ENC3250 Solutions pg 68

ENC3250 Solutions - 15.4 a ˆ because x k − x k − 1 ≥ 0 b k = 1 δ k = min[10 30,1280 = 40 ˆ y k = 990(1 × 40 = 1030 y k − 1 = min 1030

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Unformatted text preview: 15.4 a. ˆ because x ( k ) − x( k − 1) ≥ 0 b( k ) = 1 δ ( k ) = min[10 + 30,1280] = 40 ˆ y( k ) = 990 + (1 × 40) = 1030 [ )] ( y( k − 1) = min 1030, 215 − 1 = 1030 ˆ x( k − 1) = 1030 × 0.96875 = 998 b. ˆ because x ( k ) − x( k − 1) ≥ 0 b( k ) = 1 δ ( k ) = max[(0.999 × 30),10] = 29.97 € ˆ y( k ) = 990 + (1 × 29.97) = 1019.97 [ ( )] y( k − 1) = min 1019.97, 215 − 1 = 1019.97 ˆ x( k − 1) = 1019.97 × 0.96875 = 988 15.5 a. During a bust of S seconds, a total of MS octets are transmitted. A burst empties the bucket (b octets) and, during the burst, tokens for an additional rS octets are generated, for a total burst size of (b + rS). Thus, € b + rS = MS S = b/(M – r) b. S = (250 × 103 )/(23 × 106 ) ≈ 11 msec 15.6 a. b 2 (n) = a1 (n); b 1 (n) = a0 (n) ⊕ a0 (n – 2); b 0 (n) = a0 (n – 1) b. a0 (n – 2) a0 (n – 1) a0 (n) a1 (n) b 2 (n) 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 -68- 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 b 1 (n) b 0 (n) 0 0 1 1 0 0 1 1 1 1 0 0 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 ...
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This note was uploaded on 02/20/2012 for the course ENC 3250 taught by Professor Dr.timothycerner during the Fall '11 term at UNF.

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