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Maximizing survivability of acyclic transmission networks with multi-state retransmitters and vulnerable nodes Gregory Levitin Reliability Department, Planning, Development and Technology Division, Bait Amir, Israel Electric Corporation Ltd., P.O. Box 10, Haifa, 31000 Israel E-mail: [email protected] Abstract In this paper, an algorithm for optimal allocation of multi-state elements (MEs) in acyclic transmission networks (ATNs) with vulnerable nodes is suggested. The ATNs consist of a number of positions (nodes) in which MEs capable of receiving and sending a signal are allocated. Each network has a root position where the signal source is located, a number of leaf positions that can only receive a signal, and a number of intermediate positions containing MEs capable of transmitting the received signal to some other nodes. Each ME that is located in a nonleaf node can have different states determined by a set of nodes receiving the signal directly from this ME. The probability of each state is assumed to be known for each ME. Each ATN node with all the MEs allocated at this node can be destroyed by external impact (common cause failure) with a given probability. The ATN survivability is defined as the probability that a signal from the root node is transmitted to each leaf node. The optimal distribution of MEs with different characteristics among ATN positions provides the greatest possible ATN survivability. It is shown that the node vulnerability index affects the optimal distribution. The suggested algorithm is based on using a universal generating function technique for network survivability evaluation. A genetic algorithm is used as the optimization tool. Illustrative examples are presented. Keywords: transmission network, multi-state, survivability, vulnerability, optimal allocation. 1 Abbreviations ATN acyclic transmission network ME multi-state element UGF universal generating function Nomenclature S ATN survivability N total number of nodes in ATN M number of leaf nodes in ATN D number of MEs to be allocated at ATN node vulnerability probability that node survives during the system operation time (=1-) ci i-th node of ATN set of MEs i set of MEs allocated at ci ik set of nodes receiving a signal from ME located at ci when it is in state k Ki number of different states of individual ME located at node ci ~ Ki number of different states of group of MEs located at node ci K̂ i number of different states of group of MEs located at nodes c1,…,ci p d i ik probability that a signal from d-th ME located at ci reaches set of nodes ik Pd state probability distribution matrix for ME d Vi random binary vector representing set of ATN nodes receiving a signal directly from single ME located at node ci ~ Vi random binary vector representing set of ATN nodes receiving a signal directly from group of MEs located at node ci 2 random binary vector representing set of ATN nodes receiving a signal from c1 through all V̂i the MEs located at c1,…,ci Vik value of Vi at state k (vector representing the set ik) ~ Vik ~ value of Vi at state k V̂ik value of V̂i at state k V0 zero vector ~ qik ~ ~ probability that Vi is equal to Vik q̂ik probability that V̂i is equal to V̂ik uid(z) u-function corresponding to ME d located at node ci (represents probabilistic distribution of Vi) ~ U i (z) u-function corresponding to group of MEs located at node ci (represents probabilistic ~ distribution of Vi ) Û i (z) u-function corresponding to group of MEs located at nodes c1,…,ci (represents probabilistic distribution of V̂i ) u-function simplification operator , composition operators over u-functions ~ operator incorporating node vulnerability into U i (z) function for vector composition h(d) number of node in which ME d is allocated (allocation function) 1. Introduction Acyclic transmission networks (ATN) consist of a certain number of positions (nodes) in which multistate elements (MEs) capable of receiving and/or sending a signal are allocated. Each network has a root node where the signal source is located, a number of leaf nodes that can only 3 receive a signal and a number of intermediate (neither root nor leaf) nodes containing MEs capable of transmitting