Transmission Scheduling for Multi-Channel Satellite and Wireless Networks

Transcription

1 Tranmiion Scheduling for ulti-channel Satellite and Wirele Network Anand Ganti, Eytan odiano, John N. Titikli aachuett Intitute of Technology Abtract We conider a lotted ytem with N queue, and i.i.d. Bernoulli arrival at each queue during each lot. Each queue i aociated with a beam and a channel that change between on and off tate according to i.i.d. Bernoulli procee. We aume that the ytem ha K identical tranmitter ( erver ). Each erver, during each lot, can tranmit up to C 0 packet from a queue aociated with an on channel. We how that when K i arbitrary and C 0 =1,awell a when K =1and C 0 i arbitrary, a policy that aign the K erver to the on channel aociated with the K longet queue i optimal. We alo conider a fluid model under which fractional packet can be erved, for the cae K = N, and ubject to a contraint that at mot C packet can be erved in total over all of the N queue. We how that there i an optimal policy which erve the queue o that the reulting vector of queue length i ot Balanced. 1 Introduction Wirele and atellite node are often limited to a mall number of tranmitter and channel, and thee have to be allocated to uer in the face of competing demand. For example, atellite ytem employ hundred or even thouand of narrow beam over which information can be tranmitted at high data rate. Each of the downlink beam cover a different region within the atellite footprint. Data packet to be tranmitted along the different beam arrive at the atellite, either from the ground or from one of it neighboring atellite, and are tored in on-board buffer. In thi context, there i often only a limited number of tranmitter on-board the atellite, o that not all beam can be erved by the tranmitter imultaneouly. Thi give rie to a cheduling problem involving the allocation of the tranmitter to the different downlink beam. Further complicating matter i the fact that, due to weather and atmopheric condition, the tranmiion rate along the different beam varie with time; hence, the quality of the link mut be taken into account in making cheduling deciion. Similarly, a wirele bae tation typically ha far fewer channel available for tranmiion than the number of uer to be erved. Again, thi raie a nearly identical problem of allocating channel to the different uer. Thi cheduling problem ha received attention recently in the context of next generation wirele data ytem [2, 3, 4, 5, 6, 7, 8, 10]. We model the ytem a a dicrete time queueing ytem, with arrival and channel tate decribed by independent Bernoulli procee. ore pecifically, we aume that Thi reearch i upported in part by NSF Grant No. NCR and by DARPA under the Next Generation Internet Initiative

2 the number of arrival to the ith queue at time n, denoted by A i (n), are independent Bernoulli random variable, with the ame parameter for all i and n. Furthermore, we aume that the tate of the ith channel at time n, denoted by G i (n), can only take one of two value, namely, 0 or 1. We deignate 0 a an off tate, and 1 a the on tate. When the channel i in the off tate, no tranmiion i poible. When the channel i in the on tate, the channel can be utilized. Again, we aume that the G i (n) are independent and identically ditributed Bernoulli random variable, which are alo independent from the arrival procee. Finally, we let B i (n) repreent the number of packet in the ith queue at the beginning of time lot n. The atellite ha K tranmitter. Each tranmitter can only erve one queue, and can only be aigned to a queue whoe channel i on. At each lot, each tranmitter can tranmit up to C 0 packet, and the total number of tranmitted packet cannot exceed C. The parameter C 0 correpond to the power limitation on an individual tranmitter and C correpond to the power limitation on the entire atellite. Such queueing ytem, with multiple queue and tochatically varying ervice rate, have been tudied in [2] where the author how that, when K = 1(ingle tranmitter), apolicy called the Longet Connected Queue (LCQ) maximize the tability region of the ytem, and alo reult in optimal average queue lenght. Furthermore, [3] how that the LCQ policy reult in a maximal tability region under more general aumption on the arrival and channel tate procee. The model conidered in the preent work extend thi previou work in variou direction. 2 The cae of multiple erver and C 0 =1 We firt conider the cae where C = K and C 0 =1. That i, there are K erver, and each erver can erve at mot one packet during each time lot. Thi i a generalization of the ytem tudied in [2], which dealt with the cae where K =1.Wenow decribe apolicy that will be hown to be optimal in the current cenario, a well a in the cae where there i a ingle erver that can erve upto C 0 packet (ee Section 3). The Longet Connected Queue (LCQ) policy operate a follow. Conider the et of queue i that are connected, i.e., G i (n) =1. Out of that et, elect up to K queue with the larget value of B i (n), and erve min{b i (n),c 0 } packet from each one of them. Note that fewer than K queue will be elected if and only if the number of nonempty connected queue i maller than K. oreover, in the event that there are multiple queue with equal value of B i (n) competing for a tranmitter, the policy can chooe between them arbitrarily. Our main reult tate that the LCQ policy i optimal in the tochatic ordering ene [1]. The intuitive reaon i that thi policy trie to keep the queued packet pread over multiple queue and ha a better chance of avoiding idling when ome channel are off. Theorem 1 Aume that C 0 =1and C = K. Let Q(n) = N i=1 B i (n) be the total number of packet, tarting from a given initial tate, under ome policy π. Let Q L (n) be the total number of packet when the LCQ policy i ued. Then, {Q L (n)} {Q(n)}, t i.e., the proce {Q L (n)} tochatically dominate the proce {Q(n)}. A reult of thi type wa proved in [2], for the cae K =1,uing a coupling argument. Our proof for the cae of general K alo ue a coupling argument. For brevity, we only provide an outline here and refer the reader to [9] for complete proof.

