MAT 1302 – MARKOV CHAIN EXAMPLE PROF: ALISTAIR SAVAGE Suppose a city has 3 internet service providers (A, B, and C). A starts with 200 000 customers and B and C each start with 400 000 customers. Suppose that, each year, the following migration occurs .7 A D _ .3 .2 .4 BL m .4 .6 .2 - CR .2 How many customers does each company have (a) after one year? (b) after two years? (c) after many years? Solution: (a). The migration matrix is .7 .4 .2 M = .3 .2 .6 0 .4 .2 and the initial state vector is .2 x0 = .4 . .4 Therefore, .7 .4 .2 .2 0.38 x1 = M x0 = .3 .2 .6 .4 = 0.38 . 0 .4 .2 .4 0.24 So after one year, A and B have 380 000 customers each and C has 240 000 customers. 2 MAT 1302 – MARKOV CHAIN EXAMPLE, PROF. ALISTAIR SAVAGE (b). .7 .4 .2 .38 .466 x2 = M x1 = .3 .2 .6 .38 = .334 . 0 .4 .2 .24 .2 So after two years, A has 466 000 customers, B has 334 00 customers and C has 200 000 customers. (c). .61 .41 .42 M 2 = .27 .4 .3 .12 .16 .28 2 Since all the entries of M are strictly greater than zero, M is a regular stochastic matrix (we needed to check higher powers of M since M itself had a zero entry). Therefore xk approaches the unique steady-state vector q as k → ∞. To find the steady state vector, we solve M q = q ⇐⇒ (M − I)q = 0 which amounts to row reducing −.3 .4 .2 0 −3 4 2 0 1 0 − 10 0 3 .6 0 ∼ 3 −8 6 0 ∼ 0 1 −2 0 . (M − I) 0 ∼ .3 −.8 0 .4 −.8 0 0 4 −8 0 0 0 0 0 The general solution is 10 x3 3 x2 = 2x3 x3 free x1 = Switching to vector notation gives 10/3 x = x3 2 . 1 Any choice of x3 6= 0 gives an eigenvector of M with eigenvalue 1. However, we want to choose x3 so that the resulting vector is a probability vector (that is, its entries add to one). So we pick x3 to be the reciprocal of the sum of the entries −1 −1 10 19 3 x3 = +2+1 = = . 3 3 19 Therefore 10/3 10/19 3 2 = 6/19 . q= 19 1 3/19 10 Thus, in the long term A has 19 (1 000 000) ∼ = 526 316 customers, B has 3 ∼ 315 789 customers, and C has 19 (1, 000 000) = 157 895 customers. 6 (1 000 000) 19 ∼ =
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