A Markov-Based Model for Information Security Risk Assessment in Healthcare MANETs

Abstract

Information security breaches are of major concern for healthcare mobile ad-hoc networks (h-MANETs). In this paper, we propose a model that identifies and assesses risk in an h-MANET deployed by a healthcare institution in a disaster-prone region a priori by modeling the possible routes a hacker might follow to compromise a target. Our model proposes a novel method to compute the transition probability of each hop in the h-MANET. Next, it employs Markov theory to compute the maximum and minimum number of hops required to compromise the target for a given source-target <S-T> pair. It then determines the vulnerability for all the paths comprising minimum to maximum hops only for each <S-T> pair by computing their overall transition probability. Finally, our model computes the risk associated with each of these paths. Based on the calculated risk level of each path, the management can recommend an appropriate risk mitigation strategy.

Appendix

1.1 Derivation of distance between two points in a Non Orthogonal Cartesian System inclined at 60°

Fig. 10

Non Orthogonal Cartesian System inclined at 60°

Let us try to find the distance between two points P₁(x₁,y₁) and P₂(x₂,y₂) in a non-orthogonal Cartesian system where the axes are inclined at an angle of 60° using Fig. 10.

$$ {\mathrm{P}}_1\mathrm{B}={\mathrm{x}}_2-{\mathrm{x}}_1 $$

(17)

$$ {\mathrm{P}}_2\mathrm{B}={\mathrm{y}}_2-{\mathrm{y}}_1 $$

(18)

$$ {\mathrm{P}}_1\mathrm{A}={\mathrm{P}}_1\mathrm{B}+\mathrm{BA}=\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)+\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)\ \cos\ {60}^{{}^{\circ}}. $$

(19)

(using Eq. (17))

$$ {\mathrm{P}}_2\mathrm{A}={\mathrm{P}}_2\mathrm{B}\ \sin\ {60}^{{}^{\circ}}=\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)\ \sin\ {60}^{{}^{\circ}} $$

(20)

(using Eq. (18))

$$ \mathrm{BA}=\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)\ \cos\ {60}^{{}^{\circ}} $$

(21)

Now, using Pythagoras theorem,

$$ {\displaystyle \begin{array}{l}{\mathrm{P}}_1{{\mathrm{P}}_2}^2={\mathrm{P}}_1{\mathrm{A}}^2+{\mathrm{P}}_2{\mathrm{A}}^2\\ {}\kern2em ={\left[\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)+\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)\ \cos\ {60}^{{}^{\circ}}\right]}^2+{\left[\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)\ \sin\ {60}^{{}^{\circ}}\right]}^2\\ {}\kern2em ={\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2{\cos}^2{60}^{{}^{\circ}}+{2}^{\ast }{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^{\ast}\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)\ \cos\ {60}^{{}^{\circ}}+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2{\sin}^2{60}^{{}^{\circ}}\\ {}\kern2em ={\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2\left({\cos}^2{60}^{{}^{\circ}}+{\sin}^2{60}^{{}^{\circ}}\right)+{2}^{\ast }{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^{\ast }{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^{\ast}\left(1/2\right)\left[\mathrm{since},\cos\ {60}^{{}^{\circ}}=\frac{1}{2}\ \right]\\ {}\kern2em ={\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2+{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^{\ast}\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)\kern0.5em \left[\mathrm{since},\kern0.5em {\cos}^2{60}^{{}^{\circ}}+{\sin}^2{60}^{{}^{\circ}}=1\right]\end{array}} $$

(22)

[Using Eqs. (19) and (20)].