the received signal to some other nodes. The signal transmission is possible only along links between the nodes. The networks are arranged in such a way that no signal leaving a node can return to this node through any sequence of nodes (no cycles exist). Each ME located in nonleaf node can have different states determined by a set of nodes receiving the signal directly from this ME. The event that a ME is in a specific state is a random event. The probability of this event is assumed to be known for each ME and for every its possible state. All the MEs in the network are assumed to be statistically independent. The whole network is in working condition if a signal from the root node is transmitted to each leaf node. Otherwise, the network fails. (Note that it is not always necessary for a signal to reach all the network nodes in order to provide its propagation to the leaf ones). An algorithms for ATN reliability evaluation and for ME allocation maximizing the ATN reliability were suggested [1] and [2] respectively. In this paper we consider the case when the MEs allocated at the same mode are subject to a common cause failure. When a system operates in battle conditions or is affected by a corrosive medium or other hostile environment, the ability of a system to tolerate both the impact of external factors (attack) and internal causes (accidental failures or errors) should be considered. The measure of this ability is referred to as system survivability. An external factors usually cause failures of group of system elements sharing some common resource (allocated within the same protective casing, having the same power source, gathered geographically etc.). Therefore adding more redundant parallel elements will improve a system availability but will not be effective from a vulnerability standpoint without sufficient separation between elements. In ATN considered, all the MEs located at the same node can be destroyed by a single external impact. As it was shown in [2], in many cases greater ATN reliability can be achieved if some of MEs are gathered in the same node providing redundancy (in hot standby mode) and some nodes remain 4 empty, than if all the MEs are evenly distributed between all the nonleaf nodes. When the ATN nodes are vulnerable (all MEs belonging to a node can be destroyed by a single impact), the allocation providing the greatest ATN survivability can change. In order to estimate the effect of the node vulnerability on the optimal ME allocation one has to include this parameter into ATN survivability model. Consider, for example, the simplest case in which two identical MEs should be allocated within ATN with N=3, M=1. When allocated at node c1, each ME can have four states: - total failure: ME does not connect node c1 with any other node (probability of this state is p11= p21= p1); - ME connects c1 with c2 (probability of this state is p11{2}= p21{2}= p1{2}); - ME connects c1 with c3 (probability of this state is p11{3}= p21{3}= p1{3}); - ME connects c1 with both c2 and c3 (probability of this state is p11{2,3}= p21{2,3}= p1{2,3}). When allocated at node c2 the MEs can have two states: - total failure: ME does not connect node c2 with any other node (probability of this state is p12= p22= p2); - ME connects c2 with c3 (probability of this state is p12{3}= p22{3}= p2{3}); The probability that each node survives during the system operation time is . Let suppose that p1=p2. There are two possible allocations of the MEs within the ATN (figure 1): A. Both MEs are located in the first position. B. The MEs are located in the first and second positions. In case A, the ATN survives if node c1 is not destroyed and at least one of the MEs is in state {3} or {2,3} and the system survivability is SA={2(p1{3}+p1{2,3})-(p1{3}+p1{2,3})2}. 