We introduced a load-based cascade model to study the vulnerability of complex net- works under random single-node attacks, where the ER random graph with finite size was used to represent the network. We assumed that the capacity of a node is proportional to its initial load and the load of a failed node is redistributed to its neighbors according to their capacity. The average failure ratio at each step was used to quantify the damage ex- perienced by the network. A step-by-step estimation of the average failure ratio has been provided. The accuracy of such estimations was validated by numerical results. Our anal- ysis for finite-size networks revealed a phase transition phenomenon in network reactions to single-node attacks, where the average value of the failure ratio drops quickly within a short interval of the load margin. We characterized this interval by finding the critical value of the tolerance parameter at which the failure ratio takes its median value and is most sensitive to the variation of the tolerance parameter. We also derived the threshold interval within which this phase transition occurs. Our findings shed light on how to set the load margin for both robustness and efficient use of resources in designing networks resilient to random single-node attacks.

*∗*_{Reprinted from [11] “Load-Dependent Cascading Failures in Finite-Size Erd¨os-R´enyi Random Net-}
works” by D. Lv and A. Eslami and S. Cui, 2017, Network Science and Engineering, IEEE Trans on (TNSE),

REFERENCES

[1] S. Boccaletti, V. Latora, Y. Moreno, M. Chavez, and D. U. Hwang. Complex net-
*works: Structure and dynamics. Physics Reports, 424(45):175–308, Jan. 2006.*
[2] Paolo Crucitti, Vito Latora, and Massimo Marchiori. Model for cascading failures in

*complex networks. Phys. Rev. E, 69(4):045104, Apr. 2004.*

[3] I. Dobson, B.A. Carreras, and D.E. Newman. A branching process approximation to
*cascading load-dependent system failure. In Proceedings of the 37th Annual Hawaii*
*International Conference on System Sciences, pages 1–10, Hawaii, Jan. 2004.*
*[4] P. Erd¨os and A. R´enyi. On random graphs I. Publ. Math. Debrecen, 6:290–297,*

1959.

*[5] Paul Erd¨os and A R´enyi. On the evolution of random graphs. Publ. Math. Inst.*
*Hungar. Acad. Sci, 5:17–61, 1960.*

*[6] E. N. Gilbert. Random graphs. The Annals of Mathematical Statistics, 30(4):1141–*
1144, 1959.

[7] Ake J. Holmgren. Using graph models to analyze the vulnerability of electric power
*networks. Risk Analysis, 26(4):955–969, Aug. 2006.*

[8] R. Kinney, P. Crucitti, R. Albert, and V. Latora. Modeling cascading failures in the
*North American power grid. The European Physical Journal B - Condensed Matter*
*and Complex Systems, 46(1):101–107, July 2005.*

[9] Jure Leskovec and Andrej Krevl. SNAP Datasets: Stanford large network dataset collection. http://snap.stanford.edu/data, June 2014.

[10] Daqing Li, Bowen Fu, Yunpeng Wang, Guangquan Lu, Yehiel Berezin, H. Eugene Stanley, and Shlomo Havlin. Percolation transition in dynamical traffic network with

*evolving critical bottlenecks. Proceedings of the National Academy of Sciences,*
112(3):669–672, 2015.

[11] D. Lv, A. Eslami, and S. Cui. Load-dependent cascading failures in finite-size erdos-
*renyi random networks. IEEE Transactions on Network Science and Engineering,*
4(2):129–139, April 2017.

[12] Yi-Hua Ma and Dong-Li Zhang. Cascading network failure based on local load
*distribution and non-linear relationship between initial load and capacity. In Machine*
*Learning and Cybernetics (ICMLC), 2012 International Conference on, volume 3,*
pages 935–940, July 2012.

[13] Adilson E. Motter and Ying-Cheng Lai. Cascade-based attacks on complex networks.
*Phys. Rev. E, 66(6):065102, Dec. 2002.*

[14] M. E. J. Newman, S. H. Strogatz, and D. J. Watts. Random graphs with arbitrary
*degree distributions and their applications. Phys. Rev. E, 64:026118, Jul 2001.*
*[15] Mark EJ Newman. Random graphs as models of networks. arXiv preprint cond-*

*mat/0202208, Feb 2002.*

[16] Ke Sun and Zhen-Xiang Han. Analysis and comparison on several kinds of models
*of cascading failure in power system. In Transmission and Distribution Conference*
*and Exhibition: Asia and Pacific, IEEE/PES, pages 1–7, Dalian, China, Aug. 2005.*
[17] JianWei Wang and LiLi Rong. Cascade-based attack vulnerability on the US power

*grid. Safety Science, 47(10):1332–1336, Dec. 2009.*

[18] JianWei Wang and LiLi Rong. Robustness of the western United States power grid
*under edge attack strategies due to cascading failures. Safety Science, 49(6):807–812,*
July 2011.

