A Variation of Minimum Latency Problem

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(1)A Variation of Minimum Latency Problem . Yung-Hui Huang. Yaw-Ling Lin. In this paper we study the variation of the minimum latency problem(MLP) [2]. The MLP is to

(2) nd a walk tour on the graph G(V; E ) with a distance matrix d . Where d indicate the distance between v and v . Let `(v ) is the latency length of v , de

(3) ned to be the distance traveled before

(4) rst P visiting v . The minimum latency tour is to minimize =1 `(v ). In some message broadcast and scheduling problem [8] the vertex also has latency time and weight. Those problem need to extend the P objective function of the minimum latency tour as =1 `(v )w(v ). The de

(5) nition is equivalent to the MLP with no edge distance but vertex latency time and vertex weight. We give an linear algorithm for the unweighted full k-ary tree or k-path graphs, and O(n log n) time for general tree graphs. The time complexity in trees is the same as Adolphson's result; however, the algorithm given here is not only simpler, easier to understand, but also more exible and thus can be easily extended to other classes of graphs. minimum latency problem, linear ordering, broadcast network, trees, k-path graphs i;j. i. i. j. i. n i. i. i. n i. i. z. Chuan-Yi Tang. the air and save the powers of the mobile devices. However, users need to wait for the required data to appear on the broadcast channel. It follows that the more time they wait then the more power devices have to consume. Finding the minimum latency tour can thus help us in solving this kind of problem. Several variations of this problem was focused on index and data allocation in a single broadcast channel using conventional techniques [4, 5]. In general trees, this problem was solved in O(n log n) time [1]. The issue of multiple broadcast channels is addressed in [7], and the problem becomes NP-hard [6]. Further, there are discussion using prune strategies for multiple broadcast channel and heuristic answers form single broadcast channel [6]. Given a directed acyclic graph G with a set of job vertices v1 : : : v with weight w(v ), latency time t(v ) and latency edges e1 : : : e with distance d(e ). Let T be a tour which completes the job vertices in precedence order. If the vertex v wants to be completed, at least one ancestor of v must be

(6) nished before. Further, because the walk back cost is zero, T does not need to include the walk back edge and job vertex already been

(7) nished. Let the latency `(v ) of a vertex v be the sum of the edge and vertex latency time of the tour from the staringPpoint to v . We can de

(8) ne the total latency L(T ) = =1 `(v )w(v ). We say that a tour T is the minimum latency tour if L(T ) has the smallest value among all possible tours of G. In this paper, we investigate tree graphs, tree, full kary trees and k-path graphs. We propose a polynomial time algorithm for general trees to get the optimal solution in O(n log n) using a greedy manner, a linear time DFS-like algorithm for full k-ary tree and a O(n log k) time algorithm for k-path graphs. Figure 1 gives an example of the dags and its minimum latency tour.. Abstract. i;j. y. i. n. i. i. m. j. i. i. Keywords:. i. i. i. n i. 1. Introduction. i. Consider a worker (a processor,or a repairman) facing a set of jobs with precedence and the latency time to initialize before job start(latency of edges), each of jobs is a vertex with latency time and weight(latency and weight of vertices). The worker must schedule its visits so as to minimize the total latency time the jobs wait before being complete. (We assume that the walk back latency of completed jobs are zero.) This is a simple and natural combinatorial optimization problem faced in realistic world, and may be formalized as follows. In mobile environment, users retrieve information by portable devices. Since the mobile devices usually have limited power, the issue of minimization the data access latency is important. Periodic broadcasts of frequently requested data can thus reduce the traÆcs in Department. ment.. [email protected].. E-Mail:. [email protected].. This. work was partially supported by NSC 89-2213-E-126-003.. z Department. of Computer Science, National Tsing-Hua Uni-. versity, Taiwan, R.O.C. E-mail:. i. Basic Notations and Proper-. Intuitively, we would want to place the heaviest vertex at the front of visited tour as early as possible. However, at least one ancestors of the heaviest vertex must be placed before, and it might not be wise exploring these vertices too early. This simple observation suggests that we would like to explore the properties of the tree graphs, for short, in a bottom-up manner. Without loss of generality, we can add the weight w(v) and latency time t(v) of vertex v into all edges. of Computer Science and Information Manage-. Providence University.. i. ties. of Computer Science, National Tsing-Hua Uni-. versity, Taiwan, R.O.C. E-mail:. y Department. 2. i. [email protected].. 1.

