Analysis of Power Assignment in Radio Networks with Two Power Levels

Transcription

1 Analysis of Power Assignment in Radio Networks with Two Power Levels Miguel Fiandor Gutierrez & Manuel Macías Córdoba Abstract. In this paper we analyze the Power Assignment in Radio Networks with Two Power Levels paper by Paz Carmi and Matthew J. Katz, referring to the problem of the power assignment in radio networks, where each transmitter-receiver can transmit in one of two possible power levels. We will explain in further detail the points of the paper we find more difficult to understand, as well as give examples. This paper will also analyze the application of the algorithm and give a list of advantages and disadvantages of the use of the algorithm in different purpose radio based networks. Keywords. Radio Network, Power Consumption, Approximation Algorithm, NP-Hard. 1. INTRODUCTION As stated in the original paper, the assignment of power levels (corresponding to transmission ranges) to the transmitters of a radio network, so the total power used is as low as possible, is a well known and in some cases (embedded networks) is a extremely important issue. Let P be the plane containing n point (each point represents a transmitter-receiver).we will assign transmission ranges to the transmitter-receiver in P, so that the resulting transmission graph is strongly connected; in other words, there will be a directed edge between the transmitter p and the transmitter q, if and only if q is in range r p assigned to p, so there will be a directed path in G between every p E P and every q E P. And the total power consumption (the cost of the assignment of ranges) is minimized, were the total power consumption is a function of the form Σ pєp r p c and c > 0 is a constant typically between 2 and 5. This version of the power assignment problem is know to be NP-hard, Kirousis proved it to 3-dimenssional point sets and presented a 2-approximation algorithm based on the minimum spanning tree of P which is the best approximation known, and Clementi prove that the problem is np-hard for planar point sets.

2 Miguel Fiandor Gutiérrez & Manuel Macías Córdoba Outside the papers, normally is impossible to assign arbitrary power levels, and we are forced to choose between predetermined ones. The problem analyzed in this paper is the one referred in the original paper by Paz Carmi and Matthew J. Katz, in which there are two power levels the radio transmitter-receivers can be assigned with, one corresponding to short range (r 1 ), and the other one corresponding to long range (r 2 ). Since the cost of the assignment of power level to the transmitters-receivers is the form of n1r1 c + (n n1)r2 c were n 1 is the number of transmitters-receivers assigned with range r 1 and c 1 is some constant, the cost of the assignment is determined logically as the number of transmitters-receivers assigned with r 2. In the paper is proved that the power assignment problem for two different power levels is np-hard when r2 > (3 r1/2). Being m the optimal number of transmitters-receivers assigned long range, we will show a polynomial time algorithm that solves the stated problem with a (11/6)mapproximation. A immediate corollary of this result is that for any ranges r 1 and r 2, and for any c, we can compute an assignment which cost is at most (11/6) times the cost of the optimal assignment. But as we will show in section 2.2, this cost is usually lower. The algorithm that is going to be shown consist mainly in two parts, the first one will assign short and long range to the transmitter-receivers in a way that the resulting communication graph will be a tree. The second part will perform an algorithm to assign the power levels with a (4/3) approximation to the number of optimal long range assignments needed to solve the tree G. A by product of the algorithm is logically generated, in the case the initial graph is a tree the algorithm will be 4/3m, were m is the optimal solution. In the last part of the paper we will give examples of the algorithm use in real radio networks, we will analyze its advantages and disadvantages according to fault tolerance, energy consumption management. 2. AN (11/6)-APPROXIMATION In the following paragraphs we will try to explain the algorithm presented in the original paper. Let n be the number of transmitter-receivers of the set P in a plane. Each transmitter can transmit in two different power levels, low (short range r 1 ) and high (long range r 2 ). If all transmitters in P use long range the resulting communication graph is strongly connected. The problem that the algorithm is designed to solve is to assign ranges to the transmitters

