# Analysis of a Model for Computer Virus Transmission

Authors: Peng Qin
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

Computer viruses remain a significant threat to computer networks. In this paper, the incorporation of new computers to the network and the removing of old computers from the network are considered. Meanwhile, the computers are equipped with antivirus software on the computer network. The computer virus model is established. Through the analysis of the model, disease-free and endemic equilibrium points are calculated. The stability conditions of the equilibria are derived. To illustrate our theoretical analysis, some numerical simulations are also included. The results provide a theoretical basis to control the spread of computer virus.


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## Body

## 1. Introduction

With the rapid development of computer, communication, and network technology, network information system has become an important way of the development of countries and industries, amongst others. Information security has become one of the most important and challenging issues faced in the age of information sharing. The computer virus is one common information security threat.Computer virus is not only destructive, but also highly contagious. Once the virus is copied or generates variety, its speed is difficult to be controlled. Infectivity is the basic characteristic of the virus. In biology, the virus can diffuse from one organism to another. Cohen, Kephart and White pointed out that the spread mechanisms of computer viruses and biological viruses have many similarities [1, 2]. Under appropriate conditions, biological viruses can multiply quickly, and the infected organisms exhibit symptoms or even die. Similarly, the computer virus can also spread to uninfected computers from the infected computer through many kinds of ways. In some cases, the infected computers do not work and even get paralysed. Unlike biological viruses, the computer virus is a software program designed to replicate itself and spread to other machines. Computer virus enters the computer and gets executed; it will search for other programs or storage media in line with the conditions of their infection and target and then insert the code itself, achieving the purpose of self-reproduction. As long as a computer is infected (if not promptly treated), the virus will spread rapidly on this computer, in which a large number of files (usually an executable file) will be infected. The infected file has become a new source of infection and then will exchange data with other machines or over the network exposure; the virus will continue to be contagious. A computer virus can enter your computer in any number of ways, such as via mobile hard disk, via an email attachment, during file downloads from the Internet, or even upon a visit to a contaminated web site. By the time you find the virus that infected the computer, the mobile hard disk which is often used on this computer has been infected with the virus. Other computers connected to the machine network might be infected with the virus. The network has no permanent immunity to the computer virus. Therefore, there is always computer virus.Kephart et al. [2, 3] study the spread of computer virus using the biological virus model; they mainly focus on the influence of network topology on the spread of computer virus. In the homogeneous mixing nodes and only susceptible node input cases, Mishra et al. [4, 5] establish the mathematical model of Internet spread of computer virus. They analyze the propagation law of virus using threshold theory of infectious disease dynamics and predict the development trend of computer virus. Piqueira et al. [6] study the effects of equipping antivirus software on the virus in computer network. They prove the stability of the disease-free equilibrium and endemic equilibrium without incorporating new computers and removing old computers. For more studies on virus in computer network, see [7–13] and so on.In this paper, the new computers are incorporated to the network, and the old computers are removed from the computer network. Meanwhile, the computers are equipped with antivirus software on the computer network. The remainder of this paper presents the model and results. In Section2, the mathematical model on computer virus is described, and in Section 3 we obtain the equilibriums. In Section 4, the analysis of this model is derived. Numerical simulations supporting the theoretical analysis are given in Section 5. The paper ends with a conclusion and discussion in Section 6.


