. For a large number of onboard passengers, the resource contenton among multple servces should be resolved. However, dfferent from the cellular networks, the coverage provded by the nfostatons may not be seamless for a low deployment cost. As a hgh-speed tran travels along a ral lne, the wreless lnk between the nfostatons and a vehcle staton s hghly dynamc and subject to perodc dsconnectons, whch makes the resource allocaton challengng. In the lterature, the optmal on-demand broadcast schedulng s nvestgated for satelltes and cellular networks [7] [8]. The proposed algorthm can resolve the resource contenton among multple servces. However, the approaches are not applcable to an nfostaton based network snce the broadcast lnk s assumed to have a constant rate. On the other hand, the data delvery n a vehcular network wth ntermttent lnks s studed based on the moblty patterns of vehcles [9] [10]. The solutons focus on sngle servce delvery and cannot be drectly appled to mass transportaton vehcles such as hgh-speed trans. How to provde effcent resource allocaton among multple on-demand data servces n such an ntermttently connected network s stll an open ssue. In ths paper, we frst formulate the optmal resource allocaton problem by consderng the ntermttent network connectvty and mult-servce demands. In order to acheve effcent resource allocaton wth low computatonal complexty, the orgnal frame-based formulaton s transformed nto a capacty-based formulaton va tme-capacty mappng. Based on the theory of sequencng and schedulng, the transformed problem s a sngle-machne preemptve schedulng problem wth nteger generatng tmes, processng tmes, and deadlnes. An onlne resource allocaton algorthm s proposed to address the uncertantes n servce demands. The performance of the proposed algorthm s evaluated and compared wth exstng algorthms under a real hgh-speed tran schedule. II. SYSTEM MODEL In the followng, we ntroduce the tran trajectory under consderaton, and descrbe the qualty of servce (QoS) requrements of the on-demand data servces, and network resources. A. Tran Trajectory Consder a sngle trp of a tran from an orgn staton to a destnaton termnal wthn the tme duraton [T I,T O ]. A total number of H tracksde nfostatons are deployed along the ral lne. Denote Th and T h o as tme for the tran to come nto and go out of the transmsson range of the hth (h [1,,H]) nfostaton, respectvely. Snce the coverage of the nfostatons may not be seamless, we may have Th o T h+1 for 1 h H 1. Takng account the duraton of a trp, we have T I T1 and TH o T O. The nfostatons are connected to content /11/$ IEEE

2 servers n the Internet wth suffcently large bandwdth. We defne the transmsson perod and dle perod as the tme when the tran s n and out of the coverage area of an nfostaton, respectvely. B. QoS Requrements AsetS of on-demand data servces are consdered over the trp of the tran. Servce s (s S) s generated at tme G s. If servce s s delvered before ts deadlne D s, a reward ω s can be obtaned by the servce provder. We consder G s T I and D s T O, assume all other servces can be delvered to the passengers when they are off-board. Erasure codng based servce delvery s consdered [3] [9]. The nformaton data of servce s s encoded and segmented nto Q s blocks wth equal sze B. Servce s can be decoded when any Q s (Q s < Q s ) dstnct blocks are receved. The coverage of a cellular network s consdered to be seamless, so that the servce requests and acknowledgements can be delvered wth a neglgble delay. C. Network Resources Consder the MAC layer frame structure proposed n [3]. Tme s parttoned nto frames wth equal duraton T F. When the tran vsts the hth nfostaton, the number of frames s gven by K h = (Th o T h )/T F. Note that the small dfference between Th and the begnnng tme of the frst frame s omtted. The kth frame begns and ends at tmes Th +(k 