5 (1) In case B, the ATN survives either when the node c1 is not destroyed and ME located in c1 is in state {3} or {2,3}, or if both nodes c1 and c2 are not destroyed, the ME located in c1 is in state {2} and the ME located in c2 is in state {3}. The system survivability in this case is SB={p1{3}+p1{2,3}}+2p1{2}p2{3}. (2) Since p1=p2, one can rewrite the expression (2) as SB={p1{3}+p1{2,3}}+2(1-p1{3}-p1{2,3}-p1)(1-p1). (3) By comparing (1) and (3), one can decide which allocation of the elements is preferable for any given , p1 and p1{3}+p1{2,3}. Condition SASB can be rewritten as (p1{3}+p1{2,3})(1-p1{3}-p1{2,3})(1-p1{3}-p1{2,3}-p1)(1-p1) and finally as (p1{3}+p1{2,3})(1-p1{3}-p1{2,3})/(1-p1{3}-p1{2,3}-p1)(1-p1). (4) Figure 2 presents the maximal values of for whish SASB as function of variables p1 and p1{3}+p1{2,3}. Observe that for given combinations of p1 and p1{3}+p1{2,3} the values of located below the curve correspond to cases when the solution A is preferable while the values of located above the curve correspond to cases when solution A provides lower system survivability than solution B. This paper presents an algorithm for optimal allocation of MEs in ATN in which the nodes vulnerability is taken into account. The algorithm finds the allocation of arbitrary number D of MEs with given state probability distributions (depending on MEs allocation) which maximizes survivability of an ATN with given topology. Section 2 of the paper presents a description of the acyclic transmission network model. Section 3 describes the technique used for evaluating the reliability of network with given ME allocation. Illustrative examples are presented in the fourth section. 2. Model description 6 An ATN can be represented by acyclic directed graph G=(C,E) with N nodes ciC (1iN), M of which are leaf ones. The nodes are numbered in such a way that for any arc (ci,cj)E j>i and last M numbers are assigned to the leaf nodes: cN-M+1,…,cN (note that such numbering is always possible in acyclic directed graph). The existence of arc (ci,cj)E means that a signal can be transmitted directly from node i to node j. One can define for each nonleaf node ci a set of nodes i directly following ci: cji if (ci,cj)E (see Fig 3). There are D available MEs with different characteristics. Each ME has its unique number. Multistate elements located in each nonleaf node ci (1iN-M) can transmit a signal to the nodes belonging to the set i. In each state k, these elements transmit a signal to some subset ik of i (in the case of total failure, the ME cannot transmit a signal to any node: ik=). Each ME d (1dD) located at ci can have Ki different states and each state k has probability p d i ik , such that Ki pd i ik 1 . The states of all the MEs are independent. One can see that for each ME d all its k 1 possible state probability distributions depending on its location are defined by matrix Pd={ p d i ik }, 1iN-M, 1kKi. A signal can be retransmitted by the ME located at ci only if it reaches this node. All the MEs belonging to the same mode can be destroyed with probability . The system survivability S is defined as a probability that a signal generated at the root node c1 reaches all the M leaf nodes cN-M+1,…,cN. The ME allocation problem can be considered as a problem of partitioning a set of D MEs into a collection of N-M mutually disjoint subsets i (1iN-M), i.e. such that NM i , (4) i 1 i j ø, ij. 7 (5) Each set i, corresponding to ATN position ci, can contain from 0 to D MEs. The partition of the set can be represented by the vector H={h(d),1dD}, where h(d) is the number of the subset to which ME d belongs. For the given vector H one can obtain state probability distribution of each ME d allocated at node ch(d) from matrix Pd as p d h (d ) h (d ) k for 1kKh(d). For the given ATN topology (C,E) and for the given state probability distributions of the MEs Pd (1dD) and node vulnerability , the only factor influencing the ATN survivability is the allocation of its elements H. Therefore, the optimal allocation problem can be formulated as follows. Find vector H* maximizing the ATN survivability S: H*(C,E,P1,…, PD)=arg{S(H,C,E,P1,…, PD)max}. (6) The same Genetic Algorithm (GA) based procedure as described in Ref. [2] is used in this work for finding the optimal ME allocation. The GA is based on an evolutionary search in the space of possible solutions. Each time a new allocation solution is generated by the GA, its quality (ATN survivability) is determinated using the procedure described in the following section. The details of solution encoding and basic GA procedures can be foung in [2]. 