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# A Hexagonal-Tree TDMA-Based QoS Multicasting Protocol for Wireless Mobile Ad-Hoc Networks

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(3) 3. 12345678 B. A. B. D. A. F. 12345678. C. D. F. 12345678. C. E. E. 12345678. ( a) hidden-terminal problem. ( b) exposed-terminal problem. Fig. 1. Example of bandwidth calculation by the hidden-terminal and exposed-terminal problems. Y. X. B`. B. Y`. X`. Fig. 2. Example of a hexgonal-block. represent as a path from node h1 to node hk . A node B is said as a branch node if there exist more than one. disjoint paths from a three-hop neighboring node B , so that nodes B and B are called as a pair of branch 0. 0. nodes. Some terms are defined as follows.. Definition 1 Hexagonal-Block: Given a pair of two branch nodes B and B , there are two disjoint three0. hop paths (B. X Y B ) and (B X  Y  B ) between these two branch nodes, where X and X are not not 1-hop neighbor. Paths (B X Y B ) and (B X  Y  B ) forms a 1-hop neighbors andY and Y are  X Y hexagonal-block, let B as a hexagonal-block between B and B . X Y B denote   X Y For instance, a hexagonal-block B X Y B is shown in Fig. 2(a), there are two disjoint three-hop paths (B X Y B ) and (B X  Y  B ) between a pair of branch nodes B and B : In general, nodes X and X may be neighboring, and nodes Y and Y may be neighboring. The hexagonal-block gives a less stronger condition that nodes X and X are not 1-hop neighbors and nodes Y and Y are not 1-hop neighbors, as 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. illustrated in Fig. 2(a). This condition is important for the time-slot reservation scheme in order to exploit the time-slot re-use capability. Using the hexagonal-block allows us to equally split the original data packet into two sub-packets, such that.

(4) 4. B. D. A A. G. B C. E. (b). (a) F. B. H. J. D. A. G. C. I. E. (c) Fig. 3. Example of (a) an uni-path, (b) a hexagonal-block and (c) a hexagonal-twin. one sub-packet goes through (B X Y B ) and another sub-packet goes through (B X 0. 0.  Y  B ). Continually, 0. 0. the hexagonal-twin is now formally defined as follows. This hexagonal-twin is a fundamental component to construct the hexagonal-path and hexagonal-tree structure.. ". Definition 2 Hexagonal-Twin: Given two hexagonal-blocks Z= 0. Z. Z. 2 6 6 6 4. =. A. N1 N3 N2 N4. P1 P2. B P3. #. 1 N3 AN N2 N4 B.  P 1 P2 and Z = N3 B P3 C  let . 0. 3. C. 7 7 7 denote as a hexagonal-twin if a link (N 3 , B) is shared by Z and Z 0 5. For instance, a hexagonal-twin Z. 2. Z. 0. 6 6 4. = 6. A. B D. F H G I. :. 3. J. 7 7 7 is given in Fig. 3(c), there are two 5. C E B D F H J ) and (A C E G I J ) between nodes A and J: Similarly, using the hexagonal-twin, one sub-packet travels along (A B D F H J ) and another sub-packet travels along (A C E G I J ).. disjoint paths (A. Definition 3 Hexagonal-Path: A path is denoted as a hexagonal-path if each sub-path in the path may be an uni-path, a hexagonal-block, or a hexagonal-twin..

(5) 5. (a). (c). (b) Fig. 4. Examples of hexagonal-paths. In the QoS route discovery, if an uni-path exists under satisfying a given QoS requirement, the uni-path will be identified. However, if the QoS uni-path is not existed, then the hexagonal-block and hexagonal-twin are used to offer a multi-path routing scheme, which aim to increase the success rate of identifying the QoS route. For instance as shown in Fig. 3(a), an uni-path with two time slots is constructed. If we cannot find an uni-path with two time slots, then a hexagonal-block with two time slots or a hexagonal-twin with two time slots are possibly identified, as shown in Fig. 3(b) and Fig. 3(c). Further, Fig. 4(a) shows a hexagonal-path with two non-adjacent hexagonal-blocks connecting by an uni-path. Fig. 4(b) illustrates a hexagonal-path constructed by two adjacent hexagonal-blocks. Fig. 4(c) displays a hexagonal-path constructed by a hexagonal-twin. Note that, all cases are instances of the hexagonal-path. We introduce the time-slot re-use capability of a hexagonal-block and hexagonal-twin as follows.. Lemma 1 A time slot t can be used by a host X to send to another host Y without causing collision if the. following conditions are all satisfied.. 1. Slot t is not yet scheduled to send or receive in neither X nor Y:. 2. For any 1-hop neighbor Z of X , slot t is not scheduled to receive in Z: Slot t can be scheduled (or reused). to send in Z but cannot send to any other 1-hop neighbors of X (conflict with condition 1).. 3. For any 1-hop neighbor Z of Y , slot t is not scheduled to send in Z: Slot t can be scheduled (or reused) to.

