网络拓扑结构对计算机病毒传播扩散的模型模拟研究
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本文是一篇发表在《国际计算机数学杂志》上的研究论文，标题为《网络拓扑对计算机病毒传播的影响：一个建模研究》（The effect of network topology on the spread of computer viruses: a modelling study）。作者是Lu-Xing Yang和Xiaofan Yang，该研究于2016年8月19日接受作者版本在线发布，并于同年9月7日正式出版，电子期刊号为10.1080/00207160.2016.1226499。
研究关注的核心议题是网络拓扑结构如何影响计算机病毒的传播过程。网络拓扑是一个计算机科学中的关键概念，它描述了网络中节点和连接的方式，如星型、环形、网状或层次结构等。在病毒传播模型中，不同的网络拓扑可能导致病毒扩散的速度、范围和复杂性有所不同。例如，中心化网络（如星型）可能使得病毒更容易集中在某个点上爆发，而分布式网络（如无中心或网状）则可能使得病毒传播更为分散且难以预测。
这篇论文通过建立数学模型来深入分析这个问题。作者可能采用了流行病学模型，如SIR（易感-感染-康复）模型或者更复杂的模型，来模拟病毒在不同网络结构中的传播路径和动态。他们可能考虑了各种因素，如节点间的连接度、病毒的传染性、用户的防范意识等，以评估网络拓扑对病毒传播效果的影响。
论文提供了210次的阅读量，显示出其在相关领域的关注度。此外，还引用了8篇相关的研究文章，表明此研究在学术界引起了讨论和应用。通过链接doi.org/10.1080/00207160.2016.1226499，读者可以访问到原文全文。
总结来说，这篇文章对于网络安全研究人员、计算机科学家以及政策制定者来说，具有重要的参考价值，因为它提供了理论基础和实证分析，帮助理解网络拓扑如何在抵御和管理计算机病毒传播方面起到关键作用。理解这种关系对于设计更有效的网络安全策略和预防措施至关重要。
INTERNATIONAL JOURNAL OF COMPUTER MATHEMATICS 1593
• s(t): the expected percentage of susceptible nodes at time t, s(t) := S(t)/N.
• l(t): the expected percentage of latent nodes at time t, l(t) := L(t)/N.
• b(t): the expected percentage of bursting nodes at time t, b(t) := B(t)/N.Thus,s(t) + l(t) +
b(t) = 1.
• N
k
: the total number of k-degree nodes in the Internet. Then,
k
N
k
= N.
• S
k
(t): the expected number of susceptible k-degree nodes at time t.Then,
k
S
k
(t) = S(t).
• L
k
(t): the expected number of latent k-degree nodes at time t.Then,
k
L
k
(t) = L(t).
• B
k
(t): the expected number of bursting k-degree nodes at time t.Then,
k
B
k
(t) = B(t).
• s
k
(t): the expected percentage of susceptible k-degree nodes at time t, s
k
(t) := S
k
(t)/N
k
.
• l
k
(t): the expected percentage of latent k-degree nodes at time t, l
k
(t) := L
k
(t)/N
k
.
• b
k
(t): the expected percentage of bursting k-degree nodes at time t, b
k
(t) := B
k
(t)/N
k
.Thus,
s
k
(t) + l
k
(t) + b
k
(t) = 1.
• s(t) := (s
1
(t), s
2
(t), ..., s
(t)).
• l(t) := (l
1
(t), l
2
(t), ..., l
(t)).
• b(t) := (b
1
(t), b
2
(t), ..., b
(t)).
• N
(r)
: the total number of removable storage media in the Internet.
• S
(r)
(t): the expected number of susceptible removable storage media in the Internet at time t.
• I
(r)
(t): the expected number of infected removable storage media in the Internet at time t.Then,
S
(r)
(t) + I
(r)
(t) = N
(r)
.
• s
(r)
(t): the expected percentage of susceptible removable storage media in the Internet at time t,
that is, s
(r)
(t) := S
(r)
(t)/N
(r)
.
• i
(r)
(t): the expected percentage of infected removable storage media in the Internet at time t,that
is, i
(r)
(t) := I
(r)
(t)/N
(r)
.Then,s
(r)
(t) + i
(r)
(t) = 1.
We sha ll foc us on the change of S
k
(t), L
k
(t), B
k
(t), S
(r)
(t),andI
(r)
(t) (equivalently, the change of
l
k
(t), b
k
(t),andi
(r)
(t)). For that purpose, let us make the following assumptions.
(A1) The topology of the Internet is unvaried, implying that N and all N
k
are unvaried.
(A2) The total number of removable storage media in the Internet, N
(r)
,isunvaried.
(A3) A susceptible node gets infected by an infected node and becomes latent with constant
probability per unit time β
1
> 0.
(A4) A susceptible node gets infected by an infected removable storage medium and becomes latent
with constant probability per unit time β
2
> 0.
(A5) A susceptible removable storage medium gets infected by an infected node with constant
probability per unit time β
3
> 0.
(A6) A latent node becomes bursting with constant probability per unit time α>0.
(A7) An infected node gets cured and becomes susceptible with constant probability per unit time
γ
1
> 0.
(A8) An infected removable storage medium gets cured and becomes susceptible with constant
probability per unit time γ
2
> 0.
Following Yang et al. [40], the following additional assumption is also incorporated.
(A9) The probability that an edge has a latent node as one endpoint does not depend on the degree of
the other endpoint and, hence, depends only on l(t). Likewise, the probability that an edge has
a breaking-out node as one endpoint depends only on b(t).Let
1
(l(t)) and
2
(b(t)) denote
these two probabilities, respectively.
Direct probabilistic calculations yield
1
(l(t)) =
1
k
k
kp
k
l
k
(t),
2
(b(t)) =
1
k
k
kp
k
b
k
(t),(1)
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