## 2. The Dynamic Model

The model proposed here is based on the compartmental SAIR model [6–8], including an antidotal population compartment (A) representing nodes of the network equipped with fully effective antivirus programs, a susceptible compartment (S), an infective compartment (I), and a temporarily immune compartment (R). Connections between the compartments represent operational parameters of the network and their control can be used as a strategy to maintain the reliability of the whole system, even in the presence of infections (see Figure 1).Figure 1
Transfer diagram of the model.The total populationN is divided into four groups. S(t) denote noninfected computers subjected to possible infection. I(t) denote infected computers. R(t) denote removed computers due to infection or not. A(t) denote noninfected computers equipped with antivirus.The SAIR model for computer viruses propagation was proposed and can be described by(1)dStdt=C-αSASA-βSI-μS+σR,dItdt=βSI-αIAIA-δI-μI,dRtdt=δI-σR-μR,dAtdt=αSASA+αIAIA-μA,where C is influx rate, representing the incorporation of new computers to the network; β is infection rate of susceptible computers; μ is proportion coefficient for the mortality rate, not due to the virus; δ is removal rate of infected computers; σ is recovering rate of removed computers, with an operator intervention; αSA is conversion of susceptible computers into antidotal ones, occurring when susceptible computers establish effective communication with antidotal ones and the antidotal one installs the antivirus in the susceptible ones; αIA represents infected computers that can be fixed by using antivirus programs being converted into antidotal ones.LetN=S+I+R+A; then dN(t)/dt=C-μN. Let (2)Ω=S,I,R,A:S,I,R,A≤0,S+I+R+A≤Cμ.Then, it is clear that Ω is a positive invariant set. Hence, we will focus our attention only on the region Ω.But [6, 7] think the influx rate is considered to be C=0, μ=0. In fact, every day new computers can be incorporated to the network or old computers can be removed from the network. Therefore, they can not be ignored. In this paper, we analyse completely dynamical behavior of the spread of the virus in computer network with incorporating new computers and removing old computers.


## 3. The Disease-Free and Endemic Equilibrium

WhenI=0, if A=0, the disease-free equilibrium of system (1) is P1=(S1,I1,R1,A1)=(C/μ,0,0,0). If A≠0, we obtain the threshold R01=CαSA/μ2. When R01>1, the disease-free equilibrium of system (1) is P2=(S2,I2,R2,A2)=(μ/αSA,0,0,C/μ-μ/αSA).In the following, we compute the positive equilibrium; namely,I≠0.According to the third equation of system (1), we have (3)R=δσ+μI.When A=0, using the second equation of system (1), we have (4)S=δ+μβ.According to the first equation of system (1), we obtain (5)I=βC-μδ+μσ+μβμσ+δ+μ.Therefore, when the threshold R02=βC/μ(δ+μ)>1, we have the positive equilibrium (6)P3=S3,I3,R3,A3=δ+μβ,R02-1δ+μσ+μβσ+δ+μ,δσ+μI3,0.WhenA≠0, taking advantage of the second equation of system (1), we have (7)A=βS-δ+μαIA.According to the fourth equation of system (1), we have (8)I=μ-αSASαIA.Substituting (7) and (8) into the first equation of system (1), we have (9)C-αSASβS-δ+μαIA-βSμ-αSASαIA-μS+σδσ+μμ-αSASαIA=0.Namely,(10)βμ+αIAμ+σδαSAσ+μ-δ+μαSAS=CαIA+σδμσ+μ.Hence, (11)S=CαIA+σδμ/σ+μβμ+αIAμ+σδαSA/σ+μ-δ+μαSA.In order to make S>0, it must satisfy the following conditions:(12)βμ+αIAμ+σδαSAσ+μ-δ+μαSA>0.Namely, (13)R03=βμ+αIAμ+σδαSA/σ+μδ+μαSA>1.Therefore, if the threshold R03>1, we obtain (14)S4=CαIA+σδμ/σ+μδ+μαSAR03-1,I4=μ-αSAS4αIA,A4=βS4-δ-μαIA.In order to make I4,A4>0, it must satisfy (15)R03>1+CαIAσ+μ+σδμμσ+μδ+μ,R03<1+βμδ+μαSACαIAσ+μ+σδμμσ+μδ+μ.When 1+(CαIA(σ+μ)+σδμ)/μ(σ+μ)(δ+μ)<R03<1+(βμ/δ+μαSA)((CαIAσ+μ+σδμ) / μσ+μδ+μ) (namely, the threshold R04=βμ/(δ+μ)αSA>1), we have the positive equilibrium (16)P4=S4,I4,R4,A4=CαIA+σδμ/σ+μδ+μαSAR03-1,μ-αSAS4αIA,δσ+μI4,βS4-δ-μαIA.