(6) 6. Fig. 5. Time-slot re-use capability of a hexagonal-twin. receive in Z , but cannot receive from any other 1-hop neighbors of Y (conflict with condition 1). 2 3 6 6 Given a hexagonal-twin 6 4. A. B D C E. G H F I. J. 7 7 7 as illustrated Fig. 5, the time-slot re-use capability 5. on all links of a hexagonal-twin are listed.. 1. Time slot t scheduled in AB can be re-used in CE GH IJ (see Fig. 5(a)). ;!. ;! ;! ; !. 2. Time slot t scheduled in BD can be re-used in AC JH (see Fig. 5(b)). ;!. ;! ;!. 3. Time slot t scheduled in DG can be re-used in FI (see Fig. 5(c)). ;!. ; !. 4. Time slot t scheduled in EF can be re-used in HJ (see Fig. 5(d)). ;!. ;!. To prepare constructing a hexagonal-tree structure, a hexagonal-branch is defined based on the hexagonaltwin structure.. . Definition 4 Hexagonal-Branch: Given three hexagonal-blocks Z=. B P3 D  let Z Z = = N4 P4 P5 Z 2 P1 P2 6 N1 N3 C 6 6 A B P 6 3 6 4 N2 N4 D P4 P5 . Z. 00. 0. 00.   P1 P2 C  and 1 N3 Z = N3 AN B  N N BP 0. 2. 4. 3. 3 7 7 7 7 denote as a hexagonal-branch which is constructed by Z, Z 0 , and Z00 7 5 0. 00. where three links (N 3 , B), (N 4 , B) and (B, P3 ) are shared by Z, Z , and Z .. .

(7) 7. G. B. D. B. J. A. F. C. G. H. E. C. B. L. K. C. G. H. B. J. F. L. J. F. C. M. H. D. A. I. E. K. M. (b). D. A. I. E. (a). G. J. F. M. K. D. A. I. H. I. E. M. K. L. (c). L. (d). Fig. 6. Time-slot re-use capability of a hexagonal-branch. 2 6 6 6 Given a hexagonal-branch 6 6 4. A. B D C E. G H F I K L. 3. J M. 7 7 7 7 as illustrated Fig. 6, the time-slot re-use capa7 5. bility of all links of a hexagonal-branch are listed. 1. Time slot t scheduled in 5(a)).. 2. Time slot t scheduled in. AB can be re-used in CE GH , IJ , LK (see Fig. ;!. ;!. ;!. ; !. ;!. 6(a), which is same as Fig.. AC can be re-used in BD JH , IM KL (see Fig. 6(b), which is same as Fig. ;!. ;!. ;! ;! ;!. 5(b)).. 3. Time slot t scheduled in DG can be re-used in FI EK (see Fig. 6(c), which is same as Fig. 5(c)). ;!. ; ! ; ; !. 4. Time slot t scheduled in HJ can be re-used in FI LM DB EC (see Fig. 6(d), which is different with ;!. Fig. 5(d)).. ; ! ; ; ! ;! ;!.

(8) 8. 2 6. 6 Z 6 As shown in Fig. 7(d), a hexagonal-branch Z = 6 6 Z 4 0. A. 00. B D C E. 00. Z. Z. 00. 6 6 = 6 4. A. B D C E. K L. 7 7 7 7 is given. Observe 7 5. B D. A. C E. 3. F I. M. K2 L. 0. 2. J. F I. 6 Z Z 6 can be viewed as constructed by two hexagonal-twins Z = 6 4 Z 0. that Z. 3. G H. G H F I. 3. J. 7 7 7 and 5. 7 7 7 Observe that, a original packet is split into two sub-packets. From 5. M :. host A, the two sub-packet can send to host J by paths (A B D G H J ) and (A C E F I J ). Similarly,. the two sub-packet can send to host M by paths (A B D F I M ) and (A C E K L M ). This achieves the purpose of delivering the original message to two distinct hosts. This is the fundamental operation of the tree structure.. 2 6. 6 Z 6 As shown in Fig. 7(d), a hexagonal-branch Z = 6 6 Z 4 0. 00. 2 0. viewed as two hexagonal-twins Z. Z. 6 6 4. =6. A. B D C E. A. F H. B D. G I. C E. F H G I. 3. J. K L. 3. J M. 7 7 Z0 7 7 is given. The Z 00 can be 7 Z 5. 2. 7 6 7 6 7 and Z 00 = 6 5 4 Z. A. B D C E. 3. G I K L. M. A, the two sub-packet can send to host J by paths (A B D F H J ) and (A C E G I J ). Similarly, the two sub-packet can send to host M by paths (A B D G I M ) and (A C E K L M ). This achieves the purpose of delivering the original. Observe that, a original packet is split into two sub-packets. From host. message to two distinct hosts. This is the fundamental operation of the tree structure. The purpose of the tree branch is to send the same data packet into two distinct nodes under satisfying the given bandwidth requirement. Each node can continually build a sub-tree to cover all possible destination nodes. All possible cases of tree branch are illustrated in Fig. 7, and a tree branch is occurred between node. A and nodes J M . 7. Initially, Fig. 7(a) shows that node A sends packet to nodes J and M by using three uni-paths AG GJ and GM with two time-slots. If this case is failed, three other cases are then to construct ;! ; !. ; ; !. for greatly improving the success rate of constructing a QoS multicast tree. First, Fig. 7(b) illustrates a   ; ! ; ; ! B D hexagonal-block A G and two uni-paths GJ and GM with two time-slots are used. Fig. 7(c) displays. CE. an uni-path AG and two hexagonal-blocks with two time-slots are used. Finally, Fig. 7(c) shows a hexagonal;!. branch with two time-slots is utilized. With the hexagonal-path and hexagonal-branch, the hexagonal-tree is formally defined as follows. Definition 5 Hexagonal-Tree: A tree is said as a hexagonal-tree if each path of the tree is the hexagonalpath.. A possible hexagonal-tree, from source host S to destinations E and F with uni-paths is illustrated in Fig.. 7 7 7. 5.

