An Efficient Video Program Delivery algorithm in Tree Networks*

Transcription

1 3rd International Symosium on Parallel Architectures, Algorithms and Programming An Efficient Video Program Delivery algorithm in Tree Networks* Fenghang Yin 1 Hong Shen 1,2,** 1 Deartment of Comuter Science, University of Science and Technology of China, China 2 School of Comuter Science, University of Adelaide, Australia yfh2008@mail.ustc.edu.cn, hongshen@ustc.edu.cn Abstract As the roortion of video rograms is exected to grow significantly, video services will require a huge amount of Internet bandwidth in the future. In this aer, we model the video rogram lacement (VPP roblem in tree networks which sends video rograms to the requesting (demand nodes using a broadcast method. The model considers the cost for both assigning rograms to nodes and broadcasting video rograms through links. The model is formulated as an integer rogram, and its objective is to minimize the total cost of VPP in a tree network. We develo a dynamic rogramming algorithm to solve this roblem with time comlexity ( where N is the number of nodes and P is the number of video rograms. 1. Introduction With an increase in the number of internet users, video services will become more and more oular. The roortion of video contents is exected to grow significantly in the future in Internet, making otimizing delivery of video services which requires a large amount of bandwidth an increasingly imortant roblem. In video services, video rograms are broadcast to assign or demand nodes. A large bandwidth is required when video rograms are broadcast through a link, and a large storage sace is required when video rograms are stored at a node. A simle model of VOD network is roosed in [1] which ays attention to server locations and rogram assignments. A VOD network is artitioned into several levels, and the model finds out otimal server locations and rogram assignments. In this model, servers installed and rograms assigned to a single level means each node in the selected level has a server and same subset of rograms. In [2][3] some heuristic *This work is suorted by National Science Foundation of China under its General Projects funding # and Chinese Academy of Science 100 Talents Project **Corresonding author. hongshen@ustc.edu.cn algorithms are roosed for lacing video rograms in tree networks to satisfy stochastic demands subject to the constraints of storage caacity and secified blocking robabilities. Some algorithms are also roosed to allocate limited bandwidth among multile video rograms in [4][5]. Some other VOD delivery technologies are discussed in [6][7][8]. Hanan et al.[9] roose a model of VOD network to determine the otimal location of servers and otimal assignment of rograms to each of these servers in a tree network. A video rogram may be stored in several servers. In this aer, we model the video rogram lacement (VPP roblem in tree networks. We assume the servers have been installed in some nodes in a tree network. The model considers the cost for both assigning rograms to nodes and broadcasting video rograms through links. Our objective is to minimize the sum of rogram assignment cost at nodes and bandwidth cost on links. This roblem has been solved with time comlexity ( where N is the number of nodes and P is the number of video rograms as a secial case of the model roosed in [9]. We roose a new dynamic rogramming algorithm insired by [10] to solve this roblem with time comlexity ( which outerforms the algorithm roosed in [9] significantly in large-scale networks. This aer is organized as follows: in Section 2, we formulate the roblem. And the dynamic rogramming algorithm is roosed in Section 3. In Section 4 an examle is given to demonstrate our algorithm. We conclude the aer in Section Problem formulation First of all, we resent the formulation of VPP roblem in tree networks which use the broadcasting method to send video rograms to the nodes which require the rogram. It is obvious that the roblem with P rograms to assign can be decomosed into P searate roblems, one er rogram. So we can solve the roblem with one rogram to assign firstly and then imlement it for each rogram. In this aer, we /10 $ IEEE DOI /PAAP

