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M ULTICAST R ECIPIENT M AXIMIZATION 802 16 W I MAX R ELAY P ROBLEM IN 802.16 N ETWORKS 郭文興, 32 years old 郭文興, 32 years old Bachelor, National Taiwan University (1998~2002) PhD in EE, National Taiwan University (2002~2008) Advisor: Wanjiun Liao Assistant Professor, Department of EE, Yuan Ze University (2008~) O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Numerical Results Another Problem BMR Another Problem: BMR Proposed Solution Numerical Results Conclusions O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Simulation Results Another Problem BMR Another Problem: BMR Proposed Solution Simulation Results Conclusions 802 16 W I MAX 802.16 IEEE 802.16 (WiMAX) is an emerging last mile IEEE 802.16 (WiMAX) is an emerging last mile technology for broadband wireless access. Compared with 3G cellular networks and IEEE Compared with 3G cellular networks and IEEE 802.11 Wi‐Fi, it has better coverage and throughput. A typical IEEE 802.16 WiMAX network comprises a base station (BS) and multiple subscriber stations (SSs) 802 16 W I MAX 802.16 IEEE 802.16j standard introduces a new type of IEEE 802.16j standard introduces a new type of infrastructure nodes called relay stations (RSs). Specifically, RSs forward data, both upward and Specifically RSs forward data both upward and downward, over the wireless medium between the BS and the SSs to increase the system capacity and throughput. 802 16 W I MAX 802.16 802 16 W I MAX 802.16 M ULTICAST R ECIPIENT M AXIMIZATION P ROBLEM WiMAX technology is an efficient platform for multicast services. Wireless resource is relatively scarce; hence, the resource available for each wireless service is inevitably limited. With the given budget, a BS should serve as many recipients (i e SSs) recipients (i.e., SSs). When a network only consists of a BS and SSs ‐> tuning the allocated resource of the BS. Relay Networks ‐>The problem becomes much difficult because resource must be allocated between the BS and the RSs. O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Simulation Results Another Problem BMR Another Problem: BMR Proposed Solution Simulation Results Conclusions R ELATED W ORKS In [5], a reliable multicast scheme uses Code In [5], a reliable multicast scheme uses Code Division Multiple Access (CDMA) codes in WiMAX networks [6] proposes a two‐level superposition coded multicast scheme (2‐level SCM) to improve the channel efficiency. [11] considers, two kinds of nodes, (i.e., relay nodes and receiving nodes, ) but with fixed allocated energy. S IMILAR P ROBLEM : M INIMUM -E NERGY M ULTICAST (MEM) Given the channel condition of each node, the Given the channel condition of each node, the MEM problem finds the multicast tree that minimizes the total energy required to deliver the stream to a given set of subscribers. h i f b ib S IMILAR P ROBLEM : M INIMUM -E NERGY M ULTICAST (MEM) Although MEM appears to be similar to our Although MEM appears to be similar to our Multicast Recipient Maximization (MRM) scheme, they are substantially different: ~ MEM assumes that each intermediate node can forward data. ~ MEM minimizes the resource required to serve subscribers. ~ Each node is assumed to separately receive a copy from its sender in MEM. O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Simulation Results Another Problem BMR Another Problem: BMR Proposed Solution Simulation Results Conclusions N OTATIONS D EFINITION OF MRM The routing of each SS is assumed to be decided beforehand It is impractical that the whole multicast tree can be whole multicast tree can be dynamically formed and adjusted as any recipient changes its channel condition. s 0 ,1 sM,NM s0,2 s0, N0 s1 ,1 s0,3 s1,2 s1N , 1 s2,1 s 2,2 s2,N2 sM,2 sM ,1 D EFINITION OF MRM Given the channel quality , RS Given the channel quality R = [R 0 , R1 , R 2 ,,...,, R M ] , RS r I = [i0 , i1 ,..., iM ] positions , and budget , find a Γ = [ n , n ,..., n ] resource allocation that maximizes the total number of served SSs h l b f d SS ∑ D (n ) ⋅ (U (n ) − U (i )) (i.e., ), subject to the total M consumed resource m∑=0 (r(m, n ) − r(m,0) ) ≤ rbudget g budget 0 1 M M m =0 m 0 m m 0 m m im ≤ n ⎧1 Dm ( n ) = ⎨ ⎩0 otherwise m=0 ⎧⎪n − max m′ U m (n) = ⎨ im′ ≤n ⎪⎩n + U 0 (im ) otherwise D IFFICULTY OF MRM MRM is NP‐complete MRM is NP complete ~ It is NP‐hard because it can be reduced from a famous problem called integral knapsack. famous problem called integral knapsack. ~ It is NP because the performance of a single instance can be verified in polynomial time. p y A UXILIARY G RAPH For each node, the total distance (i.e. the sum of , ( weight of edges) to the root is equal to the number of resource it requires. MRM involves finding a spanning tree rooted at BS i l fi di i d S that maximizes the number of SSs covered, subject to the constraint that the total length of the tree is not greater than the resource budget. t th th b d t BS ( RS0 ) Δr0 (1) ss Δr0(2) r( 0,1) r(0, 2) r( 0 , 3 ) r(1 ,1 ) r(1, 2 ) ss RS2 RS 1 Δr0 (3) Δr0 (4) Δr1 (1) ss RSM ss ss ss ss ss Δr1 (2) ss ss ss1N1 ss A UXILIARY G RAPH 2 1 0.4 0. 