Question

Consider the following graph: V4 V1 V2 V3 V5 and consider the following process: Initially, start at v1. • At ach time step, choose one of th and move there. vertices adjacent to your current location uniformly at random, Let pi(n), p2(n), p3(n), p4(n), p5(n) be the probability your location after n time steps is v1, V2, 03, 04, or V5 respectively. So pi(0) = 1 and p2(0) = P3(0) = p4(0) = p5(0) = 0. (a) Express pi(n+1), p2(n+1), p3(n +1), p4(n +1), and p5(n+1) in terms of p1(n), p2(n), p3(n), p4(n), and p5(n). (5 marks) (b) As n gets larger, each pi(n) converges to a single value (called the steady state) which can be deter- mined by setting Pi(n + 1) = Pi(n) in the above equations. Determine the steady state probabilities for all vertices. (5 marks) (c) The distance between any two vertices is the length of the shortest path between them. What is your expected distance from vi in the steady state? (4 marks)

Answer 1

+ ļB (n) Pp (u) +Ps (4) (a) Pi (n+1) - į B (W) Pa (n+1) Pich) P3 (n+1) Pz () + Pa (n+1) (n) + Ps (n+1) P (W) + (6) Pi(n+1) - Picu) > I Pos PS 2. PuCw 2 12 O P оо O ½ => A x = x o Pu V3 o 2 'l 12o Py Ps Ps (A - 1) x = 0 RREF (A- 0) 12 1 0 -312 0 B0- 1 0 > X = t and, + [ 1414 24.1 +1]: 5 Steady state probability, t Vio lis 17 9110 P IP2 13 Py Ps is Is © expected distance: I d = 5 + 3xt & 1 x 3 10 + 2x +3x s 10 + 3 - + 3 s + 3 min - 10 s = /2 unit