3 Let u note that optimality in the tochatic ordering ene i a rather trong notion. It implie the optimality of the LCQ policy over either a finite or an infinite horizon, involving one-tep cot of the form E[f(Q(n))], where f i any nondecreaing function. The ame comment applie to Theorem 2 later in thi paper. Let Z N + be the et of nonnegative integer. We ue Π d : R N R N to denote the function that ort it argument in decreaing order. We alo let (N) = {δ { 1, 0, 1} N δ i > 0 [δ j 0, j <i]} be the et of vector whoe component are either minu one, zero, or one, with all the one appearing before all the negative one. For example, [0, 1, 0, 1, 1, 0, 1, 0, 1] belong to (9). Finally, we ue Z(N) = {b Z N + : b 1 b 2 b N } to denote the et of nonnegative vector whoe component are nonincreaing. For example, [9, 2, 2, 1] belong to Z(4). Proof Outline: We conider a ytem (to be referred to a ytem 0 ) that tart at an initial tate (b(0),g(0)) and i controlled by policy π. Wewill contruct a policy π 1 and a correponding ytem 1 on the ame probability pace, by appropriately coupling the arrival, ervice, and channel tate procee, o that π 1 act imilar to LCQ at n =0,and o that Q 1 (n) Q(n) atall time n, for every ample path. We will ue a upercript 1 to denote variou quantitie, uch a total queue length or channel tate, under policy π 1 in ytem 1. Both ytem tart with the ame initial tate (b(0),g(0)). If π behave like LCQ at time 0, we let π 1 be equal to π, and we couple the two ytem o that they evolve in an identical fahion. In thi cae, the inequality Q 1 (n) Q(n) hold with equality for all n. Suppoe now that at time 0, π behave differently from LCQ. We contruct ytem 1 and policy π 1 a follow. At time 0, we have the ame channel tate g(0) in both ytem. We apply an LCQ policy to ytem 1. After ubtracting the packet withdrawal at time 0, we arrange the queue length of ytem 0 and ytem 1 in decreaing order We then couple any arrival to the ith longet queue of ytem 0 to an arrival to the ith longet queue of ytem 1. Let β(n) be the vector of queue length in ytem 0, arranged in decreaing order, o that β 1 (n) β 2 (n) β N (n). Similarly define β 1 (n) for ytem 1. Let δ(n) = β(n) β 1 (n) The following lemma (whoe proof we omit) how that δ(1) (N), a reult we will ue in decribing the coupling procedure for n 1. The firt part of the lemma how that if we firt ubtract the packet withdrawal under either policy, and then ort the reulting queue length vector, the difference belong to the et (N). The econd part etablihe that after coupling the arrival at time 0, a decribed earlier, the vector δ(1) belong to (N). Lemma 1 (a) Let b Z(N) and g {0, 1} N. Let r be afeaible packet withdrawal vector, and let r be the LCQ control, in tate (b, g). Then Π d (b r) Π d (b r) (N). (b) Let e {0, 1} N,b, b Z(N),b b (N). Then Π d (b + e) Π d ( b + e) (N). We now decribe the coupling and the policy π 1 at poitive time n. Cae I: Suppoe that there exit ome i uch that δ i (1) = 1 and β 1 i (1) = 0. Since δ(1) (N) (Lemma 1), we mut have δ(1) 0, i.e., the jth longet queue in ytem 0