1 u m (n) = U m ( n ) − U m ( n − 1) Δ rm ( n ) E NVELOPE F UNCTION * ′ U ( n ) − U * m m ( n − 1) u m ( n) = max S ( m′.n′ ) ∈ST(m, (m n) r( m ′, n ′) − r( m , n −1) * * * Um (n) = U m ( n − 1) + u m ( n ) Δrm ( n ) E NVELOPE T REE * um ( n) = * U m (n′) − U m (n − 1) S ( m′.n′ ) ∈ST(m,n) r( m ′, n ′) − r( m, n −1) max * * * Um (n) = U m (n − 1) + u m ( n)Δrm (n) If nodes are allocated based on the envelope tree instead of g , g the original tree, the budget can be more utilized. P ROPOSED DSS S CHEME P ROPOSED DSS S CHEME P ERFORMANCE A NALYSIS DSS has polynomial‐time DSS has polynomial time complexity. complexity. * ∀ m, n, . Um (n) ≥ U m (n) ∀ S ( m ′, n ′ ) ∈ ST ( m , n ) u m* ( n ) ≥ u m* ′ ( n ′) , . (i.e., nodes farther from U m* ( n) the BS have smaller . The envelope tree if concave) concave) P ERFORMANCE A NALYSIS Γ op _ R : the optimal solution to the envelop tree : the optimal solution to the envelop tree Γ op : the optimal solution to the original tree Γ DSS = Γ op _ R P MRM ( Γ op ) − P MRM ( Γ ≤ P R − MRM ( Γ = P R − MRM ( Γ = max {U ∀ m ,n * m op _ r DSS DSS ) ) − P MRM ( Γ ) − P MRM ( Γ (n ) − U m ( n )} ) DSS DSS ) O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Simulation Results Another Problem BMR Another Problem: BMR Proposed Solution Simulation Results Conclusions S IMULATION S ETTINGS 1 The required resource is represented as d . The required resource is represented as . a ~ d is the distance between the sender and receiver ~ a is the channel attenuation factor that is usually between 2 and 4. S IMULATION I In the first simulation, M In the first simulation, M and N and N are fixed at 5 and are fixed at 5 and 100 respectively. Since the network topologies are generated Since the network topologies are generated randomly, we conduct 100 runs for each setting. S IMULATION I S IMULTAION I The trends of both figures are very similar, The trends of both figures are very similar, meaning that DSS provides same satisfying performance under different radio conditions. Given different budgets, DSS always outperforms GD. S IMULATION II rbudget = 20000 20000 a=2 We tune the value of N from 100 to 1000 to observe the impact of recipient number on the performance. performance Again, each point on the curves represents the average of 100 simulation runs. f 100 i l ti S IMULATION II S IMULATION II The number of served recipients is linearly The number of served recipients is linearly proportional to the node number. This is because more SSs result in higher density. This is because more SSs result in higher density Given the same amount of allocated resource, more SSs would be served. As N increases, the performance of DSS always stays close to that of OP, while the difference between OP and GD increases. O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Simulation Results Another Problem BMR Another Problem: BMR Proposed Solution Simulation Results Conclusions I NTRODUCTION – A N E XAMPLE The number of each link in (a) refers to the required resource. As shown in (b), the conventional approach conventional approach looks for the route that consumes the least resources for each SS independently. I NTRODUCTION – A N E XAMPLE However, as shown in (c), the resource requirement can be requirement can be further reduced to 5 if the two SSs receive the stream from the h f h BS directly. S YSTEM M ODEL AND P ROBLEM S PECIFICATION RS0 : the BS RSy , : the yth RS 1≤ y ≤ Y SS n , : the nth SS 1≤ n ≤ N Y, N: the number of RS and SS respectively rn , y : the required resource for multicast from y to n ry : the required resource of RSy ra y : the resource allocated to RS y RA = [ra 0 , ra1 ,...raY ] : the resource allocation rtotal : the total resource budget S YSTEM M ODEL AND P ROBLEM S PECIFICATION ry r1 r 2 RS 1 rY RS y RS2 RS Y rn , 0 rn ,1 rn, 2 SS1 SS2 rn , y rn,Y SS n rN ,Y SSN S YSTEM M ODEL AND P ROBLEM S PECIFICATION Find that maximizes RA = [ra 0 , ra1 ,...raY ] N Y n =1 y =0 ∑ min((1, ∑ U (ra y − rn, y )U (ra 0 − ry )) Subject to Y ~ ∑ ra y ≤ rtotal y =0 ~ 0 ≤ ra y ≤ rtotal O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Simulation Results Another Problem BMR Another Problem: BMR Proposed Solution Simulation Results Conclusions for n=1 to N, rn,0 is put in a queue q; q is sorted in decreasing order of r ; ra0DRA ← rtotal ; for y=1 to Y, ra yDRA ← 0 ; RAmax ← RADRA ; do q′ ← q ; DRA are popped from q′ ; all elements larger than ra0 do for y=1 to Y, RAtemp ← RADRA ; Δr ← r0temp − q′[0] ; r0temp ← q ′[0] ; r ytemp ← r yanswer + Δ r ; temp max if P ( RA ) > P ( RA ) then RAmax ← RAtemp ; end if; end for; q ′[0] is popped from the queue; ratemp loop until ( q′[0] < 1≤ y≤Ymax ) or ( q′ is empty); y , ra >0 DRA ← RAmax ; if RA max ≠ RA DRA then RA else l exit it do; d end if; loop; return RADRA ; n, 0 temp y P ERFORMANCE E VALUATION – S IMULATION S ETTING The coordinate of the BS is The coordinate of the BS is set as (0,0). In each run, the positions of , p all SSs are randomly distributed in a circular area whose center and radius are whose center and radius are (0,0) and 100 respectively. All the RSs are placed All the RSs are placed randomly in circular region whose radius ranges from 40 to 60 such that each RS is 40 to 60 such that each RS is located between the BS and some SSs Range of SSs Range of RSs 0 10 60 40 RS (0,0) O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Simulation Results Another Problem BMR Another Problem: BMR Proposed Solution Simulation Results Conclusions S IMULATION I In the first simulation, we compare the In the first simulation, we compare the performance of the allocations of the three schemes under different resource budgets. The value of N and Y are fixed at 100 and 5 respectively, while the value of is tuned in the range 0 to 1,000,000. Each point is averaged over 100 random node placements. S IMULATION I Available Resouce v.s. Served SSs (Y=5, N=100, a=2) Number of Serrved SSss 100 80 60 ra^BS 40 ra^DRA 20 ra^fixed 0 0 2000 4000 6000 r_total 8000 10000 S IMULATION I More SSs are served as the budget increases More SSs are served as the budget increases ra^DRA achieves a better performance than ra BS and ra fixed, indicating that the ra^BS and ra^fixed indicating that the participation of RSs allows more SSs to be served, and thus improves resource utilization. Although RSs are also used in rs^fixed, it does not yield an obvious performance gain over rs^BS. S IMULATION II rtotal Next, we fix the value of N, and at 100 and Next, we fix the value of N, and at 100 and 500,000 respectively, and tune the number of RSs (i.e., Y) from 0 to 5 to observe the impact of RS density on performance. RS d i f Again, each point is averaged over 100 runs. S IMULATION II v s Served SSs Number of RSs v.s. (N=100, a=3, r_total=500000) Numberr of Serveed SSs 100 80 60 ra^BS 40 ra^DRA 20 ra^fixed ra fixed 0 0 1 2 Y 3 4 5 S IMULATION II Given the same budget, the performance of Given the same budget, the performance of ra^DRA improves as Y grows, while that of ra^BS remains unchanged because RSs are not used. Although RSs also relay the stream for SSs in ra^fixed, a significant increasing trend and performance gain over are not observed because finding the minimal route independently does not achieve high resource utilization in not achieve high resource utilization in multicast.. O UTLINE Multicast Recipient Maximization Problem Related Works and Similar Problems Proposed Solution Proposed Solution Simulation Results Another Problem BMR Another Problem: BMR Proposed Solution Simulation Results Conclusions C ONCLUSION We study the multicast maximization problem in y p 802.16j WiMAX relay networks. MEM aims to maximize the number of recipients in a fixed network topology fi d k l We prove that the maximization has the performance very approximate to the optimum very approximate to the optimum MRM aims to maximize the number of recipients in a dynamic network topology dynamic network topology The simulation results show that the performance is satisfying R EFERENCE Wen Hsing Kuo and Jeng Wen‐Hsing Jeng‐Farn Farn Lee, Lee, “Multicast Multicast Recipient Maximization in IEEE 802.16j WiMAX Relay Networks,” IEEE Transactions on Vehicular Technology, vol. 1, pp.335‐343, Jan. 2010. T h l l 1 335 343 J 2010 Wen‐Hsing Kuo and Jeng‐Farn Lee, "Multicast Routing Scheme for Recipient Maximization in Wireless Relay Networks," IEEE Transactions on Vehicular Technology, Vol. 8, Issue 19, pp.4002‐ Vehicular Technology, Vol. 8, Issue 19, pp.4002 4011, Oct. 2010.