3. ATN survivability estimation based on a universal generating function The procedure used in this paper for network reliability evaluation is based on the universal generating function (also called u-function) technique, which was introduced in [3] and was applied for ATN reliability evaluation in [1,2]. In this paper, the approach is extended by incorporating node vulnerability into the ATN model. The basic UGF technique is described in brief in the following sections. 3.1. u-function for individual MEs The UGF (u-function) of a discrete random variable X is defined as a polynomial 8 u (z) K qkzXk , (7) k 1 where the variable X has K possible values and qk is the probability that X is equal to Xk. In order to represent random sets of ATN nodes that receive a signal, we modify the UGF by replacing the random value X with the random binary vector V={v(1)…v(N)} such that v(j) corresponds to node cj. Consider a multistate element d located at position ci. In each state k (1k<Ki), the ME provides a signal transmission from ci to a set of nodes ik. In order to represent the set ik, we determine vector Vik as follows 1, c j ik vik ( j) . 0 , c j ik (8) The polynomial u id (z) Ki pd i ik z Vik (9) k 1 represents all the possible states of the ME located at ci by relating the probabilities of each state k to the value of a random vector Vi (representing set ik) in this state. Note that the absence of any ME at position ci implies that no connections exist between ci and any other position. This means that any signal reaching ci cannot be retransmitted in this node. In this case, the corresponding u-function takes the form ui0(z)= z V0 , (10) where v0(j)=0 for 1jN. 3.2. u-function for group of MEs allocated at the same position (node) Consider two MEs n and m allocated at the same position ci. Assume that first ME is in state s and second one is in state g. The probability of this composition of states is p n i is p m i ig . A 9 signal generated by the two MEs reaches all the nodes belonging to set isig. This set can be represented by vector Vis Vig, where the operator (logical OR) for two arbitrary vectors A and B is defined as follows: a ( j) b( j) 0 if a ( j) b( j) 0, (1jN). 1 otherwise (11) In order to obtain the u-function of a subsystem consisting of two MEs n and m located at the same position ci, a composition operator is introduced. This operator determines the u-function for a group of MEs using simple algebraic operations on the individual u-functions of the MEs. The composition operator for a pair of MEs n and m takes the form: Ki (u in (z), u im (z)) ( p i ik z Ki Ki p k 1f 1 n Vik k 1 n m i ik p i if z Ki , p m i if z Vif ) f 1 Vik Vif (12) . The resulting polynomial relates probabilities of each of the possible combinations of states of the two MEs (obtained by multiplying the probabilities of corresponding states of each ME) with vectors representing sets of nodes reseiving the signal in the given combination of states. One can see that the operator satisfies the following conditions: {u1 (z),..., u t (z), u t 1 (z),..., u v (z)} {{u1 (z),..., u t (z)}, {u t 1 (z),...., u v (z)}} (13) for arbitrary t. Therefore, it can be applied in sequence to obtain the u-function for an arbitrary group of MEs allocated at ci: d i (u id (z)) ~ Ki ~ ~qik z Vik , (14) k 1 ~ where K i is a number of different states of the group of MEs allocated at ci, ~ qik is the probability ~ that only the nodes belonging to the set represented by the vector Vik receive the signal directly from ci. One can consider the group of MEs allocated at ci as a single ME with state distribution (14). 10 3.3. Incorporating node vulnerability into corresponding u-function When MEs are gethered within a single node having vulnerability , this means that all these elements can be destroyed with probability . The probabilities ~ qik in (14) relate only to reliability of MEs (internal ATN reliability). In order to incorporate external destructive factors into the model the probabilities ~ qik should be considered as conditional probabilities that a signal from MEs located at ci reaches a set of nodes ik under assumption that the node ci survives external attack during the ATN operation time. Unconditional probability that a signal from MEs located at ci reaches the set of