Class 20: Robustness Cascades PartI
Prof. Boleslaw Szymanski
Prof. Albert-László Barabási
Dr. Baruch Barzel, Dr. Mauro Martino
Network Science: Robustness Cascades
2015
A SIMPLE STORY (3):
Thex
Network Science: Introduction 2012
ROBUSTNESS IN COMPLEX SYSTEMS
Complex systems maintain their basic functions even under errors and failures
cell  mutations
There are uncountable number of mutations and other errors in our cells, yet, we do not notice their
consequences.
Internet  router breakdowns
At any moment hundreds of routers on the internet are broken, yet, the internet as a whole does not
loose its functionality.
Where does robustness come from?
There are feedback loops in most complex systems that keep tab on the
component’s and the system’s ‘health’.
Could the network structure affect a system’s robustness?
Network Science: Robustness Cascades
2015
ROBUSTNESS
Could the network structure affect a system’s robustness?
node failure
How do we describe in quantitave terms the breakdown of a
network under node or link removal?
~percolation theory~
Network Science: Robustness Cascades
2015
PERCOLATION THEORY
p= the probability that a node is occupied
Increasing p
Critical point pc: above pc we have a spanning cluster.
Network Science: Robustness Cascades
2015
PERCOLATION: CRITICAL EXPONENTS
p – play the role of T in thermal phase transitions
Order parameter:
P∞~(p-pc)β
Correlation length:
ξ~|p-pc|-ν
P∞
probability that a node (or link) belongs to the  cluster
mean distance between two sites on the same cluster
Cluster size:
S~|p-pc|-γ
Average size of finite clusters
•β, ν, γ: critical exponents—characterize the behavior near the phase transition
•The exponents are universal (independent of the lattice)
•pc depends on details (lattice)
•ν and γ are the same for p>pc and p<pc
•For ξ and S take into account all finite clusters
Network Science: Robustness Cascades
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I:
Subcritical
<k> < 1
II:
Critical
<k> = 1
III:
Supercritical
<k> > 1
IV:
Connected
<k> > ln N
N=100
<k>
<k>=0.5
<k>=1
<k>=3
<k>=5
ROBUSTNESS
Could the network structure contribute to robustness?
node failure
How do we describe in quantitave terms the breakdown of a
network under node removal?
~percolation theory~
Network Science: Robustness Cascades
2015
ROBUSTNESS: INVERSE PERCOLATION TRANSITION
Remove
nodes
Unperturbed
network
Remove
nodes
Giant Component
Persists
P
1
Network
Collapses
P∞:probability that a node
belongs to the giant
component
f: fraction of removed nodes.
0
fc
f
Network Science: Robustness Cascades
2015
Damage is modeled as an inverse percolation process
f= fraction of removed nodes
Graph
Component
structure
S
fc
f
(Inverse Percolation phase transition)
Network Science: Robustness Cascades
2015
BOTTOM LINE: ROBUSTNESS OF REGULAR NETWORKS IS WELL UNDERSTOOD
f=0: all nodes are
part of the giant
component, i.e.
S=N, P∞=1
P
1
0
fc
f
0<f<fc: the network is
fragmented into many clusters
with average size S~|p-pc|-γ
f>fc: the network
collapses, falling into
many small clusters;
there is a giant component;
the probability that a node
belongs to it: P∞~(p-pc)β
giant component
disappears
Network Science: Robustness Cascades
2015
ROBUSTNESS: OF SCALE-FREE NETWORKS
The interest in the robustness problem has three origins:
Robustness of complex systems is an important problem in many areas
Many real networks are not regular, but have a scale-free topology
In scale-free networks the scenario described above is not valid
Albert, Jeong, Barabási, Nature 406 378 (2000)
Network Science: Robustness Cascades