(9) 9 B. D J. J A A. F. F C. M. (a). G. (b). G. J. B. H. D. A. F. L. H. J. D. A. I. E. K. M. E. F. C. I. M. E. K. M. (c). L. (d) Fig. 7. Four cases of hexagonal-branchs. 8, note that a hexagonal-branch appears between node B and nodes. C D.. Observe that, the uni-path with. satisfied QoS requirement will be identified with the higher priority during constructing the multicast tree. In the following, the construction of the hexagonal-tree will be presented. III. T HE H EXAGONAL -T REE TDMA-BASED Q O S M ULTICAST ROUTING P ROTOCOL We first give an overview of our proposed protocol, a hexagonal-tree TDMA-Based QoS multicast protocol, in a MANET. The proposed protocol mainly constructs a hexagonal-tree to perform the on-demand QoS multicast routing operation. The designed protocol is achieved by the hexagonal-branch/twin identification and hexagonal-tree construction phases. The hexagonal-branch/twin identification phase identifies the hexagonal-branch/twin in a MANET. The hexagonal-tree construction phase constructs the hexagonal-tree by merging hexagonal-paths from a source to all destinations..

(10) 10. Destination E C. Source S. A. B. D. F. Destination. Fig. 8. Example of a hexagonal-tree. A. Phase 1: Hexagonal-Branch/Twin Identification A.1 Hexagonal-Branch Identification This phase searches for a hexagonal-branch under a given bandwidth requirement. Given a hexagonal2. 6 6 6 branch 6 6 4. A. G H. B D. F I. C E. 3. J M. 7  7 7 7, which is constructed by three hexagonal-block 7 5.    B D G H AC E F , DF I J ,. K L  F I E K L M : Node F contains the following data structure if we attempt to successfully identify a. . and. hexagonal-branch. Observe that each node keeps local information of all one/two/three-hop neighboring nodes.. nodex : Let nodex A B C D E F G H I J K L M and all one- or two-hop neighbors of A B C D E F G H I J K L M . A list of sending activities of node x : Node nodex records on which time slots nodex will have sending activities. If y l1   lk ] denotes nodex send to node y on time slots l1  lk : That is, a list of y l1   lk ] is maintained. A list of receiving activities of node x : Node nodex records on which time slots nodex will have receiving activities. If y l1   lk ] denotes nodex receive from node y on time slots l1   lk  a list of y l1   lk ] 2 f. . g. f. g. . . f. . g. f. . g. . . . . is maintained. Observe that this section only discuss how to successfully identify a hexagonal-branch to exploit the time-. slot re-use capability. Observe that, all records are collected into node F , which indicates that the time slot reservation of paths between. A to J and A to M are determined by node F .. This is the main overhead of. our proposed protocol. From the simulation results in Section IV, the improved success rate of identifying. QoS on-demand multicast tree can successfully cover this extra-overhead. Let Free time slot (  ) denote ; !.  to  Send time slot (  ) denote the time slots have been used to send from  to  and Receive time slot ( ) denote the time slots have been used to receive from  ; !. the time slots can be used to send from. ; !.

