On Component Order Edge Reliability and the Existence of Uniformly Most Reliable Unicycles

Transcription

1 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel On Component Order Edge Reliability and the Existene of Uniformly Most Reliable Uniyles DANIEL GROSS Seton Hall University Dept of Math and CS South Orange, NJ LAKSHMI ISWARA Stevens Institute Dept of Mathematial Sienes Hoboken, NJ L. WILLIAM KAZMIERCZAK Stevens Institute Dept of Mathematial Sienes Hoboken, NJ KRISTI LUTTRELL Stevens Institute Dept of Mathematial Sienes Hoboken, NJ CHARLES SUFFEL Stevens Institute Dept of Mathematial Sienes Hoboken, NJ JOHN T. SACCOMAN Seton Hall University Dept of Math and CS South Orange, NJ Abstrat: Let G be a graph with n nodes and e edges, where the nodes are perfetly reliable and the edges fail independently with equal probability ρ. A failure state exists if the surviving edges indue a graph having all omponents of order less than a preassigned threshold value k. The unreliability of G,U k (G; ρ), is the probability of a failure state and a graph G is k-uniformly most reliable (k-umr) over a lass of graphs if and only if U k (G; ρ) U k (H; ρ) for all 0 < ρ < 1 and all H in the same lass as G. If U k (G; ρ) U k (H; ρ) for all 0 < ρ < 1 and all H in the same lass as G then G is k- uniformly least reliable (k- ULR). In this paper we show that K 1,n is the unique tree that is k-umr over all trees with k n. We also show that the uniyle U x s, i.e. K 1,n with an added edge, is uniquely 3-UMR over all graphs having n 5 nodes and e = n edges. We extend this study for 4 k n and prove that U x s is the unique k-umr uniyle. In the last two setions we give the neessary onditions for a graph to be k-umr and show that there exists a range of values of k for whih a k-umr uniyle does not exist. Key Words: Component Order edge reliability, uniformly most reliable, unreliability, uniyle 1 Introdution In [1] we introdued the omponent order edge reliability model. We repeat some of the elementary notions pertinent to this model for the sake of ompleteness as well as disuss other fundamentals. Unless otherwise noted, we follow the notation of []. Consider a senario of a network whih is modeled by an undireted graph G having n nodes and e edges in whih nodes are perfetly reliable but edges fail independently of one another, all with the same probability ρ (0, 1). A threshold value < k < n is given and the surviving subgraph remaining after the failure of edges is said to be an operating state provided it ontains at least one omponent of order k or more. It is a failure state provided eah omponent has order at most k 1 and the subset of failing edges that produe the failure state is referred to as a failure set. The probability that, at a snapshot of time, the network is in an operating state is referred to as the k-omponent order edge reliability and is denoted by R k (G; ρ) while U k (G; ρ) = 1 R k (G; ρ), the k-omponent order edge unreliability, is just the probability that the network is in a failure state. As a onsequene of the underlying assumption of independent failure of edges, all with the same probability ρ (0, 1), the probability of a failure set for whih a speifi set of i edges fail is just ρ i (1 E-ISSN: Issue 9, Volume 1, September 013

2 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel ρ) e i. Thus if f i (G) denotes the number of failure sets of size i, then U k (G; ρ) = i f i (G)ρ i (1 ρ) e i. Now the smallest number of edges required to fail in order to produe a failure state is referred to as the k-omponent order edge onnetivity of G and is denoted by (G) (see [3],[4],[5],[6] for studies of λ (k) (G)). Thus f i (G) = 0 for all i < λ (k) (G). Furthermore, assuming e k, a set of edges of size i e (k ), leaves a surviving subgraph having at most k edges upon failure, and hene a failure state, so f i (G) = ( e )fori e (k ). i Thus, (G) e (k ) and we may write U k (G; ρ) = e i=λ (k) (G) f i (G)ρ i (1 ρ) e i, where f i (G) = ( e ) i for all e (k ) i e. Of ourse, if λ k (G), then U k (G; ρ) = e i=e (k ) (e i )ρi (1 ρ) e i. Thus any suh graph will have the minimum value of