(9) A. is iterative select a maximum ratio cost edge and shrinking with is parent or output the selected subtour until all vertex is be output; totally it will spend O(n) time to obtain a Fibonacci heap. Second, we uses O(log n) time to extract maximum and decrease the ratio cost of the parent edge exactly n times. The output result is then the minimum latency tour. First we show when to glue two inseparable subtours into one:. B. C. D. E. F. G. 1. Weight of vertices w(vi). A. B. C. D. E. F. G. 30. 0. 45. 50. 42. 65. 101. a and b be G, and a is the parent of b.. Lemma 3.1 Let. two inseparable subtours. of. Further,. b has the largest ab is insep-. ratio cost among all others. It follows that arable subtour.. 2. Work Time of vertices t(vi) is zero. 3. All edge delay d(ej) is 1. By contradiction, assume that there is a nonempty subsequence of edges of the trees such that L(a b) < minfL( ab); L(ab )g. First we assume c( ) < c(ab). Note that d(a); d(b) > 0 and c(a) c(b), we have c( ) < c(ab) w(a) + w(b) < d(a) + d(b) d(a)c(a) + d(b)c(b) < d(a) + d(b) d(a)c(b) + d(b)c(b) d(a) + d(b) = c(b) (1) then L(ab ) L(a b) = L(b ) L( b) = d(b)[c( )d( )] d( )[c(b)d(b)] = d(b)d( )[c( ) c(b)] By equation (1), it follows that L(ab ) < L(a b), contradicting the assumption. Similarly, for the other case that c( ) c(ab), we get that L( ab) < L(a b), which also contradicts the assumption. So, if c(a) c(b) then we can shrink a and b into inseparable subtour ab. Based on Lemma 3.1, we claim that if an inseparable subtour has the maximum ratio cost among all others. It can glue with its parent. Proof.. Result: T = {A,C,E,G,F,B,D} L(T) = 1*30+2*45+3*42+4*101+5*65+6*0+7*50 = 1325. Figure 1: The minimum latency tour of dags which incident to v. Let w(e) = w(v) and d(e) = d(e)+ t(v). Given a tree T , we say a subtour T = he ; : : : ; e i is inseparable if and only if there exists a minimum latency tour on T containing this subsequence T contiguously. That is, T T , and all adjacent edges of T are also adjacent to each other in T . Let d(T ) denote the sum of distances d(e)'s for each edge e in T , and w(T ) donate the sum of weights w(e)'s. Further, we de

(10) ne the ratio cost of T by c(T ) = (( )) . By de

(11) nition, each edge e of G is initially inseparable subtour and have c(e) = (( )) . The idea here is to iteratively glue the child and parent vertices and

(12) nd its minimum latency tour for the new sub-graph. Taking advantage of the nice properties of trees, we can glue the inseparable subtour that has the maximum ratio cost with its parent such that edges of the trees are grouped into a new inseparable subtour. If an inseparable subtour T 0 is the root edge and has maximum ratio cost then T 0 can be output directly. From above idea, we found that there are some interesting properties of trees. Section 3 introduce the O(n log n) algorithm for tree graph, and provides some properties for trees. Those properties are also used in Section 4 to provide a DFS-like algorithm for the special case "full k-ary tree". The algorithm in Section 4 is O(n). After the tree and its special case, we discuss another case called "k-path graphs" in Section 5. k. i. j. k. k. k. k. k. k. K. w Tk. k. d Tk. w e d e. k. k. 3. Lemma 3.2 If. T0. k. is inseparable subtour with maxi-. mum ratio cost among all others, It can glue with its parent or if. T0. k. is the root edge, then. T0. k. can be output. directly.. First from the Lemma 3.1, we can say that the inseparable subtour has maximum ratio cost must be the maximum child of its parent and can always be shrink with its parent. Second, if T 0 k is a root edge and inseparable with maximum ratio cost. According to Lemma 3.1, no nonempty subsequence can separate T 0 . Further T 0 is the root edge, so we can easily claim that all other nonempty subsequence will be traversed after T 0 is completed. Proof.. Algorithm for General Trees. In this section we prove some property of general tree graphics. According to those properties an O(n log n) time algorithm for tree naturally evolves.The idea here. k. k. k. 2.