(11) 11. to  . Observe that, Send time slot(  ) = Receive time slot(): For instance as illustrated in Fig. 9(a), ; !. ; !. Send time slot(LK ) = f1 2g and Receive time slot(KL) = f1 2g: ;!. ;!. 2. 6 6 6 Given a hexagonal-branch 6 6 4. A. B D C E. G H F I K L. 3. J M. 7 7 7 7 as shown in Fig. 9(a), 7 5. AB. ;!. and. AC. ;!. cannot. share same time slots, since our scheme is a multi-path (two-path). Based on Fig. 6(a), we do know. that Free time slot (AB ) can be same with Free time slot (CE ), Free time slot (GH ), Free time slot (IJ ), ;!. ;!. Free time slot (LK ), but time slots scheduled in. ;!. ; !. AB cannot be used in FD: Similarly, from Fig. 6(b), Free time slot (AC ) can be same with Free time slot (BD ), Free time slot (IM ), Free time slot (KL), Free time slot (JH ), but time slots scheduled in AC cannot be used in FE etc. Let the bandwidth requirement be  , we do the following reservation operations; (1) we try to allocate the  time slots to AB CE , GH , and IJ (see Fig. 6(a)), (2) we then try to allocate the 2 time slots to AC BD 2 IM and KL (see Fig. 6(b)), (3) we then attempt to allocate 2 time slots to HJ FI and LM (see Fig. 6(c)), (4) we then attempt to allocate 2 time slots to DG FI and EK (see Fig. 6(d)), (5) we finally try to allocate  2 time slots to DF and EF: If any one is failed, then the time slot reservation for a hexagonal-branch is failed. Assume that the bandwidth requirement is  , the formal reservation procedure is given herein. A1. Repeatedly reserve 2  same time slots and  distinct time slots on AB CE , GH and IJ , for  = 0 ;!. ;!. ;!. ;!. ;!. ;!. ;! ;! ;!. ;!. ;!. ;!. ;!. ;!. ; !. ;! ;!. ;!. ;! ; !. ;! ; !. ;!. ; ; !. ; ; !. ;!. ;! ;! ;!. ; !. ;. to 2 , until the reservation is successful, and update the possible nodes’ records of sending and receiving. activities. Otherwise, if the reservation is failed, then the reservation for a hexagonal-branch is failed, and exit the procedure (see Fig. 6(a)). A2. Repeatedly reserve.  2;.  same time slots and  distinct time slots on AC , BD, IM , and KL, for  = 0 ;! ;! ;!. ;!. to 2 , until the reservation is successful, and update the possible nodes’ records of sending and receiving. activities. Otherwise, if the reservation is failed, then the reservation for a hexagonal-branch is failed, and exit the procedure (see Fig. 6(b)). A3. Repeatedly reserve.  2;.  same time slots and  distinct time slots on DG, FI , and EK , for  ;!. ; !. ; ; !. = 0. to 2 , until the reservation is successful, and update the possible nodes’ records of sending and receiving activities. Otherwise, if the reservation is failed, then the reservation for a hexagonal-branch is failed, and exit the procedure (see Fig. 6(c)). A4. Repeatedly reserve to.  2,.  ; 2.  same time slots and  distinct time slots on HJ , FI , and LM , for  ;!. ; !. ; ; !. = 0. until the reservation is successful, and update the possible nodes’ records of sending and receiving. activities. Otherwise, if the reservation is failed, then the reservation for a hexagonal-branch is failed, and exit the procedure (see Fig. 6(d)). A5. Repeatedly reserve.  2;.  same time slots and  distinct time slots on DF. ;!. and. EF , for  ;!. = 0 to 2 ,. until the reservation is successful, and update the possible nodes’ records of sending and receiving activities. Otherwise, if the reservation is failed, then the reservation for a hexagonal-branch is failed, and exit the procedure..

(12) 12. 12 G. G. H. H 34. 34 B. B. J. D. J. D. 12 A. F. 12. A. I. F. I. 12 C. E. C. M. K. E. M. K. L. L 12. 12. (a). G. (b). 12. G. H 34. B. 34. B. A. F. 12. J. D 12 F. I 34. C. E. 56. 34. 34 C. 34. A. I. 34. 34. 12. 12. 12. H. 56. J. D. 12. M. 12. E. M. 56 K. 34. K. L. 12. 12. B. 34. G. H. 78. 56. F. 78 56. 34. B. 12. 12 5678. F. I 34. 1112 C. 12. 78 34 12. J. 9 10. 34 M. K. 78 34. D. A. E 56. 34. 12. I 34. C. H. 56. 12. 12. 12. 34 J. D. A. L. (d). (c). G. 34 12. L. (e) Fig. 9. Time-slot reservation for a hexagonal-branch. E. M 56. 78 K. 34 12. (f). L.

(13) 13. 1 2 G. G. H. H 3 4. 3 4 B. B. J. D. J. D. 1 2 A. F. 1 2. A. I. F. I. 1 2 C. C. E. E. (a). G. 1 2. (b). G. H 3 4. B. 3 4. B. A. F. 3 4. 3 4. J. D 1 2. 1 2. 1 2. 1 2. H. 5 6. J. D. 1 2. A. I. 5 6. F. I. 3 4. 3 4 C. 1 2. C. E. 1 2. E. (d). (c). G. 1 2. 7 8. 5 6 B. 3 4. H 3 4 J. D 1 2. 1 2 A. F 3 4. 5 6. I. 7 8 C. 1 2. E. (e) Fig. 10. The time slot reservation for a hexagonal-twin. For instance as shown in Fig. 9(a), let Send time slot( JH )=f3 4g and Send time slot (LK )=f1 2g before ;!. ;!. constructing the hexagonal-branch. Let bandwidth requirement  be 4 slots. Fig. 9(b) shows that slots f1 2g. AB CE , GH and IJ for A1 step. Fig. 9(c) displays that slots 3 4 are successfully allocate to AC , BD , IM , and KL for A2 step. Fig. 9(d) displays that slots 5 6 are successfully allocate to DG, FI , and EK for A3 step. Fig. 9(e) illustrates that slots 7 8 are successfully allocate to HJ , FI , and LM for A4 step. Fig. 9(f) illustrates that slots 9 10 and 11 12 are respectively allocate to DF and EF for A5 step. Finally, a hexagonal-branch with four time slots is successfully constructed. are successfully allocate to. ;!. ;!. ;! ;! ;!. ;!. ; !. f. ;!. f. ;!. ; !. ; ; !. f. ;! ; !. f. ;!. g. ; ; !. ;!. g. f. g. g. g.