the k-omponent order edge reliability among all graphs on n nodes and e edges for all ρ (0, 1). Now it is lear that if ρ is held fixed then, as the olletion of all graphs on n nodes and e edges is finite, there is a graph on n nodes and e edges having minimum k-omponent order edge reliability among all those in the olletion. However, as has been shown for k = n, there may be no one graph in a olletion that minimizes the unreliability for all ρ (0, 1)[7]. If a graph exists in a speified olletion of graphs, that minimizes U k (G; ρ) for all G in the olletion and all ρ (0, 1), it is referred to as a k-uniformly most reliable graph (k-umr) in the olletion. It has been shown that if e ( n ) n, then K n minus a mathing is n UMR over the lass of all graphs having n nodes and e edges [8, 9]. If e = n 1, then all trees have the same n-omponent order edge unreliability, i.e., 1 (1 ρ) n. On the other hand it was shown that if e = n, then C n is the unique uniyle whih is n UMR over all uniyles on n [10]. In this work we shall show that K 1,n is the unique tree that is k UMR over all trees and all k n. We shall also show that there exists a range of values of k for whih a k UMR uniyle exists and a range for whih no k UMR uniyle exists. Graphs with λ k = e (k ) As was noted above, any graph G in the olletion of all graphs having n nodes and e edges, where n k and e k ; having = e (k ) has k-omponent order edge unreliability equal to U k (G; ρ) = e i=e (k ) (e i )ρi (1 ρ) e i Now any graph on n nodes and e edges with e (k 1) has stritly larger unreliability than G so we have the following theorem: Theorem 1 If n k, e k and e (k 1), then G is k-umr over all graphs having n nodes and e edges. In the remainder of this setion we determine those ases for whih = e (k ). First note that if e = k, then λ k = 0 = e (k ) and U k (G; ρ) = 1, so all suh G on n nodes and e edges, where n k, have the same k-omponent order reliability. Another trivial situation ours when k =. In this ase = e = e (k ) and U k (G; ρ) = ρ e for all G on n nodes and e edges where n > k =. Next suppose e = k 1. Note that if G = T k (n k)k 1 where T k is a tree on k nodes, then = 1 = (k 1) (k ) = e (k ). Otherwise, = 0 = e (k 1). Hene, every G = T k (n k)k 1 where T k is a tree on k nodes has = e (k ) and no other graph on n k 3 nodes and e = k 1 edges does. Note that when n = k this subsumes the fat that trees on n nodes have (T ) = 1 = e (k ). The next logial ase is e = k 3. Suppose G has n nodes and e edges where n k = e 3 and assume = e (k ). Then f e () = 0 and every subset of k 1 edges must indue a tree on k nodes. Choose one, say T k, and let the remaining edge be denoted by x. Now at least one endpoint of x must lie in T k or else the removal of an edge in T k and the addition of x yields a disonneted subgraph with k 1 edges. Suppose the addition of x to T k yields another tree. If that tree has a path with three edges, the removal of an internal edge of that path yields a disonneted subgraph having k edges. Hene, if T k with x added is a tree, then it must be K 1,k. Otherwise, onsider the possibility that the addition of x to T k yields a uniyle. If that uniyle ontains a pendant edge its removal yields a subgraph of order k 1 ontaining k 1 edges, whih ontradits f e () = 0. E-ISSN: Issue 9, Volume 1, September 013

3 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel Thus, in this ase, T k with x added is just C k. In summary, if = e (k ) = k (k ) =, then G = K 1,k (n k)k 1 or G = C k (n)k 1. Observe that, if n = k + 1, then in the first ase G = K 1,n, while in the seond ase G = C n. Finally, suppose e k and assume λ (k) = e (k ). As in the previous ase, we hoose T k, a tree on k nodes, from the set of e edges. Suppose T k has two independent edges so that the addition of an edge to T k annot produe K 1,k.Thus eah of the remaining edges when added to must produe C k. As there are at least two additional edges, this is impossible. Thus, T k = K 1,k and eah of the additional edges, when added to T k,must yield K 1,k. Hene G = K 1,e follows. Observe that in this ase if n = e + 1, then G = K 1,n. We summarize those findings in our next theorem. Theorem Consider n k, and let G be a graph on n nodes and e edges. 