*[19] Duncan J. Watts. A simple model of global cascades on random networks. Proceed-*
*ings of the National Academy of Sciences, 99(9):5766–5771, Apr. 2002.*

[20] ZhiXi Wu, Gang Peng, WenXu Wang, Sammy Chan, and Eric WingMing Wong.
*Cascading failure spreading on weighted heterogeneous networks. Journal of Statis-*
*tical Mechanics: Theory and Experiment, 2008(05):P05013, May 2008.*

APPENDIX A

SOME PROOFS FOR THEOREMS, COROLLARIES, LEMMAS

*Lemma 2.3.1 Consider a random single-node attack applied to G(n, p). Let node a be*
*the attacked node, and node e be an arbitrary node in V\{a}. Let Pd* be the probability

*that the shortest path from e to a has length d; Pr{Bd} be the probability that at least one*

*path from e directly through a node in V\{a ∪ e} to a has a length which is less than or*
*equal to d. Then E[|Vd|], d ≥ 1, the average size of Vd*, is simply

*E[|Vd|] = (n − 1)Pd,*

*where Pd, d≥ 1, can be obtained recursively as*

*P*1 *= p,*
*P*2 = (1*− p)(1 − (1 − p*2)*n−2),*
*Pd* = (1*− p) Pr{Bd} −*
*d−1*
∑
*j=2*
*Pj, d > 2.*

In the numerical calculation, we assume that Pr*{Bd}, d > 2, can be approximated recur-*

sively as
Pr*{Bd} ≈ 1 − ((1 − p) + p · (1 −*
*d−1*
∑
*j=1*
*Pj*))*n−2.*

*proof 4 (Proof of Lemma 2.3.1) Suppose that we randomly pick a node a from the ER*
*graph G(n, p), where any two nodes are connected with probability p. For an arbitrary*
*node e* *∈ V \a, we define a family of probabilities as Pd* = Pr*{the shortest path between*

*nodes e and a has length d}, P[i,j]* = Pr*{the shortest path between nodes e and a has*

*length within [i, j]}, satisfying*

*P[i,j]* =

*j*

∑

*d=i*

*Pd,* *0 < i≤ j.*

*Since Pd* *is the same for all the nodes in V\a, the expectation of |Vd| over all possible*

*topologies can be calculated as*

*E[|Vd|] = (n − 1)Pd.* (5.1)

*We use induction to obtain Pd. We first have P*1 *= p. When d≥ 2, we find Pdas*

*Pd* =
*P[2,d]* *d = 2*
*P[2,d]−*
∑*d−1*
*j=2Pj* *d > 2*
*,* (5.2)

*where P[2,d]* *remains to be found. The event “the shortest distance from e to a is within*

*[2, d]” is true if the following two independent events happen at the same time: A =“e is*
*not directly connected to a” and Bd=“at least one path from e directly through a node in*

*V\{a ∪ e} to a has a length which is less than or equal to d”. It can be easily seen that*
Pr*{A} = 1 − p. Then P[2,d], d* *≥ 2 can be obtained as*

*P[2,d]* = Pr*{A} Pr{Bd}*

Pr*{A}=1−p*

*−−−−−−−→ = (1 − p) Pr{Bd}.* (5.3)

*Now we aim to obtain Pr{Bd}, d ≥ 2. Given d = 2, node e can go through any node*

*happens if any of these paths is connected. Therefore,*

Pr*{B*2*} = 1 − (1 − p*2)*n−2.* (5.4)

*Combining (5.2), (5.3) and (5.4) we have*

*P*2 = (1*− p)(1 − (1 − p*2)*n−2).* (5.5)

*When d > 2, substituting (5.3) into (5.2), we have*

*Pd* = (1*− p) Pr{Bd} −*
*d−1*

∑

*j=2*

*Pj, d > 2.* (5.6)

*Combining P*1 *= p, (5.5) and (5.6) yields Lemma 1.*

*Now we aim to verify the approximation*

Pr*{Bd} ≈ 1 − ((1 − p) + p · (1 −*
*d−1*

∑

*j=1*

*Pj*))*n−2, d > 2.* (5.7)

*Since directly obtaining Pr{Bd} is complicated, we first focus on its complement ¯Bd=“the*

*lengths of the (n−2) shortest paths from e directly through a node in V \{a∪e} to a are all*
*greater than d”. Therefore, Pr{Bd} can be obtained through 1 − Pr{ ¯Bd}. Now our goal*

*is to approximate Pr{ ¯Bd}. For notational simplicity, let l*1*,· · · , ln−2* *denote the lengths*

*of the (n− 2) shortest paths from e directly through a node in V \{a ∪ e} to a, such that*
Pr*{ ¯Bd} is the joint probability that l*1*,· · · , ln−2are greater than d, which can be expressed*

