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Unformatted text preview: 228 Chapter 16 16.1 Use (16132) with r = 1. This gives p n = n n ! p , n 1 n p , 1 < n m = n p , n m Thus m X n =0 p n = p m X n =0 n = p (1 m +1 ) 1 = 1 = p = 1 1 m +1 and hence p n = 1 1 n +1 n , n m, 6 = 1 and lim 1, we get p n = 1 m + 1 , = 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 ) k (1 ) n k = ( n + 1)(1 ) 2 n , n = 0 , 1 , 2 , (i) where = / . 229 (b) When the two queues are merged, the new input rate = + = 2 . Thus from (16102) p n = ( / ) n n ! p = (2 ) n n ! p , n < 2 2 2 2! ( 2 ) n p = 2 n p , n 2 . Hence X k p k = p (1 + 2 + 2 X k =2 k ) = p (1 + 2 + 2 2 1 ) = p 1 ((1 + 2 )(1 ) + 2 2 ) = p 1 (1 + ) = 1 = p = 1 1 + , ( = / ) . (ii) Thus p n = 2(1 ) n / (1 + ) , n 1 (1 ) / (1 + ) , 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 ) n = (1 ) 2 X n =2 ( n 1) n 2 = (1 ) 2 X k =1 k k 1 = (1 ) 2 1 (1 ) 2 = 2 (1 ) . (iv) Since n 1 ( t ) = X + Y we have L 1 = E { n 1 ( t ) } = E { X } + E { Y } = 2 L 1 = 2 2 1 (v) For L 2 we can use (16106)(16107) with r = 2. Using (iii), this gives L 2 = p r (1 ) 2 = 2 (1 ) 2 1 + (1 ) 2 = 2 3 1 2 = 2 2 1 1 + ! < 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 , = = m, , 1 = i,i +1 = ( m i ) , i,i 1 = i 231 i,i = [( m i ) + i ] , i = 1 , 2 , ,m 1 m,m = m,m 1 = m. Substituting these into (1663) text, we get mp = p 1 (i) [( m i ) + i ] p i = ( m i +1) p i 1 +( i +1) p i +1 , i = 1 , 2 , ,m 1 (ii) and mp m = p m 1 . (iii) Solving (i)(iii) we get p i = m i ! + ! i + ! m i , i = 0 , 1 , 2 , ,m 16.4 (a) In this case p n = 1 1 1 = 1 ! n p , n < m 1 1 1 2 2 p , n m = n 1 p , n < m m 1 1 n m +1 2 p , n m, where X n =0 p n = p " m 1 X k =0 k 1 + m 1 1 2 X n =0 n 2 # = p " 1 m 1 1 1 + 2 m 1 1 1 2 # = 1 232 gives p = 1 m 1 1 1 + 2 m 1 1 1 2 !...
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This note was uploaded on 10/01/2009 for the course ECE 2521 taught by Professor Jacobs during the Fall '09 term at Pittsburgh.
 Fall '09
 Jacobs

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