4 ha no fewer packet than the jth longet queue in ytem 1, for every j. Wethen let the channel tate, control, and arrival for the jth longet queue in ytem 1 be the ame a for the jth longet queue in ytem 1, at all poitive time n. Weobtain Q 1 (1) Q(1), and the coupling alo preerve thi inequality for all time n. Cae II: Suppoe now that for every i atifying β 1 i (1) = 0, we mut alo have δ i (1) = 0, which implie β i (1) = 0. We map the channel tate of the ith longet queue in ytem 0 to the the ith longet queue in ytem 1. If π erve a packet from the ith longet queue in ytem 0, we have β i (1) > 0, which implie that β 1 i (1) > 0, and we let π 1 erve a packet from the ith longet queue in ytem 1. To couple the arrival, we again ubtract the packet withdrawal, arrange the queue length in ytem 0 and 1, repectively, in decreaing order, and couple any arrival to the ith longet queue of ytem 0 to an arrival to the ith longet queue of ytem 1. The next lemma (whoe proof i omitted) tate that δ(2) (N), which we can then ue to repeat the coupling at time 1, for time n =2, 3,... Lemma 2 Let e {0, 1} N,b, b Z(N),b b (N). Then Π d (b e) Π d ( b e) (N). For n =2, 3,..., either the coupling ha already been determined (if at ome prior time the condition for Cae I wa atified), or ele we repeat the above coupling procedure and ue the fact δ(n 1) (N) δ(n) (N) (Lemma 1 and 2). We now how that Q 1 (n) Q(n) for all n. Thi tatement clearly hold when n =0. We have alo etablihed it for Cae I. For Cae II, we have already noted that we can alway withdraw a many packet in ytem 1 a in ytem 0, which implie our aertion for all n. We have etablihed that for every policy π, there exit a policy π 1 that agree with the LCQ policy at time 0, and uch that Q 1 (n) Q(n), for all n. By repeating the contruction, we obtain policie π 1,π 2,... uch that the policy π k agree with LCQ until time k 1, and uch that Q k (n) Q k 1 (n 1) Q 1 (n) Q(n), for all n. Thi tatement, in conjunction with a tandard characterization of tochatic dominance [1] complete the proof. 3 The ingle erver cae We now conider a variant of the problem in which there i a ingle erver. Thi i another extenion of the problem tudied in [2], where the erver could only erve one packet per lot. In our generalization, the erver can erve up to C 0 packet during the ame lot, but they mut all be taken from the ame queue. Without lo of generality, we aume that if a queue with B i (n) packet i erved, then the number of packet withdrawn i equal to min{b i (n),c 0 }. Wehave the following reult, which i again proved uing a coupling argument. Theorem 2 Aume that K =1.LetQ(n) = N i=1 B i (n) be the total number of packet, tarting from a given initial tate, under ome policy π. Let Q L (n) be the total number of packet when the LCQ policy i ued. Then, {Q L (n)} t {Q(n)}.