*as*

Pr*{ ¯Bd} = Pr{*
*n*∩*−2*

*i=1*

*To obtain (5.8), we start from analyzing the joint probability that arbitrary two path*
*lengths from l*1*,· · · , ln−2* *are greater than d. Let us randomly pick two nodes v*1*, v*2 *∈*

*V\{a ∪ e} and consider paths e → v*1 *99K a and e → v*2 *99K a, where “99K” stands for*

*the shortest path between the two nodes. Let l*1 *denote the length of e* *→ v*1 *99K a and*

*l*2 *denote the length of e* *→ v*2 *99K a. In ER random graph, e is connected to each node*

*independently with probability p. We can list the following three scenarios regarding the*
*connectivity between e and v*1*, v*2*:*

*1. e is not directly connected to v*1*, such that l*1 =*∞. In this case, l*1*and l*2*will not be*

*affected by each other, such that they are independent.*

*2. e is not directly connected to v*2*. Similarly, l*1 *and l*2*are independent.*

*3. e is directly connected to both v*1 *and v*2*. In this case l*1 *and l*2 *become dependent.*

*Let C*1 *=“Scenario 3): e is directly connected to both v*1*and v*2*”. According to scenarios*

*1) and 2), we have l*1 *⊥ l*2 *| ¯C*1*. Now we aim to calculate the joint probability Pr{l*1 *>*

*d, l*2 *> d}. Given d > 2, Pr{l*1 *> d, l*2 *> d} can be rewritten according to the law of total*

*probability:*
Pr*{l*1 *> d, l*2 *> d}*
= Pr*{l*1 *> d, l*2 *> d| C*1*} Pr{C*1*}+*
Pr*{l*1 *> d, l*2 *> d| ¯C*1*} Pr{ ¯C*1*}*
= Pr*{l*1 *> d, l*2 *> d| C*1*}p*2+
Pr*{l*1 *> d, l*2 *> d| ¯C*1*}(1 − p*2)
*l1⊥l2| ¯C1*
*−−−−−→ = Pr{l*1 *> d, l*2 *> d| C*1*}p*2+ (5.9)
Pr*{l*1 *> d| ¯C*1*} Pr{l*2 *> d| ¯C*1*}(1 − p*2*),*

*where Pr{l*1 *> d| ¯C*1*} can be obtained by*

Pr*{l1> d| ¯C*1*} = (Pr{l1* *> d} − Pr{l1* *> d| C1} Pr{C1})/ Pr{ ¯C*1}

= (Pr*{l1* *> d} − Pr{l1* *> d| C1}p*2*)/(1− p*2*).* (5.10)

*Similar to (5.10), we can obtain Pr{l*2 *> d| ¯C*1*}. Substituting them into (5.9), we obtain*

Pr*{l*1 *> d, l*2 *> d} = Pr{l*1 *> d, l*2 *> d| C*1*}p*2+

(Pr*{l*1 *> d} − Pr{l*1 *> d| C*1*}p*2)

*·(Pr{l*2 *> d}− Pr{l*2 *> d| C*1*}p*2*)/(1− p*2*),*

*When p*2* _{≈0, cancel p}*2

*−−−−−−−−−−−→ ≈ Pr{l*1 *> d} Pr{l*2 *> d}.* (5.11)

*The approximation (5.11) holds when p*2 *is small, which is true under typical values*
*of p in networks of typical sizes, e.g., n* *≥ 20. In addition, some simulations were con-*
*ducted to test the accuracy of approximation (5.11), and the results support our analysis.*
*The following steps describe how the simulations were conducted: in a network of size n,*
*arbitrarily four nodes a, e, v*1*, and v*2*were selected. Then we counted l*1*and l*2*in 100, 000*

*realizations of ER random graph. Based on the numerical results of l*1 *and l*2*, we calcu-*

*lated Pr{l*1 *> d, l*2 *> d} and Pr{l*1 *> d} Pr{l*2 *> d}. Under varied values of n, p and d,*

*these two probabilities are always approximately equal, which indicates the assumption is*
*valid even in small networks with relatively larger p, e.g., n = 20, p = 0.3. Partial results*
*are shown in Table 5.1.*

*in (5.8). First we rewrite (5.11) as*

Pr*{l*1 *> d, l*2 *> d}*

= Pr*{l*1 *> d| l*2 *> d} Pr{l*2 *> d}*

*≈ Pr{l*1 *> d} Pr{l*2 *> d},*

*Since Pr{l*2>d*}̸=0, cancel Pr{l*2>d*} on both sides*

*−−−−−−−−−−−−−−−−−−−−−−−−−−−→*

Pr*{l*1 *> d| l*2 *> d} ≈ Pr{l*1 *> d}.* (5.12)

*Since an ER random graph is homogeneous network and each node has an identical*
*statistical property, result (5.12) also applies to other path lengths l*1*,· · · , ln−2. In a*

*finite-size network G(n, p), according to the chain rule, the joint probability Pr{ ¯Bd} =*