(13) Algorithm. k -ary Trees We say that a tree T is a full k-ary tree if each internal node of T has exactly k children. Within the given tree T , we distinguish two kinds of vertices: let I = fi1; i2 ; : : : ; ij jg be the set of internal nodes and D = fd1 ; d2 ; : : : ; dj j g be the set of leaf nodes. In this section, we provide a linear time algorithm for full kary tree T where each internal edge has a constant weight and latency w(I ); d(I ) and each leaf edge has w(D); d(D). First we consider the case that c(I ) c(D). This case is trivial because we can easily traverse all internal node by any order and visit every leaf nodes sequentially. Clearly, it only need O(n). If c(I ) < c(D), than we must to show two things, one is every subtree of fullary tree is inseparable so we can propose a top-down algorithm. Another is if a subtree have more internal node, it ratio cost must small than subtree have less internal node. Form this two result, we can easily provide a linear time algorithm and prove its correctness. Let w = x; w = y; d(I ) = u; d(D) = v.. 4. (T ). TreeMLT. A tree T . Output: Minimum latency tour. Data Structure: Consist each edge e(u; v ) and vertex v pair into one data node. Each node E = (e; v) have the following data: d: The latency time of e and v. w: The weight of v. cr: The cost ratio of inseparable subtour form E . parent: The parent edges. next: The next node of inseparable subtour. Initial: Build each edge and vertex pair. Let each ( ) E :cr , and set all other variable by it de

(14) nition. ( ) Input:. I. D. w e. i. d e. Build a Fibonacci heap F for all E by the key E :cr. Step 2: If F is empty then stop the job else let e is the node returned from the extract-max from F . Step 3: If e is the root edge than output the inseparable subtour starting from e following the e:next link to get all member for output. Continue the Step 2. Step 4: If e is not root edge, then we modify the parent's edge e0 :w e0:w0 + e:w, e0 :d e0 :d + e:d and re-compute the e0 :cr 0 . Continue the step 2. Step 1:. i. I. i. c. I. and. c. and. D. c <c. D.. I. k-ary. tree. T. Any sub-tree. with ratio cost. T0 i. of full. k-ary. tree T is inseparable.. Since T is a full k-ary tree ,every subtrees are de

(15) ned recursively. If T has i internal nodes T has j internal nodes, then T has ki i + 1 leaf and T has kj j + 1 respectively. We want to prove that for any sub-tree T 0 is inseparable, then (ki i + 1)y + ix c(T 0 ) lim !1 (ki i + 1)v + iu +x y = ky (2) kv + u v Proof.. i. j. i. j. i. e :d. TreeMLT. i. Figure 2: Algorithm for

(16) nding minimum latency tour in tree.. i. Since c(I ) < ++ < c(D), we can conclude that c(I ) < c(T 0 ) for all i. Let T 0 be a subtree of T , and the root of T 0 be . Let T 0 have m inseparable unit T 0 1 ; T 0 2 :::T 0 , where T 0 denote a subtree with a internal nodes. Assume that if i < j c(T ) > c(T ). We prove the induction basis by shrinking the with an inseparable unit T 0 1 with maximum ratio cost within fT 0 1 ; T 0 2 ; : : : ; T 0 g. After the shrinking process, we have (ka a + 1)y + (a1 + 1)x ky + x y c( T 0 1 ) = 1 1 (ka1 a1 + 1)v + (a1 + 1)u kv + u v (3) Since k 2, and c(I ) < c(D), we can easily prove that Equation(3) holds. Again, now we have c( T 0 1 ) + and c(T 0 ) ++ 82 j m; we can + iterative shrinking the next subtrees. Let A = a1 + a2 + : : : + a .Assume that the following equation is hold. c( T 0 1 :::T 0 2 ) kA 2 A 2 + m 2)y + (A 2 + 1)x = ((kA A 2 + m 2)v + (A 2 + 1)u 2. Theorem 3.3 The minimum latency tour can be found in. D. Lemma 4.1 Given a full. e :w. End of. Algorithm for Full. ky. x. y. kv. u. v. n. i. O(n log n) time for tree case.. n. a. From the Lemma 3.1 and Lemma 3.2, we can propose an algorithm with O(n log n) time. The algorithm is shown in Figure 2. Note that Step 1 of the algorithm takes O(n) time to build a Fibonacci heap structure. Clearly Step 2 will be execute O(n) times. Because every time the Step 2 execute, one node been deleted. Each time Step 2 execute will perform an Extract-Max action. This action need O(log n) time. Further, Step 3 can be done in O(n) time, too. The reason is every node will be output only once. Finally, Step 4 perform an increase key of Fibonacci heap. The key increasing action only cost O(1). Base on the analysis above, we can prove that the minimum latency tour of tree case can be found in O(n log n) time. Proof.. a. n. am. i. ai. ai. aj. a. a. a. am. a. a. ky. kv. x. u. y. aj. v. ky. x. y. kv. u. v. m. m. a. 3. am. m. m. m. m. m. m.

