Suppose that server 1 serves at rate 4 while server 2

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Unformatted text preview: in with transition probability q (i, j ) r(i, j ) = i goes through the same sequence of states as Xt but stays in each one for on unit of time. Let VA = min{t : Xt 2 A} be the time of the first visit to A. Then h(i) = Pi (X (VA ) = a) satisfies h(a) = 1, h(b) = 0 for b 2 A {a} and X h(i) = r(i, j )h(j ) for i 62 A. j The expected hitting time g (i) = Ei VA satsifies g (a) = 0 for a 2 A and g (i) = 1 i + X r(i, j )g (j ) j for i 62 A. One can also work directly with the transition rates. In the first case the equation with the same boundary conditions (h(a) = 1, h(b) = 0 for b 2 A {a} and for i 62 A) is X Q(i, j )h(j ) = 0 for i 62 A i In the second case, if if we let R be the part of the Q-matrix where i, j 62 A then g = R 11 where 1 is a column vector of all 1’s. P Stationary distributions. A stationary distribution has i ⇡ (i) = 1 and satisfies ⇡ pt = ⇡ for all t > 0, which is equivalent to ⇡ Q = 0. To solve these equations mechanically, we replace the last column of Q by all 1’s to define a matrix A and then ⇡ will be the last row of...
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