nodes ik is therefore equal to (1-) ~ qik = ~ qik . If the node ci is destroyed the empty set of nodes i (corresponding to zero vector V0) is reached from ci independently of states of MEs. Therefore the u-function of MEs located at vulnerable node ci takes form ~ K ~ K i i ~ ~ ~ U i (z) ( ~ qik z Vik ) ~ qik z Vik z V0 . k 1 (15) k 1 Note that one of states in (14) corresponds to the total failure of all the MEs and, therefore, the expression (14) can be rewritten as ~ Ki ~qik z k 1 ~ Vik ~ qi1z V0 ~ Ki ~ ~qik z Vik . k 2 In this case operator takes form ~ Ki ~ ~ Ki ~ ( ~ qik z Vik ) ~ qik z Vik ( ~ qi1 )z V0 . k 1 k 2 3.4. u-function for the entire ATN Assume that a signal generated by MEs located at c1 in state s reaches c2 (c21s which corresponds to ~v1s (2)=1). If the MEs located at c2 are in state g, the signal generated at c2 reaches 11 all the nodes belonging to 2g. Therefore, when the first group of MEs is in state s and the second group is in state g, the set of nodes receiving the signal is 1s2g. This set can be represented by ~ ~ vector V1s V2 g . If a signal generated at c1 at some state s does not reach c2 (c21s which corresponds to ~v (2)=0), the group of MEs located at c2 cannot retransmit the signal in any of its states and, 1s therefore, MEs located at c2 don't affect the state of the ATN. The set of nodes receiving the signal ~ remains 1s represented by the vector V1s . In the general case of arbitrary states of the two groups of MEs, one can use the following function to determine the random vector V̂2 representing the set of nodes receiving the signal from c1 directly or through c2: ~ ~ ~ V1, ˆ (V V , V ) ~ 2 1 2 ~ V1 V2 , ~ v1 (2) 0, ~ v (2) 1. (16) 1 To represent all the possible combinations of states of the two groups of MEs, one has to relate the corresponding probabilities (obtained by multiplying the probabilities of corresponding states of each ME group) with the values of the random vector V̂2 in these states. For this purpose, we introduce a composition operator over u-functions of groups of MEs located at c1 and c2: ~ K ~ K ~ 1 ~ 2 ~ ~ V Û 2 (z) ( U1 (z), U 2 (z)) ( ~ q1s z V1s , ~ q 2g z 2 g ) ~ ~ K1 K 2 ~q1s ~q 2g z s 1 ~ ~ ( V1s , V2 g ) s 1g 1 g 1 K̂ 2 q̂ 2k z (17) ˆ V 2k k 1 The resulting polynomial Û 2 (z) represents the probabilistic distribution of the possible values of the random vector V̂2 corresponding to set of nodes receiving the signal from c1 directly or ~ ~ through the MEs located at c2. The random vector V̂2 can have K̂ 2 K1K 2 different values. The probability of each state k of group of MEs located at c1 and c2 is ~ q 2k . 12 Consider a random vector V̂i representing a set of nodes receiving the signal directly from c1 or through the MEs located at c2,…, ci. It can easily be seen that the addition of the MEs located at ˆ , ci+1 changes the set of nodes receiving the signal in such a way that the random vector V i 1 representing this new set, takes the form: V ˆ , i ~ ˆ ˆ ,V ) V ( V ˆ i 1 i i 1 ~ Vi Vi 1, v̂ i (i 1) 0, (18) v̂ i (i 1) 1. Let Û i ( z ) be the u-function representing probabilistic distribution of V̂i . Since node ci+1 cannot receive the signal from any node cm with m>i+1, the probability that the signal generated at c1 reaches ci+1 is completely determined by Û i ( z ) . Therefore, we can obtain a recursive expression for the u-function representing the distribution of ATN states: K̂ i ˆ ~ Û i 1 (z) ( Û i (z), U i 1 (z)) ( q̂ ik z Vik , ~ K̂ i K i 1 q̂ ik ~qi 1 f z k 1 ~ ˆ (Vik , Vi 1 f ) k 1 f 1 ~ K i 1 f 1 ~ Vi 1 f ~ qi 1 f z ) K̂ i1 q̂ i 1 k z ˆ V i 1 k ) , (19) k 1 ~ where K̂ i 1 K̂ i K i 1 . ~ Note that for any Û i ( z ) and U i1 (z) =ui+1 0(z) (corresponding to empty position ci+1) Ûi 1 (z) ( Ûi (z), u i 1 0 (z)) Ûi (z) . (20) One can obtain the u-function representing the distribution of the ATN states when all the ˆ MEs are considered (or, equivalently, the probabilistic distribution of random vector V NM ) applying the Eq. (19) in sequence for i=1, i=2,…, i=N-M-1. Summing probabilities q̂ N M k for all the states k in which v̂ N M k ( j) 1 for N-M+1jN, one obtains the probability that the signal reaches all the leaf nodes, which is equal to ATN reliability index. Using the simplification technique described in [1] and [2], one can drastically simplify polynomials Û i ( z ) for 1iN-M using the following operator ( Û i ( z ) ) which zeroes 13 v̂ik (1) ,…, v̂ik (i) for 1k K̂ i in each term of Û i ( z ) , removes all the terms in which V̂ik contain only zeros and collects like terms in the resulting polynomial. 