(17) +x y ky kv + u v. where T 0 1 :::T 0 2 has m 2 subtrees of T 0 . By induction, we must show that after shrink T 0 and c( T 0 1 :::T 0 2 ). The c( T 0 1 :::T 0 1 ) ++ a. am. am. a. am. c( T 0 1 :::T 0 m 2 T 0 m 1 ) kA 1 A 1 + m 1)y + (A = ((kA A 1 + m 1)v + (A 1 ky + x y kv + u v a. (T ) Input: A full k -ary tree with constant internal node weight x, latency time u and leaf node weight y, latency time v where < . Output: Minimum latency tour. Data Structure: Each node of Tree have a value V :internal indicate the internal node number of subtrees rooted by V , and a priority queue V :queue. Initial: V :internal =0 and V :queue is empty. Step 1: Use a recursive function to compute internal nodes number of each sub-trees. And keep it in V :internal. Step 2: Use a radix sort function to sort all node by V :internal. Step 3: By the sorting result R, scan R form minimum to maximum and add itself into the priority queue of its parent. Step 4: Use a depth-

(18) rst search and use priority queue V :queue of each node to choose next child to be visited. Output the chose node.. n. am. a. Algorithm. (4). a. ky. x. y. kv. u. v. a. m. m. m. 1. m. m. m. 1. 1. + 1)x + 1)u. ky. x. y. kv. u. v. I. D. I. i. j. i. i. i. I. End of kTreeMLT. j. D. I. i. D. i. i. i. According to Lemma 4.1,the c(T ) can be evaluated reasonable with full subtree node. If i < j then c(T ) c(T ) ix (kj j + 1)y + jx = (((kiki ii++1)1)vy ++iu (ki j + 1)v + ju )(j i) = ((ki i + 1)(vyu+ iuxv )((ki j + 1)v + ju) >0 Since c < c and i < j , then c(T ) > c(T ). Thus, if i < j and c < c then c(T ) > c(T ). Proof.. i. i. D. i. v. i. k-ary tree T with ratio cost c < c . Denote the sub-tree of T with nodes by T . It follows that c(T ) > c(T ) if and. y. i. Lemma 4.2 Given a full. c and c i internal i < j.. x. u. i. Because c(T 1 ) > ++ and k 2, we can prove that a root of any subtree can iterative shrink all subtree. Further, we can say any subtree of full k-ary tree is inseparable. am. kTreeMLT. Figure 3: Algorithm for

(19) nding minimum latency tour in full k-ary tree. walk back cost, we can always choose a best subpath to visit and then choose another subpath without any extra cost. Those observations construct the skeleton of k-Path algorithm. To simplify the discussion below, we give some de

(20) nitions about the k-path graph G. Let P be a path of G and = h 1 ; 2 : : : ; i be a path partition of P . Where the is the i-th path partition of . By the de

(21) nition in Section 2, each has a ratio cost called c( ). We call a subpath = is Right-Skew if and only if the ratio cost for the pre

(22) x is always smaller than the ratio cost of the remaining subpath . We also call a is Decreasing Right-Skew if each is right-skew and c( ) > c( ) for all i; j and i < j .. j. j. d. Theorem 4.3 The minimum latency tour can be found in linear time, if. i. G is full k-ary tree with constant ratio. cost.. a. a. According lemma 4.1, we can propose an algorithm with linear time. The algorithm is shown in Figure 3. Note that Step 1 of the algorithm takes O(n) time to compute internal node number of each subtree. Clearly Step 2 and Step 3 can be done in O(n) time. Step 4 is a depth

(23) rst search that traverse each node and output it in O(n) time [3]. Proof.. 5. b. b. i. i. Observation 5.1 Let a+b . If then 2. a>b. c>. a; b; c be >b 2. j. the real numbers and. c+a. Here we

(24) rst show that all single path P can be partition into Decreasing Right-Skew.. Algorithm for k-Path Graphs. In this section, we will discuss some properties of the kpath graph and propose an O(n log k) time algorithm. If k is a constant, the problem is linear time solvable. A graph G(V; E ) is called the k-path if G is a star graph which central vertex has degree k. To solve this kind of graph, we have some observations in k-path. First, after the starting point been visited, G split into k paths. Next, because we need not to compute the. Lemma 5.2 Let. P. be a path, there exist a partition. . is Decreasing Right-Skew.. Let P = v1 ; v2 : : : v . The valid partition can be found by the following step. First, for all i from 1 to n, compute the c(v1 : : : v ). Assume c(v1 : : : v ) be the maximum ratio cost among all i. Let 1 = v1 : : : v . Next, eliminate the pre

(25) x v1 : : : v from P and Proof.. n. i. j. 4. j. j.