3 Analysis of Power Assignment in Radio Networks with Power Levels Figure Transmitters-Receivers network The algorithm that is going to be shown, will in polynomial-time assign ranges to the transmitters in P, such as the number of transmitters assigned long range is at most (11/6)m, were m is the number of transmitters that are assigned long range by OPT. For the set P of transceivers-receivers we will set all of them to short range. Next we will draw an edge between every pair of transceivers-receivers p and q which Euclidean distance is smaller than the short range distance, p-q < r1.we will get a (undirected) graph G, of components, being each component a subset of P. Figure Graph G of components. Now we draw an edge between two components C 1 and C 2 of G if there exist transmitters p Є C 1 and q Є C 2, such that p, q r2. See Figure

4 Miguel Fiandor Gutiérrez & Manuel Macías Córdoba Figure Graph G of connected components We can easily obtain a 2-approximation algorithm. By computing a minimum spanning tree of G,and,for each edge (C 1, C 2 ) of the tree, assign long range to two transmitters p Є C 2 and q Є C 2, such that p, q r2. The range assignment algorithm explained in the paper consists in two parts. In the first one we will repeatedly find a cycle (between components) in G and reduce it to a single component by assigning long range to one of the transceivers-receivers in each of the components of the cycle. The second part begins when there are no more cycles in G, the graph G will be a tree, and more transmitters will be assigned long range in order to finish our task. The following paragraphs will describe the two parts in detail: The first part of the algorithm will continue until there are no cycles in G. For the cycle C 1, C 2, C l, C 1 (be l 3). We will assign long range to a transmitter-receiver in C 1 that can reach C 2, then assign long range to a transmitter-receiver in C 2 that can reach C 3,etc. L transmitters will have now being assigned with long range. After doing this, any pair of transmitters in the union C = C 1 U U C l can talk with each other mainly trough other transmitter in C. Now we can merge these l components in the union C, reducing the number of components of G by l-1. After doing so, we forget that some of the transmitters-receivers in C have been set to long range and update the edges of G accordingly, see Figure 2.4. Figure Reducing the Cycle C 1, C 2, C 3, C 4, C 1 to the single component C When the first part of the algorithm have finished there will be no cycles in G (G will be a tree)

5 Analysis of Power Assignment in Radio Networks with Power Levels In the second part of the algorithm the 4/3-approximation algorithm for a tree, shown in section 3, will be applied to G in order to complete the range assignment task. Now we will show, as in the original paper, that the overall number of transmitters that were assigned long range is bounded by 11/6 m. Theorem 1. The range assignment algorithm (described above) computes an (11/6)-approximation in polynomial time. Proof. In the first part of the algorithm while there are still cycles in G, they will be replace by a single component in G. Note, we assume that this cycles are of length three as this is the worst case for the analysis. Let i be the number of cycles found during the first part of the algorithm. Let k be the number of components in G, at the beginning of the first part of the algorithm, just before the loop starts. Them m, the number of transmitter-receivers assigned long range by OPT, is at least k, since in initial component at least one of the transmitters-receivers has to be assigned long range. During the first part of the algorithm 3i (3 components in each cycle by i cycles) transmitters have been assigned long range, and the number of components in G at the end of this first part of the algorithm are k 2i. As it was explained above, when the first part of the algorithm finishes G is a tree, to further the analysis we have to distinguish between two different cases: Case 1: i > k/2 m/3, in this case instead of performing the second stage, we proceed in the most trivial way (for the porpoise of the analysis only) and assign long range to 2(k - 2i -1) transmitters. In other words, for each edge in G connecting two components for example the edge between C 1 and C 2, we assign long range to any transmitter-receiver in C 1 that can reach C 2 and vice versa. The total number of transmitters-receivers assigned long range is bounded by: 3 i + 2(k -2i -1) 2k i 2k (k/2 m/3) = 3k/2 m/3 11/6m 3 i + 2(k -2i -1) : number of transmitters-receivers assigned long range in the first part of the algorithm plus the ones in the second part. i > k/2 m/3 m is at least k Case 2: i k/2 m/3. In this case the second part of the algorithm is applied and assign long range to at most 4/3 m tree transmitters-receivers, where m tree is the number of long range assignments needed to solve the tree G. Obviously m tree m, so the number of transmitters-receivers assigned long range in the second part of the algorithm is at most 4/3 m. The total number of transmitters-receivers assigned long range is bounded by: 3i + 4/3m 3(k/2 m/3) + 4/3m = 3/2k + m/3 11/6m As in booth cases we were able to achieve an upper bound of 11/6m, we can reach the conclusion that our range assignment algorithm computes an 11/6-approximation