1. If e k, then λ (k) (G) = 0, and equals e (k ) when e = k. Also, U k (G; ρ) = 1 for all ρ (0, 1).. If k =, then (G) = e = e (k ), and U k (G; ρ) = ρ e for all ρ (0, 1). 3. If e k 1, then (G) = 1 = e (k ) if and only if G = T k (n k)k 1, where T k is a tree on k nodes. In the event that n = k, then every tree on n nodes has λ k (T ) = 1 and U k (T ; ρ) = 1 (1 ρ) n. 4. If e = k 3,then (G) = = k (k ) if and only if G = K 1,k (n (k + 1))K 1 or G = C k (n k)k 1. Also, U k (G; ρ) = 1 (1 ρ) k k(ρ)(1 ρ) k. In the event that n = k, G = C n is the unique graph on n nodes with e = n edges having λ n = and therefore, the unique suh graph whih is n-umr. If n = k + 1 (i.e. k = n 1), then K 1,n is the unique graph on n nodes with e = n 1 edges having λ (n) = and therefore, the unique suh graph whih is n- UMR. 5. If e k + 1 4,then λ k (G) = e (k ) if and only if n e + 1 and G = K 1,e [(n (e + 1))K 1 ]. Also as was observed previously, U k (G; ρ) = e i=e (k ) (e i )ρi (1 ρ) e i. In the event that e = n 1 k + 1 4, the unique graph having λ (k) (G) = e (k ) = n k + 1 is K 1,n. It is, therefore, the unique k-umr graph having n nodes and e = n 1 edges, and U k (G; ρ) = n i=n k+1 (n i )ρi (1 ρ) e i. We onlude this setion with three orollaries of the last theorem. Corollary 3 1. If e = n 1, n k and k, then, K 1,n is k- UMR over all graphs with n nodes and e = n 1 edges.. If e = n = k, then K 1,n is the unique k-umr tree on n nodes. 3. If e = n 1 k + 1 4, then K 1,n is k-umr over all graphs with n nodes and e = n edges. Corollary 4 If e = n 1 k + 1 4, then (G) e (k 1). Proof: By Theorem (5), (G) = e (k ) fores e n 1. Corollary 5 If e = n = k, then C n is the unique graph in the lass of all graphs having n nodes and e = n edges with (G) = = e (k ) and is therefore also the unique n-umr graph in this lass. 3 k-umr Uniyles exist for 3 k n We begin with preliminary observations. First observe that by Corollary 4, (U) e () = n () for eah uniyle U on n nodes. With n = + k, we obtain (U) + 1. Next, it is easy to see that (Us x ) = + 1 for k 3, where Us x denotes the star K 1,n with edge x added between two leaf nodes. Sine f i (U) = ( n ) i for eah i n (k ), for eah U it follows that we an prove Us x to be k-umr by establishing the inequality f n0 +1(Us x ) f n0 +1(U) for all uniyles U Us x. Furthermore, if either (U) < (Us x ) and f n0 +1(Us x ) f n0 +1(U) or (U) (Us x ) and f n0 +1(Us x ) < f n0 +1(U) for eah uniyle U Us x, then Us x is seen to be uniquely k-umr over the lass of uniyles on n nodes. The final result of a preliminary nature is given in our first proposition. Proposition 6 If U Us x is a uniyle on n nodes, then U onsists of two node disjoint trees T 1 and T eah having at least two nodes, joined by two edges x and y (see Figure 3.1). E-ISSN: Issue 9, Volume 1, September 013

4 Proof: Let U be a uniyle different from U x s having unique yle v 1, v,..., v l, v l so that U onsists of the yle together with l node disjoint trees T 1, T,..., T l rooted at v 1, v,..., v l, respetively (see Figure 3.). If l 4, then setting x = v 1 v and y = v 3 v 4 establishes the laim. If l = 3, then, as U U x s,at least two of the trees, say T 1 and T, have two or more nodes eah. Then setting x = v 1 v and y = v 1 v 3 establishes the laim. Thus for eah n 7, U(Us x ; ρ) < U(G; ρ) for all ρ (0, 1). Next suppose that (G) 3 and let node u have degree equal to (G). Now eah edge not inident at u forms pair of independent edges with at least (G) of the edges inident at u. Hene f n (G) (n (G))( (G) ). The parabola y = (n )( ) opens downward with axis of symmetry = n + 1. Also, the y values at = 4 and = n are equal to n 8. Thus, if 4 then f n (G) n 8 > n 3, for n 6. Now (G) n = (Us x ) sine the removal of Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel n edges from G leaving an independent pair yields a failure state. Thus when 4 it follows that U(Us x ; ρ) < U(G; ρ) for all ρ (0, 1). Finally, we onsider the ase (G) = 3. Let node u have degree equal to (G) = 3 and N(u) = v 1, v, v 3. As there are at least seven edges