5 For the purpoe of the proof of thi theorem, we expand the et of policie to include policie that may add extra packet to queue. We let Φ a be the expanded et of policie. We let Φ na ( no addition ) be the et of all policie that never add extra packet. We ay that a policy ha the LCQ property at time t if at that time it can only erve packet from a longet connected queue. We let L τ be the et of all policie that have the LCQ property for all time t τ, and which do not add any packet before time τ. We will ue a(n) to denote arrival, g(n) to denote channel tate, b(n) to denote queue ize, and h i (n) todenote packet added to queue i, attime n. Proof Outline: The core of the proof conit of the following two tep. 0. We tart with a policy π in L τ 1 Φ na acting on a queueing ytem with initial tate (b(0),g(0)). We call thi ytem 0. Such a policy ha the LCQ property until time τ 1, and never introduce additional packet. 1. We contruct a new policy π in L τ (i.e., π ha the LCQ property until time τ, and doe not introduce additional packet before time τ), and a correponding ytem 1 on the ame probability pace, o that π dominate π, inthe ene that Q 1 (n) Q(n) for all n. (We ue a upercript of 1 to denote variou quantite in ytem 1 and no upercript to denote related quantitie in ytem 0.) We now decribe the contruction of π. Before time τ, welet π be identical to π, and let arrival and channel tate be the ame for both ytem. In particular, b(τ) =b 1 (τ). At time τ, welet the tate of each channel be the ame for both ytem. Step 1.1. Policy π at time τ. If π erve a longet connected queue at time τ, we let π do the ame, and et a 1 (τ) =a(τ), h 1 (τ) =0. In thi cae, we have b 1 (τ +1)=b(τ + 1). Suppoe now that π chooe to erve at time τ ome queue which i not a longet connected queue. Without lo of generality, let u aume that π erve queue 2 and that queue 1 i a longet connected queue. Thu, b(τ) iofthe form b(τ) = for ome Z+ N 2 and >m.weditinguih three cae. (i) Suppoe that C 0 m<. Policy π remove C 0 packet from queue 2. We let policy π remove C 0 packet from queue 1. We alo let h 1 (τ) =0anda 1 (τ) =a(τ). The reulting configuration are of the form b(τ +1)= m C 0 m, + a(τ), b 1 (τ +1)= C 0 m + a(τ). (ii) Suppoe that m< C 0.Inthi cae, π drive m down to zero. We let π erve queue 1, alo driving it down to zero. Furthermore, π add packet to queue 2 to drive it up to, i.e., h 1 2(τ) = m. Wealo let a 2 (τ), if i =1, a 1 i (τ) = a 1 (τ), if i =2, a i (τ), otherwie.

6 The reulting configuration are of the form b(τ +1)= 0 + a(τ), b 1 (τ +1)= Thu, b 1 (τ +1)and b(τ +1)are permutation of each other. 0 + a 1 (τ). (iii) Suppoe finally that m<c 0 <.Inthi cae, π drive m down to zero. We let π erve queue 1 and remove C 0 packet. We alo let π add C 0 m packet to queue 2, i.e., h 1 2(τ) =C 0 m, driving it up to C 0. The reulting configuration are of the form b(τ +1)= 0 + a(τ), b 1 (τ +1)= C 0 C 0 + a(τ). Thi complete the decription of policy π at time τ. Wewill now contruct π, for time t>τ,o that at any time one of the following three relationhip hold: (i) b(t) =b 1 (t). (ii) b(t) and b 1 (t) differ only in their firt two component, and b 1 (t) =b 1 2(t), b 2 (t) =b 1 1(t). (iii) b(t) and b 1 (t) differ only in their firt two component, and there exit a poitive integer k uch that for either (i, j) =(1, 2) or for (i, j) =(2, 1), we have b j (t) min{b 1 1(t),b 1 2(t)} max{b 1 1(t),b 1 2(t)} b i (t), b 1 i (t) =b i (t) kc 0, b 1 j(t) =b j (t)+kc 0, We will ue b 1 (t) b(t) to indicate that b(t) and b 1 (t) are related in one of the above three way. Note that our contruction of π at time τ guarantee that b 1 (τ +1) b(τ + 1). Step 1.2. Policy π at time t>τ. We now contruct the policy π for time t > τ. We proceed recurively. For time τ +1, thi i already accomplihed. Suppoe that π ha been defined up to ome time t 1 and that b 1 (t) b(t). We conider three cae, which correpond to the three cae in the definition of the relation. Cae (i): If b(t) =b 1 (t), we let the channel tate, arrival, and control be the ame for both ytem, which enure that b 1 (t +1)=b(t +1)and b 1 (t +1) b(t +1). Cae (ii): Suppoe that b 1 (t) iobtained from b(t) bypermuting the firt two component. For queue i / {1, 2} we let the channel tate, arrival, and control be the ame for both ytem. For queue 1 and 2, we let channel tate, arrival, and control for queue 1 in ytem 0 be the ame a for queue 2 in ytem 1, and vice vera. Then, the lat N 2 component of b 1 (t +1)and b(t +1)are equal, wherea the firt two remain permutation of each other. In particular, b 1 (t +1) b(t +1). Cae (iii): We finally conider the remaining cae (iii) in the definition of. Without lo of generality, we aume that the firt component of b(t) iatleat a large a the