Pr*{*∩*n _{i=1}−2li*

*> d} can be expanded as*

Pr*{ ¯Bd} = Pr{*
*n*∩*−2*
*i=1*
*li* *> d}*
=
*n*∏*−2*
*i=1*
Pr*{li* *> d|*
*i−1*
∩
*j=1*
*lj* *> d},*
(5.12)
*−−−→ ≈*
*n*∏*−2*
*i=1*
Pr*{li* *> d}.* (5.13)

*In order to find (5.13), we need to have Pr{li* *> d}, i = 1, · · · , n − 2. Assume l*1 *is the*

*length of the path through node v*1*, so l*1 *> d happens when either e is not connected to v*1

*(l*1 =*∞) or e is connected to v*1 *but the distance between v*1 *and a is greater than d− 1.*

*Because ER random graph is a homogeneous network and l*1*,· · · , ln−2* *have the identical*

*statistical property, Pr{li* *> d}, i = 1, 2, · · · , n − 2, can be obtained by*

*Substituting (5.14) into (5.13), we have*
Pr*{ ¯Bd} ≈ ((1 − p) + p · (1 −*
*d−1*
∑
*j=1*
*Pj*))*n−2, d > 2,* (5.15)

*Thus, the approximation in (5.7) is justified.*

Table 5.1: Simulation results to test the dependence of two path lengths.
Network *d* Pr*{l*1 *> d, l*2 *> d} Pr{l*1 *> d} · Pr{l*2 *> d}*

*G(100, 0.05)* 4 *0.9287* *0.9286*

10 *0.9054* *0.9051*

*G(20, 0.3)* 4 *0.8265* *0.8270*

8 *0.4926* *0.4922*

*Theorem 3.2.1 Consider a random single-node attack applied to G(n, p). We assume*
*the conditional distribution of k(V*1) given*|V*1*| = x is approximately normal with mean µ*

*and variance σ*2_{, where}

*µ = x + x(x− 1)p + (n − x − 1)xp,*
*σ*2 *= (2x(x− 1) + (n − x − 1)x)p(1 − p).*

*Then E[f*1], i.e., the average failure ratio at step 1, can be approximated as

*E[f*1] *≈*
1
*n*
(
1 +
*n−1*
∑
*x=1*
*x·*
(
*n− 1*
*x*
)
*px*(1*− p)n−1−x*Φ(
*x*
*α−1* *− µ*
*σ* )
)
*,*

*proof 5 (Proof of Theorem 3.2.1) Since E[f*1*] can be obtained as E[f*1] = * _{n}*1

*(1 + E[|F*1

*|]),*

*it is enough to find E[|F*1*|]. Based on Corollary 1, we have the failure condition for V*1 *as*

*k(V*1*) <* *k(V0 _{α−1}*)

*. According to the failure condition, the distribution of|F*1

*| can be expressed*

*as*

Pr*{|F*1*| = x}*

= Pr*{k(V*1*) <*

*k(V*0)

*α− 1* *| |V*1*| = x} · Pr{|V*1*| = x},* (5.16)

*and expectation of|F*1*| can be obtained by the law of total probability as*

*E[|F*1*|] =*
*n−1*
∑
*x=1*
*x· Pr{k(V*1*) <*
*k(V*0)
*α− 1* *| |V*1*| = x} · Pr{|V*1*| = x},* (5.17)

*where* *|V*1*| is the number of nodes in V*1*, which obeys a binomial distribution. That is,*

*|V*1*| ∼ B(n − 1, p) and*
Pr*{|V*1*| = x} =*
(
*n− 1*
*x*
)
*px*(1*− p)n−1−x.* (5.18)

*Now to find (5.17), we need to calculate the conditional distribution of k(V*1*) given*

*|V*1*| = x. The links adjacent to nodes in V*1 *can be divided into three categories: edges*

*between V*0*and V*1*, within V*1*, and between V*1*and V*2*, denoted by the setsE(V*0*, V*1*),E(V*1*),*

*andE(V*1*, V*2*), respectively. Such partition of edges is illustrated in Fig. 5.1. We then have*

*k(V*1) = *|E(V*0*, V*1)*| + 2|E(V*1)*| + |E(V*1*, V*2)*|.* (5.19)

*Figure 5.1: Partition of edges within and adjacent to V*1: *E(V*0*, V*1) = *{1, 2, 3}, E(V*1) =

*{4}, E(V*1*, V*2) = *{5, 6, 7, 8}.*

*can be easily seen that|E(V*0*, V*1)*| = x, while |E(V*1)*| and |E(V*1*, V*2)*| depend on connec-*

*tivity of nodes. In a ER random graph, each pair of nodes are connected with a probability*
*p independent of other pairs [6]. Thus,* *|E(V*1)*| follows a binomial distribution B(*