## 4. Stability of Equilibrium

### 4.1. Stability of the Disease-Free Equilibrium

The Jacobin matrix of system (1) at the disease-free equilibrium P1 is (17)JP1=-μ-βCμσ-CαSAμ0βCμ-δ-μ000δ-σ+μ0000CαSAμ-μ.The characteristic equation of J(P1) is given by (18)λ+μλ+l1λ+σ+μλ+l2=0,where l1=-βC/μ+δ+μ and  l2=-CαSA/μ+μ. Therefore, we have l1>0 if and only if R02=βC/μ(δ+μ)<1, and l2>0 if and only if R01=CαSA/μ2<1. It follows from the Routh-Hurwitz criterion that the eigenvalues have negative real parts if and only if R01<1 and R02<1. Hence, the disease-free equilibrium P1 of model (1) is locally asymptotically stable if R01<1 and R02<1 and unstable if R01>1 or R02>1.IfR01>1, the disease-free equilibrium P1 is unstable, and system (1) exhibits the other disease-free equilibrium P2.IfR02>1, the disease-free equilibrium P2 is unstable, and system (1) exhibits the positive equilibrium P3.The Jacobin matrix of system (1) at the disease-free equilibrium P2 is (19)JP2=-μ-αSAA2-βS2σ-αSAS20βS2-αIAA2-δ-μ000δ-σ+μ0αSAA2αIAA20αSAS2-μ.Since αSAS2-μ=0, the characteristic equation of J(P2) is given by (20)λ+σ+μλ-βS2-αIAA2-δ-μλ2+μ+αSAS2λ+μαSAS2=0.By calculating, we obtain that the characteristic equation has the four eigenvalues: -(σ+μ),  βS2-αIAA2-δ-μ, -μ,  and -αSAA2. When R01>1,  -αSAA2<0. We have βS2-αIAA2-δ-μ<0 if and only if R05=(β+αIA)μ2/(CαSAαIA+αSAδ+μ)<1. Hence, the disease-free equilibrium P2 of model (1) is locally asymptotically stable if R01>1 and R05<1. The disease-free equilibrium P2 of model (1) is unstable if R01>1 and R05>1.Theorem 1.
(1) IfR01<1 and R02<1, the disease-free equilibrium P1 of system (1) is locally asymptotically stable. If R01>1 or R02>1, the disease-free equilibrium P1 of system (1) is unstable.
(2) IfR01>1, the disease-free equilibrium P1 is unstable, and system (1) exhibits the other disease-free equilibrium P2.
(3) IfR02>1, the disease-free equilibrium P1 is unstable, and system (1) exhibits the positive equilibrium P3.
(4) IfR01>1 and R05<1, the disease-free equilibrium P2 is locally asymptotically stable. If R05>1, the disease-free equilibrium P2 is unstable.