(14) 14. Destination nodes E. D. Destination nodes. Destination nodes. D. D. E. E. C. B. B1. C. C1. C2. B2. B. B1. C2. Release C. C1. Release B2. Source nodes. B1. B2. B. B1. Release. C2. B2. A. Source nodes. (b). C. C1. Release. A. Source nodes. (a). C2. B. A. A. E. D. Release C1. Destination nodes. Source nodes. (c). (d). Fig. 11. Four merging steps of the hexagonal-tree. A.2 Hexagonal-Twin Construction This phase searches for a hexagonal-twin if given bandwidth requirement is  . A hexagonal-twin Z. Z. 0. or. 0. Z without the hexagonal-block Z or Z : The operation is simZ Z ilar with the identification of the hexagonal-branch. We omit the details herein. For instance as shown in Fig.. Z. 00. is constructed from a hexagonal-branch Z. 0. 00. 00. 10(a), let Send time slot(JH )=f3 4g before constructing the hexagonal-twin. Let bandwidth requirement  ;!. AB CE , GH and IJ (same as A1 step, and see Fig. 5(a)). Fig. 10(c) displays that slots 3 4 are successfully allocate to AC and BD (same as A2 step, and see Fig. 5(b)). Fig. 10(d) displays that slots 5 6 are successfully allocate to DG and FI (same as A3 step, and see Fig. 5(c)). Fig. 10(e) illustrates that slots 7 8 are successfully allocate to EF and HJ (see Fig. 5(d)). Finally, a hexagonal-twin with four time slots is successfully constructed. be 4 slots. Fig. 10(b) shows that slots f1 2g are successfully allocate to f. ;! ;! ;!. ; !. ;!. ;!. g. ;!. f. ; !. g. ;!. f. g. ;!. B. Phase 2: Hexagonal-Tree structure The hexagonal-tree construction phase is divided into three operations.. 1: The hexagonal-path. discovery: Based on identified hexagonal-branches and hexagonal-twins, many. hexagonal-paths from a source to a given set of destinations are constructed (see Fig. 11(a)).. 2: The hexagonal-path reply : Destination receives the route-request packet from a source, and replies a. route-reply packet to the source (see Fig. 11(b)). 3:. The hexagonal-tree construction:. Multiple hexagonal-paths are received from all destinations, the. hexagonal-tree is finally established in the source (see Fig. 11(c)). Finally, the source sends data packet according to the determined the hexagonal-tree as shown in Fig. 11(d). B.1 The Hexagonal-Path Discovery After identifying possible hexagonal-branches and hexagonal-twins, we now describe how to construct the. hexagonal-path. Let 1 . 2  3. .  i](b) denote as a hexagonal-path, where i can be an uni-path, or a.

(15) 15. A2. A4. A3. A1. D1. Destination. D2. Destination. B3. Source. B1. S. B2. B4 C1. C2. C3. C4. C6. C5. Fig. 12. The hexagonal-tree discovery. hexagonal-block, or a hexagonal-twin, and b represents as the hexagonal-path bandwidth. The hexagonal-path discovery operation is given.. (C 1) Source node initiates and floods a hexagonal-path REQUEST-packet into MANET. If there is an uni-. path 1 satisfying the required bandwidth, then a hexagonal-path  1 ] is constructed, where 1 indicates an uni-path satisfying the bandwidth requirement.. (C 2) If no uni-path exists, we will further check if one or more hexagonal-branches exist. If a hexagonal-. branch exists, then a hexagonal-path 1 ] is constructed, where 1 is a hexagonal-branch.. (C 3) If both of steps C 1 and step C 2 are failed, we further check if a hexagonal-twin or a hexagonal-block. exists or not. If a hexagonal-twin or a hexagonal-block exists, then a hexagonal-path  1 ] is constructed,. where 1 is a hexagonal-twin or a hexagonal-block.. (C 4) Repeatedly perform C 1, C 2 and C 3 steps until a hexagonal-path arriving to one of destination nodes,. then a hexagonal-path 1  2  3      k ](b) is established.. For instance, many hexagonal-paths, from source S to destination nodes D 1 and D 2, are illustrated in Fig.. 12. B.2 The Hexagonal-Path Reply Each destination node replies all possible hexagonal-path  i  i.  i 2  1] to the source node. For instance, some hexagonal-paths are replied, from destination nodes D 1 and D 2 to source S , as illustrated in Fig. 13.. 1. ;. ;. .

(16) 16. A5. A4. A3. A2. D1. Destination. D2. Destination. B3. Source. B1. S. B2. B4 C1. C2. C3. C2. C3. C4. Fig. 13. The hexagonal-tree reply. B.3 The Hexagonal-Tree Construction This study mainly constructs the hexagonal tree which is modified from the spiral-fat-tree on-demand multicast(SOM) protocol . Observe that, the construction of the multicast tree can use the results of , , , , . In this work, all spiral-path of the spiral-fat-tree  are replaced with the hexagonal-path for the purpose of providing the QoS-extension multicast protocol. Given that two hexagonal-paths  1 . 2  p p+1  k ](b) and 1  2  p p+1  k ](b) have the same hexagonal sub-path  1  2   p](b) we denote (1 2  p p+1  k ](b) 1 2 p p+1  k ](b)) = 1  2   p](b): Further, let 1 2   p](b s) denote as s hexagonal-paths with same 1  2   p ](b): We also denote 1  2   p ](b) = p to be the shared hexagonal-path length. Given each pair of s and s replied hexagonal-paths, among many replied hexagonal-paths from destination . 0. . 0. \. . . 0. . . 0. . . j. . j. 0. nodes to the source node, we have the merging cirterion to construct the hexagonal-tree according to values of p s and b.. (D 1) If the s replied hexagonal-paths with the maximum values of p, s and b, then these s hexagonal-paths. are merged together with the highest priority.. (D 2) If there are s and s replied hexagonal-paths with the same values of p and s, then s hexagonal-paths 0. with the greater value of b are merged together.. (D 3) If there are s and s replied hexagonal-paths with the same values of b and s, then s hexagonal-paths 0. with the greater value of p are merged together.. (D 4) If there are s and s replied hexagonal-paths with the same values of b and p, then s hexagonal-paths 0. with the greater value of s are merged together.. (D 5) If there are s and s replied hexagonal-paths with the same values of p, then s hexagonal-paths with 0.