(_{x}

2

)
*, p)*
*when x* *≥ 2 and |E(V*1)*| = 0 when x = 1. |E(V*1*, V*2)*| follows a binomial distribution*

*B((n* *− x − 1)x, p). Under typical settings, |E(V*1)*| is much smaller than |E(V*1*, V*2)*|.*

*For example, in G(100, 0.05), given* *|V*1*| = 5, we have E[|E(V*1)*|] = 0.25, whereas*

*E[|E(V*1*, V*2)*|] = 94 × 0.25. Therefore, 2|E(V*1)*| + |E(V*1*, V*2)*| ≈ |E(V*1)*| + |E(V*1*, V*2)*|,*

*which follows B(*1_{2}*x(x−1)+(n−x−1)x, p). This binomial distribution is approximately*
*normal since (n− x − 1)x · p and (n − x − 1)x · (1 − p) are both greater than 5 under*
*typical settings in networks with practical-sizes (We usually have np≥ ln n in a connected*
*graph G(n, p) [5]). Therefore, k(V*1*) is approximately normal given|V*1*| = x. Let µ and*

*σ*2 *denote the conditional mean and variance of k(V*1*) given|V*1*| = x, respectively. They*

*can be obtained by*

*µ = x + x(x− 1)p + (n − x − 1)xp,*
*σ*2 *= (2x(x− 1) + (n − x − 1)x)p(1 − p).*

*Such that Pr{k(V*1*) <* *k(V0 _{α}_{−1}*)

*| |V*1

*| = x} in (5.17) can be approximated as*Pr

*{k(V*1

*) <*

*k(V*0)

*α− 1*

*| |V*1

*| = x} ≈ Φ(*

*x*

*α−1*

*− µ*

*σ*

*).*(5.20)

*After substituting (5.18) and (5.20) into (5.17), we obtain*

*E[|F*1*|] ≈*
*n−1*
∑
*x=1*
*x·*
(
*n− 1*
*x*
)
*px*(1*− p)n−1−x*Φ(
*x*
*α−1* *− µ*
*σ* *).*

*By definition, E[f*1] = * _{n}*1

*(1 + E[|F*1

*|]), which yields Theorem 1.*

*Theorem 3.3.1 Consider a random single-node attack applied to G(n, p). We assume,*
*1. We only consider the failures propagating in the forward direction; i.e., at step t,*

*only the nodes in V\ ∪t _{d=0}−1*

*Vd*are considered as potential nodes to fail.

*2. The set Ft−1*is considered as a large virtual node that redistributes its load to its alive

*neighbors at step t with the rule defined in (3).*

*3. n is large enough such that the variance of| ˆVt| is small and | ˆVt| can be approximated*

*by E[| ˆVt|].*

*4. E[| ˆVt| | |Ft−1| = E[|Ft−1|]] is applied to approximate E[| ˆVt|].*

5. Given*| ˆVt| = E[| ˆVt|], Lt(Ft−1) and (α−1)L*0( ˆ*Vt*) are independent and approximately

*normal. Lt(Ft−1*) has conditional mean ˜*µ = E[|Tt−1|](n − 1)p and unknown condi-*

tional variance ˜*σ*2_{. (α}_{− 1)L}

0( ˆ*Vt*) has conditional mean ˆ*µ = (α− 1)(n − 1)E[| ˆVt|]p*

and conditional variance ˆ*σ*2 * _{= (α}− 1)*2

_{(n}− 1)E[| ˆ_{V}*t|]p(1 − p). Φ(µ*˜*−ˆµ _{σ}*

_{ˆ}) is applied

*Then an estimate of the average failure ratio E[ft] for step t≥ 2 is obtained recursively as*
*E[ft*] *≈*
1
*n*Φ(
˜
*µ− ˆµ*
ˆ
*σ* *)E[| ˆVt|] + E[ft−1],*
where
*E[| ˆVt|] = (n −*
*t−1*
∑
*d=0*
*E[|Vd|]) · (1 − (1 − p)E[|Ft−1|]),*
*E[|Tt−1|] = nE[ft−1],*
*E[|Ft−1|] = n(E[ft−1*]*− E[ft−2]),*

*E[|V*0*|] = 1 by definition and E[|Vd|], d ≥ 1 are given by Lemma 2.3.1.*

*proof 6 (Proof of Theorem 2) After step 1, E[ft] depends on random variables|V*1*|, |V*2*|, · · · , |Vt−1|,*

*as well as k(V*1*), k(V*2*),· · · , k(Vt−1). However, finding the joint distribution of all these*

*random variables is very difficult. Therefore, we need to make several necessary sim-*
*plifying assumptions and approximations to obtain a closed-form result, as listed in the*
*theorem. In this proof, we will first use these assumptions and approximations to derive*
*the approximation of E[ft], and then justify all the assumptions and approximations point-*

*by-point in the end of the proof.*

*According to the assumption 1): we only consider the failures propagating in the for-*
*ward direction, i.e., at step t only the nodes in V\ ∪t _{d=0}−1*