(26) V1. V2. V3. V4. V5. V6. constant time. The time complexity of Algorithm 5 is linear. To prove the correctness of Algorithm 5, assume the result path partition is = h 1 ; 2 : : : ; i and there exist c( ) < c( ); i < j or is not rightskew. Base on the shrink condition in Step 2,4,5. Each must be right-skew. If c( ) < c( ); i < j then c( ) > c( ) when V = . If c( ) c( +1 ) then will shrink the +1 into and set the V as the new and repeat the step until the c( ) > c( ) or c( ) > c( +1 ). If c( ) > c( ) contradict the assumption. If c( ) > c( +1 ) the V will move to and then c( ) > c( ) also contradict the assumption.. V7. d. Figure 4: The example of the path partition. Algorithm. i. j. j. i. j. i. j. j. i. j. i. i. i. i. id. i. j. j. Assume in optimal solution 1 placed in front of 1 . We compare the case which 1 be placed in front of 1 . ` ` c( 1 ) ` ` c( 1 ) = c( 1 ) c( 1 ) >0 Since c( 1 ) > c( 1 ), we can change the position between 1 and 1 to get better solution. Contradiction the original assumption. Proof.. j. i. i. j. j. j. i. i. i j. i. j. i. End of. j. j. i. j. j. i. i. i. k paths. We denote the partition = h 1 ; 2 : : : ; i. If c( 1 ) > c( 1 ) then in optimal solution 1 must placed in front of 1 . as. i. j. j. j. j. Lemma 5.4 Given. j. j. j. j. j. . j. j. j. j. j. i. j. j. j. j. j. j. j. n. j. j. j. n. i. i. i. A path P = v1 ; v2 : : : ; v . Output: A path partition. Initial: i n 1; j n; V v . Step 1: Set v V and set v as the parent of v . Step 2: if v has no parent and c(v ) c(v +1 ) then shrink v +1 to v . Step 3: if v has no parent then stop the algorithm and return the partition. Step 4: If c(v ) c(v ) then shrink v to v and set V vv. Step 5: If c(v ) > c(v ) and c(v ) c(v +1 ) shrink v +1 to v and set V v v +1 . Step 6: If c(v ) > c(v ) and c(v ) > c(v +1 ) set V v. Step 7: Goto Step 1. j. i. i. (T ). pathPartition. Input:. j. i. i. pathPartition. Figure 5: Algorithm to partition path. j. j. j. j. Theorem 5.5 If. G. is a. k-path graph, the O(n log k) time.. minimum. latency tour can be done in. repeat the

(27) rst step to

(28) nd the next sub-path . In this phrase we can easily prove that c( ) > c( ) because if c( ) c( ) then c( ) > c( i) (Observation 5.1) contradict the

(29) rst step. Further, because c( ) > c( ) for all i; j , where i < j has been proved. From the Observation 5.1, if 1 is not the right-skew then there exist a pre

(30) x of 1 has larger ratio cost. Contradict the assumption of

(31) rst step. The 1 is right-skew. If +1 is not the rightskew then may concatenate the pre

(32) x of +1 to enlarge the ratio cost of . Contradict the assumption of

(33) rst step. The +1 is right-skew. Thus, we can prove that given a path P , we always can partition it into Decreasing Right-Skew. Algorithm to prove Lemma 5.2 costs O(n2 ) times. In Algorithm 5, we will use a bottom-up manner to reduce the time complexity to linear time. i. i. j. i. j. i. i. i. i. id. cording to the Lemma 5.4, we can initial build a heap from 1 ; : : : ; 1 node. And iterative extract the maximum right-skew form heap and insert the ( +1) into heap. After the iterative we get a sequence of walk tour. From the Lemma 5.4, we prove the output walk tour is a minimum latency tour for G. . j. i. k. ij. i. i. Base on Lemma 5.3, we can partition all the. Proof.. k path into several Decreasing Right-Skews in linear time. We denote as = h 1 ; 2 : : : ; i. Then ac-. j. i. i j. i. 6. i. Lower Bound. In this section, we prove that (n log n) and (n) are lower bounds for general tree , full k-ary tree with constant weight, and k-path graphs respective. Theorem 6.1 Lower bound of the variation minimum. (n log n) even for binary tree. Proof. Given a set of nonnegative real values fN0 ; N1 ; : : : ; N g, we construct a binary tree illustrated by Figure 6. Note that all internal nodes and one leaf node have a very large weight, say 1. Because w(D +1 ) = 1, we