3.5. Algorithm for determination of ATN reliability Using the UGF technique described above, one can obtain the ATN reliability for the given set of parameters ( p d i ik , ik) 1iN-M, 1kKi, 1dD and the given ME allocation H applying the following procedure, which is convenient for numeric implementation: 1. Determine vectors Vik corresponding to sets ik for the positions c1,…,cN-M using rule (8). 2. For each ME d located at position h(d) determine the u-function uh(d)d(z) using expression (9) with probabilities p d h (d ) h (d ) k for each state k. 3. Obtain u-functions for group of MEs allocated at each nonempty node ci using expression ~ (14) and operator (12). For empty nodes j assign U j (z) u j0 (z) , where uj0(z) is defined in (10). ~ 4. Obtain U i (z) for each node ci using operator (15) over u-functions for group of MEs allocated at this node. ~ 5. Assign Û1 (z) U1 (z) . ~ 6. Apply expression Û i 1 (z) (( Û i (z)), U i 1 (z)) for i=1,2,…,N-M-1 in sequence using operator (17) and operator described in the previous section. 7. Simplify polynomial Û N - M (z) using operator and obtain the ATN reliability R as the coefficient of the term of ( Û N - M (z) ) in which v̂ N-M(j)=1 for all N-M+1jN. Note that in the general case, the resulting polynomial contains 2M-1 terms. Therefore, the suggested method can be applied for ATNs with moderate values of M. 4. Illustrative examples 4.1. Simple analytical example of ATN reliability determination Consider the ATN presented in section 1 (Figure 1). The u-functions of the MEs are 14 u11(z)=p11z000+p11{2}z010+p11{3}z001+p11{2,3}z011 , u12(z)=p21z000+p21{2}z010+p21{3}z001+p21{2,3}z011 if the MEs are located at c1 and u21(z)=p12z000+p12{3}z001, u22(z)=p22z000+p22{3}z001 if the MEs are located at c2. When both MEs are allocated at node c1 (case A): (u11(z), u12 (z)) p11p21z000+(p11{2}p21+p21{2}p11+p11{2}p21{2})z010+ (p11{3}p21+p21{3}p11+p11{3}p21{3})z001+ (p11{2,3}+p21{2,3}-p11{2,3}p21{2,3}+p11{2}p21{3}+p11{3}p21{2})z011, ~ U1 (z) {(u11 (z), u12 (z))} (+p11p21)z000+ (p11{2}p21+p21{2}p11+p11{2}p21{2})z010+(p11{3}p21+p21{3}p11+p11{3}p21{3})z001+ (p11{2,3}+p21{2,3}-p11{2,3}p21{2,3}+p11{2}p21{3}+p11{3}p21{2})z011, ~ U 2 (z) {u 20 (z)} (+)z000=z000. Following steps 5 and 6 of the algorithm 3.6. one obtains ~ Û1 (z) U1 (z) , ( Û1 ( z ) )=(p11{2}p21+p21{2}p11+p11{2}p21{2})z010+ (p11{3}p21+p21{3}p11+p11{3}p21{3})z001+ (p11{2,3}+p21{2,3}-p11{2,3}p21{2,3}+p11{2}p21{3}+p11{3}p21{2})z011. ~ Û2 (z) ((Û1(z)), U2 (z)) ((Û1(z)), u 20 (z)) (Û1(z)). ( Û 2 (z)) (p11{3}p21+ p21{3}p11+p11{3}p21{3}+p11{2,3}+ p21{2,3}-p11{2,3}p21{2,3}+p11{2}p21{3}+p11{3}p21{2})z001. If the MEs are identical ( p1i ik p 2i ik pi ik ) ( Û 2 (z)) (2p1{3}p1+(p1{3})2+2p1{2,3}-(p1{2,3})2+2p1{2}p1{3})z001. 15 The system reliability is equal to the coefficient corresponding to the single term of the polynomial: SA={2p1{3}p1+(p1{3})2+2p1{2,3}-(p1{2,3})2+2p1{2}p1{3}}={2p1{3}(1-p1{2}-p1{3}-p1{2,3})+ (p1{3})2+2p1{2,3}-(p1{2,3})2+2p1{2}p1{3}}={2p1{3}+2p1{2,3}-(p1{3})2-(p1{2,3})2-2p1{3}p1{2,3}}= {2(p1{3}+p1{2,3})-(p1{3}+p1{2,3})2}. Consider case B in which first ME is located at c1 and second one is located at c2: ~ U1 (z) {u11 (z)} (+p11)z000+p11{2}z010+p11{3}z001+p11{2,3}z011, ~ U 2 (z) {u 22 (z)} (+p22)z000+p22{3}z001. ~ Û1 (z) U1 (z) , ( Û1 ( z ) )=p11{2}z010+p11{3}z001+p11{2,3}z011, ~ Û 2 (z) (( Û1 (z)), U 2 (z)) (p11{2}z010+p11{3}z001+p11{2,3}z011,(+p22)z000+p22{3}z001)= p11{2}(+p22)z010+{p11{3}(+p22)+2p11{3}p22{3}}z001+ {p11{2,3}(+p22)+2p11{2,3}p22{3}+2p11{2}p22{3}}z011= p11{2}(+p22)z010+{p11{3}+2p11{3}(p22+p22{3})}z001+ {p11{2,3}+2p11{2,3}(p22+p22{3})+2p11{2}p22{3}}z011= p11{2}(+p22)z010+{p11{3}+2p11{3}}z001+{p11{2,3}+2p11{2,3}+2p11{2}p22{3}}z011. ( Û 2 (z) )={(p11{3}+p11{2,3})+2(p11{3}+p11{2,3}+p11{2}p22{3})} z001. Taking into account that the MEs are identical one obtains the ATN reliability SB=(p1{3}+p1{2,3})+2(p1{3}+p1{2,3}+p1{2}p2{3}). Since =1-, SB=(1-)(p1{3}+p1{2,3})+2(p1{3}+p1{2,3}+p1{2}p2{3})=(p1{3}+p1{2,3})+2p1{2}p2{3}. 