2 formulate the roblem with one rogram and then solve it by using a dynamic rogramming method. Consider a tree (,, where V is the set of nodes in a tree network and is the set of links. There are N+1 nodes in the network, i.e. +1=. For simlicity, we label the nodes as 0,1,2,, from to to bottom and left to right. Node 0 which is the root of the tree is the original server and stores the rogram. We also label link using the label of its end node. For examle, lin k 1 in Fig.1 is the lin k which connects node 0 and node 1. There are N links, ={1,2,,}. In the VPP roblem, servers have been installed in many nodes and the rogram can only be assigned to the nodes which have a server installed in it. We will use some notations as follows: whether node want to access the rogram or not. I.e. =1 if node wants to visit the rogram and 0 otherwise. cost of assigning the rogram to node.this cost includes storage and rocessing cost of the rogram at node. If there is no server installed in it, =. ( cost of the bandwidth required for transmitting the rogram on link. ( decision variable. =1 if the arent of node needs to broadcast the rogram through link, 0 otherwise. To ensure each node s requirement can be met, we will give a assignment strategy which is the set of nodes to which the rogram is assigned to determine value for each link. Because the value of is fixed, can only be determined by assignment strategy only. To avoid confusion, we use, as the decision variable to identify the link which should be or not be under the assignment strategy. The method to determine the value of, by using the assignment strategy will be given in theorem 1. Suose is the set of nodes which stores the rogram, the total cost of assigning and broadcasting the rogram as follows: ( = +, ( where the first art of equation is the assignment cost and the second is the broadcast cost. Our objective is to find an assignment strategy A subject to: ( = {(} \{} This roblem can be solved in olynomial time ( for an N-nodes tree as a secial case of SLPAM using the method roosed in [9]. In this aer we roose a new dynamic rogramming algorithm whose time comlexity is (. Figure 1. A tree network with nodes are labeled from to to bottom and left to right, and the bandwidth required for broadcasting rogram has been labeled near the link, and the nodes with shadow mean a server is not installed in it 3. Dynamic rogramming aroach We use a bottom-u dynamic rogramming aroach en-lightened by [10] to get a solution of a tree network. Let denote the subtree of whole network with as its root node and as the set of children nodes of, e.g. in Fig.2 is the subtree which is enclosed by dotted line and = {, }. is the arent node of and the rogram is assigned or broadcast to node. We use (, to denote the extended subtree including. For examle the tree shown in Fig.2 can be denoted as (,. Definition 1 (otimal solution of (, is a feasible solution, (,, = + \, ( is the total cost in (, under the lacement strategy. The otimal solution of (, is which satisfies (,, =(,, ={ +, ( } \ We denote the value of otimal solution as (, = (,,. Definition 2 (otimal solution of (, By changing the cost of link to, we get the extended tree (,. The otimal solution of (, is which satisfies (,, ={ (,,} =min { +, ( +, \( {} } 4

3 w1 w w Figure 2. Definition of (, where is the subtree with w as its root node and is the subtree with as its root node and has the same definition We denote the value of otimal solution of (, as (, = (,,. The link should not be used under the strategy in (, obviously. All of the feasible solutions of (, can be divided into two kinds: using link to broadcast the rogram or not. In next theorem, we will rove that the best solution which doesn't use link to broadcast the rogram is the otimal solution of (,. Theorem 1. For each, we definite (, = {, (,\{}} and (, = { (,\{}, (,} where (, is the set of nodes on the ath that connects nodes and and. We have ={ =0} (, which is equivalent to, =0 for each and the minimum total cost of solutions in is (, with the otimal solution. Proof: Assume, for each (,, all access request of nodes in can be satisfied by the nodes in and the rogram shouldn t be broadcasted to the trees { }. According to the definitions of (, and (,, we can conclude that (, = \( (,. Because (, =0which means that all of the nodes in (, don t want to access the rogram, the rogram shouldn t be broadcast to (,. And (, (, =, so the rogram shouldn t be broadcast to which means, =0. If a set F, we have (, 0which means there is at least a node in (, which wants to access the rogram, because there is no node that can satisfy it in, the rogram has to be broadcast into subtree using the link, so, =1. (,, = + ( +0 \( {} w2 = + ( +0 ( \( {} = + ( \ =(,, Based on the above analysis, (,, = (,, for each. We define =2 \. In (, (,, = +, ( +1 \ {} (,,, according this inequality, we know that, and (,, = (,, (,, = (,, which means the is the otimal solution in that the total cost of (, is (,. Thus the theorem is roved. Lemma 1. If and, then =,, we have. Proof: Firstly, we will rove that (, =(,. For each (,, we have (,,, (,\{}, (,\{} Because is a rerequisite,, (,\{}, ((, \{} {}, (,\{} where. And because (,, (,, (, \{}, (,\{}, (,\{} (, thus (, =(,. Then we rove (, =(,. (, (,\{}, (, Because for each,(, =, (, (,\{}, (, (,\{}, (, (, \{}, (, (, It is obvious that. Because (, (, and (, =0, (, =0. As mentioned above,. Hence, the lemma is roved. As reviously mentioned, the best solution in is the otimal solution of (,, according to the 5