(34) rst shrink fI0 ; I1 ; : : : ; I ; D +1g into one inseparable supper node. After this, a comparison sorting is required for determining the output sequence. Lemma 5.3 Algorithm 5 is correct and the time complexity is. latency problem is. O(n).. Consider the Algorithm 5. The V denote the current working pointer initially at the end vertex of P . Every times from Step 1 to Step 6 will cause the V move to v1 at least one vertex or shrink at least one vertex. After the linear step, the path will be partition into Decreasing right-skew. Because each step only cost Proof.. n. n. n. 5. n.

(35) Those problems still remain open. Of course, the ultimate open problem is whether the variant of minimum latency problem can be eÆciently solved in dags.. N0 N1. References. N2. [1] D. Adolphson and T. C. Hu. Optimal linear ordering. SIAM J. Appl. Math., 25:403{423, 1973. [2] A. Blum, P. Chalasani, D. Coppersmith, B. Pulleyblank, P. Raghavan, and M. Sudan. The minimum latency problem. Proc. 26th Annu. ACM Sympos. Theory Comput., pages 163{171, 1994. [3] T. Corman, C. Leiserson, and R. Rivest. Introduction to Algorithms. MIT Press, 1990. [4] T. Imielinski, S. Viswanathan, and B.R. Badrinath. Power eÆcient

(36) ltering of data on air. 4th Inter-. 8. Nn. Figure 6: The worst case for general tree graph of fN0; N1 ; : : : ; N g. Note that (n log n) time is the lower bound for comparison sorting. Thus, we can say the lower bound of the minimum access latency problem of general tree is (n log n) time. n. national Conference on Extending Database Tech-. Theorem 6.2 Algorithm kTreeMLT is optimal in terms. [5]. Obviously the lower bound is (n) since the size of the output set is exactly the input size n for the full k-ary tree. It follows that algorithm kTreeMLT is optimal for the full k-ary tree. Proof.. 7. , pages 245{258, March 1994. T. Imielinski, S. Viswanathan, and B.R. Badrinath. Data on air: Organization and access. IEEE Trans. on Knowledge and Data Engineering, 9(3):353{372, May 1997. Shou-Chih Lo and Arbee L.P. Chen. Index and data allocation in multiple broadcast channels. IEEE Interantion Conference on Data Engineering 2000, pages 293{302, 2000. N. Shivakumar and S. Venkatasubramanian. Energy-eÆcient indexing for information dissemination in wireless systems. ACM, Journal of Wireless and Nomadic Application, 1996. M. Chen P. Yu and K. Wu. Indexed sequential data broadcasting in wireless mobile computing. IEEE noloy. of time complexity.. [6]. Concluding Remarks. In this paper, we de

(37) ne a variation of original minimum latency problem. In the new problem, the weight and latency cost of vertices been de

(38) ned additionally and we ignore the walk back cost to make the problem more realistic in some real problem. This variation is only de

(39) ned in trees and some simpler graphs, like full k-ary tree. In this paper, we propose three versions of algorithms for di erent graphs. Let n denote number of vertices in the given graph. We

(40) nds an O(n log n) time algorithm for general trees with weighted vertices and weighted edges, and an O(n) time algorithm for full k-ary tree with constant ratio costs of each vertex. For general case, previous results [1] show that O(n log n) is optimal, but our algorithm seems more intuitive and easily to implement. Two topics concerning the minimum latency problem will be addressed in the future. One is the problem in dags, we believe that is NP or NP Hard, another is to consider an eÆcient on line algorithm to immediately re ect the weight change. For the case of direct acyclic graph, we didn't

(41) nd an polynomial time algorithm yet. It will be interesting to know whether the time complexity of the algorithm on dags can be solved in polynomial time. Another interesting problem is about the starting point. In our de

(42) nition, the starting point is given and never change. If the starting point is online changed. Can we get the new minimum latency tour eÆcient? If the starting point is unde

(43) ned, where the best starting point is?. [7]. [8]. Interantional Conference on Distributed Computing Systems. 6. , pages 124{131, 1997..

(44)