{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Papoulis ch. 16

# Papoulis ch. 16 - 228.1 Use(16-132 with r = 1 This gives p...

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 228 Chapter 16 16.1 Use (16-132) 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 (16-102) 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 (16-106) with r = 1) E { X } = L 1 = ∞ X n =2 ( n- 1) p n 230 where p n is an in (16-88). 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 (16-106)-(16-107) 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 non-zero 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 (16-63) 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 !...
View Full Document

{[ snackBarMessage ]}

### Page1 / 22

Papoulis ch. 16 - 228.1 Use(16-132 with r = 1 This gives p...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online