1)T F and Th + kt F, respectvely. We defne the capacty of the kth frame (A h,k ) as the maxmum number of blocks that can be delvered from the hth nfostaton to the vehcle staton wthn ths frame. A round-robn scheduler s appled when multple trans are present n the coverage area of an nfostaton. If the kth frame wthn the hth nfostaton s not allocated to the vehcle staton under consderaton, we have A h,k =0. Snce the resource allocaton s frame based, we update the servce generatng tmes and deadlnes as G s Th + T F (Gs Th)/T F, for T h G s Th o (1) D s Th + T F (Ds Th)/T F, for T h D s Th. o (2) Tran-to-tran communcatons are not consdered n ths work because of an extremely short contact duraton and a hghly dynamc wreless channel when two hgh-speed trans move n opposte drectons. For a vehcle staton, the buffer space s consdered to be unlmted, and the data transmsson rate to a passenger devce s assumed to be suffcently large. Therefore, a data block can be successfully delvered to a passenger devce f t s delvered to the vehcle staton. Snce the schedule of a hgh-speed tran s hghly stable, the system parameters n terms of tran trajectory and network resources can be obtaned n advance. However, a servce demand (wth parameters G s,d s,ω s,q s ) s not known a pror untl the servce request s receved at tme G s. III. PROBLEM FORMULATION The objectve n servce provsonng s to maxmze the total reward of delvered servces over a trp of the tran. Defne x h,k,s as the number of blocks delvered to the vehcle staton durng the kth frame wthn the hth nfostaton for servce s. Then the resource allocaton varable over the trp of the tran s gven by X = x h,k,s h 1,,H},k 1,,K h },s S}. (3) For a specfc X, defne ψ X,s as a delvery ndcator of servce s whch equals 1 f servce s s delvered before ts deadlne, and 0 otherwse. We have 1, f H Kh ψ X,s = x h,k,s = Q s 0, f H Kh x (4) h,k,s <Q s. Note that we do not consder the case H Kh x h,k,s > Q s. The reason s that, for erasure codng based servce delvery, the resources are underutlzed by delverng more than Q s blocks for servce s. The optmal resource allocaton problem s formulated as (P1) max X subject to ω s ψ X,s (5) x h,k,s 0, h 1,,H}, k 1,,K h }, s S (6) H K h x h,k,s Q s, s S (7) x h,k,s =0, f G s Th + kt F or D s Th +(k 1)T F, h 1,,H} k 1,,K h }, s S (8) x h,k,s A h,k, h 1,,H}, k 1,,K h }. (9) Constrant (6) mples that negatve resource allocaton s not allowed. Constrant (7) s n accordance wth (4). Constrant (8) states that the blocks of a servce can only be scheduled after the generatng tme and before the deadlne. Wth constrant (9), the number of blocks that can be delvered to the vehcle staton durng the kth frame wthn the hth nfostaton s lmted by the capacty of the frame (A h,k ). Problem P1 s a mxed nteger programmng (MIP) problem whch cannot be solved effcently [11]. Because of the mass transportaton functonalty of hgh-speed trans, a large amount of servces may be generated durng a sngle trp of a tran, whch results n a prohbtve computatonal complexty. The dffculty of analyzng problem P1 stems from the fact that the servces are not contnuously schedulable over the tme axs because of the lnk dsconnectons. Moreover, the data transmsson rate from nfostatons to a vehcle staton (n terms of A h,k ) s not a constant. Therefore, the duraton to complete each servce s dependent on the tme when the servce s scheduled. IV. PROBLEM TRANSFORMATION In order to smplfy problem P1, we consder a transformaton n ths secton whch vrtually maps the tme nto a cumulatve capacty. Here we defne the cumulatve capacty