2015
ROBUSTNESS OF SCALE-FREE NETWORKS
Scale-free networks do not appear to
break apart under random failures.
Reason: the hubs.
The likelihood of removing a hub is small.
1
S
0
1
f
Albert, Jeong, Barabási, Nature 406 378 (2000)
Network Science: Robustness Cascades
2015
MALLOY-REED CRITERIA: THE EXISTENCE OF A GIANT COMPONENT
A giant cluster exists if each node is connected to at least two other nodes.
The average degree of a node i linked to the GC, must be 2, i.e.
Bayes’ theorem
P(ki|i <-> j): joint probability that a node has degree ki and is connected to nodes i and j.
For a randomly connected network (does NOT mean random network!) with P(k):
i can choose between N-1 nodes to
link to, each with probability 1/(N-1).
I can try ki times.
< k2 >
kº
=2
<k>
κ>2: a giant cluster exists
κ<2: many disconnected clusters
Network Science: Robustness Cascades
Malloy, Reed, Random Structures and Algorithms (1995); Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
2015
Probability Distribution Function
(PDF)
Apply the Malloy-Reed Criteria to an Erdos-Renyi Network
Discrete Formulation
Continuum Formulation
-binomial distribution-
-Poisson distribution-
æ N -1ö k
(N -1)-k
P(k) = ç
÷ p (1- p)
è k ø
P(k) = e
-<k>
< k >k
k!
< k >= (N -1) p
< k >=< k >
< k 2 >= p(1- p)(N -1) + p2 (N -1)2
< k 2 >=< k > (1+ < k >)
s k = (< k 2 > - < k >2 )1/ 2 = [ p(1- p)(N -1)]1/ 2
s k = (< k 2 > - < k >2 )1/ 2 =< k >1/ 2
Network Science: Robustness Cascades
2015
Apply the Malloy-Reed Criteria to an Erdos-Renyi Network
A giant cluster exists if each node is connected to at least two other nodes.
< k2 >
kº
=2
<k>
κ>2: a giant cluster exists;
κ<2: many disconnected clusters;
< k >=< k >
< k 2 >=< k > (1+ < k >)
s k = (< k 2 > - < k >2 )1/ 2 =< k >1/ 2
< k 2 > < k > (1+ < k >)
kº
=
= 1+ < k >= 2
<k >
<k >
< k >=1
Malloy-Reed; Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
Network Science: Robustness Cascades
2015
< k2 >
kº
=2
<k>
S
Graph
Component
structure
RANDOM NETWORK: DAMAGE IS MODELED AS AN INVERSE OERCOLATION PROCESS
fc
f= fraction of removed nodes
kc : <k>=1
f
<k>
Network Science: Robustness Cascades
(Inverse percolation phase transition)
2015
BREAKDOWN THRESHOLD FOR ARBITRARY P(k)
Problem: What are the consequences of removing a fraction f of all nodes?
At what threshold fc will the network fall apart (no giant component)?
Random node removal changes
the degree of individual nodes [k  k’ ≤k]
the degree distribution [P(k)  P’(k’)]
A node with degree k will loose some links and become a node with degree k’ with probability:
æ k ö k -k'
(1 - f ) k'
ç ÷f
è k'ø
Remove k-k’
links, each with
probability f
k'£ k
Leave k’ links
untouched, each
with probability 1-f
The prob. that we had a k
degree node was P(k), so
the probability that we will
have a new node with
degree k’ :
ækö
P'(k') = å P(k)ç ÷ f k -k' (1 - f ) k'
è k'ø
k =k'
¥
Let us asume that we know <k> and <k2> for the original degree distribution P(k)
 calculate <k’> , <k’2> for the new degree distribution P’(k’).
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
Network Science: Robustness Cascades
2015
BREAKDOWN THRESHOLD FOR ARBITRARY P(K)
ækö
P'(k') = å P(k)ç ÷ f k -k' (1 - f ) k'
è k'ø
k =k'
¥
¥
¥
Degree distribution after we removed f fraction of nodes.
¥
¥
k!
k -k'
k'
< k' > f = å k' P'(k') = å k' å P(k)
f
(1 - f ) = å
k'!(k
k')!
k' =0
k' =0 k =k'
k' =0
The sum is done over
the triangel shown in
the right, so we can
replace it with
¥
< k' > f = å
k' =0
k=[k’, ∞)
¥
¥
k(k -1)!
å P(k) (k' -1)!(k - k')! f
k -k'
(1 - f ) k'-1(1 - f )
k =k'
¥
åå
k' =0 k =k'
¥
=å
k =0
k
å
k' =0
k’
¥
¥
k
k(k -1)!
(k -1)!
k -k'
k'-1
å P(k) (k' -1)!(k - k')! f (1 - f ) (1 - f ) =å(1 - f )kP(k) å (k' -1)!(k - k')! f k -k' (1 - f )k' -1 =
k =k'
k =0
k' =0
¥
æ k -1ö k -k'
k'-1
åå(1 - f )kP(k) åçè k' -1÷ø f (1 - f ) =å(1 - f )kP(k) = (1 - f ) < k >
k' =0 k =0
k' =0
k =0
k
¥
k
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
Network Science: Robustness Cascades
2015
BREAKDOWN THRESHOLD FOR ARBITRARY P(K)
ækö
P'(k') = å P(k)ç ÷ f k -k' (1 - f ) k'
è k'ø
k =k'
¥
Degree distribution after we removed f fraction of nodes.
¥
< k' > f =< k'(k' -1) - k' > f = å k'(k' -1)P'(k')- < k' > f
2
k' =0
k=[k’, ∞)
The sum is done over
the triangel shown in
the right, i.e. we can
replace it with
¥
< k'(1 - k') > f = å
k' =0
¥
¥
åå
k' =0 k =k'
¥
=å
k =0
k
å
k' =0
k’
¥
¥
k
k(k -1)(k - 2)! k -k'
(k - 2)!
k -2'
2
2
å P(k) (k' -2)!(k - k')! f (1 - f ) (1 - f ) =å(1 - f ) k(k -1)P(k) å (k' -2)!(k - k')! f k -k' (1 - f )k' -2 =
k =k'
k =0
k' =0
¥
æ k - 2ö k -k'
k' -2
åå(1 - f ) k(k -1)P(k) åçè k' -2÷ø f (1 - f ) =å(1 - f )2 k(k -1)P(k) = (1 - f )2 < k(k -1) >
k' =0 k =0
k' =0
k =0
k
¥
k
2
< k'2 > f =< k'(k' -1) - k' > f = (1 - f )2 (< k 2 > - < k >) - (1 - f ) < k >= (1 - f )2 < k 2 > + f (1 - f ) < k >
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
Network Science: Robustness Cascades
2015
BREAKDOWN THRESHOLD FOR ARBITRARY P(K)
Robustness: we remove a fraction f of the nodes.
At what threshold fc will the network fall apart (no giant component)?
Random node removal changes
the degree of individuals nodes [k  k’ ≤k)
the degree distribution [P(k)  P’(k’)]
< k' > f = (1 - f ) < k >
kº
< k'2 > f = (1 - f )2 < k 2 > + f (1 - f ) < k >
Breakdown threshold:
fc = 1 -
< k' 2 > f
< k' > f
=2
κ>2: a giant cluster exists
κ<2: many disconnected clusters
1
k2
1
-1
k
f<fc: the network is still connected (there is a giant cluster)
0
f>fc: the network becomes disconnected (giant cluster vanishes)
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
S
fc
Network Science: Robustness Cascades
f
2015
ROBUSTNESS OF SCALE-FREE NETWORKS
Scale-free networks do not appear to
break apart under random failures.
Reason: the hubs.
The likelihood of removing a hub is small.
1
S
0
1
f
Albert, Jeong, Barabási, Nature 406 378 (2000)
Network Science: Robustness Cascades
2015
ROBUSTNESS OF SCALE-FREE NETWORKS
fc = 1 -
Scale-free networks do not appear to break apart
under random failures. Why is that?
<k
m
>= (g -1)K min
g -1
K max
òk
K min
< k m >=
m -g
[
k2
k
K max
(g -1)
-1
m -g +1
dk =
K gmin
k
[
] K min
(m - g +1)
(g -1)
-1
K gmin
K max m -g +1 - K min m -g +1
(m - g +1)
1
-1
K max = K min N
1
g -1
]
(2 - g ) K max 3-g - K min 3-g
2 -g
g > 3: k =
=
K
(3 - g ) K max 2-g - K min 2-g
3 - g min
< k 2 > (2 - g ) K max 3-g - K min 3-g
=
<k>
(3 - g ) K max 2-g - K min 2-g
(2 - g ) K max 3-g - K min 3-g
2 -g
-2
3 >g > 2: k =
=
K max 3-g Kgmin
2-g
2-g
(3 - g ) K max
- K min
3 -g
(2 - g ) K max 3-g - K min 3-g
2 -g
2 > g > 1: k =
K
2-g
2-g =
(3 - g ) K max
- K min
3 - g max
Network Science: Robustness Cascades
2015
ROBUSTNESS OF SCALE-FREE NETWORKS
1
fc = 1 k -1
ì
K min
ï
<k >
2 -g
-2
íK max 3-g K gmin
k=
=
<k>
3 -g ï
K max
î
g >3
3 >g >2
2 > g >1
2
K max = K min N
1
g -1
γ>3: κ is finite, so the network will break apart at a finite fc that depens on Kmin
γ<3: κ diverges in the N ∞ limit, so fc  1 !!!
for an infinite system one needs to remove all the nodes to break the system.
For a finite system, there is a finite but large fc that scales with the system size as:
Internet: Router level map, N=228,263; γ=2.1±0.1;
AS level map, N= 11,164; γ=2.1±0.1;

κ=28
κ=264

k @1 - CN
-
3-g
g -1
fc=0.962
fc=0.996
Network Science: Robustness Cascades
2015
NUMERICAL EVIDENCE
Scale-free random graph with
P ( k )  Ak  , with k  m ,...K
1
if   3
 2
m 1
 3
1
fc  1 
if 2    3
  2   2 3 
m K
1
3
  2 .5
fc  1 
K  400
  3 .5
K  25
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
Infinite scale-free networks with   3 do not break down under
random node failures.
Network Science: Robustness Cascades
2015
SIZE OF THE GIANT COMPONENT DURING RANDOM DAMAGE –WITHOUT PROOF-
S: size of the giant component.
(i) γ>4: S≈f-fc (similar to that of a random graph)
(i) 3>γ>4: S≈(f-fc)1/(γ-3)
(i) γ<3: fc =0 and S≈f1+1/(3-γ)
R. Cohen, D. ben-Avraham, S. Havlin,
Percolation critical exponents in scale-free networks
Phys. Rev. E 66, 036113 (2002);
See also: Dorogovtsev S, Lectures on Complex Networks, Oxford, pg44
Network Science: Robustness Cascades
2015
INTERNET’S ROBUSTNESS TO RANDOM FAILURES
failure
attack
Internet
1
fc = 1 k -1
R. Albert, H. Jeong, A.L. Barabasi, Nature 406 378 (2000)
Internet: Router level map, N=228,263; γ=2.1±0.1;
AS level map, N= 11,164; γ=2.1±0.1;
κ=28

fc=0.962
κ=264

fc=0.996
Internet parameters: Pastor-Satorras & Vespignani, Evolution and Structure of the Internet: Table 4.1 & 4.4
Network Science: Robustness Cascades
2015
Achilles’ Heel of scale-free networks
1
Attacks
Failures
  3 : fc=1
S
(R. Cohen et al PRL, 2000)
0
fc
Albert, Jeong, Barabási, Nature 406 378 (2000)
f
1
Network Science: Robustness Cascades
2015
Attack threshold for arbitrary P(k)
Attack problem: we remove a fraction f of the hubs.
At what threshold fc will the network fall apart (no giant component)?
Hub removal changes
the maximum degree of the network [Kmax  K’max ≤Kmax)
the degree distribution [P(k)  P’(k’)]
A node with degree k will loose some links because some of its neighbors will vanish.
Claim: once we correct for the changes in Kmax and P(k),we are back to the robustness problem.
That is, attack is nothing but a robusiness of the network with a new Kmax and P(k).
fc
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
Network Science: Robustness Cascades
f
2015
Attack threshold for arbitrary P(k)
Attack problem: we remove a fraction f of the hubs.
the maximum degree of the network [Kmax  K’max ≤Kmax) `
If we remove an f fraction of hubs, the maximum degree changes:
K max
ò P(k)dk =
f
'
K max
K max
ò P(k)dk = (g -1)K
'
K max
æ K min ög -1
ç
÷ = f
è K'max ø
g -1
min
g -1 g -1 1-g
g
k dk =
K min (K max - K'1max )
1-g
K max
ò
-g
'
K max
K' max = K min f
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
1
1-g
As K’max ≤Kmax
we can ignore
the Kmax term
 The new maximum degree after
removing f fraction of the hubs.
Network Science: Robustness Cascades
2015
Attack threshold for arbitrary P(k)
Attack problem: we remove a fraction f of the hubs.
the degree distribution changes [P(k)  P’(k’)]
A node with degree k will loose some links because some of its neighbors will vanish.
Let us calculate the fraction of links removed ‘randomly’ , f’, as a consequence of we removing f
K
fraction of hubs.
ò kP(k)dk
max
0
K max
ò kP(k)dk
f '=
'
K max
<k>
=
1
1 g -1 g -1 2-g
-1
(g -1)K gmin
k1-g dk =
K (K
ò
<k>
< k > 2 - g min max
'
K max
1 g -1 g -1 2-g
f '= K K f
< k > 2 - g min min
< k m >= -
(g -1)
m
K min
(m - g + 1)
K max
2-g
1-g
1 g -1
=K f
< k > 2 - g min
f '= f
2-g
1-g
(g -1)
K
(2 - g ) min
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
< k >= -
as K’max ≤Kmax
1 g -1 g -1 2-g
g
- K'2K K'
max ) = < k > 2 - g min max
2-g
1-g
For γ2, f’1, which means that even
the removal of a tiny fraction of hubs will
destroy the network. The reason is that
for γ=2 hubs dominate the network
Network Science: Robustness Cascades
2015
Attack threshold for arbitrary P(k)
Attack problem: we remove a fraction f of the hubs.
At what threshold fc will the network fall apart (no giant component)?
Hub removal changes
1
1
the maximum degree of the network [Kmax  K’max ≤Kmax) K'
max  K min f
2
the degree distribution [P(k)  P’(k’)]
A node with degree k will loose some links because some of its neighbors will vanish. f '  f 1
Claim: once we correct for the changes in Kmax and P(k), we are back to the robustness problem.

That is, attack is nothing but a robustness of the network with a new K’max and f’.
1
f ' 1
 '1
K min

2  
3
 2

K
 max K min
3 
K max

 k '2 
 k2 


'


 k'
(1  f c )  k  1  f c
 3
3 2
2   1
Cohen et al., Phys. Rev. Lett. 85, 4626 (2000).
fc
2
1
2 
 3

2
K min f c 1 1


3 
Network Science: Robustness Cascades
2015