### 4.2. Stability of the Endemic Equilibrium

In the following, we study the stability of the endemic equilibriums.The Jacobin matrix of system (1) at the endemic equilibrium P3 is (21)JP3=-βI3-μ-βS3σ-αSAS3βI3βS3-δ-μ0-αIAI30δ-σ+μ0000αSAS3+αIAI3-μ.Since βS3-δ-μ=0, the characteristic equation of J(P3) is given by (22)λ-αSAS3+αIAI3-μFλ=0,where (23)Fλ=λ3+a1λ2+a2λ+a3,a1=βI3+μ+σ+μ>0,a2=βI3+μσ+μ+β2S3I3>0,a3=β2S3I3σ+μ-βσδI3=βI3δ+μσ+μ-δσ>0.It is easy to calculate a1a2-a3>0. Hence, the eigenvalues of F(λ)=0 have negative real parts.Therefore, if the eigenvalues of the characteristic equation inP3 want to have negative real parts, only αSAS3+αIAI3-μ<0. Namely, (24)R02<1+αSAσ+δ+μδ+μβμ/αSAδ+μ-1αIAδ+μσ+μ=R07.According to Routh-Hurwitz criterion, the epidemic equilibriumP3 is locally asymptotically stable if 1<R02<R07 and unstable if R02>R07.System (1) has the following limiting system: (25)dItdt=βCμ-I-R-AI-αIAAI-δI-μI,dRtdt=δI-σR-μR,dAtdt=αSACμ-I-R-AA+αIAAI-μA.In the following, we prove the stability of the endemic equilibriumP4 by the limit system. The Jacobin matrix of system (25) at the endemic equilibrium P4 is (26)JP4=βS4-βI4-αIAA4-μ-δ-βI4-βI4-αIAI4δ-σ+μ0-αSAA4+αIAA4-αSAA4αSAS4-αSAA4+αIAI4-μ.Because βS4-αIAA4-μ-δ=0,  αSAS4+αIAI4-μ=0, the above matrix becomes (27)JP4=-βI4-βI4-βI4-αIAI4δ-σ+μ0-αSAA4+αIAA4-αSAA4-αSAA4.The characteristic equation ofJ(P4) is given by (28)λ3+b1λ2+b2λ+b3=0,where (29)b1=βI4+σ+μ+αSAA4>0,b2=αSAβI4A4+αSAA4σ+μ+βI4σ+μ-αSAA4-αIAA4βI4+αIAI4+δβI4,b3=αSAβI4A4σ+μ-δαSAA4βI4+αIAI4-σ+μαSAA4-αIAA4βI4+αIAI4+δαSAβI4A4.Because of αIAI4=μ-αSAS4, we have (30)b2=αSAβI4A4+αSAA4σ+μ+βI4σ+μ-αSAA4-αIAA4βI4+αIAI4+δβI4=αSAA4σ+μ+βI4σ+μ-αSAA4αIAI4+αIAA4βI4+αIAI4+δβI4=αSAA4σ+μ+βI4σ+μ-αSAA4μ-αSAS4+αIAA4βI4+αIAI4+δβI4=αSAA4σ+βI4σ+μ+αSAA4αSAS4+αIAA4βI4+αIAI4+δβI4>0.b3=αSAβI4A4σ+μ-δαSAA4βI4+αIAI4-σ+μαSAA4-αIAA4βI4+αIAI4+δαSAβI4A4=αIAA4I4σ+μβ+αIA-δαSAA4αIAI4-σ+μαSAαIAA4I4.It is easy to calculate b1b2-b3>0. Therefore, in order to make b3>0, only there is R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)>1. Hence, if the threshold R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)>1, the endemic equilibrium P4 is locally asymptotically stable.Theorem 2.
(1) If1<R02<R07, the positive equilibrium P3 is locally asymptotically stable. If R02>R07, the positive equilibrium P3 is unstable.
(2) If1+(CαIA(σ+μ)+σδμ)/μ(σ+μ)(δ+μ)<R03<1+(βμ/δ+μαSA)((CαIAσ+μ+σδμ) / μσ+μδ+μ) and R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)>1, the positive equilibrium P4 is locally asymptotically stable. If R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)<1, the positive equilibrium P4 is unstable.


## 4.1. Stability of the Disease-Free Equilibrium

The Jacobin matrix of system (1) at the disease-free equilibrium P1 is (17)JP1=-μ-βCμσ-CαSAμ0βCμ-δ-μ000δ-σ+μ0000CαSAμ-μ.The characteristic equation of J(P1) is given by (18)λ+μλ+l1λ+σ+μλ+l2=0,where l1=-βC/μ+δ+μ and  l2=-CαSA/μ+μ. Therefore, we have l1>0 if and only if R02=βC/μ(δ+μ)<1, and l2>0 if and only if R01=CαSA/μ2<1. It follows from the Routh-Hurwitz criterion that the eigenvalues have negative real parts if and only if R01<1 and R02<1. Hence, the disease-free equilibrium P1 of model (1) is locally asymptotically stable if R01<1 and R02<1 and unstable if R01>1 or R02>1.IfR01>1, the disease-free equilibrium P1 is unstable, and system (1) exhibits the other disease-free equilibrium P2.IfR02>1, the disease-free equilibrium P2 is unstable, and system (1) exhibits the positive equilibrium P3.The Jacobin matrix of system (1) at the disease-free equilibrium P2 is (19)JP2=-μ-αSAA2-βS2σ-αSAS20βS2-αIAA2-δ-μ000δ-σ+μ0αSAA2αIAA20αSAS2-μ.Since αSAS2-μ=0, the characteristic equation of J(P2) is given by (20)λ+σ+μλ-βS2-αIAA2-δ-μλ2+μ+αSAS2λ+μαSAS2=0.By calculating, we obtain that the characteristic equation has the four eigenvalues: -(σ+μ),  βS2-αIAA2-δ-μ, -μ,  and -αSAA2. When R01>1,  -αSAA2<0. We have βS2-αIAA2-δ-μ<0 if and only if R05=(β+αIA)μ2/(CαSAαIA+αSAδ+μ)<1. Hence, the disease-free equilibrium P2 of model (1) is locally asymptotically stable if R01>1 and R05<1. The disease-free equilibrium P2 of model (1) is unstable if R01>1 and R05>1.Theorem 1.
(1) IfR01<1 and R02<1, the disease-free equilibrium P1 of system (1) is locally asymptotically stable. If R01>1 or R02>1, the disease-free equilibrium P1 of system (1) is unstable.
(2) IfR01>1, the disease-free equilibrium P1 is unstable, and system (1) exhibits the other disease-free equilibrium P2.
(3) IfR02>1, the disease-free equilibrium P1 is unstable, and system (1) exhibits the positive equilibrium P3.
(4) IfR01>1 and R05<1, the disease-free equilibrium P2 is locally asymptotically stable. If R05>1, the disease-free equilibrium P2 is unstable.