3 at tme t (t [T I,T O ]) as the summaton of the capactes of all frames wthn [T I,t]. The problem transformaton conssts of two steps,.e., tme-capacty mappng and capacty-based formulaton. Here problem P1 s referred to as the frame-based formulaton. The proofs of the Lemmas and Theorem 1 are omtted because of space lmtaton. A. Tme-Capacty Mappng For a specfc trp of the tran, the maxmum number of blocks that can be delvered to the vehcle staton s lmted by H Kh A h,k. [ Defne a tme-capacty mappng functon f(t) :[T I,T O ] 0, 1,, H ] Kh A h,k whch maps tme t to the cumulatve capacty. Based on the nformaton of tran trajectory and network resources, we have the followng lemma. Lemma 1: The value of f(t) s gven by f(t) = (t T h t )/T F j=1 A ht,j + h t 1 l=1 ht l=1 Kl j=1 A l,j, otherwse Kl j=1 A l,j, f h, T h t T o h (10) where h t = arg max h T h t }. Intutvely, more blocks can be potentally delvered to the vehcle staton as tme t ncreases. Ths property s nherent for f(t) and stated by the followng lemma. Lemma 2: f(t) s a non-decreasng functon wth respect to t (t [T I,T O ]). Based on Lemma 1 and Lemma 2, the followng theorem holds. Theorem 1: For problem P1, constrant (8) s equvalent to a capacty-based constrant gven by x h,k,s =0, f G c s k h 1 K l A h,j + A l,j, j=1 l=1 j=1 k 1 h 1 K l or Ds c A h,j + A l,j, j=1 l=1 j=1 h 1,,H}, k 1,,K h }, s S. (11) where G c s = f (G s ) and D c s = f (D s ). B. Capacty-Based Problem Formulaton By replacng constrant (8) n problem P1 wth constrant (11), we can obtan problem P2. Snce all constrants of problem P2 are defned based on the number of blocks, we can smplfy problem P2 by ntroducng a capacty-based formulaton. By defnton, we can verfy f(t I ) = 0 and f(t O ) = H h=1 Kh k=1 A h,k. Then the set T I,T O,G s,d s s S} of tme ndces can be represented by a set C of unque cumulatve capactes, gven by C = f (G s ),f(d s )} f (T I ),f(t O )} } H K h = G c s,ds} c 0, A h,k. (12) Let C = N +1 (N 1) and c n (1 n N +1)bethe cardnalty and elements of set C, respectvely. Wthout loss of generalty, we consder an ascendng order of the elements n C,.e., c 1 <c 2 < <c N+1. We partton the trp of the tran nto N non-overlapped vrtual perods accordng to the cumulatve capactes n C. Wthn the nth vrtual perod (defned by [c n +1,c n+1 ]), no servce s generated or expred snce all servce generatng tmes and deadlnes are consdered n the calculaton of set C. Therefore, for a feasble resource allocaton, changng the sequence of servce schedulng wthn a vrtual perod does not affect the servce delvery performance. Ths property s stated by the followng lemma. Lemma 3: Consder a feasble resource allocaton varable X wth four elements x h1,k 1,s 1, x h1,k 1,s 2, x h2,k 2,s 1, x h2,k 2,s 2. Suppose x h1,k 1,s 1,x h2,k 2,s 2 1, x h1,k 1,s 2,x h2,k 2,s 1 0. All blocks of the two frames (.e., the k 1 th frame wthn the h 1 th nfostaton and the k 2 th frame wthn the h 2 th nfostaton) belong to the same vrtual perod, whle the two frames are not dentcal. Construct another resource allocaton varable X by replacng the elements x h1,k 1,s 1,x h1,k 1,s 2,x h2,k 2,s 1,x h2,k 2,s 2 n X wth x h1,k 1,s 1 1, x h1,k 1,s 2 +1, x h2,k 2,s 1 +1, x h2,k 2,s 2 1 and keepng all other elements unchanged. Then we have the same feasbltes and objectve functon values for X and X. Based on Lemma 3, the optmal resource allocaton problem can be smplfed by consderng the total number of blocks scheduled for each servce wthn each vrtual perod. Defne y n,s as the number of blocks delvered to the vehcle staton for servce s wthn the nth vrtual perod. Then the resource allocaton varable over the trp of the tran s gven by Y = y n,s n 1,,N},s S}. (13) The delvery ndcator s gven by 1, f N n=1 η Y,s = y n,s = Q s 0, f N n=1 y (14) n,s <Q s. Then problem P2 can be transformed nto a capacty-based formulaton as follows (P3) max ω s η Y,s (15) Y subject to y n,s 0, n 1,,N}, s S (16) N y n,s Q s, s S (17) n=1 y n,s =0, f G c s c n+1 +1 or D c s c n, n 1,,N} (18) y n,s c n+1 c n, n 1,,N}(19) where constrant (19) states that the number of blocks that can be delvered to the vehcle staton durng the nth vrtual perod s lmted by c n+1 c n.