in G, at least one of the n - 3 edges not inident at u is inident at most one node in N(u). Eah of the remaining n - 4 edges not inident at u forms a pair of independent edges with at least one edge inident at u. Thus f n (G) n > n 3 = λ (3) (U x s ). We proeed now to the ase k = 3 and n 7. In this ase + 1 = n and we shall prove that Us x is uniquely 3-UMR not only over all the uniyles but also over all graphs having n nodes and e = n edges provided n 7. Theorem 7 The uniyle U x s is uniquely 3-UMR over all graphs having n 7 nodes and e = n edges. Proof: Consider a graph G on n 5 nodes with e = n edges and maximum degree (G). If G Us x then (G) n. Suppose (G) ; then, sine the sum of the degrees of the nodes in G is n, either G = C n or G is a disjoint union of yles. Note that, in eah ase, for n 7, f n (G) = n(n 3) > n 3 = f n (U x s ). Sine f n (G) n it follows that U(U x s ; ρ) < U(G; ρ) for all ρ (0, 1) and the proof is omplete. To omplete the ase k = 3 we examine the situations e = n = 5 and e = n = 6 in turn. First, if n = 5 and (G) = 4 then G = Us x, and if (G) =, then G = C 5. Now (C 5 ) = 3 but f 3 (C 5 ) = 4 > = f 3 (Us x ). If (G) = 3, let deg(u) = 3. If one of the two edges not inident at u is inident at only one node in N(u) then f 3 (G) 3. Otherwise G = (K 4 x) K 1 and has λ (3) (G) = 3, f 3 (G) =. Thus Us x and (K 4 x) K 1 are the 3-UMR graphs on n = 5 nodes. Next suppose n = 6. We know if (G) 4 and G Us x, then U(Us x ; ρ) < U(G; ρ) for all ρ (0, 1). If (G) = then G = C 6 or G = K 3. If G = C 6, then (C 6 ) = 3 but f 4 (C 6 ) = 9 > 3 = f 4 (U x s ). If G = K 3, then (K 3 ) = 4 but f 4 (K 3 ) = 9 > 3 = f 4 (U x s ) If (G) = 3 then let deg(u) = 3 and observe that if one of three edges not inident at u is inident with at most one node in N(u) then suh an edge forms at least two pairs of independent edges with edges inident at u. The other two form at least one pair with an edge inident at u so f 4 (G) 4 > 3 = f 4 (U x s ). Otherwise, G = K 4 K 1, whih has λ (3) (G) = 4, = λ (3) Thus Us x (Us x ) f 3 (G) =. and K 4 K 1 are the 3-UMR graphs with e = n = 6. The preeding analysis leads to the following orollary. E-ISSN: Issue 9, Volume 1, September 013

5 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel Corollary 8 The graph U x s is the unique 3-UMR uniyle for n 5. Next we proeed to the general ase: 4 k ;. We begin by proving an important vulnerability result in the ontext of uniyles. We shall show that in all instanes, save one, Us x is the unique uniyle with = + 1. All others have = or less, thereby indiating that Us x is the unique most invulnerable uniyle subjet to system failure. Initially, we deal with the ase where the yle of the uniyle U Us x has length l 4 and in preparation for that result we require the next lemma. Lemma 9 If and k 4, then (C n ). Proof: We know that λ (k) (C n ) = n = [5]. But n k 0 if and only if k, whih is the ase when k 4. Theorem 10 If k 4, and U is a uniyle with yle length l 4, then (U). Proof: We shall employ indution on beginning with base ase =. First, if l = n, so that U = C n, the result follows by Lemma 9. If 4 l n 1, then referring to Proposition 6 there is a T i, say, T 1, with order at least two. Removal of the edges v 1 v l and v v 3 leaves two omponents, one onsisting of T 1 and T together with edge v 1 v and the other onsisting of T 3,..., T l together with the path v 3, v 4,..., v l. The first omponent ontains at least two edges, so if the seond does as well, eah will ontain at most k edges, as the number of remaining edges is k. Hene in this event (U) = =. Suppose the seond omponent onsists of one edge v 3 v 4, i.e. l = 4, (see Figure 3.3). Sine n = k + 6, V (T 1 ) + V (T ) 4. In the event that V (T 1 ) 3 but V (T ) = 1, removal of v 1 v and v 1 v 4 yields a failure state while if V (T 1 ), V (T ), removal of v 1 v and v 3 v 4 yields a failure state. Hene if = and k 4, λ (k) (U) = =. Our indution hypothesis is that if = m 1, where m 3, and U has yle length l where 4 l n, then (U) = m 1, for k 4. Now onsider a uniyle with = m and yle length l where 4 l n and suppose U = C n ; then λ (k) (U) by Lemma 9. If U C n