## 4.2. Stability of the Endemic Equilibrium

In the following, we study the stability of the endemic equilibriums.The Jacobin matrix of system (1) at the endemic equilibrium P3 is (21)JP3=-βI3-μ-βS3σ-αSAS3βI3βS3-δ-μ0-αIAI30δ-σ+μ0000αSAS3+αIAI3-μ.Since βS3-δ-μ=0, the characteristic equation of J(P3) is given by (22)λ-αSAS3+αIAI3-μFλ=0,where (23)Fλ=λ3+a1λ2+a2λ+a3,a1=βI3+μ+σ+μ>0,a2=βI3+μσ+μ+β2S3I3>0,a3=β2S3I3σ+μ-βσδI3=βI3δ+μσ+μ-δσ>0.It is easy to calculate a1a2-a3>0. Hence, the eigenvalues of F(λ)=0 have negative real parts.Therefore, if the eigenvalues of the characteristic equation inP3 want to have negative real parts, only αSAS3+αIAI3-μ<0. Namely, (24)R02<1+αSAσ+δ+μδ+μβμ/αSAδ+μ-1αIAδ+μσ+μ=R07.According to Routh-Hurwitz criterion, the epidemic equilibriumP3 is locally asymptotically stable if 1<R02<R07 and unstable if R02>R07.System (1) has the following limiting system: (25)dItdt=βCμ-I-R-AI-αIAAI-δI-μI,dRtdt=δI-σR-μR,dAtdt=αSACμ-I-R-AA+αIAAI-μA.In the following, we prove the stability of the endemic equilibriumP4 by the limit system. The Jacobin matrix of system (25) at the endemic equilibrium P4 is (26)JP4=βS4-βI4-αIAA4-μ-δ-βI4-βI4-αIAI4δ-σ+μ0-αSAA4+αIAA4-αSAA4αSAS4-αSAA4+αIAI4-μ.Because βS4-αIAA4-μ-δ=0,  αSAS4+αIAI4-μ=0, the above matrix becomes (27)JP4=-βI4-βI4-βI4-αIAI4δ-σ+μ0-αSAA4+αIAA4-αSAA4-αSAA4.The characteristic equation ofJ(P4) is given by (28)λ3+b1λ2+b2λ+b3=0,where (29)b1=βI4+σ+μ+αSAA4>0,b2=αSAβI4A4+αSAA4σ+μ+βI4σ+μ-αSAA4-αIAA4βI4+αIAI4+δβI4,b3=αSAβI4A4σ+μ-δαSAA4βI4+αIAI4-σ+μαSAA4-αIAA4βI4+αIAI4+δαSAβI4A4.Because of αIAI4=μ-αSAS4, we have (30)b2=αSAβI4A4+αSAA4σ+μ+βI4σ+μ-αSAA4-αIAA4βI4+αIAI4+δβI4=αSAA4σ+μ+βI4σ+μ-αSAA4αIAI4+αIAA4βI4+αIAI4+δβI4=αSAA4σ+μ+βI4σ+μ-αSAA4μ-αSAS4+αIAA4βI4+αIAI4+δβI4=αSAA4σ+βI4σ+μ+αSAA4αSAS4+αIAA4βI4+αIAI4+δβI4>0.b3=αSAβI4A4σ+μ-δαSAA4βI4+αIAI4-σ+μαSAA4-αIAA4βI4+αIAI4+δαSAβI4A4=αIAA4I4σ+μβ+αIA-δαSAA4αIAI4-σ+μαSAαIAA4I4.It is easy to calculate b1b2-b3>0. Therefore, in order to make b3>0, only there is R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)>1. Hence, if the threshold R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)>1, the endemic equilibrium P4 is locally asymptotically stable.Theorem 2.
(1) If1<R02<R07, the positive equilibrium P3 is locally asymptotically stable. If R02>R07, the positive equilibrium P3 is unstable.
(2) If1+(CαIA(σ+μ)+σδμ)/μ(σ+μ)(δ+μ)<R03<1+(βμ/δ+μαSA)((CαIAσ+μ+σδμ) / μσ+μδ+μ) and R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)>1, the positive equilibrium P4 is locally asymptotically stable. If R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)<1, the positive equilibrium P4 is unstable.