6 Miguel Fiandor Gutiérrez & Manuel Macías Córdoba As stated above, the cost of the algorithm is n1r1 c + (n n1)r2 c,were n 1 are the transmitters-receivers assigner range r 1 and c 1 is a constant between 2 and 5. An immediate corollary of Theorem 1 is that for any ranges r 1 and r 2 and for any c, we can compute and assignment of at least 11/6 cost of the optimal assignment. We will show below that usually the cost of the assignment that our algorithm makes is lower than this upper bound Algorithm proof example In order to give a clear and easy example of the behavior of the algorithm shown in this paper, we have chosen a communication network which fulfils the requirements of the case used to analyze the algorithm in the previous section. We start with the following 18 transmitter-receiver nodes in the plane, and the two power levels referring to the range distance r 1 and r 2, r 1 << r 2. In the firs step we set the range of the transmitters to r 1 and we draw an edge that connects every pair of nodes which Euclidean distance is less than r 1. The nodes that fulfill this condition are the pairs: (T 1a, T 1b ), (T 2a, T 2b ), (T 3a, T 3b ), (T 4a, T 4b ), (T 5a, T 5b ), (T 6a, T 6b ), (T 7a, T 7b ), (T 8a, T 8b ), (T 9a, T 9b )

7 Analysis of Power Assignment in Radio Networks with Power Levels Each connected nodes, will form a component of the graph G (C 1, C 2, C 3, C 4, C 5, C 6, C 7, C 8, C 9,). Now we draw an edge between two components C a and C b of G if there exist a transmitter p Є C a and q Є C b, such that p, q r2 and we obtain the graph of connected components G

8 Miguel Fiandor Gutiérrez & Manuel Macías Córdoba At this point we start the algorithm. As we can see the first cycle we encounter is the one composed by the components C 1, C 2, C 3. As the algorithm says we assign long range to any transmitter-receiver in C 1 which Euclidian distance with a transmitter from C 2 is less than r 2 and so on. In total we assign long range to three transmitters. Next we group the components C 1, C 2 and C 3 in a new component C 10 and substitute it in G. The next cycle we found is the one composed by C 10, C 4, C 5. We follow the steps taken before and we generate the component C 11 = C 10 U C 4 U C

9 Analysis of Power Assignment in Radio Networks with Power Levels We continue with the algorithm, and we select the next cycle C 11, C 7, C 8 in this iteration of the loop. We follow the steps taken before and we generate the component C 12 = C 10 U C 7 UC 8. The resulting graph has three component and cero cycles. The first part of the algorithm is finished. We have assigned long range to 9 transmitters-receivers (three in each iteration). Continuing with the analysis done in the section 2, we see that i > k/2 m/3, so as case one says, we assign long range to four more transmitter-receivers. Two for the communication between C 12 and C 6, plus two between C 12 and C 9. Analyzing the results we get: Total number of transmitters assigned long range: 3i + 4 = 13. m = 9. K = 9. i = 3. 3 i + 2(k -2i -1) 11/6m. => 13 < 16.5 This fits in the formulas stated in the sections above. Where: i = number of iterations in the first part of the algorithm m = optimal solution of the problem. 2.2 The cost for ranges 1 and d. This is the particular case we approximate defining the range of the two possible power levels as, r1 = 1 and r2 = d. Thus, we have the following theorem, - 9 -