*Vd*

*are considered as potential*

*nodes to fail, we define the set of target nodes ˆVtat step t as the nodes in V\ ∪td=0−1* *Vdcon-*

*nected to Ft−1, as illustrated in Fig. 5.2. Since each node in V\ ∪td=0−1* *Vdhas probability*

*p to be connected with a node in Ft−1* *independently,* *| ˆVt| obeys a binomial distribution*

*B(n−*∑*t _{d=0}−1*

*|Vd|, 1 − (1 − p)|Ft−1|). The target nodes will receive redistributed load from*

*Figure 5.2: At step t, Ft−1* redistributes its load to set ˆ*Vt. Vt−1* is the set of nodes with a

*shortest distance t− 1 to node a; Ft−1* *is the set of nodes failing at step t− 1; ˆVt*is the set

*of target nodes at step t, defined as the nodes in V\ ∪t _{d=0}−1*

*Vdconnected to Ft−1*.

*Furthermore, instead of analyzing the load redistribution for every node in Ft−1, we*

*make the assumption 2): the set Ft−1is considered as a large virtual node that redistributes*

*its load to its alive neighbors at step t. This assumption makes the analysis mathematically*
*tractable. The problem now becomes “a single node redistributing its load to its neigh-*
*bors”. Similar to step 1, by applying Corollary 1 to this equivalent setting, the failure*
*condition for ˆVtcan be found as*

*Lt(Ft−1) + L*0( ˆ*Vt) > αL*0( ˆ*Vt),* (5.21)

*and E[|Ft|] can be obtained as*

*E[|Ft|] =*
∑
*z*
*z Pr*
{
ˆ
*Vtfails| | ˆVt| = z*
}
Pr*{| ˆVt| = z}.* (5.22)

*Recall our goal is to obtain E[ft] through*

*where E[ft−1] is estimated in the previous step analysis. Now we aim to find E[|Ft|] in*

*(5.22). However, it is difficult to find the exact distribution of| ˆVt| in (5.22) as it requires*

*the joint distribution of|V*1*|, |V*2*|, · · · , |Vt−1| and k(V*1*), k(V*2*),· · · , k(Vt−1). According to*

*the approximation 3): E[| ˆVt|] is applied to approximate | ˆVt|, we have*

*E[|Ft|] ≈ Pr*
{
ˆ
*Vtfails| | ˆVt| = E[| ˆVt|]*
}
*E[| ˆVt|],* (5.23)

*where E[| ˆVt|] depends on random variable |Ft−1|:*

*E[| ˆVt|] =*

∑

*y*

*E[| ˆVt| | |Ft−1| = y] · Pr{|Ft−1| = y}.* (5.24)

*And by ˆVt’s definition, we have*

*E[| ˆVt| | |Ft−1| = y] = (n −*
*t−1*

∑

*d=0*

*E[|Vd|]) · (1 − (1 − p)y).* (5.25)

*Now we aim to find E[| ˆVt|]. To avoid finding Pr{|Ft−1| = y} and the summation in*

*(5.24), we make the approximation 4): E[| ˆVt| | |Ft−1| = E[|Ft−1|]] is applied to approxi-*

*mate E[| ˆVt|]. This approximation leads to*

*E[| ˆVt|] ≈ (n −*
*t−1*
∑
*d=0*
*E[|Vd|]) · (1 − (1 − p)E[|Ft−1|]),* (5.26)
*where*
*E[|Ft−1|] = n(E[ft−1]),*

*and E[ft−1] is given by the previous step analysis; E[|V*0*|] = 1 by definition, and E[|Vd|], ∀d ≥*

*1 are given by Lemma 1. We now estimate the probability Pr*
{

ˆ

*Vtfails| | ˆVt| = E[| ˆVt|]*

}

*in (5.23). According to the failure condition (5.21), Pr*
{

ˆ

*Vtfails| | ˆVt| = E[| ˆVt|]*

}

*obtained by*
Pr
{
ˆ
*Vtfails| | ˆVt| = E[| ˆVt|]*
}
= Pr*{Lt(Ft) > (α− 1)L*0( ˆ*Vt*)*| | ˆVt| = E[| ˆVt|]}.* (5.27)

*According to assumption 5), the above probability can be approximated by*

Pr*{Lt(Ft) > (α− 1)L*0( ˆ*Vt*)*| | ˆVt| = E[| ˆVt|]} ≈ Φ(*
˜
*µ− ˆµ*
ˆ
*σ* *),* (5.28)
*where*
˜
*µ = E[|Tt−1|](n − 1)p,*
ˆ
*µ = (α− 1)(n − 1)E[| ˆVt|]p,*
ˆ
*σ*2 *= (α− 1)*2*(n− 1)E[| ˆVt|]p(1 − p).*