then U has a pendant node and edge. Remove the pendant node and edge obtaining a uniyle Û with n 1 = m 1 + k nodes and yle length l. The indution hypothesis fores (Û) m 1 so λ(k) (U) m = for k 4. Our next theorem deals with the ase l = 3 and inludes the one exeptional ase previously mentioned. Theorem 11 Consider the lass of all uniyles on n = +k nodes when and l = 3, k 4. If U 6 is the uniyle where V (T 1 ) = V (T ) = V (T 3 ) = (see Figure 3.3), then λ (4) (U 6 ) = 3 = + 1. Otherwise, i.e., if U Us x, U 6 then (U). Proof: First observe that if U has six nodes but is not equal to Us x or U, then, without loss of generality, V (T 1 ) = 3, V (T ) =, V (T 3 ) = 1(see Figure 3.4). The set {v 1 v, v 1 v 3 } is a failure set so λ (4) (U) = =. We omplete the proof by indution on starting with the base ase =. Sine we have already onsidered =, k = 4, it remains to assume n 7 so that by the Pigeonhole Priniple, some E-ISSN: Issue 9, Volume 1, September 013

6 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel V (T i ), say V (T 1 ) 3. But U Us x implies, without loss of generality, that V (T ) and it follows that {v 1 v, v 1 v 3 } is a failure set sine eah of the two omponents it leaves upon removal has at most k edges. Our indution hypothesis simply states that if U is a uniyle on n nodes, different from Us x with yle length l = 3 and = m, then (U) where k 4. Now onsider a uniyle U on n nodes with yle length l = 3 and = m 3. The Pigeonhole Priniple fores one V (T i ), say V (T 1 ) 3 and, sine U Us x, another, say V (T ). Again, as n 7, either 1. V (T 1 ) 3, V (T ), and V (T 3 ) or. V (T 1 ) 3, V (T ) 3, and V (T 3 ) = 1 or 3. V (T 1 ) 4, V (T ), and V (T 3 ) = 1. In the first ase remove a pendant node and its edge from T 3, in the seond ase from T and in the third ase from T 1, thereby arriving at a uniyle Û on n 1 = m+k nodes with yle length l = 3 and different from Us x, U 6. Hene λ (k) (U) λ (k) (Û)+1 m+ 1 = m = and the proof is omplete. In the remainder of this setion we prove that Us x is the unique k-umr uniyle for 4 k n. Sine (U) λ (k) (Us x ) for all uniyles with only one exeption, i.e. n = 6, =, k = 4, and e (k ) = +, it is only neessary to prove that f 3 (U 6 ) > f 3 (Us x ) in the exeptional ase and that f n0 +1(U 6 ) f n0 +1(Us x ) for the other ases. Let s onsider n = 6, = and k = 4 to begin with. The uniyles in this ase are shown in Figure 3.3. It is easy to see that U 1, U, U 3, U 4 and U 5 all have λ (4) = = and f 3 4 while f 3 (Us x ) = 4. On the other hand, f 3 (U 6 ) > 4 = f 3 (Us x ). Now, for the remaining ases, we begin with the observation that if U Us x then by Proposition 6 we an represent U by two trees T 1 and T of orders n 1 and n, respetively, joined by two edges x and y (See Figure.1). We begin the analysis of this ase under the assumption that n 1 k and n k and establish a lower bound on the failure sets of size + 1 whih inlude x and y. Of ourse, the removal of + 1 edges leaves a total of k 1 edges so that if x and y are removed and 1 additional edges are removed so that at least one edge from T 1 and one edge from T remain then a failure state is obtained. Thus, the number of failure sets of size + 1 whih inlude x and y, denoted by f x,y +1, satisfies f x,y k +1 i=1 ( n 1 1 )( n 1 i k 1 i ) = ( n 1 + n ) ( n 1 1 k 1 k 1 ) (n 1 k 1 ). But n 1, n k implies that ( n 1 )+(n ) is maximized when n 1 = k and n = or vie versa. Thus f x,y +1 (U) (n k 1 ) ( 1 k 1 ) 1. Next we prove that there exist ( n 4 ) additional failure sets that inlude at most one of the edges x and E-ISSN: Issue 9, Volume 1, September 013