10 Miguel Fiandor Gutiérrez & Manuel Macías Córdoba Theorem 2. If r 1 = 1 and r 2 = d, then one can compute a range assignment whose cost is at most 11d c / 6d c + 5 times the cost of an optimal assignment. For d = 2 we get a (44/29)-approximation, if c = 2, and a (22/17)-approximation, if c = 1. Proof. The cost of an optimal algorithm is d c m + 1 (n m) = n + (d c 1)m, where m is the number of transmitters assigned long range. We apply both our algorithm and the naive algorithm which assigns range d to all the transmitters, depending on the variable a, where a = n/m, n is the total number of transmitters in the problem. We distinguish between two cases. Case 1: a 11/6. In this case we use the naive algorithm whose cost is d c n. The ratio between the cost of the naive algorithm and the cost of an optimal algorithm is: d c n / (n + (d c 1)m) d c n / (n + (d c 1)(6/11)n) = d c / ((6/11)d c + (5/11)) In the left side of the constraint we have the cost of the naive algorithm, so in the right side we have the cost of an optimal algorithm. To calculate the ratio m has been replaced by (6/11)n in the right side, because the optimal value is obtained when m = (6/11)n. And it is demonstrated that every possible solution in case 1, left side constraint, will be upper bounded by d c / (6/11)d c + (5/11), the right side. Case 2: a 11/6. In this case we run our algorithm whose cost is: d c (11/6)m + 1 (n (11/6)m) = n + (11/6)(d c 1)m The ratio between the cost of our algorithm and the optimal one is: n + (11/6) (d c 1)m / (n + (d c 1)m) = a + (11/6)(d c 1) / a + (d c 1) (11/6)d c / d c + 5/6 = d c /(6/11)d c + (5/11)

11 Analysis of Power Assignment in Radio Networks with Power Levels Also in this case we got a d c / (6/11)d c + (5/11) -approximation on the cost. Thus we can check that in both cases for d = 2 we get a (44/29)-approximation, if c = 2, and a (22/17)-approximation, if c = A (4/3)-APPROXIMATION FOR A TREE OF COMPONENTS Now we explain the case where the components graph G is cycle free, i.e., where G is a tree, so a (4/3)-approximation algorithm is given. G is the graph obtained at the end of the first step of the general algorithm above. So, the (4/3)-approximation algorithm forms the second step of the initial algorithm and it works as follows. Fig 3.1 Firstly is picked an arbitrary component in G to be the root of G. Given a component C in G, we talk about its children components or leafs, and about the parents and neighbours. For connecting the different components C, we need to assign long range to some of the transmitters that belong to it, with the purpose of connect such component C to its children. So at least one of the transmitters in C assigned long range can reach C, and also at least one of these transmitters can reach the parent of C. A neighbour C of C is satisfied if at least one of the transmitters in C that can reach it when assigned long range, finally it has long range. Fig 3.2 The second step is start with the leaf components. We have two types of components leaves and internals. We assign long range to any transmitter in C that can reach the parent of C using long range. Finishing with all the leaf components of particular C, we will consider next group of leaf components until we finish the leaves

12 Miguel Fiandor Gutiérrez & Manuel Macías Córdoba Then we will start considering the internal components in the same way it has been done before. Now we show that this builds a (4/3)-approximation algorithm with the following definitions: C be the internal component that is about to be considered. X C is the number of children of C. m C is the number of long range assignments (transmitters in C) needed to satisfy all children of C. m C is the number of long range assignments (transmitters in C) assigned by OPT. Clearly for each child C of C, we must assign long range to at least one of the transmitters in C that can reach C. We have two possible cases, if the m C transmitters satisfying the children of C can be chosen so that one of them also satisfies the parent of C, then m C = m C, otherwise we would have to choose a new transmitter in C to satisfy its parent, so, m C = m C + 1. The following inequalities are immediate: Σ C m C = m, where m is the overall number of long range assignments assigned by OPT, and Σ C X C = k 1 < m, where k is the number of components in G. We will assign long range to at most (1/3)X C + m C transmitters in C. Summing over all components in G we obtain: ΣC ((1/3)XC + mc) 1/3m + m = 4/3m Notice that for each transmitter p in C there are a number of children of C that would be satisfied if p were assigned long range, that is called d p (the degree of p). This value has to be updated each time a transmitter q in C is selected to long range, this demands d q = 0. Because increasing the range to q we can satisfy children that were counted in any d p value. of X C, There is a last step of the algorithm of the original paper. Depending on the value

