n j 1 with D j n i n j otherwise T jk n j n j 1 T jk n k n k n k 1 and T jk n i

# N j 1 with d j n i n j otherwise t jk n j n j 1 t jk

• University of Waterloo
• STAT 333
• Notes
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• 280
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= n j - 1 with ( D j n ) i = n j otherwise ( T jk n ) j = n j - 1 , ( T jk n ) k = n k = n k + 1 , and ( T j,k n ) i = n j otherwise 4.5. QUEUEING NETWORKS 207 Note that if n j = 0 then D j n and T jk n have - 1 in the j th coordinate, so in this case q ( n, D j n ) and q ( n, T jk n ) = 0. In equilibrium the rate at which probability mass leaves n is the same as the rate at which it enters n . Taking into account the various ways the chain can leave or enter state n it follows that the condition for a stationary distribution πQ = 0 is equivalent to π i λ i = j = i π j q ( j, i ) which in this case is π ( n ) K k =1 q ( n, A k n ) + K j =1 q ( n, D j n ) + K j =1 K k =1 q ( n, T jk n ) = K k =1 π ( A k n ) q ( A k n, n ) + K j =1 π ( D j n ) q ( D j n, n ) + K j =1 K k =1 π ( T jk n ) q ( T jk n, n ) This will obviously be satisfied if we have π ( n ) K k =1 q ( n, A k n ) = K k =1 π ( A k n ) q ( A k n, n ) (4.32) and for each j we have π ( n ) q ( n, D j n ) + K k =1 q ( n, T jk n ) = π ( D j n ) q ( D j n, n ) + K k =1 π ( T jk n ) q ( T jk n, n ) (4.33) Taking the second equation first, if n j = 0, then both sides are 0, since D j n and T jk n are not in the state space. Supposing that n j > 0 and filling in the values of our rates (4.33) becomes π ( n ) φ j ( n j ) q ( j ) + K k =1 p ( j, k ) = π ( D j n ) λ j + K k =1 π ( T jk n ) φ k ( n k + 1) p ( k, j ) (4.34) 208 CHAPTER 4. MARKOV CHAINS IN CONTINUOUS TIME The definition of q implies q ( j ) + K k =1 p ( j, k ) = 1, so filling in the proposed formula for π ( n ) the left-hand side of (4.33) is π ( n ) φ j ( n j ) = K i =1 c i r n i i ψ i ( n i ) · φ j ( n j ) = K i =1 c i r ˆ n i i ψ j n i ) · r j = π n ) · r j where ˆ n = D j n has ˆ n j = n j - 1 and ˆ n i = n i for i = j . To compute the right-hand side of (4.33) we note that ( T jk n ) i = ˆ n i for i = k and ( T jk n ) k = ˆ n k + 1 = n k + 1 so π ( T jk n ) = π n ) · r k φ k ( n k + 1) Since D j n = ˆ n , we can rewrite the right-hand side of (4.33) as = π n ) λ j + π n ) K k =1 r k p ( k, j ) = π n ) · r j where the last equality follows from (4.27): λ j + k r k p ( k, j ) = r j . At this point we have verified (4.33). Filling our rates into (4.32) and noting that π ( A k n ) = π ( n ) r k k ( n k + 1) we want to show π ( n ) K k =1 λ k = π ( n ) K k =1 r k φ k ( n k + 1) · φ k ( n k + 1) q ( k ) (4.35) To derive this, we note that summing (4.27) from j = 1 to K and interchanging the order of summation in the double sum on the right gives K j =1 r j = K j =1 λ j + K k =1 r k K j =1 p ( k, j ) = K j =1 λ j + K k =1 r k - K k =1 r k q ( i ) since K j =1 p ( i, j ) = 1 - q ( i ). Rearranging now gives K k =1 r k q ( k ) = K j =1 λ j This establishes (4.35), which implies (4.32), and completes the proof. 4.6. CLOSED QUEUEING NETWORKS 209 4.6 Closed Queueing Networks At first, the notion of N customers destined to move forever between K servers may sound like a queueing hell that might be the subject of a “Far Side” cartoon. However, as the next two examples show, this concept is useful for applications. Example 4.26. Manufacturing system. The production of a part at a factory requires two operations. The first operation is always done at machine 1. The second is done at machine 2 or ma- chine 3 with probabilities p and 1 - p after which the part leaves the system. Suppose that the factory has only a limited number of palettes, each of which holds one part. When a part leaves the sys-  #### You've reached the end of your free preview.

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