7 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel y. There are two senarios to onsider dependent on deg(u ), where x = u 1 u. Assume without loss of generality that y = uv where u V (T 1 ), v V (T ) but v u. Now if deg(u ) 3 let w be the edge inident on u that lies on the unique path of T from u to v and let z be any other edge inident on u in T. Now if x and w are removed together with 1 additional edges not inluding y and z a failure state results, sine y and z lie in separate omponents (see Figure 3.6(a)). As there are ( n 4 ) of these sets the laim is established in this ase. The other senario involves the ase where deg(u ) =. Let w be the edge of T with endpoint u and onsider removing 1 edges inluding y and w but not x or z where z is an arbitrary but fixed edge of T different from w. Realize edge z exists sine n k 4 (see Figure 3.6(b)). A failure state results sine x and z lie in separate omponents. Here too there are ( n 4 ) suh sets. Thus f n0 +1(U) ( n 4 k 1 ) + ( n 4 1 ) ( 1 k 1 ) 1. Consider D = f n0 +1(U) f n0 +1(U x s ) = ( n k 1 )+( 1 k 1 )+( n 4 1 ) ( n 3 1 ) ( n ) = (n )! (k 1)!( 1)! + (n 4)! (k 3)!( 1)! (n 3)! (k )!( 1)! (n 3)! (k 4)!( + 1)! [( 1 k 1 )+1] = (n 4)! (k 1)!( + 1)! [(n )(n 3)( +1)( )+()(k )( +1)( ) (n 3)()( +1)( ) (n 3)()(k )(k 3)] [( ) + 1]. Now, ( 1 k 1 ) = ( 1)! ( k)!(k 1)! = (n 4)! ( + 1)!( ) ( k + 1)! ( + 1)!(k 1)! (n 4)(n 5) ( ) (n 4)! ( + 1)!(k 1)! ( + 1)( )( 1)( ) and it follows that D (n 4)! ( + 1)!(k 1)! [(+1)( )( )( 3)+ ()(k )( +1)( ) ()(n 3)( +1)( ) (n 3)()(k )(k 3) ( +1)( )( )( )] 1 = (n 4)! ( + 1)!(k 1)! [(n 3)(k 1)(( + 1)( ) (k )(k 3))] 1 = ( n 3 1 ) ( n ) 1 = ( n k ) (n 3 k 4 ) 1 > 0 beause the binomial oeffiients stritly inrease as the subset size inreases toward the midpoint of the set size. Next we onsider the ase where 3 n 1 k 1, n n 1. As in the previous ase, f x,y +1 (U) ( +k ) (n ). But here, n n 3 so f x,y +1 (U) ( +k ) (+k 4 ). Also, as the argument used in the previous ase applies whether y is inident at either u 1 or u, we an show that there exists ( +k 4 ) additional failure sets so that f x,y +1 (U) (+k ). Thus, D = f x,y x,y +1 (U) f +1 (U x) s ( +k ) (+k 3 ) ( +k 3 +1 ) = (n 3 ) (n 3 k 4 ) > 0 for basially the same reason as in the previous ase. Finally, suppose that n 1 =. In this ase, n = n so f x,y +1 (U) (+k 3 ). Here we laim that there are at least ( +k 4 ) additional failure sets for all possible senarios but one (see Figure 3.7). Indeed, if this is the ase then f x,y x,y +1 (U) f +1 (U x) s E-ISSN: Issue 9, Volume 1, September 013

8 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel ( + k 3 1 ) + ( + k 4 1 ) ( + k 3 1 ) ( + k ) = ( + k 4 1 ) ( + k ) = ( + k 4)! (k 3)!( + 1)! [ (k 5) (k 3) ] > 0 if k. To prove the laim we onsider all possible senarios as shown below in Figure 3.7 and show that for (a) through (d) the laim holds. We verify f x,y +1 (U) f x,y +1 (U x) s 0 diretly for the senario shown in (e). In (a) T must ontain a path from u to some node u of length at least two, sine U Ux. s Then by hoosing failure sets, inluding x and z but not y and w and failure sets inluding y and z but not x and w we obtain ( +k 4 ) additional failure sets. In (b) realize that either T or T ontains an edge, say T. Then failure sets ontaining x and z but not y and w together with failure sets ontaining w and z but not x and y yield an additional failure sets. In () we may onsider failure sets inluding x and w but not y and z and failure sets inluding y and z but not x and w to obtain the laim. As regards (d), failure sets (i) inluding x and z but not y and w and (ii) inluding y and z but not x and w establish the laim. As for (e) observe that if either x and y are inluded but z isn t or if x and z are inluded but y isn t, then the number of failure sets is at least ( +k 4 x,y x,y ). Thus f +1 (U) f +1 (U x) s = ( n 3 ) ( n 3 +1 ) = (n 3 k ) (n 3 ) k 4 > 0 and the proof is omplete. E-ISSN: Issue 9, Volume 1, September 013

9 4 Neessary onditions for uniform optimality Throughout this setion the overriding assumption is that 4 k + 1 n e so that by Corollary 4, (G) e (k 1) and we may write U k (G; ρ) = e () f i(g)ρ i (1 ρ) e i + e i= (G) i=e (k ) (e i )ρi (1 ρ) e i.. Observe that e f e is the number of subtrees of G with k nodes. The theorem of this setion desribes onditions on the oeffiients in the unreliability expression given above for determining when U k (G 1 ; ρ) < U k (G ; ρ) for all suffiiently small ρ and also when the inequality holds for all suffiiently large ρ. Theorem 1 Suppose G 1 and G have n nodes and e edges where 4 k + 1 n e. Then 1. if (G 1 ) > (G ), then there exists ρ 0 (0, 1) suh that U k (G 1 ; ρ) < U k (G ; ρ) for all ρ (0, ρ 