4.2. Optimal ME allocation using GA based algorithm Consider an ATN with N=10 and M=3 presented in Figure 4. The list of possible states of MEs (represented by sets ik) when they are located at positions c1,…,c7 and corresponding probabilities 16 pi ik is presented in Table 1. The system fails if a signal generated at position c1 does not reach at least one of positions from the set {c8,c9,c10}. In [2] the optimal allocation of identical MEs (maximizing the system reliability) was obtained for =0. These solutions were compared with ones obtained by the suggested algorithm for =0.1. Table 2 presents ME allocation solutions obtained by the GA for D=6 and D=14. This table contains numbers of identical MEs located at each position and resulting ATN survivability. When =0 the solutions which are optimal for =0.1 provide system survivability very close to survivability of ATN with allocation optimal for =0. When =0.1 the solutions which are optimal for =0.1 provide greater survivability then solutions, which are optimal for =0 (survivability increase is 2.4% for D=6 and 5.7% for D=14). Figure 5 presents the survivability of ATNs with the obtained ME allocations as function of single node survivability . One can see that the greater the node vulnerability, the greater difference between ATN survivability provided by ME allocations optimal for =0 and =0.1. In order to solve the allocation problem for nonidentical MEs we modify the ME state probabilities in the following way: p d i ik (d) pi ik for ik and pdi=1-(d)+(d) pi, where pi ik are presented in Table 1 and (d)=1.02-0.02d for 1dD=7, (d is number of ME). Such a modification provides unique state distribution for each element being allocated in each position. The optimal ME allocation solutions for =0 and =0.1 are presented in Table 3 which contains lists of MEs located in each position. Figure 6 presents the survivability of ATNs with the obtained allocations of nonidentical MEs as function of single node survivability . The same increase of difference between ATN survivability provided by ME allocations optimal for =0 and =0.1 with the growth of node vulnerability can be observed. Note that in the optimal solutions obtained for the vulnerable nodes, the number of occupied nodes is greater than in the solutions for =0. Indeed, by increasing ME separation the system tries to compensate its vulnerability. 17 References [1] Levitin G. Reliability evaluation for acyclic consecutively-connected networks with multistate elements, Reliability Engineering and System Safety, 2001, vol. 73, pp. 137-143. [2] Levitin G, Optimal allocation of multi-state retransmitters in acyclic transmission networks, to appear in Reliability Engineering and System Safety. [3] I. A. Ushakov, Universal generating function, Sov. J. Computing System Science, vol. 24, No 5, 1986, pp. 118129. Figure Captions Figure 1: Two possible allocations of MEs in the simplest ATN. Figure 2: Decision curves for comparison of the two possible allocations of MEs in the simplest ATN. Figure 3: Fragment of ATN. Figure 4: ATN for the numerical example. Figure 5: ATN survivability as function of single node survivability for different allocations of identical MEs. Figure 6: ATN survivability as function of single node survivability for different allocations of nonidentical MEs. 18 Table 1. Probabilistic state distribution of the system MEs. i 1 2 3 4 5 6 7 ik pi ik {2,3,4} {2,3} {3,4} {2} {3} {4,6,8} {4,6} {4,8} {6,8} {4} {6} {8} {4,5} {4} {5} {6,7,10} {6,7} {6,10} {7,10} {6} {7} {10} {6,7} {6} {7} {8,9} {8} {9} {9,10} {9} {10} 0.75 0.10 0.08 0.02 0.01 0.04 0.65 0.08 0.05 0.08 0.05 0.02 0.05 0.02 0.85 0.06 0.04 0.05 0.62 0.08 0.06 0.02 0.05 0.05 0.07 0.05 0.83 0.04 0.07 0.06 0.8 0.06 0.10 0.04 0.60 0.35 0.02 0.03 19 Table 2. Allocation solutions obtained for ATN with identical MEs D 6 14 Solution for 0.0 0.1 0.0 0.1 1 2 1 3 2 2 1 2 2 3 1 Positions 4 5 2 2 5 3 1 S(=0.0) S(=0.1) 6 2 1 3 2 7 1 1 3 0.8693 0.8628 0.9993 0.9965 0.6337 0.6491 0.7882 0.8333 Table 3. Allocation solutions obtained for ATN with different MEs Solution for 0.0 0.1 1 2,4 4,7 2 3 3 - Positions 4 5,6,7 2,5 5 - 20 6 1,3 1 7 6 S(=0.0) S(=0.1) 0.8868 0.8816 0.6464 0.6640