4 Algorthm 1 Resource Allocaton Algorthm Input: k, h, G c s,ds,ω c s,q s,q r h,k,s (s Sg h,k ) Output: x h,k,s, Q r h,k,s (s Sg h,k ) 1: Intalze x h,k,s =0, Q r h,k,s = Qr h,k,s, for s Sg h,k ; 2: Intalze m = A h,k ; 3: whle m 0do 4: S A = s s S g h,k, Q r h,k,s > 0,Dc s Q r h,k,s 5: + k j=1 A h,j + h 1 } Kl l=1 j=1 A l,j m ; 6: f S A = then 7: break; 8: end f 9: U s calculaton, for s S A ; 10: s = arg max A U s }; 11: Qr h,k,s Q r h,k,s 1; 12: m m 1; 13: x h,k,s x h,k,s +1; 14: end whle 15: return x h,k,s, Q r h,k,s (s Sg h,k ) V. RESOURCE ALLOCATION ALGORITHM Based on the problem transformaton, the servces are contnuously schedulable over the cumulatve capacty axs, and the duraton to complete each servce (Q s ) s not dependent on the tme when the servce s scheduled. Accordng to the theory of sequencng and schedulng, problem P3 defnes a snglemachne preemptve schedulng problem wth nteger generatng (or release) tmes (G c s), processng tmes (Q s ), and deadlnes (Ds) c (formal notaton: 1 G c s, preempton ω s η Y,s ) [12] [13] [14], whle the tme ndces s vrtually transformed nto the cumulatve capactes. By approxmatng the servce reward to ntegers, problem P3 can be solved by a pseudopolynomal algorthm [15]. However, for on-demand data servces, snce the servce demands are not known a pror, onlne algorthm should be desgned to acheve effcent resource allocaton. As the tran moves from the orgn staton to the destnaton termnal, the onlne algorthm allocates the network resources to multple servces frame-by-frame. Consder the kth frame wthn the hth nfostaton, denote the number of blocks allocated to servce s (s S g h,k ) as x h,k,s, where S g h,k = s s S, G s Th + kt } F represents the set of already generated servces. The resource allocaton algorthm s detaled n Algorthm 1, where Q r h,k,s and Q r h,k,s are the numbers of remanng blocks of servce s before and after the resource allocaton, respectvely. Note that the algorthm needs to be performed only when S g h,k. For a newly generated servce s,.e., S g h,k, f h =1,k =1 s S g h,k \ Sg h 1,K h 1, f h 1,k =1 (20) S g h,k \ Sg h,k 1, otherwse we have Q r h,k,s = Q s. Otherwse, for a servce s generated n one of the prevous frames, we have Qr Q r h,k,s = h 1,Kh 1,s, f k =1 Q r h,k 1,s, otherwse. (21) Note that n (20) and (21), k =1corresponds to the frst frame n a transmsson perod. In step 4, S A represents the set of servces whch are avalable for schedulng and can possbly be delvered before ther deadlnes. In step 9, U s represents the utlty of servce s. We consder two knds of utltes n ths work,.e., Smth rato and exponental capacty [14]. For the Smth rato based resource allocaton, we have U s = ω s /Q s, whle for the exponental capacty based resource allocaton, we have ( ) ln max g h,k U s = ω s 1 Q Q r h,k,s 1 s max g Q. (22) s h,k The compettve factors of the Smth rato and exponental capacty based algorthm are 2 max Q s } and max Q s }/ ln(max Q s }), respectvely [14]. The complexty of Algorthm 1 s O (max h,k A h,k } S ). VI. NUMERICAL RESULTS In order to evaluate the performance of the our proposed resource allocaton algorthm, we consder a real tran schedule based on the Huhang hgh-speed ralway (also known as the Shangha-Hangzhou hgh-speed ralway) [16] [17]. The schedule s obtaned n February Snce no moblty trace s avalable, we consder a synthetc tran moblty model proposed n [18]. Each tran moves at a constant speed when travelng from one staton to another. When a tran leaves (arrves at) a staton, t accelerates (decelerates) accordng to a constant acceleraton (deceleraton). For smplcty, we consder the deceleraton equals to the negatve value of the acceleraton (α). A typcal value for the acceleraton of a hghspeed tran s gven by α =0.4m/s 2 [19]. We consder the schedule of the hgh-speed tran G7302 from Hangzhou to Shangha