13 Analysis of Power Assignment in Radio Networks with Power Levels If XC 2, then we solve C optimally, that is, we find a minimum subset of transmitters in C that can reach all children of C and can also reach its parent (when assigned long range). We can do this since in this case mc 3. Otherwise, as long as the number of unsatisfied children is at least 3 and there exists a transmitter of degree at least 3, we assign long range to any transmitter q and update the degrees of all transmitters in C accordingly. By assigning long range to q we satisfy at least 3 of the children of C. Since for each of these 3 children, OPT assigns long range to one of their transmitters so that it can reach C, we charge the assignment to q to these 3 assignments of OPT. Thus in this loop we have used at most 1/3(XC x) long range assignments to transmitters in C, where x 0 is the number of remaining unsatisfied children of C. All this process justifies the following theorem, Theorem 3. If the components graph G is a tree, one can compute a range assignment that is a (4/3)-approximation in polynomial time. In order to clarify the ideas and demonstrate the Theorem 3 we show the next example, 3.1 Example 1. We start with five components that build a graph G free of cycles, what means it generates a tree T. See Fig Now we pick an arbitrary component in G to be the root of a new tree T. The component chosen as root is C We have to start with the leaf components, in this case C 5, we assign long range to any transmitter in C 5 that can reach its parent, C 4. Here could come the first difference with OPT, in case there is a transmitter in C 4 that can reach at the same time C 5 and its parent, C 2, when it is assigned with long range. OPT would choose a transmitter in C 5 that could reach such transmitter in C

14 Miguel Fiandor Gutiérrez & Manuel Macías Córdoba 4. We repeat this step with components C 2, C 3, and C 4. This last case, C 4, is when XC 2 when so is solve optimally and trivially, only has to be chosen the transmitters in C 4 that can reach all children, C 5, and also its parent C 2, at most 3 transmitters. 5. After this, we continue with the last component C 2, where the unsatisfied children are at least 3, and we simulate the case there exists one or more transmitters with degree at least 3 (in case with a smaller degree it is solved in a similar way). Founded all transmitters in C with degree equal 3; those are assigned with long range. Here OPT would difference from the algorithm choosing only one transmitter since a set of transmitters reach the same transmitters, and their degree is at least 3. It is not the case, but after this loop only can left transmitters with degree at most 2, and it would be solve optimally, as in step NP-HARDNESS As any computing problem is of huge interest to know about its NP-Hardness (non-deterministic polynomial-time hardness). In the original paper is shown that the problem of finding an optimal range assignment for a given set P of points in the plane, that represent transmitters-receivers, is NP-Hard, when any two ranges r 1 and r 2 subject to r2 > (3/2) r1. Even in the special case where is assigned short range to every component, we obtain a graph G, and then G consist on a central component C that is connected to k orbit components. This special case is also NP-Hard

15 Analysis of Power Assignment in Radio Networks with Power Levels Now we explain how the original paper qualifies this problem as NP-Hard. First it is used a reduction from minimum vertex cover in planar cubic graphs, PGC, where each of the nodes has degree at most 3. If a PCG = (V, E) is a planar cubic graph, a vertex cover for PCG is a subset U of V, such for each edge (v 1, v 2 ) Є E, either v 1 Є U or v 2 Є U. So, they use this reduction of the problem whose NP-Hardness is justified to its first reference. The original paper uses a second reduction, from the min vertex cover-4 to min vertex cover-3 where is demonstrated that is APX-Hard in its references [1] and [9]. The process continues transforming the planar cubic graph PGC = (V, E) in a rectangular grid of size O( V 2 ), what is showed by Valiant in the original paper reference [13]. Then each node v Є V corresponds to some grid vertex, and each edge (v1, v2) Є E corresponds to a rectilinear path of grid edges, whose endpoints are the grid vertices corresponding to v 1 and v 2. V 1 V