7 econd component. In particular, for ome m and, with m<, for ome poitive integer k, and for ome Z N 2 +,wehave b(t) = m, b 1 (t) = kc 0 m + kc 0 The ret of the argument will be different, depending on whether we have m + kc 0 kc 0 ( Type I ) or m + kc 0 > kc 0 ( Type II ). Type I: Suppoe that m + kc 0 kc 0.Wecouple the channel tate and arrival by letting g 1 (t) =g(t) and a 1 (t) =a(t). (a) If π erve queue 1, bringing it down to C 0,policy π alo remove C 0 packet from queue 1. (Thi i poible becaue kc 0 m + kc 0 C 0. The reulting configuration i b(t +1)= C 0 m + a(t), b 1 (t +1)=. (k +1)C 0 m + kc 0 + a(t), and we have b 1 (t+1) b(t+1). (b) If π remove u packet from queue 2 (note that either u = C 0 if m C 0,oru = m otherwie), then π remove the ame number of packet from queue 2. (Thi i done by removing C 0 packet and then adding h 1 2(t) =C 0 u packet.) The reulting configuration i b(t +1)= m u + a(t), b 1 (t +1)= kc 0 m u + kc 0 + a(t). (c) Finally, if π erve ome queue j>2, we let π do the ame. In all ubcae, we have b 1 (t +1) b(t + 1). Type II: Suppoe now that m + kc 0 > kc 0.Welet g 1 1(t) =g 2 (t), g 1 2(t) =g 1 (t), and g 1 j (t) =g j (t) for j>2. That i, we couple the channel tate for queue 1 under policy π to that for queue 2 under policy π, and vice vera. For all other queue, channel tate coincide under the two policie. (a) Suppoe that π erve ome queue j>2. Then, π remove the ame number of packet from the ame queue, which i poible becaue b 1 j(t) =b j (t) and g 1 j (t) =g j (t). We then let a 1 (t) =a(t). (b) Suppoe that π erve queue 1 (in particular, g 1 (t) =1), bringing it down to C 0.Wethen let π erve queue 2, bringing it down to m +(k 1)C 0. Thi i poible becaue g 1 2(t) =g 1 (t) =1and b 1 2(t) =m + kc 0 C 0. We then let a 1 (t) =a(t). The reulting configuration i b(t +1)= C 0 m + a(t), b 1 (t +1)= kc 0 m +(k 1)C 0 Notice that m m +(k 1)C 0,becaue k i poitive. In particular, + a(t). m min{ kc 0,m+(k 1)C 0 } max{ kc 0,m+(k 1)C 0 } C 0. We then ee that b 1 (t +1) b(t +1). (If k =1,wehave cae (i) in the definition of ; if k>1, we have cae (iii).)