4 next theorem, we can get the otimal solution of (,. Theorem 2. Let = (, + and B= (, +. For the otimal solution value of (,, we have (, ={,B} And {} B = Proof: Assume = { } and = { }. Clearly is the set of all feasible solutions of (,. Consider the value of : (1 If =0. When, and F, so, = 0. \ = ( \ = ( \ {} = (,, \(. then = +, ( +0 \ {} = +, ( \ = + \{} +, ( \ = + +, ( \( = + +, ( = + (,, + (, ( if = for each, (,, = + (,, meaning = { (,,, } with solution =( {} when, we can divide it into two cases: and \. If, then and, =0. We have (,, = (,, according to Lemma 1 and Theorem 1, (,, = (,,, so (,, = (,, (, = + (, (,, = + (, =B, if and only if = for each, If \, then and, =1. we have (,, =, + (,, (,, So the otimal solution among can be achieved when = and B={ (,,, } with solution = Hence when =0, if B, (, = with solution =, else if >, (, =B with solution =. (2 If =1. It is obvious that = (, + with solution =( {} if it just like in the case =0. When, because, (,and =. So, =1and for each, we have (,, =, + (,, = + (,, (, + We still define B = (, +. In this case, (, = with solution =. This is consistent with the fact that when =1 and, to guarantee node can access the rogram, node has two chooses: store the rogram in node or let node to broadcast the rogram to, because ( =, node should store the rogram in its own storage. Hence, the theorem is roven. According to Theorem 2, we notice that if node has a demand, the otimal solution of (, is ( {} which means the rogram must be assigned to node. 6

5 The otimal solution of (, can be achieved by comaring the otimal solution of (, and the solutions which must use link to broadcast the rogram. To rove theorem 3, we give Lemma 2 firstly. Lemma 2. =2 \ and = ( + (,, then there is a feasible solution =, such that (,, =if and only if =1,. Proof: for each, (,, =, ( + (,,, ( + (, ( + (, It is obvious that if and only if, =,,(,, =. Because = 2 \, we know that, =1. (1 If =1, to satisfy the requirement of node, we have to broadcast the rogram by link, so, =, =1. (2 If,, let =. By the roof of lemma 1, we know that (, = (, and (, = (,. Because there is at last one,, we conclude that (, 0. Then we can have and, =1=,,. (3 If,, we still let =, and by the same method we can have (, = 0 which means, so, =0. For (1(2(3, we know that (,, = in (1(2 and (,, in (3, which means this theorem is correct. Theorem 3. Let = ( + (,. The otimal solution of (, is (, = < (, and (, = { (,,}. Proof: we can div ide all feasible solutions of (, into two kinds: and, just like mentioned in theorem 1. (1 If (,, according to above discussion, we have (,, (,. Obviously (, = (, and the otimal solution of (, is. (2 If < (,. Let =. If,, and according to the roof of lemma 2, we know that, =0, so F and = ( + (, > (,. Because is a feasible solution and the total cost of is (,, we conclude that > (,, (,. But < (, and (,,, so (, which is contrary to >(,. Based on the above discussion, we know that, if < (,, and according to lemma 2, there is a feasible solution = whose total cost is. As a result, we conclude that (, = with as its otimal solution. For (1(2, The theorem is roved. According to theorem 3, we can conclude that, for the whole tree network, the minimum value of the total cost of assigning and broadcasting the rogram is (, and the otimal solution is where is the original server. According to the above theorems, we roose a dynamic rogramming algorithm as follows which runs in a bottom-u order. Algorithm 1: Ste 1. Initialization: = ; = ;, ( =0;, ( =0, for each \{}where (w is the arent of and s is the root of whole tree. Ste 2. Running the algorithm in ost-order traversal. For =, 1,,1do = +, (; B= +, (; = ( +, (; if Bthen =( {} ;, ( =; else = ;, ( =B; endif (,( +=, (; if, then = ;, ( = (, (; else = ; 7