(17) 17. Destination nodes. Destination nodes. A. A. B. B. D. C. E. C. Merging. D. Merging. E. F. Source nodes. Source nodes. (a). (b). Destination nodes. Destination nodes. A. A. F. B. B D. C. C. Merging. D. E. E. F. Source nodes. Merging. Source nodes. (c). (d) Fig. 14. The merging examples. the greater value of b are merged together.. (D 6) If there are s and s replied hexagonal-paths with the same values of s, then s hexagonal-paths with 0. the greater value of b are merged together.. (D 7) If there are s and s replied hexagonal-paths with the same values of b, then s hexagonal-paths with 0. the greater value of s are merged together.. (D 8) If there are s and s replied hexagonal-paths with the different values of p, b and s then s hexagonal0. paths with the greater value of b are merged together.. For example, Fig. 14(a) and Fig. 14(b) are instance of the D3 case, Fig. 14(b) and Fig. 14(c) are instance of D7 case, Fig. 14(c) and Fig. 14(d) are instance of D2 case, and Fig. 14(b) and Fig. 14(d) are instance of D6 case. IV. E XPERIMENTAL R ESULTS A simulator is developed by C++ to evaluate the performance of our approach. The simulation parameters in the simulator are given below. . The number of mobile hosts is ranging from 20 to 50..

(18) 18. . The number of time slots of the data frame is assumed to be 16 slots.. . The bandwidth requirement are 2, 4, 6 and 8 time slots.. . The mobility is ranging from 10 to 40 Km/hr.. . The message length is ranging from 1 to 4 Mbit/s.. . The radio transmission range is 200 m.. . The transmission rate is 5Mbits/sec. To examine the effectiveness of our approach, we compare our proposal with two well-known MANET. multicast routing protocols, which are AODV  and ODMRP . Observe that, AODV and ODMRP do not offer the QoS capability. In addition, Tseng  designed a TDMA-based QoS uni-path routing protocol. To make a fair comparison, we provide the QoS-extension AODV and ODMRP protocols by integrating with Tseng’s TDMA-based QoS uni-path routing result . In this simulation, these two integrated results are denoted as AODV(+Tseng) and ODMRP(+Tseng), respectively. The simulator is simulated in a 10001000 m 2 area. The duration of each time slot of a time frame is assumed to be 5 ms, and the duration of a control slot is 0.1 ms. Source and destination are selected randomly. Once a QoS request is successful, a time slots is reserved for all the subsequent packets. The reservation is released when either the data transmission process is finished or the link is broken. A packet is dropped if the packet stays in a node exceeds the maximal queuing delay time, which is setting to four frame lengths (328 ms). The performance metrics of our simulation are given below. . Success Rate (SR):. The number of successful QoS route requests divided by the total number of QoS. route requests from source to destination. . . Overhead (OH ): The number of packets of transmitted packets, including the control and data packets. Latency(LT ): The interval from the time the multicast was initiated to the time the last host finishing its. multicasting.. It is worth mentioning that an efficient QoS routing protocol is achieved by with high success rate SR, and. low overhead. OH and low LT .. In the following, we illustrate the performance of. SR, OH and LT. from. several prospects. A. Performance of Success Rate (SR) The observed results of the performance of success rate vs. number of mobile hosts and network bandwidth are illustrated in Fig. 15. The low success rate of our approach will be obtained if the bandwidth requirement is large. Four kinds of effects are illustrated. 1a) Effects of number of hosts: Each value in Fig. 15(a) is obtained by assuming the bandwidth requirement (slot number) is 2, the number of destinations is 2 and the mobility is 10 Km/hr. From Fig. 15(a), the. greater number of hosts is, the higher SR value will be obtained. For instance, if number of hosts is 20, then. SR values of AODV(+Tseng), ODMRP(+Tseng) and our protocol are 12%, 14%, and 17%, respectively. If number of hosts is 50, then SR values of AODV(+Tseng), ODMRP(+Tseng) and our protocol are 62%, 69%, and 81%, respectively. This indicates that our protocol acquires the better success rate than other two schemes under various number of hosts..