*By substituting (5.28) and (5.26) into (5.23), we find E[|Ft|]. Then E[ft] can be ob-*

*tained as E[ft] = E[|Ft|]/n + E[ft−1], which yields Theorem 2.*

*Now we show the point-by-point justifications of all assumptions and approximations*
*used:*

*1. We only considered failures propagating in the forward direction, i.e., at step t, we*
*only considered the nodes in V\ ∪t _{d=0}−1*

*Vdas potential nodes to fail.*

*This assumption will be discussed in Section 3.4.*

*2. The set Ft−1* *is considered as a large virtual node that redistributes its load to its*

*alive neighbors at step t with the rule defined in (3).*

3 3.5 4 4.5 5 5.5 6 Degree 0.8 1 1.2 1.4 1.6 1.8 2 2.2

Received Load V2 data

Linear Regression

Figure 5.3: Example of load redistribution. White numbers located inside of circles are degrees, and numbers outside of circles are received load amounts.

*ematically analyzable. Without this assumption, we would need to have the joint*
*distribution of all node degrees in Ft−1, as well as the link connections between*

*Ft−1* *and ˆVt, which is analytically complicated, especially when t is large.*

*This assumption is appropriate in ER random graph because such graph is homo-*
*geneous by construction. In a typical realization of ER graph, the loads of nodes*
*in Ft−1* *have small variation. After the nodes in Ft−1* *distribute their loads to their*

*alive neighbors according to their degrees, a neighbor node with a higher degree*
*tends to receive more load, and vice versa. Such an example is shown in Fig. 5.3.*
*The given partial network is a typical realization of ER random graph. Nodes in*
*V*1 *redistribute their loads to their neighbors in V*2*. The scatter plot of the received*

*load amounts and degrees in V*2 *is also shown in Fig. 5.3. The linear correlation of*

*the received load amounts and degrees in V*2 *is 0.9556, indicating a strong linear*

*relationship, which matches the assumed case.*

*3. n is large enough such that the variance of| ˆVt| is small and | ˆVt| can be approximated*

*By definition of target nodes, we have*
*| ˆVt| ∼ B(n −*
*t−1*
∑
*d=0*
*|Vd|, 1 − (1 − p)|Ft−1|),*
*with variance*
*V ar = (n−*
*t−1*
∑
*d=0*
*|Vd|)(1 − (1 − p)|Ft−1|*)(1*− p)|Ft−1|*
*= E[| ˆVt|] · (1 − p)|Ft−1|.*

*For t* *≥ 2 and before the stage of steady state, we have |F _{t−1}| ≫ 0, (1 − p)|Ft−1|≈ 0*

*and V ar* *≈ 0 under typical settings of finite networks. For example, given p = 0.06,*
*|Ft−1| = 100 and E[| ˆVt|] = 40, V ar = 0.0822. With Chebyshev’s inequality, we*

*have Pr{|| ˆVt| − 40| > 3 ×*

*√*

*0.0822} ≤* 1_{9}*. We can see that| ˆVt| stays close to its*

*mean with high probability such that it can be approximated by its mean, as long as*
*network’ size is large enough (still finite). In addition, for the asymptotic case, we*
*also have (1− p)|Ft−1|→ 0 and V ar → 0 with |F*

*t−1| → ∞.*

*4. E[| ˆVt| | |Ft−1| = E[|Ft−1|]] is applied to approximate E[| ˆVt|].*

*For notational simplicity, let Y =* *|Ft−1|. Now our goal is to show that E[| ˆVt|] ≈*

*E[| ˆVt| | Y = E[Y ]]. According to (5.24), we have*

*E[| ˆVt|] =*

∑

*y*

*E[| ˆVt| | Y = y]P (Y = y),* (5.29)

*where E[| ˆVt| | Y = y] can be rewritten as a function of y:*

*Figure 5.4: Histogram of Lt(Ft−1*) and fit normal distribution. Skewness and kurtosis

indicate that this distribution is quite symmetric, not heavily tailed.

*with c = (n−*∑*t _{d=0}−1*

*E[|Vd|]) as a constant. Combining (5.29) with (5.30), we have*

*E[| ˆVt|] = E[f(Y )].* (5.31)

*We now aim to show that E[f (Y )]* *≈ f(E[Y ]). First we look at the derivative of*
*f (y):*

*f′(y) =* *−c(1 − p)y*ln(1*− p),*

*where c is a positive constant, (1* *− p)y* _{< 1, and ln(1}_{− p) ≈ 0 under typical}

*settings in finite networks (where p is a small number). So f′(y) is a very small*
*positive number. For example, given c = 10, p = 0.08, and y = 30, we have*
*f′(y) =−10 · 0.92*30*· ln(0.92) = 0.0683, which is close to zero. Since f′ _{(y) is close}*

*to zero and f (y) is approximately constant over y, E[f (Y )] can be approximated*
*by:*