6 , ( =; endif (,( +=, (; endfor = = ; return A ; Time Comlexity What we need to calculate is (, 0,, and the comlexity of our algorithm is ( where N is the number of nodes in a tree network. Because the algorithm is just for one rogram, we have to imlement it once for each rogram, and the total time comlexity is ( for rograms, which is significantly lower than O(N2P of algorithm in [9]. 4. Examle In this section, we give an examle to show the imlementation of our algorithm. The tree network of 13 nodes, labeled as 0-12, as shown in Fig. 1. Because the imlementation er rogram is similar, we just give the imlementation of rogram. Nodes that want to access rogram have been marked as P under it, for examle, in Fig. 2 nodes 2, 3, 5, 7, 8, 9, 11, 12 want to access rogram. The broadcast cost through link also is marked in this link. The cost of assigning rogram to any of the nodes is a =3 for i= 2,3,4,6,7,8,9,10,11. And a = for others because there is no server installed in them. We now demonstrate the imlementation of our algorithm for this examle. The algorithm runs in bottom-u order, so it starts with leaf nodes 7, 8, 9, 10, 11, 12. For node 12, =,B = and =4+ 0=4, so G (12,6 =min{, } =,B = {12}, G(12,6 =min{g (12,6,4} =4,A =. the results of nodes 7, 8, 9, 10, 11, 12 are resented in Table 1. Table 1. The results of leaf nodes 7, 8, 9, 10, 11, 12 Node (, ( (, ( labels Notice that the result of node 10 is secial comaring with other nodes, because node 10 doesn t want to access the rogram. Consider the node 6, =3+G(12,6 +G(11,6 +G(10,6 =10,B= G (12,6 +G (11,6 +G (10,6 = +3 =,=4+3+4=11, so G (6,2 =min{10, } = 10, B = {11,6},G(6,2 =min{10,11} =10,A = B ={6,11}. The results of second level nodes are shown in Table 2. Table 2. The results of 3, 4, 5, 6 Node (, ( (, ( labels , ,11 Then we will give the rocess details of nodes 1 and 2. For node 1, = +3+8=,B = 3+9 = 12,=4+3+8=15,so G (1,0 =min{,12} = 12, B = {3,4},G(1,0 =min{12,15} =12,A = B ={3,4}. And for node 2, =3+2+10= 15, B =, = =16,so G (2,0 = min{,15} =15,B = {2,6,11},G(2,0 = min{15,16} =15,A =B = {2,6,11}. So according to our algorithm, the otimal solution of the examle is {2, 3, 4, 6, 11}. 5. Conclusions The dynamic rogramming algorithm roosed in [9] solves the server location roblem in VOD networks with t ime co mle xity Ο(N 3. This cost is excessively high, making the network administrator unable to determine server locations and reassign rograms (requires time comlexity Ο(N P frequently In this aer, we roose a dynamic rogramming algorithm with time comlexity Ο(NP which allows more frequent rogram reassignment for otimizing erformance and hence is more suitable for solving the video rograms assignment roblem in large-scale networks. 6. References [1] D. Dammicco, U. Mocci, and F. U. Bordoni, Otimal server location in VOD networks, In Proc. IEEE GLOBECOM, Nov. 3-8, 1997, vol. 1, [2] R.H. Hwang and P.H. Chi, Fast otimal video lacement algorithms for hierarchical video-on-demand systems, IEEE Trans. Broadcast, vol.47, no.44, , Dec [3] J.D. Ryoo and S. S. Panwar, File distribution in networks with multimedia storage servers, Networks, vol.38, no.3, Oct.2001, [4] C. Y. Lee, Y. P. Moon and Y. J. Cho, A lexicograhically fair allocation of discrete bandwidth for multicast traffics, Comut. Oer. Res. vol.31, no.14, Dec.2004,

7 [5] H. Luss, An equitable bandwidth allocation model for video-on-demand networks, Netw. Satial Econ., vol.8, no.1, Mar. 2008, [6] K. A. Hua, M. A. Tantaoui, and W. Tavanaong, Video delivery technologies for large-scale deloyment of multimedia alications, Proc. IEEE, vol.92, no.9, Se. 2004, [7] B. Li and J. Liu, Multirate video multicast over the Internet: An overview, IEEE Netw., vol.17, no.1, Jan. 2003, [8] S. Ramesh, I. Rhee, and K. Guo, Multicast with cache (Mcache: An adative zero-delay video-on-demand service, IEEE Trans. Circuits Syst. Video Technol., vol.11, no.3, Mar. 2001, [9] H. Luss, Otimal content distribution in video-ondemand tree networks, IEEE Trans. ON SYSTEM, MAN, AND CYBERNETICS, vol.40, Jan. 2010, no.1. [10] Shihong Xu and Hong Shen, An O(nh Algorithm for Dual-Server Coordinated En-Route Caching in Tree Networks, PDCAT 2006: