Workshop ”Games and Optimization”, Saint-Etienne
Optimal design and defense of networks under link attacks
Bravard C.
a, Charroin L.
b, Touati C.
caUniversit´e Grenoble-Alpes.
bUniversit´e de Lyon and Ecole Normale Sup´erieure de Lyon.
cInria, Montbonnot Saint-Martin.
June 2016
Motivation
Networks represent a crucial feature in our society, and are of particular interest in di↵erent fields such as military defense, telecommunication or computer networks.
Networks can be damaged by natural disasters or intelligent attacks.
Attacks can a↵ect links (roads, communications flows, ...),
and may disconnect a network.
Illustration
Consider a firm which consists of several production units (nodes of the network).
Each production unit produces a part of the product and the pieces are assembled by a given production unit.
The links of the network allow the parts of the product to be transferred among the units.
If one unit is not connected to the rest of the units, its
production cannot be transferred and the product has no
value.
Illustration
During the Second World War, the production units for the weapons (nodes) were buried, so they were impossible to target, and attacks had to target the roads (links) in order to destroy the production process of the enemy.
Therefore, the issue was to design a network of
communication between the production units that the enemy
could not disconnect.
Our Goal
Our goal is to examine how to design and protect the network in an optimal way, such that the network remains connected after an intelligent link attack.
We say that a network is designed and protected in an optimal way
when the costs associated with the design and the protection of
the network are minimized.
Structure of the Game
We consider a two-stage game with two players: a Designer (D) and an Adversary (A).
Stage 1. The Designer moves first and chooses both a set of protected, and a set of non-protected links. Protected links cannot be removed by the attacks of the Adversary.
Stage 2. After observing the protected network (strategy)
formed by the Designer, the Adversary attacks the network by
allocating attacks to specific links. Since the attacks are
costly, the Adversary has an incentive to attack at most k
links.
Main Assumptions
Creating protected and non-protected links is costly for the Designer.
The benefits obtained by the Designer at the end of the game depends on the connectivity of the residual network, that is the network obtained after the attack of the Adversary.
If the residual network is connected, then the Designer wins
the game: her benefits are equal to 1 and the benefits of the
Adversary are 0.
If the residual network is not connected, then the Adversary wins the game: her benefits are equal to 1 and the benefits of the Designer are 0.
The payo↵ obtained by the players are equal to the di↵erence
between their benefits and the costs associated with their
strategies.
Equilibrium Concept
We are interested in the Sub-game Perfect Equilibrium (SPE) of the two-stage game.
We assume that the cost of protected links and non-protected links
are sufficiently low: the Designer has some profitable strategies
which allow the residual network to be connected.
Main Results
First, we provide for each number of protected links, the minimal number of non-protected links that the Designer has to form in order to avoid the Adversary to disconnect the network.
Second, we establish that three polar non-empty networks
may arise in equilibrium in the benchmark model.
Equilibrium Networks
1
A minimal (k + 1, n)-link-connected network which contains no protected links.
2
A minimal (1, n)-link-connected network which contains n 1 protected links.
3
A network g which contains one protected link and
d(n 1)(k + 1)/2e non-protected links.
Extensions
We study some variations of the game.
1
We consider that D cannot create as many protected links as in the benchmark model.
2
We consider situations where each protected link has a
probability ⇡ to be removed when it is attacked by A.
Extensions: main results
In the framework where the number of protected links available for D is limited, we show that for intermediate relative costs (cost of a protected link / cost of a
non-protected link), the optimal strategy of D consists in designing a network which contains both protected links and non-protected links.
In the framework where protected links are removed by attacks
with some probabilities, we provide conditions under which
the results obtained in our benchmark model are preserved.
Literature
Dziubinski and Goyal, GEB, 2013. major di↵erences:
The Adversary attacks nodes in the DG’s framework while she attacks links in our framework;
In our framework, the Designer wins the game if every node is able to communicate with any other node in the residual network.
In the DG’s framework, the Designer wins the game if the residual network is connected regardless of the number of nodes removed by the Adversary.
Thus, our setting is based on the complementarity of nodes
while DG assume that nodes are substitutable.
Literature
Hoyer and De Jaegher, 2010, consider a framework where the Designer has to shape the network and form enough links in order to resist the attacks. In this framework, the Designer cannot protect specific parts of the network. The authors study the optimal way to design a network under link or node deletion with various cost levels.
Haller, 2015, extends the model of Hoyer and De Jaegher by
adding the possibility for two nodes to be connected by more than
one link. In that case, it is harder for the Adversary to disconnect
the network.
Literature
Additionally to economic, several fields investigate problems close to the one we deal with.
In an early graph theoretic work, Harary exhibits a family of (k, n)-node-connected networks with a total number of links that is minimal.
Cunningham studies network security and considers a model where the Designer allocates a di↵erent number of defense units to each link.
A defended link has a level of resistance that depends on the number of defense units the Designer has allocated to it.
The Adversary allocates attack units to remove a link.
A link is removed if more attack units than defense units have
been allocated to this link.
Model Setup: Protected Network
For any integer n 4, let N = { 1, . . . , n } and E(N) be the set of unordered pairs of N.
The elements of N are referred to as nodes while those of E (N) are called links.
We introduce the notion of protected network as a triplet g = (N, E
P, E
NP) with E
P✓ E (N), E
NP✓ E (N) and E
P\ E
NP= ; .
We call protected links the elements of E
Pand non-protected links the elements of E
NP.
To simplify notation, we let p = #E
P.
Model Setup: Players and Strategies
The players are the Designer (D) and the Adversary (A).
We consider a two-stage game where D plays first and A moves at the second stage.
Given N, a strategy S
Dfor D, is identified with a protected network (N, E
P, E
NP).
A strategy for the Adversary A is a mapping that assigns to each protected network g a subset of links E
A(g ) of E(g ).
A chooses to attack some links of g :
S
A: g = (N, E
P, E
NP) 7! E
Awith E
A✓ E(g ).
Model Setup: Residual network and benefits
The strategy S
Dplayed at the first stage is a protected network.
Then, the attack of A leads to a second protected network of the form g
R= (N, E
P(S
D), E
NP(S
D) \ E
A). which we call residual network.
The benefits of D are given by (g
R) =
⇢ 1, if g
Ris connected,
0, otherwise. (1)
Model Setup: Network and Costs
Attacking a link has a unitary cost c
A. Therefore, the cost of a strategy S
Aof the Adversary is
c
A(S
A) = c
A#E
A,
where c
A2 [1/(n 3), 1). Note that the cost is less than 1 if and only if the Adversary attacks less than k = b 1/c
Ac links.
Both protected and non-protected links are costly.
Each protected link has a strictly positive cost c
P> 0 and each non-protected link has a strictly positive cost c
L> 0, c
P> c
L. The cost of a strategy S
Dof the Designer is thus:
c
D(S
D) = c
P#E
P(S
D) + c
L#E
NP(S
D),
with c
P< 1/(n 1) and c
L< 1/( d n(k + 1)/2 e ).
Sub-game Perfect Equilibrium (SPE).
A SPE is a pair (S
?D, S
?A(S
D)) that prescribes the following strategic choices. At Stage 2, A plays a best response S
?A(S
D) to S
D:
S
?A(S
D) 2 argmax
SA✓ENPD(N)[EPD(N)
{⇧
A(S
D, S
A)} .
Note that S
?A(S
D) ✓ E
NPsince attacks cannot remove protected
links.
Let g
?R(S
D) be the residual network obtained when D plays strategy S
Dand A plays a best response to S
D, that is S
?A(S
D).
Given the best response outcome g
?R(S
D), D achieves payo↵
(g
?R(S
D)) c
D(S
D) when choosing S
?D.
By noting G(N) the set of all protected networks with nodes in N, then, at stage 1, D plays S
?Dsuch that
S
?D2 argmax
SD2G(N)
{ ⇧
D(S
D, S
?A(S
D)) } .
Specific strategies.
Let S
p,kDbe the set of strategies where D designs a network that (i) cannot be disconnected by k attacks, (ii) contains p protected links, and (iii) contains a minimal number of non-protected links.
Strategies which belong to the same set induce the same costs for
D.
Specific strategies.
The strategies associated with two specific values of p are:
S
0,kDis the set of strategies where D designs a network which contains no protected link, and which is a minimal
(k + 1, n)-link-connected network. For instance (k + 1, n)-Harary-networks belong to S
0,kD.
S
nD 1,kis the set of strategies where D designs a network
which only contains protected links, and such that (N, E
P, ; )
is a tree.
(i) (2,5)-Harary-network (ii) (3,5)-Harary-network (iii) (4,5)-Harary-network
Figure 1: Examples of (, n)-Harary-networks.
Specific strategies. We define specific strategies that play a crucial role in the rest of the paper.
(N,
;,;) is induced by the strategy where
Dforms the
emptynetwork. Let
Sp,kDbe the set of strategies where
Ddesigns a network that (i) cannot be disconnected by
kattacks, (ii) contains
pprotected links, and (iii) contains a minimal number of non-protected links. Given
N, let
SDp,k(N) be the set of protected networks induced by strategies in
Sp,kD. Strategies which belong to the same set induce the same costs for
D. The strategies associated with two specific values ofpare:
• SD0,k
is the set of strategies where
Ddesigns a network which contains no protected link, and which is a minimal (k + 1, n)-link-connected network. For instance, (k + 1, n)-Harary-networks belong to
SD0,k(N).
• SDn 1,k
is the set of strategies where
Ddesigns a network
SD(N) = (N, E
P, ENP) which only contains protected links (E
NP=
;) and such that (N, E
P,;) is a tree.
143 Model Analysis
Our first result provides, for any number of protected links, the minimal number of non-protected links that the Designer has to form in order to avoid the Adversary to disconnect the network.
To establish the first result, for any pair (n, k)
2N⇥J1, n 3
K, we set
p1(k, n) and
p2(k, n) as follows:
8>
>>
>>
><
>>
>>
>>
:
= (3k + 5)
28n(k + 1)
p1(k, n) = 4
n3
k5
p8
⌫
+ 1 if 0, and 1 otherwise,
p2
(k, n) =
⇠
4
n3
k5 +
p8
⇡
1 if 0, and 1 otherwise.
When no confusion is possible, we let
p1=
p1(k, n) and
p2=
p2(k, n).
Proposition 1 Suppose thatA attacks exactlyklinks in an optimal way. Letn1
(p, k) =
l(n p)(k+1)]2
m andn2
(p, k) = (n 2p)
k+ 1
n 2p2 1 . For SD2Sp,kD ,#E
DNP=
8>>>
<
>>
>:
n1
(p, k),
forp2J0, n 2
K\Jp1(k, n), p
2(k, n)
K, n2(p, k),
forp2J0, n 2
K\Jp1(k, n), p
2(k, n)
K,0,
forp=
n1.
(5)
14Indeed, if (N, EP,;) is a tree andENP=;, thenSD(N)2SnD 1,k(N) cannot be disconnected bykattacks. Otherwise that is if (N, EP,;) contains a cycle the network is not connected unless #EDNP>0. Therefore, that network does not satisfy condition (iii) and thus is not inSDn 1,k(N).
7
Main Results. Example I
Suppose N = { 1, . . . , 10 } , k = 6 and p = 5. We describe a strategy S
Dwhere #E
NP(S
D) = n
1(5, 6). Consider the networks S
Dthe protected network such that g
0, g
00, g
000are all subnetworks of S
D, with E
P(S
D) = E
P(g
000) and
E
NP(S
D) = E
NP(g
0) [ E
NP(g
00). Note that subnetwork g
0is a
complete network and g
00is a (3, 5)-Harary-network. Finally, we
observe that each component in (N, E
P, ; ) is incident to at least 7
non-protected links and there is no possibility for A to disconnect
S
Dwith 6 attacks.
So, ifD formspprotected links andAattacksk links, then the optimal cost function associated with the pair
(p, k)
isC?
(p, k) = #E
NPD cL+
pcP, withSD2SDp,k.(6)
Proof
The proof is given in Appendix.
⇤Let us provide the intuition of Proposition 1. If
Dforms
n1 protected links, then there exists a set of strategies for
Dthat allows to resist
kattacks without non-protected links,
SnD 1,k. Otherwise, let
Dform
p2J0, n 2
Kprotected links and build a protected network
SD(N ) = (N, E
P, ENP) in
Sp,kD(N).
First, (N, E
P,;) is acyclic. Indeed, if
SD(N ) contains a cycle, then there exists a protected link, say
ij, that can be removed without altering the fact that
SD(N ) resists
kattacks. Hence, it is be possible for
Dto remove the protected link
ijand replace a non-protected link
i0j0by the protected link
i0j0, and so reduce the number of non-protected links.
Second, consider the following sequence of networks:
g0= (N,
;,;) and for any
` 2 J1, p
K,
g`= (N, E
P(g
` 1)
[ij,;) for some
ij 2 EP\EP(g
` 1). Hence
gp=
SD(N ). Since there is no cycle in (N, E
P,;), then, for any
`, the extra link ofEP(g
`) allows to merge two components of
EP(g
` 1). Since
g0has
ncomponents, then by an immediate recurrence, (N, E
P,;) has exactly
n pcomponents.
Third, observe that each component of (N, E
P,;) has to be incident to at least
k+ 1 non-protected links, otherwise
Acan disconnect the protected network
SD(N ) with
kattacks. Since there are
n psuch components,
n1(p, k) is the minimal number of non-protected links to form. In the following example, we illustrate that there exist some values of
pand
kfor which
n1(p, k) non-protected links are sufficient to resist
kattacks.
Example 1
Suppose
N=
J1, 10
K,
k= 6 and
p= 5. We describe a strategy
SDwhere #E
DNP=
n1(5, 6). Consider the networks of Figure 2 and
SD(N ) the protected network such that
g0,
g00,
g000are all subnetworks of
SD(N ), with
EDP=
EP(g
000) and
ENPD=
ENP(g
0)
[ENP(g
00). Note that subnetwork
g0is a complete network and
g00is a (3, 5)-Harary-network. Finally, we observe that each component in (N, E
P,;) is incident to at least 7 non-protected links and there is no possibility for
Ato disconnect
SD(N) with 6 attacks.
9 3 1
5
7
(i) Subnetworkg0
10 4
2
6
8
(ii) Subnetworkg00
1 2 5 6
3 4 7 8
9 10
(iii) Subnetworkg000= (N, EP,;)
Figure 2: Subnetworks associated with Example 1: the solution graph is (N, E
P, E
NP) with E
NP= E
NP(g
0) [ E
NP(g
00).
Recall that each component of (N, E
P,;) has to be incident to at least
k+ 1 non-protected links in order to resist
kattacks. This fact implies that there exist some situations where networks in
Sp,kD(N) contains
n2(p, k) non-protected links. We illustrate these situations through the following example.
Example 2
Suppose
N=
J1, 10
K,
k= 7, and
p= 2. We assume that
Dforms a protected network
SD(N) in
Sp,kD(N ), with a protected link between nodes 1 and 2 and between nodes 3 and 4. Then
SD(N) contains 8 components. We denote by
Cthe component which contains nodes 1 and 2, and by
C0the component which contains nodes 3 and 4. Each of these components has to be incident to 8 non-protected links. Networks
g0and
g00given in Figure 3 are subnetworks of
SD(N ), with
EPD=
EP(g
0) and
ENPD=
ENP(g
0)
[ENP(g
00). Each node
a2J5, 10
Kcan form at most 5 non-protected links with each other. Hence each of them has to form at least 3 non-protected links with nodes in
J1, 4
K. Altogether, nodes 5, 6, 7, 8, 9 and 10 have to form a total of 18 non-protected links with nodes 1, 2, 3 and 4. As for
8
Main Results. Example II
Suppose N = { 1, . . . , 10 } , k = 7, and p = 2. We assume that D forms a protected network with a protected link between nodes 1 and 2 and between nodes 3 and 4. Then S
Dcontains 8
components.
components
Cand
C0, they should be incident to a total of 16 non-protected links. Hence,
Dcannot form a protected network with
n1(2, 7) protected links that resists
kattacks. More precisely, a network that cannot be disconnected by
Awith 7 attacks has to contain 33 non-protected links while
n1(2, 7) = 32.
5 6 7 8 9 10
1 2 3 4
(i) Subnetworkg0= (N, EP, E0NP)
5 6 7
8
9 10
(ii) Subnetworkg00= (N,;, ENP00 )
Figure 3: Subnetworks associated with Example 2. The solution is (N, E
P, E
NP) with E
NP(g) = E
NP0[ E
NP00. In Example 2, we have assumed that
Duses protected links to form two components of size two in (N, E
P,;). Now, assume instead that (N, E
P,;) has a unique component. For instance,
Dforms pro- tected links between nodes 1 and 2 and between nodes 2 and 3. Then, each node
a2J4, 10
Kcan form at most 6 non-protected links with other nodes in
J4, 10
K. Consequently, since
k= 7 each node
a2J4, 10
Khas to form at least 2 non-protected links with nodes in
J1, 3
K. The component which contains nodes 1, 2 and 3 is incident to 14 non-protected links. It follows that
Dforms 21 + 14 = 35 non-protected links instead of 33 links in Example 2. This example illustrates how
Ddesigns (N, E
P,;) in order to minimize the costs of forming links.
We now generalize Example 2 to provide some intuition for
p1(k, n) and
p2(k, n). Consider a pro- tected network in
Sp,kD(N ) where each component of (N, E
P,;) contains either one or two nodes. There are thus
n2p components of size 1 and
pcomponents of size 2. We observe that components which contain one node need to be incident to at least
k+ 1 non-protected links. Since the number of links between components which contain 1 node is at most (n 2p)(n 2p 1)/2, then the total number of non-protected links between components which contain one node and those which contain two nodes is at least equal to (n 2p)((k + 1) (n 2p 1)/2). Moreover, to minimize the number of links, the total number of non-protected links incident to components which contain two nodes should be equal to (k + 1)p. Equation (n 2p)((k + 1) (n 2p 1)/2) = (k + 1)p is satisfied for
x1and
x2. Since
p1(k, n) =
bx1c+ 1 and
p2(k, n) =
dx2e1, the number of non-protected links required to resist
kattacks is given by
n2(p, k) when
p2Jp1(k, n), p
2(k, n)
K.
We now characterize the SPE according to the costs of forming protected and non-protected links.
Proposition 2 Let the payo↵functions be given by Equations
(3)
and(4), and let (S
?D, S?A(S
?D(N)))
be an SPE.151. Suppose thatn(k
+ 1) = 0 mod 2.
(a) If ccP
L <⇣
n n 1
⌘ k+1
2 , thenS?D2SnD 1,k andS?A
(S
?D(N)) =
;. (b) If⇣n n 1
⌘ k+1 2 < ccP
L, then,S?D2SD0,k andSA?
(S
?D(N )) =
;. 2. Suppose thatn(k+ 1) = 1 mod 2.
15The case of equality follows the same pattern, that is:
1. Ifn(k+ 1) = 0 mod 2 and ccP
L =⇣
n n 1
⌘ ⇣k+1 2
⌘
, thenSD? 2SDn 1,k[SD? andS?A(S?D(N)) =;.
2. Suppose thatn(k+ 1) = 1 mod 2.
(a) If ccP
L =⇣n 1
n 2
⌘ ⇣k+1
2
⌘, thenSD? 2SDn 1,k[SD1,k andS?A(S?D(N)) =;.
(b) If k+22 =ccP
L, thenSD? 2SD0,k[S1,kD andSA?(S?D(N)) =;.
9
Main Results. Proposition 1
To establish the first result, for any pair (n, k) 2 IN ⇥ {1, . . . n 3}, 8 >
> >
> >
> <
> >
> >
> >
:
= (3k + 5)
28n(k + 1) p
1(k , n) = b 4 n 3 k 5 p
8 c + 1 if 0, and 1 otherwise,
p
2(k , n) = d 4 n 3 k 5 + p
8 e 1 if 0, and 1 otherwise.
Proposition
Suppose that A attacks k links in an optimal way. Let n
1(p , k) = d
(n p)(k+1)]2
e and
n
2(p , k) = (n 2p) ⇣
k + 1
n 2p2 1⌘
. For S
D2 S
p,kD,
#E
NP(S
D) = 8 >
> >
<
> >
> :
n
1(p, k), p 2 [0, . . . n 2] \ [p
1(k, n), p
2(k, n)], n
2(p, k), p 2 [0, n 2] \ [p
1(k, n), p
2(k, n)],
0, for p = n 1.
Main Results. Proposition 2
Proposition
Let (S
?D, S
?A(S
?D)) be a SPE.
1
Suppose that n(k + 1) = 0 mod 2.
1
If
ccPL
< ⇣
n n 1
⌘
k+12
, then S
?D2 S
nD 1,kand S
?A(S
?D) = ; .
2
If ⇣
n n 1
⌘
k+1 2<
ccPL
, then, S
?D2 S
0,kDand S
?A(S
?D) = ; .
2
Suppose that n(k + 1) = 1 mod 2.
1
If
ccPL
< ⇣
n 1 n 2
⌘
k+12
, then, S
?D2 S
nD 1,kand S
?A(S
?D) = ; .
2
If ⇣
n 1 n 2
⌘
k+1 2<
ccPL
<
k+22, then S
?D2 S
1,kDand S
?A(S
?D) = ; .
3
If
k+22<
ccPL
, then S
?D2 S
0,kDand S
?A(S
?D) = ; .
Observe that for any
p02J1, n 2
K\Jp1, p2Kand any ˜
p2Jp1, p2K, abs(s
0,p˜)
<abs(s
0,p0). Similarly, for any
p02J1, n 2
K\Jp1, p2K, abs(s
0,p0)
<abs(s
0,n 1). Moreover, for any
p2J1, n 2
K, abs(s
0,n 1)
<abs(s
n 1,p0).
Suppose
cP/cL>abs(s
0,n 1). Then, costs of forming links with a strategy in
S0,kDare lower than the costs of forming links with a strategy in
SDn 1,k. Moreover,
cP/cL>abs(s
0,n 1)
> cP/cL>abs(s
0,p0)
>abs(s
0,˜p), with
p0 2J1, n 2
K\Jp1, p2Kand ˜
p2 Jp1, p2K. It follows that the costs of forming links are minimized for strategies in
S0,kD. Conversely, suppose
cP/cL<abs(s
0,n 1). Then
cP/cL<abs(s
n 1,p0), for
p0 2 J1, n 2
K. It follows that the costs of forming links are minimized for strategies in
SDn 1,k. Finally, note that abs(s
0,n 1) =
⇣n n 1
⌘ k+1 2
.
Average
number of non-protected links
p
p1 p˜ p2
(d
0,˜p)
(d
0,p0)
(d
0,n 1)
(d
n 1,p0)
p0 n 2 n 1
Figure 4: Intuition of Proposition 2 when n(k + 1) = 0 mod 2.
We now consider the second part of Proposition 2:
n(k+ 1) = 1 mod 2. The intuition is similar to the situation where
n(k+ 1) = 0 mod 2 except for protected networks which belong to
SD0,k(N) and
SD1,k(N). Consequently, we focus only on three sets of protected networks:
S0,kD(N),
S1,kD(N) and
SnD1,k(N ). In Figure 5, the slope
s0,1of (d
0,1), can be interpreted as the number of non-protected links that the protected link allows to save between a protected network in
SD0,k(N) and a protected network in
S1,kD(N). The same interpretation is valid for the slope
s1,n 1of (d
1,n 1), which relates a protected network in
SD1,k(N) and a protected network in
SnD 1,k(N), and for the slope
s0,n 1of the line (d
0,n 1), which relates a protected network in
S0,kD(N ) and a protected network in
SDn 1,k(N ).
Suppose that
cP/cL <abs(s
1,n 1). Then,
cP/cL<abs(s
0,n 1), and networks in
SDn 1,k(N ) have a minimal link formation cost. Conversely, suppose that
cP/cL>abs(s
1,n 1). If
k < n3,
16then there are two possibilities. If
cP/cL >abs(s
0,1), then protected networks in
SD0,k(N) minimize the cost of forming links. If
cP/cL<abs(s
0,1), then protected networks in
S1,kD(N ) minimize the cost of forming links. Finally, note that abs(s
0,1) = (k + 2)/2, and abs(s
1,n 1) =
⇣n 1 n 2
⌘ k+1 2
.
We now compare the results obtained in our framework where
Aattacks links and the results obtained in DG’s framework where
Aattacks nodes (Proposition 1 in [9]). Recall that in DG’s paper, the non-
16Fork=n 3, k+22 =⇣n 1
n 2
⌘ ⇣k+1
2
⌘, and thus the case 2.(b) of Proposition 2 never occurs.
11
Average
number of non-protected links
p
1 n 1
(d
1,n 1) (d
0,1)
0
(d
0,n 1)
Figure 5: Intuition of Proposition 2 when n(k + 1) = 1 mod 2 and k < n 3.
empty networks formed by
Dat equilibrium are either a star network with a protected central node, or a minimal (k + 1, n)-node-connected network without any protection. Hence, (S
0,kD ,;) is an SPE when the cost of links (non-protected links in our case) is sufficiently low relatively to the cost of protection in both frameworks.
However, when
Aattacks nodes,
Duses at most one protected node at equilibrium. The role played by the protections is di↵erent since in DG, if
Dbuilds a star network, one protection is sufficient to protect the network and resist any attack of
A. By contrast, in our framework,Dmay use more than one protected link in an SPE: indeed, when the cost of protected links is sufficiently low relative to the cost of non-protected links,
Ddesigns a (1, n)-link-connected network which contains only protected links. Protecting a network under link-attack is thus more costly than protecting a network under node- attack. This is because our framework calls for the survival of every node, a requirement which does not hold in the DG’s framework.
4 Limited number of protected links and imperfectly pro- tected links
In this section, we consider two potential types of restrictions on the protection of the network for the Designer. First, we consider the case where
Dcan only use a limited number of protected links and we focus on situations where this number is smaller than
p2(k, n). Then, we consider situations where links are imperfectly protected and can be removed by the Adversary with some a priori known probability
⇡2
(0, 1).
4.1 Limited number of protected links
We examine a situation where the maximal number of protected links, ¯
p, that Dcan form is strictly smaller than
n1. More precisely, we are interested in the situation where ¯
p2Jp1(k, n), p
2(k, n)
K.
17Proposition 2 establishes that for
n(k+ 1) = 0 mod 2, there exists no SPE in which
Duses both protected and non-protected links. In contrast, when ¯
p2Jp1, p2K, there exist situations where the SPE are of the form (S
D,;) with
SD 2 SpD1 1,k. Note that networks in
SpD1 1,k(N) contain both protected and non-protected links when
p1 >1.
18The following proposition gives a condition on the values of the parameters upon which such situations arise. To simplify the analysis, we restrict our attention to situations where
k+ 1 is even.
17This interval is non empty ifp2> p1which implies that >0, that isn <(3k+5)8(k+1)2.
18Indeed, note thatp14n8 + 1n 1 sincen 4 and thereforeSDp
1 1,k always contains non-protected links.
12
Limited number of protected links and imperfectly protected links
Two potential types of restrictions on the protection of the network for the Designer.
D can only use a limited number of protected links and we focus on situations where this number is smaller than p
2(k , n).
Situations where links are imperfectly protected and can be removed by the Adversary with some a priori known
probability ⇡ 2 (0, 1).
Main Results. Proposition 3
Proposition
Assume that n(k + 1) = 0 mod 2. Assume further that p
2p
12 and that the maximal number of protected links is
¯
p 2 [p
1, p
2]. Let (S
?D, S
?A(S
?D)) be a SPE. There exists " > 0 such
that if (k + 1)/2 " < c
P/c
L< (k + 1)/2, then S
?D2 S
pD1 1and
S
?A(S
?D)) = ;.
Imperfectly protected links
We now assume that each protected link has a probability
⇡ 2 (0, 1) to be removed when it is attacked by A.
Let g = (N, E
P, E
NP) be an (imperfectly) protected network, and E
Aan attack over the links of g .
In the benchmark model, g
Ris obtained by removing the links of E
NPthat are targeted by A, i.e. g
R= (N, E
P, E
NP\ E
A).
Now, a realization of the attack is a subnetwork of g
Rof the
form g
⇢= (N, E
⇢, E
NP\ E
A) with E
⇢✓ E
P\ E
A.
Let g
⇢be a realization and (g
⇢| g
R, E
A) be the probability that g
⇢is realized given g
Rand E
A. We have
(g
⇢|g
R, E
A) = Y
ij2EP(g⇢), ij2EP\EA
(1 ⇡) Y
ij2E/ P(g⇢), ij2EP\EA
⇡.
The expected benefits obtained by D , IE (g
R, E
A), when she chooses strategy g and A chooses E
Ais
IE (g
R, E
A) = X
g⇢=(N,E⇢,ENP\EA)
(g
⇢| g
R, E
A) (g
⇢).
The expected payo↵s obtained by D, IE ⇧
D, is the di↵erence between the expected benefits and the costs of forming protected and non-protected links:
IE ⇧
D= IE (g
R, E
A) c
D(S
D). (2) Finally, the expected payo↵s obtained by A are
IE ⇧
A= 1 IE (g
R, E
A) c
A#E
A. (3)
Recall that c
A1/(n 3).
5 1 2
3
4
(i)gand attacked links (EA, in dashed lines)
5 1 2
3
4
(ii) SubnetworkgR
5 1 2
3
4
(iii) A possible realizationg%1
5 1 2
3
4
(iv) The other realizationg%2
Figure 7: Networks of Example 4. Thick lines represent protected links and dashed lines represent links that are targetted by A.
Let
g%be a realization and (g
%|gR, EA) be the probability that
g%is realized given
gRand
EA. We have
(g
%|gR, EA) =
Yij2EP(g%), ij2EPD\EA
(1
⇡) Yij /2EP(g%), ij2EPD\EA
⇡.
The expected benefits obtained by
D,(g
R, EA), when she builds a protected network
gand
Achooses
EAis
(g
R, EA) =
Xg%=(N,E%,ENPD \EA)
(g
%|gR, EA) (g
%).
We assume that the costs incurred by
Dwhen she chooses a strategy are given by Equation (2). The expected payo↵s obtained by
D, ⇧D, is the di↵erence between the expected benefits and the costs of forming protected and non-protected links:
⇧D
= (g
R, EA)
cD(S
D). (7) Finally, the expected payo↵s obtained by
Aare
⇧A
= 1 (g
R, EA)
cA#E
A.(8) Recall that
cA1/(n 3).
Proposition 4 Let the payo↵functions be given by Equations
(7)
and(8). Suppose that
⇡< cA. Then, results provided in Proposition 2 are preserved.Proof
Let
S?D(N) = (N, E
P, ENP) and let
S?A(S
?D(N)) =
EAwith
EA= (E
PA, EANP), where
EPA✓EPand
EN PA ✓ENP.
If
Acan disconnect the protected network
S?D(N ) = (N, E
P, ENP) with strategy
EA= (
;, ENPA), then since attacks are costly, her best response is to not attack any protected links, i.e.
EPA=
;.
If
Acannot disconnect the protected network
S?D(N ) = (N, E
P, ENP) with strategy (
;, ENPA), then the strategy (E
PA, EANP) should disconnect the network with a strictly positive probability (otherwise
Awould not be playing a best response). The highest probability to disconnect network (N, E
P, ENP) occurs when the deletion of any protected link implies that (N, E
P, ENP) is disconnected. This probability is 1 (1
⇡)#EPA. Since 1
⇡2(0, 1), by Taylor’s expansion 1 (1
⇡)#EPA=
P1`=0
#EPA
`
( 1)
`+1⇡`. Then by Leibniz’s rule on alternating series 1 (1
⇡)#EPA⇡#EPA. Hence, strategy
EAdisconnects the network with a probability lower or equal to
⇡#EPA. So, the expected benefits associated with
EAare at most
⇡#EPAwith a cost of at least
cA#E
AP. Hence, if
EPA 6=
;, then the payo↵ associated to strategy
EAis (strictly) negative and thus is not a best response. Therefore
EPA=
;.
Since no optimal strategy of
Atargets any protected link, the situation is equivalent to the one
examined in Proposition 2.
⇤Proposition 4 examines situations where probability
⇡is low relative to
cA, the cost of attacking links.
We now examine other situations through an example.
15
Main Results. Proposition 4
Proposition
Suppose that ⇡ < c
A. Then, results provided in Proposition 2 are
preserved.
(i) 1 8cL (ii) 1 6cL cP if⇡<2cA
1 ⇡ 6cL cP otherwise.
(iii) 1 5cL 2cP if⇡2<2cA
1 ⇡2 5cL 2cP otherwise.
(iv) 1 3cL 3cP if⇡<2cA
1 ⇡ 3cL 3cP otherwise.
(v) 1 4cP if⇡(1 ⇡/2)< cA
1 ⇡ 4cP if⇡(1 ⇡)< cA⇡(1 ⇡/2)
(1 ⇡)2 4cP ifcA⇡(1 ⇡)
(vi) 1 cL 4cP if⇡<2cA
1 ⇡ cL 4cP otherwise.
(vii) 1 5cP if⇡2<2cA
1 ⇡2 5cP otherwise.
Figure 8: Networks of Example 5: n = 5, k = 2. Optimal strategies for D for di↵erent numbers of protected links and when A attacks (up to) 2 links. The captions of the figures give the Designer’s expected benefit.
Example 5