*where y∗* *is an arbitrary point within f (y)’s domain. Let us pick y∗* *= E[Y ] such*
*that we have E[f (Y )]* *≈ f(E[Y ]), where E[Y ] is obtained from the previous step*
*analysis. According to the simulation result, Y usually has a symmetric distribu-*
*tion and E[Y ] is the median of Y . Since f (Y ) is a monotonic increasing function,*
*f (E[Y ]) must be the median of f (Y ). Thus f (E[Y ]) is a reasonable approximation*
*of E[f (Y )]. Based on (5.31), we have*

*E[| ˆVt|] = E[f(Y )] ≈ f(E[Y ]).*

*According to the definition of f (y) in (5.30), we have*

*E[| ˆVt|] ≈ f(E[Y ]) = E[X | Y = E[Y ]]*

*= E[| ˆVt| | |Ft−1| = E[|Ft−1|]].*

*5. Given* *| ˆVt| = E[| ˆVt|], Lt(Ft−1) and (α* *− 1)L*0( ˆ*Vt) are independent and approxi-*

*mately normal. Lt(Ft−1) has conditional mean ˜µ = E[|Tt−1|](n−1)p and unknown*

*variance. (α− 1)L*0( ˆ*Vt) has conditional mean ˆµ = (α− 1)(n − 1)E[| ˆVt|]p and*

*conditional variance ˆσ*2 * _{= (α}_{− 1)}*2

_{(n}_{− 1)E[| ˆ}_{V}*t|]p(1 − p). Φ(µ*˜*−ˆµ _{σ}*

_{ˆ}

*) is applied as an*

*approximation of Pr{Lt(Ft−1) > (α− 1)L*0( ˆ*Vt*)*| | ˆVt| = E[| ˆVt|]}.*

*For notational simplicity, let R*1*and R*2*denote random variables Lt(Ft−1) and (α−*

*1)L*0( ˆ*Vt), conditional on| ˆVt| = E[| ˆVt|]}, respectively. R*1 *has mean ˜µ and variance*

˜
*σ*2_{. R}

2*has mean ˆµ and variance ˆσ*2*. Thus, obtaining the probability Pr{Lt(Ft−1) >*

*(α− 1)L*0( ˆ*Vt*)*| | ˆVt| = E[| ˆVt|]} is equivalent to obtaining Pr{R*1 *> R*2*}.*

*First we look at R*1*, which does not depend on* *| ˆVt| and it equals the summation*

*over the initial loads in Tt−1. Since all the nodes’ initial load amounts are i.i.d.*

*However,* *|Tt−1| is itself a random variable with an unknown distribution. Only*

*E[|Tt−1|] is obtained from the previous step’s analysis. Thus, ˜µ = E[|Tt−1|](n − 1)p*

*is known, while ˜σ*2 _{and the distribution of R}

1 *remain unknown. To get an idea of*

*what distribution R*1*looks like, we show a histogram of Lt(Ft−1) from the simulation*

*results in Fig. 5.4, which has the following settings: n = 100, p = 0.06, α = 1.1, t =*
*4, and sample size is 50, 000. The case where no failures are triggered at step 1*
*was excluded in the histogram since no failures will happen at step 2 either in this*
*scenario. From Fig. 5.4, we see that R*1 *is approximately normal.*

*Then we look at R*2*. Given that| ˆVt| = E[| ˆVt|], we have R*2 *∼ B((n − 1)E[| ˆVt|], p),*

*which is approximately normal because E[| ˆVt|] ≫ 0 before the steady state, such*

*that (n− 1)E[| ˆVt|]p and (n − 1)E[| ˆVt|](1 − p) are both greater than 5 under typical*

*settings in a practical-size network. Given α* *− 1 is a constant, (α − 1)L*0( ˆ*V*0)

*is also approximately normal. R*2*’s mean ˆµ and variance ˆσ*2 *can be obtained as*

ˆ

*µ = (α− 1)(n − 1)E[| ˆVt|]p and ˆσ*2 *= (α− 1)*2*(n− 1)E[| ˆVt|]p(1 − p).*

*Note R*1 *and R*2 *are independent because L*0( ˆ*Vt) and Lt(Ft−1) will not affect each*

*other’s distribution with given| ˆVt|. Then,*

Pr*{R*1 *> R*2*} = Pr{R*2*− R*1 *< 0}*

*≈ Φ(√µ*˜*− ˆµ*
˜

*σ*2_{+ ˆ}* _{σ}*2

*),*(5.32)

*where ˜σ*2 *remains unknown. We notice that Pr{R*1 *> R*2*} does not depend on ˜σ*2 *in*

*the following three cases:*

*(a) ˜µ− ˆµ ≫ 0 and (5.32) ≈ 1,*
*(b) ˜µ− ˆµ ≪ 0 and (5.32) ≈ 0,*
*− ˆµ = 0 and (5.32) = 0.5.*