0 );. if (G 1 ) = (G ), and i 0 is the smallest index suh that f i0 (G 1 ) f i0 (G ), then f i0 (G 1 ) < f i0 (G ) implies there exists ρ 0 (0, 1) suh that U k (G 1 ; ρ) < U k (G ; ρ) for all ρ (0, ρ 0 ); 3. if i 1 is the largest index, neessarily at most e (k 1), suh that f i1 (G 1 ) f i1 (G ), then f i1 (G 1 ) < f i1 (G ) implies there exists ρ 1 (0, 1) suh that U k (G 1 ; ρ) < U k (G ; ρ) for all ρ (ρ 1, 1). Proof: 1. Observe that + f λ (k) e i= (G )+1 U k (G ; ρ) U k (G 1 ; ρ) = ρλ(k) (G ) (1 ρ) e λ(k) (G ) (G ) (f i (G ) f i (G 1 ))ρ i (1 ρ) e i. Set p = ρ 1 ρ so that ρj (1 ρ) e j = (1 ρ) e p j for all j. Thus U k (G ; ρ) U k (G 1 ; ρ) = (1 ρ) e p λ(k) (G ) [f (k) λ (G ) (G )+ e (f i (G ) f i (G 1 ))p i λ(k) (G ) ]. i=λ (k) (G )+1 Hene there exists a p 0 suh that if p (0, p 0 ) then the quantity in the brakets is positive. But ρ = p 1+p 0 is an inreasing funtion of p so there exists ρ 0 suh that ρ (0, ρ 0 ) implies p (0, p 0 ) and the result follows.. In this ase U k (G ; ρ) U k (G 1 ; ρ) = (1 ρ) e p i 0 [f i0 (G ) f i0 (G 1 ) + e i=i 0 +1 (f i(g ) f i (G 1 ))p i i 0 ], and the result follows as in the previous argument. 3. Observe that U k (G ; ρ) U k (G 1 ; ρ) = i 1 i=0 (f i (G ) f i (G 1 ))ρ i (1 ρ) e i = ρ i 1 (1 ρ) e i 1 i 1 i=0 [(f i (G ) f i (G 1 ))ρ i i 1 (1 ρ) i1 i ] = ρ i 1 (1 ρ) e i 1 i 1 i=0 [(f i (G ) f i (G 1 ))( 1 ρ ρ )i i 1 ]. Set p = 1 ρ ρ so that U k (G ; ρ) U k (G 1 ; ρ) = ρ i 1 (1 ρ) e i 1 i 1 i=0 [(f i (G ) f i (G 1 ))p i 1 i ]. Now if i 1 = 0 then f 0 (G ) = 1 and f 0 (G 1 ) = 0 and U k (G ; ρ) U k (G 1 ; ρ) = ρ i 1 (1 ρ) e i 1 ((f i1 (G ) f i1 (G 1 ) + i 1 i=0 [(f i(g ) f i (G 1 ))ρ) i i 1 (1 ρ) i i 1 ]). Thus there exists p 1 1 suh that if p (0, p 1 ) then the quantity in the brakets is positive. As ρ = 1 1+p is dereasing on (0, p 1] there exists ρ 1 suh that ρ (ρ 1, 1) implies p (0, p 1 ) and the result follows. The following orollary yields neessary onditions for a graph to be UMR over all graphs in a given olletion (neessarily having the same numbers of nodes and edges). Corollary 13 If C is a olletion of graphs, all with the same number of nodes n and the same number of edges e, where 4 k + 1 n e, and G is k-umr over C then 1. (G) is maximum over C;. f (k) λ (G) is minimum over all H C having (H) = (G); 3. G has the minimum value of f e () over C (or equivalently, has the maximum number of subtrees of order k). Proof: Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel 1. (H) > (G) where H C, then, by Theorem 1(1), there exists ρ 0 (0, 1] suh that U k (H; ρ) < U k (G; ρ) for all ρ (0, ρ 0 ), whih ontradits G being k-umr over C.. If (H) = (G) = λ but f λ (H) < f λ (G) then, by Theorem 1(), there exists ρ 0 E-ISSN: Issue 9, Volume 1, September 013

10 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel (0, 1] suh that ρ (0, ρ 0 ) implies U k (H; ρ) < U k (G; ρ), whih ontradits G being k-umr over C. 3. If f e () (H) < f e () (G), then as i 1 = e (k 1), we obtain a ontradition to G being k- UMR by Theorem 1(3). We apply this Corollary in our next setion in showing that there exists a range of values of k for whih no k-umr graph on n nodes with e = n exists. 5 On the Existene of k-umr Uniyles for Large Values of k Relative to n As we have shown in Setions and 3, k-umr uniyles exist whenever 3 k n. In this setion we note that the yle C n is n-umr and prove that (n - 1)- UMR uniyles exist. The somewhat surprising result that if n < k n, then k-umr uniyles do not exist is also established here. To begin, the fat that C n is the unique n-umr uniyle was established in [10] as λ (n) = λ, the lineonnetivity. The ase k = n 1 is a bit more ompliated and is the subjet of our next theorem. edges of C l exept for those adjaent to a node on the yle of degree two and every pair of edges not on the yle are failure sets. Consider the paraboli funtion f(x) = ( x ) (x)+(k+1 x ) = x (k+4)x+k +k+ = x (k )x+ k +k+ (where the binomial oeffiients have the obvious interpretation when x not an integer) whih has a unique minimum at x = k + 1. Sine U k is the only uniyle with f (U) = f( k + 1) when k is even, the result in (1) follows immediately. In the event that