(19) 19. 100%. 100% AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 90%. 80%. 70%. 70%. 60%. 60%. Success Rate. Success Rate. 80%. AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 90%. 50% 40% 30%. 50% 40% 30%. 20%. 20%. 10%. 10%. 0%. 0% 20. 30. 40. 50. 2. (a). 6. 8. (b). 100%. 100% AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 90% 80%. 80%. 70%. 70%. 60%. 60%. 50%. 50%. 40% 30% 20%. AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 90%. Success Rate. Success Rate. 4. Bandwidth Requirement (slot number). Number of hosts. 40% 30% 20% 10%. 10%. 0%. 0% 10. 20. 30. 40. 2. 3. 4. 5. Number of destinations. Mobility (Km/hr). (c). (d). Fig. 15. Performance of success rate vs. effect of (a) number of hosts, (b) bandwidth requirement, (c) mobility, and (d) number of destinations. 1b) Effects of bandwidth requirement: Each value in Fig. 15(b) is obtained by assuming the number of hosts is 50, the number of destinations is 2 and the mobility is 10 Km/hr. Fig. 15(b) illustrates that the higher bandwidth requirement is, the lower. SR value will be obtained.. For instance, if bandwidth requirement. SR values of AODV(+Tseng), ODMRP(+Tseng) and our protocol are 62%, 68%, and 79%, respectively. If bandwidth requirement is eight, then SR values of AODV(+Tseng), ODMRP(+Tseng) and. is two, then. our protocol are 10%, 14%, and 32%, respectively. This indicates that our protocol acquires the better success rate than other two schemes even in various bandwidth requirement. 1c) Effects of mobility: Each value in Fig. 15(c) is obtained by assuming the number of hosts is 50, the number of destinations is 2 and bandwidth requirement is 2. Fig. 15(c) indicates that the higher mobility is, the lower. SR value will be.. For instance, if the mobility is 10 Km/hr, then SR values of AODV(+Tseng),.

(20) 20. ODMRP(+Tseng) and our protocol are 64%, 71%, and 79%, respectively. If the mobility is 40 Km/hr, then. SR values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 13%, 16%, and 20%, respectively. This indicates that our protocol acquires the better success rate than other two schemes under various mobility. 1d) Effects of number of destinations: Each value in Fig. 15(d) is obtained by assuming the number of hosts is 50, the mobility is 10 Km/hr and bandwidth requirement is 2. Fig. 15(d) indicates that the greater number of destinations is, the lower. SR value will be.. For instance, if the number of destination is two, then. SR. values of AODV(+Tseng), ODMRP(+Tseng), and our protocol,are 75%, 78%, and 80%, respectively. If the number of destination is five, then. SR values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are. 21%, 32%, and 41%, respectively. This indicates that our protocol acquires the better success rate than other two schemes under various number of destinations. B. Performance of OverHead (OH ) The observed results of the performance of overhead vs. number of mobile hosts and network bandwidth are illustrated in Fig. 16. The low overhead of our approach will be obtained if the bandwidth requirement is large. Four kinds of effects are illustrated. 2a) Effects of number of hosts: Each value in Fig. 16(a) is obtained by assuming the bandwidth requirement is 2, the number of destinations is 2, message length is 1M and the mobility is 10 Km/hr. From Fig. 16(a),. the greater number of hosts is, the higher OH value will be obtained. For instance, if number of hosts is 20,. then OH values of AODV(+Tseng), ODMRP(+Tseng) and our protocol are 243, 312, and 411, respectively. If number of hosts is 50, then OH values of AODV(+Tseng), ODMRP(+Tseng) and our protocol are 1534, 1836, and 2410, respectively. This indicates that our protocol acquires the higher overhead than other two schemes. 2b) Effects of bandwidth requirement: Each value in Fig. 16(b) is obtained by assuming the number of hosts is 20, the number of destinations is 2, message length is 1M and the mobility is 10 Km/hr. Fig. 16(b) shows. that the higher bandwidth requirement is, the stable OH value will be obtained. For instance, if bandwidth. requirement is 2, then OH values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 286, 336, and. 409, respectively. If bandwidth requirement is eight, then OH values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 289, 337, and 410, respectively. This indicates that our protocol acquires the higher overhead than other two schemes under various bandwidth requirement. 2c) Effects of mobility: Each value in Fig. 16(c) is obtained by assuming the number of hosts is 20, the number of destinations is 2, message length is 1M and bandwidth requirement is 2. Fig. 16(c) indicates that the higher mobility is, the higher. OH value will be.. For instance, if the mobility is 10 Km/hr, then. OH. values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 243, 276, and 386, respectively. If the mobility is 40 Km/hr, then OH values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 856, 965, and 1121, respectively. This indicates that our protocol acquires the higher overhead than other two schemes under various mobility. 2d) Effects of number of destinations: Each value in Fig. 16(d) is obtained by assuming the number of hosts is 20, the mobility is 10 Km/hr, message length is 1M and bandwidth requirement is 2. Fig. 16(d) indicates.

