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Unformatted text preview: 228 Chapter 16 16.1 Use (16132) with r = 1. This gives p n = ρ n n ! p , n ≤ 1 ρ n p , 1 < n ≤ m = ρ n p , ≤ n ≤ m Thus m X n =0 p n = p m X n =0 ρ n = p (1 ρ m +1 ) 1 ρ = 1 = ⇒ p = 1 ρ 1 ρ m +1 and hence p n = 1 ρ 1 ρ n +1 ρ n , ≤ n ≤ m, ρ 6 = 1 and limρ → 1, we get p n = 1 m + 1 , ρ = 1 . 16.2 (a) Let n 1 ( t ) = X + Y , where X and Y represent the two queues. Then p n = P { n 1 ( t ) = n } = P { X + Y = n } = n X k =0 P { X = k } P { Y = n k } = n X k =0 (1 ρ ) ρ k (1 ρ ) ρ n k = ( n + 1)(1 ρ ) 2 ρ n , n = 0 , 1 , 2 , ··· (i) where ρ = λ/μ . 229 (b) When the two queues are merged, the new input rate λ = λ + λ = 2 λ . Thus from (16102) p n = ( λ /μ ) n n ! p = (2 ρ ) n n ! p , n < 2 2 2 2! ( λ 2 μ ) n p = 2 ρ n p , n ≥ 2 . Hence ∞ X k p k = p (1 + 2 ρ + 2 ∞ X k =2 ρ k ) = p (1 + 2 ρ + 2 ρ 2 1 ρ ) = p 1 ρ ((1 + 2 ρ )(1 ρ ) + 2 ρ 2 ) = p 1 ρ (1 + ρ ) = 1 = ⇒ p = 1 ρ 1 + ρ , ( ρ = λ/μ ) . (ii) Thus p n = 2(1 ρ ) ρ n / (1 + ρ ) , n ≤ 1 (1 ρ ) / (1 + ρ ) , n = 0 (iii) (c) For an M/M/ 1 queue the average number of items waiting is given by (use (16106) with r = 1) E { X } = L 1 = ∞ X n =2 ( n 1) p n 230 where p n is an in (1688). Thus L 1 = ∞ X n =2 ( n 1)(1 ρ ) ρ n = (1 ρ ) ρ 2 ∞ X n =2 ( n 1) ρ n 2 = (1 ρ ) ρ 2 ∞ X k =1 k ρ k 1 = (1 ρ ) ρ 2 1 (1 ρ ) 2 = ρ 2 (1 ρ ) . (iv) Since n 1 ( t ) = X + Y we have L 1 = E { n 1 ( t ) } = E { X } + E { Y } = 2 L 1 = 2 ρ 2 1 ρ (v) For L 2 we can use (16106)(16107) with r = 2. Using (iii), this gives L 2 = p r ρ (1 ρ ) 2 = 2 (1 ρ ) ρ 2 1 + ρ ρ (1 ρ ) 2 = 2 ρ 3 1 ρ 2 = 2 ρ 2 1 ρ ˆ ρ 1 + ρ ! < L 1 (vi) From (vi), a single queue configuration is more efficient then two separate queues. 16.3 The only nonzero probabilities of this process are λ , = λ = mλ, λ , 1 = μ λ i,i +1 = ( m i ) λ, λ i,i 1 = iμ 231 λ i,i = [( m i ) λ + iμ ] , i = 1 , 2 , ··· ,m 1 λ m,m = λ m,m 1 = mμ. Substituting these into (1663) text, we get mλp = μp 1 (i) [( m i ) λ + iμ ] p i = ( m i +1) p i 1 +( i +1) μp i +1 , i = 1 , 2 , ··· ,m 1 (ii) and mμp m = λp m 1 . (iii) Solving (i)(iii) we get p i = ˆ m i ! ˆ λ λ + μ ! i ˆ μ λ + μ ! m i , i = 0 , 1 , 2 , ··· ,m 16.4 (a) In this case p n = λ μ 1 λ μ 1 ··· λ μ 1 = ˆ λ μ 1 ! n p , n < m λ μ 1 λ μ 1 ··· λ μ 1 λ μ 2 ··· λ μ 2 p , n ≥ m = ρ n 1 p , n < m ρ m 1 1 ρ n m +1 2 p , n ≥ m, where ∞ X n =0 p n = p " m 1 X k =0 ρ k 1 + ρ m 1 1 ρ 2 ∞ X n =0 ρ n 2 # = p " 1 ρ m 1 1 ρ 1 + ρ 2 ρ m 1 1 1 ρ 2 # = 1 232 gives p = ˆ 1 ρ m 1 1 ρ 1 + ρ 2 ρ m 1 1 1 ρ 2 !...
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 Fall '09
 Jacobs
 Unit Circle, Trigraph, Emoticon, pn, k=0, di e−λ Ti

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