16 Miguel Fiandor Gutiérrez & Manuel Macías Córdoba Finally, its become with the conclusion that an optimal solution for the range assignment problem corresponds to a minimum vertex cover for the graph PCG. Theorem 4. Let r 1 and r 2 be any two ranges, such that r2 > (3/2) r1. Then the problem of finding an optimal range assignment (where r 1 and r 2 are the two possible ranges) for a given set P of points in the plane is NP-Hard. 5. Advantages and Disadvantages We have focused our study on the main cases where the power consumption problem is considered a high important issue. If we look at embedded networks (mainly consist on a collection of small devices known as motes connected each other via a wireless communication) the power consumption issue can be considered critical, as the motes carries small batteries. So, applying the algorithm in order to minimize the long range assignment to motes also saves energy to the network. We talked about power consumption as the main problem of the studied network, but this is not the only problem can arise in it, a radio network has to deal with other problems like, weak fault tolerance or low bandwidth. There are different kinds of networks with different requirements like Bluetooth or Wi-Fi. When we are in charge of developing or managing a network we would have to take a decision between all possible algorithms and the different features they give to the computed network. The algorithm exposed has an (11/6)-approximation factor to the optimal solution in the two range radio assignment. As a longer range assignment becomes with a higher power consumption, this 11/6 factor can be considered as an energy save feature of the network. When we talk about fault tolerance we mean the possibility that our network never split when one or more transmitters come down, disconnecting two components or dividing one in two. If the first step of the algorithm has finished and the resulting graph G is just one component (cycles have been founded in each iteration), between each component only exists one path. This is the worst case for computing a reliable high fault tolerance network, and also with a high bandwidth. The bandwidth feature is quite difficult to analyze, and also this kind of networks are not usually known by a high payload that goes through them. But this does not mean they do not exist, an example is the increasing of pictures and videos sharing between mobile phones, what demands a higher bandwidth transmission. The weakness in this algorithm can come in the connections between components (those made by long range transmitters). If a transmitter t 1 needs to communicate with another t 2, which is allocated in a different component it will be use the path made by the long range transmitters. So, only one path could not be a desired solution. This study introduce much more complexity and other fields of the communication problems that not belong to original paper, we only pretend to present the

17 Analysis of Power Assignment in Radio Networks with Power Levels possible weakness points can become with extremely optimal solution to the power assignment problem. 6. Problems Related Power Assignment For Symmetric Communication InWireless Sensor Networks. omp/proceedings/iscc/2006/2588/00/2588toc.xml&doi= /iscc E. Althaus, G. Cˇalinescu, I. I. Mˇandoiu, S. Prasad, N. Tchervenski, and A. Zelikovsky. Power Efficient Range Assignment in Ad-hoc Wireless Networks. Proc. IEEE Wireless Communications and Networking Conference (WCNC), , W. Liang. Constructing Minimum-Energy Broadcast Trees in Wireless Ad Hoc Networks. Proc. 3rd ACM International Symposium on Mobile Ad Hoc Networking and Computing, , I. Papadimitriou and L. Georgiadis. Energy-aware Broadcasting in Wireless Networks. Proc. Modeling and Optimization in Mobile, Ad Hoc and Wireless Networks (WiOpt), , P.-J. Wan, G. Cˇalinescu, X.-Y. Li, and O. Frieder. Minimum-Energy Broadcast Routing in Static Ad Hoc Wireless Networks. ACM Wireless Networks 8: ,