8 (c) Suppoe, finally, that π erve queue 2 and remove u = min{m, C 0 } packet, bringing it down to m u. In particular, g1(t) 1 =g 2 (t) =1. If kc 0 C 0, then, π remove C 0 packet from queue 1, bringing it to (k +1)C 0, and add h 1 2(t) =C 0 u packet to queue 2, bringing it to m +(k +1)C 0 u. We then let a 1 (t) =a(t). The reulting configuration i b(t +1)= m u + a(t), b 1 (t +1)= (k +1)C 0 m u +(k +1)C 0 + a(t). We then have b 1 (t +1) b(t +1) (cae (iii) in the definition of ). It remain to conider the cae where kc 0 <C 0.Inthat cae, we have m kc 0 C 0,othat π drive queue 2 down to zero. Policy π, drive queue 1 from kc 0 down to zero, and it add enough packet to queue 2 to drive it up to. We then let a 1 1(t) =a 2 (t),a 1 2(t) =a 1 (t) and a 1 j(t) =a j (t), for j>2. The reulting configuration i b(t+1) = 0 + a 1 (t) a 2 (t) [a 3 (t),...,a N (t)] T, b 1 (t+1) = 0 + a 2 (t) a 1 (t) [a 3 (t),...,a N (t)] T and we have b 1 (t +1) b(t +1)(cae (ii) in the definition of ). At thi point, we have completed the recurive contruction of π. The new policy π dominate π, ha the LCQ property until time τ, and doe not add any packet before time τ. The remainder of the proof of Theorem 2, which we omit, proceed a follow. We modify the policy π to obtain yet another policy ˆπ in L τ Φ na (in particular, ˆπ never add extra packet), and which dominate π. We then repeat the preceding tep for τ =0, 1,...,toobtain a policy with the LCQ property at all time and with no added packet, and which dominate the policy that we tarted with. 4 The cae of one erver per queue (K = N) In thi ection, we aume that packet can be erved from all connected queue imultaneouly, ubject to a bound C on the total number of packet that can be erved during one time lot. If we further aume that C =1,weare back to the model in [2], and the LCQ policy i optimal. The cae of general C i an open problem. However, if we lightly modify the model to allow erving a noninteger number of packet from each queue, we can how that a generalization of LCQ, which we call the ot Balanced policy i optimal. Suppoe that at time n we have (B(n),G(n)) = (b, g), for ome vector b and g. Let U(b, g) denote the et of feaible vector of packet withdrawal, when the ytem i in tate (b, g), i.e, { U(b, g) = u R N + if g i =0then u i =0; } N u i C. i=1 The ot Balanced policy chooe a u U(b, g) othat i u i i a large a poible, and in addition, o that it minimize max (b i u i ). i:g i =1,

9 For example, if b =[5, 4, 3, 2, 6, 1], g =[1, 1, 1, 1, 0, 0], and C =2,amot balanced policy will let u =[1.5, 0.5, 0, 0, 0, 0], reulting in the configuration b u =[3.5, 3.5, 3, 2, 6, 1]. It i not hard to how that the mot balanced policy i uniquely defined. Finally, we let F be the et of all function f : R N + Rthat are convex, nondecreaing, and ymmetric (permutation invariant). Theorem 3 Let B(n) be the vector of queue ize at time n. For any function f F and for every n 0, the ot Balanced policy minimize E[f(B(n))]. Proof Outline: The proof ue a dynamic programming argument. Let Vn (b, g) bethe leat poible value of E[f(B(n))], tarting from the initial tate (b, g) attime zero. We then have V0 (b, g) =f(b), and V n (b, g) = min w U(b,g) a p A (a)p G ( g)v n 1(b + a w, g), g for every poitive n. Here, p G ( g) ithe probability that the next vector of channel tate i g, and p A (a) ithe probability that the vector of queue arrival i a. We ue the above equation and induction to how that the function Vn belong to F, for all n. The convexity and ymmetry of V (n), together with the form of the above dynamic programming equation, can be ued to how that more balanced configuration are alway preferable. Uing Theorem 3, it i eaily een that the mot balanced policy i optimal for a wide variety of performance criteria, uch a a dicounted um of the E[f(Q(n))] over a finite or infinite horizon, or an undicounted um over a finite horizon. Furthermore, the theorem cover the important pecial cae of f(b) =b b N (total queue length). 5 Dicuion Let u remark that Theorem 1 and 2 remain valid for certain generalization of the model. For example, alway auming that the arrival A i (n) are independent and identically ditributed, the Bernoulli aumption can be relaxed. Intead, it uffice to aume that the random variable A i (n) can be expreed a a um of independent Bernoulli random variable, with poibly different parameter. The pecial cae of equal parameter allow the A i (n) tohave abinomial ditribution. For another generalization, we can relax the aumption that the ditribution of the A i (n) ithe ame for different time n. For example, we can aume that it i Bernoulli with a parameter which i itelf random and independent at different time (a determinitically changing parameter i a pecial cae). In the ame pirit, the probability that a channel i on can alo vary with time. It turn out that the reult alo remain valid in certain ituation where the arrival at different queue (repectively, the channel tate) at a given time are dependent, a long a the dependence i ymmetric, in the ene that it lead to permutation-invariant ditribution. Reference [1] D. Stoyan, Comparion ethod for Queue and other Stochatic odel, John Wiley, 1983.

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