## 5. Numerical Simulation

In this section, we will perform a series of numerical simulations to verify the mathematical analysis. In Figure2, when R01<1 and R02<1, the disease-free equilibrium P1 is asymptotically stable, where the initial values are S=50, I=20, R=1, and A=1. The parameters are C=1, αSA=0.00045, β=0.05, μ=0.05, σ=0.8, αIA=0.0025, and δ=0.96.Figure 2
R
01
<
1 and R02<1. The disease-free equilibrium P1 of system (1) is asymptotically stable.
(a)
(b)
(c)
(d)
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(f)IfR01>1 and R05<1, Figure 3  signifies that the disease-free equilibrium P2 of system (1) is asymptotically stable. The initial values are S=50, I=20, R=1, and A=1. Choose the parameters as follows: C=6, αSA=0.0045, β=0.1, μ=0.05, σ=0.8, αIA=0.0025, and δ=0.8.Figure 3
R
01
>
1 and R05<1. The disease-free equilibrium P2 of system (1) is asymptotically stable.
(a)
(b)
(c)
(d)
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(f)If1<R02<R07, the positive equilibrium P3 of system (1) is asymptotically stable (see Figure 4). The initial values are S=50, I=20, R=1, and A=1. Choose the parameters as follows: C=1, αSA=0.00045, β=0.2, μ=0.05, σ=0.8, αIA=0.0025, and δ=0.96.Figure 4
1
<
R
02
<
R
07. The positive equilibrium P3 of system (1) is asymptotically stable.
(a)
(b)
(c)
(d)
(e)
(f)If1+(CαIA(σ+μ)+σδμ)/μ(σ+μ)(δ+μ)<R03<1+(βμ/δ+μαSA)(CαIAσ+μ+σδμ / μσ+μ(δ + μ)) and R06=(σ+μ)(β+αIA)/αSA(σ+δ+μ)>1, the positive equilibrium P4 of system (1) is asymptotically stable (see Figure 5). The initial values are S=50, I=20, R=1, and A=1. Choose the parameters as follows: C=10, αSA=0.00045, β=0.2, μ=0.05, σ=0.8, αIA=0.0025, and δ=17.Figure 5
1
+
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δ  +  μ)αSA)((CαIA(σ+μ)+σδμ)/μσ+μδ+μ), and R06=(σ+μ)(β+αIA) / αSA(σ+δ+μ)>1. The positive equilibrium P4 of system (1) is asymptotically stable.
(a)
(b)
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## 6. Conclusion

This paper mainly considers the incorporation of new computers to the network, the removal of old computers from the network, the computer equipped with antivirus software, and so forth. They affect the spread of the virus. The model for computer virus transmission is established. Through the analysis of the model, two disease-free and two positive equilibriums are obtained. The stability conditions of the equilibriums are derived.Through the qualitative analysis of computer virus propagation model, mastering the virus prevention and control technology is very necessary. In the meantime, computer users have been advised to update their security settings. We can strengthen the knowledge of the computer virus spread (e.g., enhance the user’s information security awareness) in a timely manner to install antivirus software or fix bugs, to minimize the impact on network computer virus.



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