Hongqao as an example. The duraton of the trp (T O T I ) s 45 mnutes. For the wreless channel condton, we use a typcal settng for a hgh-speed tran [3]. The MAC layer frame structure for blnd nformaton ranng scheme s consdered as an example. The sze of each erasure codng block s B = 240 bts. There are 40 nfostatons deployed along the ral lne to support on-demand data servces. The overall servce arrval rate to the tran s represented by λ. The number of blocks of each servce (Q s ) s unformly dstrbuted wthn [50000, ], correspondng to the sze of a servce wthn [1.5, 15] Mbytes. The relatve deadlne (D s G s ) of each data servce s exponentally dstrbuted wth average value of 2 mnutes. The reward of each servce (ω s ) s unformly dstrbuted wthn [1, 10]. In addton to the proposed resource allocaton algorthm, we consder three exstng algorthms for comparson,.e., frst-n-frst-out (FIFO), earlest due date (EDD), and RAPID

5 Total reward of delvered servces Exponental Capacty Smth Rato Modfed RAPID 2 Modfed RAPID 1 FIFO EDD λ (servce/s) Fg. 1: Total reward of delvered servces versus λ. [10]. For the FIFO and EDD algorthms, the servces are scheduled accordng to an ascendng order of ther generatng tmes and deadlnes, respectvely. Snce moblty nformaton s requred by the RAPID algorthm to calculate the utlty functon, we modfy the algorthm, refer to modfed RAPID 1, by replacng the tme ndces wth cumulatve capactes based on the tme-capacty mappng. Moreover, snce the RAPID algorthm does not support servce dfferentaton, we further modfy t, refer to modfed RAPID 2, by multplyng the reward of each servce to ts utlty functon. Note that the modfed RAPID 2 algorthm, whch schedules the avalable servces one-by-one, can be consdered as a non-preemptve [12] verson of our proposed Smth rato based resource allocaton algorthm. The total reward of delvered servces versus λ s shown n Fg. 1. The standard devatons are also shown for reference. As we can see, the total reward s low for FIFO and EDD algorthms snce they do not consder the tran moblty and the QoS requrements. Although the EDD algorthm performs well when most servces can be delvered before ther deadlnes, ts performance degrades as the traffc load (n terms of λ) ncreases [20]. For a larger λ, the total rewards acheved by the modfed RAPID algorthms and our proposed algorthm mprove snce more servces can be potentally scheduled. The modfed RAPID 1 algorthm performs better than the FIFO and EDD algorthms snce the moblty nformaton of the tran s taken nto account. By further consderng the reward of each servce, the modfed RAPID 2 algorthm outperforms the modfed RAPID 1 algorthm because of the servce dfferentaton. However, ts performance s nferor to the newly proposed algorthm snce the total reward acheved by a non-preemptve scheduler s upper-bounded by the total reward acheved by a preemptve scheduler [21]. Although the compettve factor of the exponental capacty based algorthm s much hgher than that of the Smth rato based algorthm, the performance mprovement s not obvous snce a worst case scenaro for the algorthm happens wth a low probablty for the hgh-speed tran applcatons. VII. CONCLUSIONS In ths paper, we study the on-demand data delvery to hgh-speed trans va tracksde nfostatons. The optmal resource allocaton problem s formulated and transformed nto a sngle-machne preemptve schedulng problem wth nteger generatng tmes, processng tmes, and deadlnes. Snce the servce demands are unknown a pror, an onlne resource allocaton algorthm s proposed based on the Smth rato and exponental capacty, respectvely. As demonstrated by the smulaton results, our proposed algorthm can mprove the total reward of delvered servces over a trp of a tran as compared wth the exstng approaches such as FIFO, EDD, and RAPID. 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