(21) 21. 3000. 700 AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 2500. AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 600 500. 2000 Overhead. 400 Overhead. 1500 1000 500. 300 200 100. 0. 0 20. 30. 40. 50. 2. 4. 6. 8. Bandwidth Requirement (slot number). Number of hosts. (a). (b). 1400. 1200 AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 1000. AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 1200 1000. 800 Overhead. Overhead. 800 600 400 200. 600 400 200 0. 0 10. 20. 30. 2. 40. Mobility (Km/hr). 3. 4. 5. Number of destinations. (c). (d). Fig. 16. Performance of overhead vs. effect of (a) number of hosts, (b) bandwidth requirement, (c) mobility, and (d) number of destinations. that the greater number of destinations is, the higher. OH. value will be. For instance, if the number of. destination is two, then OH values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 235, 314 and. 413, respectively. If the number of destination is five, then OH values of AODV(+Tseng), ODMRP(+Tseng). and our protocol are 724, 863 and 998. This indicates that our protocol acquires the higher overhead than other two schemes under various number of destinations. C. Performance of Latency (LT ) The observed results of the performance of latency vs. number of mobile hosts and network bandwidth are illustrated in Fig. 17. The high overhead of our approach will be obtained if the bandwidth requirement is large. Four kinds of effects are illustrated..

(22) 22. 1.2. 0.45 AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 1. 0.35. 0.8. 0.3 Latency (sec). Latency (sec). AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 0.4. 0.6 0.4. 0.25 0.2 0.15 0.1. 0.2 0.05 0. 0 20. 30. 40. 50. 2. 4. 8. (b). (a) 0.9. 0.8 AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 0.7. AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 0.8 0.7 Latency (sec). 0.6 Latency (sec). 6. Bandwidth Requirement (slot number). Number of hosts. 0.5 0.4 0.3. 0.6 0.5 0.4 0.3. 0.2. 0.2. 0.1. 0.1. 0. 0 10. 20. 30. 40. 2. 3. Mobility (Km/hr). 4. 5. Number of destinations. (c). (d) 0.9 AODV (+Tseng) scheme ODMRP (+Tseng) scheme Our protocol. 0.8. Latency (sec). 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1. 2. 3. 4. Message length (M). (e) Fig. 17. Performance of latency vs. effect of (a) number of hosts, (b) bandwidth requirement, (c) mobility, (d) number of destinations, and (e) message length.

(23) 23. 3a) Effects of number of hosts: Each value in Fig. 17(a) is obtained by assuming the bandwidth requirement is 2, the number of destinations is 2, message length is 1M and the mobility is 10 Km/hr. From Fig. 17(a), the. greater number of hosts is, the higher LT value will be obtained. For instance, if number of hosts is 20, then. LT values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 0.39, 0.31, and 0.22, respectively. If number of hosts is 50, then LT values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 0.91, 0.82, and 0.75, respectively. This indicates that our protocol acquires the lower latency than other two schemes under various number of hosts. 3b) Effects of bandwidth requirement: Each value in Fig. 17(b) is obtained by assuming the number of hosts is 20, the number of destinations is 2, message length is 1M and the mobility is 10 Km/hr. Fig. 17(b) illustrates that the higher bandwidth requirement is, the lower. LT. value will be obtained. For instance, if. bandwidth requirement is two, then LT values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 0.39, 0.31, and 0.22, respectively. If bandwidth requirement is eight, then. LT. values of AODV(+Tseng),. ODMRP(+Tseng), and our protocol are 0.09, 0.07, and 0.06, respectively. This indicates that our protocol acquires the lower latency than other two schemes even in various bandwidth requirement. 3c) Effects of mobility: Each value in Fig. 17(c) is obtained by assuming the number of hosts is 20, the number of destinations is 2, message length is 1M and bandwidth requirement is 2. Fig. 17(c) indicates that the higher mobility is, the higher LT value will be. For instance, if the mobility is 10 Km/hr, then LT values. of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 0.37, 0.31, and 0.22, respectively. If the mobility is 40 Km/hr, then LT values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 0.68, 0.61, and 0.55, respectively. This indicates that our protocol acquires the lower latency than other two schemes under the various mobility. 3d) Effects of number of destinations: Each value in Fig. 17(d) is obtained by assuming the number of hosts is 20, mobility is 10 Km/hr, message length is 1M and bandwidth requirement is 2. Fig. 17(d) indicates that the. greater number of destinations is, the low LT value will be. For instance, if the number of destination is two,. then LT values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 0.37, 0.3, and 0.23, respectively. If the number of destination is five, then LT values of AODV(+Tseng), ODMRP(+Tseng), and our protocol. are 0.75, 0.67, and 0.49, respectively. This indicates that our protocol acquires the lower latency than other two schemes under the various number of destinations. 3e) Effects of message length: Each value in Fig. 17(e) is obtained by assuming the number of hosts is 20, mobility is 10 Km/hr, the number of destinations is 2 and bandwidth requirement is 2. Fig. 17(e) indicates that. the greater number of destinations is, the lower LT value will be. For instance, if the number of destination is two, then. LT. values of AODV(+Tseng), ODMRP(+Tseng), and our protocol are 0.37, 0.22, and 0.15,. respectively. If the number of destination is five, then LT values of AODV(+Tseng), ODMRP(+Tseng), and. our protocol are 0.75, 0.67, and 0.56, respectively. This indicates that our protocol acquires the lower latency than other two schemes under various message lengths.. Apparently, it is desirable to have a high SR and low LT . The higher the SR is, the lower the LT will be.. It is always beneficial to adopt our protocol as demonstrated by the simulation results..

(25)    +7

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