k is odd onsider the problem of minimizing f(x) when x is onstrained to be an integer. Then the minimum value of f(x) ours only when x = k+1 or x = k+3. As f (U k ) = f(k+1 ) = f(k+3 ) = f (U k ), () follows and the proof is omplete. Theorem 14 If e = n = k + 1 5, then 1. when k is even, the uniyle U k onsisting of k pendant edges all attahed to a single node of the yle C k +1 is the unique (n - 1)-UMR uniyle on n nodes (see Figure 5.1(a));. when k is odd, the uniyle U k k+1 onsisting of pendant edges all inident on a single node of the yle C k+1 and the uniyle U k onsisting of pendant edges all attahed to a single node of the yle C k+3 are the only two (n - 1)-UMR uniyles on n nodes (see Figure 5.1(b)). Proof: Suppose that U is a uniyle on n nodes with yle length l and let l be the number of nodes on the yle of degree equal to two. We laim that (U). Indeed, if l = 3 then there exists a node on the yle C 3 of degree at least three, so removal of the two edges of C 3 adjaent to suh a node yields a failure state. If l 4, then removal of two independent edges of C l yields a failure state. Next realize that f (U) ( l ) l +( k+1 l ) sine every pair of λ (n) The final result of this setion establishes the nonexistene of k-umr uniyles for a range of large values of k. E-ISSN: Issue 9, Volume 1, September 013

11 Daniel Gross, Lakshmi Iswara, L. William Kazmierzak, Kristi Luttrell, John T. Saoman, Charles Suffel Theorem 15 If e = n = + k, and k 4, then a k-umr uniyle does not exist when k > n Proof: First reall from Theorem 10 and Theorem 11 that (U) for U Us x with the sole exeption of n = 6, k = 4 and =, where (U 6 ) = 3 = +1. As (Us x ) = +1, it follows that (U) < (Us x ) whenever n 4, k 4 and. Hene if a k-umr uniyle exists it must be Us x, by Corollary 13(1), and in this ase, f n0 +1(U) f n0 +1(Us x ) for all uniyles U on n nodes, by Corollary 13(3). But we shall see that if k satisfies the ondition of the theorem the uniyle U 4, having n 4 pendant edges all inident on a single node of C 4 (see Figure 5.) has a smaller value of f n0 +1 than Us x, thereby proving that a k-umr uniyle doesn t exist. Indeed so that f n0 +1(U x s ) = ( + k ) + 3( + k 4 1 ) +( + k 4 ) f n0 +1(U x s ) f n0 +1(U 4 ) = 1! [( + k 4)( + k 5) (k 1)] [(k )(k 3) + ( k + 3) ( + 1)] ( + k 4)!!(k )! (k ( + 5)k ). Now the expression in the seond set of brakets is positive if k > n Of ourse if then the ondition requires that k > > 7 whih is onsistent with the proviso that n 7. This onludes the proof. Corollary 16 follows immediately from Theorem 15 after diret substitution of numerial values. Corollary 16 If =, then when k 8, no k-umr uniyle exists. Referenes: [1] Lakshmi Iswara, L. Kazmierzak, Kristi Luttrell, C. Suffel, D. Gross, J.T. Saoman, On Component order Edge Reliability and Uniform Optimality, Cong. Numer., Volume 1 (01) pp [] G. Chartrand, L. Lesniak and P. Zhang, Graphs and Digraphs (Fifth Edition), Chapman & Hall/CRC,Boa Rato11. [3] F. Boesh, D. Gross, J.T. Saoman, L. Kazmierzak, C. Suffel and A. Suhartomo, A Generalization of an Edge-Connetivity Theorem of Chatrand, Networks, Volume 54, Issue (009) pp [4] F. Boesh, D. Gross, L.W. Kazmierzak, C. Suffel, A. Suhartomo, Component order edge onnetivity an introdution, Congr. Numer. 178 (006), pp [5] F. Boesh, D. Gross, L. Kazmierzak, C. Suffel and A. Suhartomo, Bounds for the Component Order Edge Connetivity, Cong. Numer., 185(007), pp [6] D. Gross, L. Kazmierzak, J.T. Saoman, C. Suffel, A. Suhartomo, On Component Order Edge Connetivity of a Complete Bipartite Graph, ARS Combinatoria, (to appear). [7] W. Myrvold, K.H. Cheung, L.B. Page and J.E. Perry, Uniformly-Most Reliable Networks Do Not Always Exist, Networks, Volume 1(1991) [8] A.K.Kelmans, On Graphs with Randomly Deleted Edges, Ata Mathematia Hungaria, Volume 37, issue 1-3(1981) [9] A. Satyanarayana, L. Shoppmann, C. Suffel, A Reliability-improving GraphTransformation with Appliations to Network Reliability, Networks, Volume (199) [10] F. Boesh, X. Li, C. Suffel, On the Existene of Uniformly Optimally Reliable Networks, Networks, Volume 1, Issue (1991), pp E-ISSN: Issue 9, Volume 1, September 013