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HW1.sols.09 - Pentium D part(b Arithmetic Means Chip Memory...

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Computer Architecture Homework 1 Due: Friday, September 11, 2009 All problems are in the text. 1. Problem 1.5, page 57, parts (a) and (c) part (a): (14 KW) / (79W + 2.3 W + 7.0 W) = (14KW /88.3 W) = 158 servers part (c): Failure rate = [1/ (9 * 10 6 )] + [ 8 * (1/4500)] + [1 / (3 * 10 4 )] = [1 + 8 * 2000 + 300 ] / [9 * 10 6 ] = [16301] / [9 * 10 6 ] MTTF = 1/(Failure rate) = [9 * 10 6 ] / [13601] = 522 hours 2. Problem 1.7, page 58 part (a): 50% part (b): Power New (V * 0.5) 2 * (F * 0.5) = = 0.5 3 = 0.125 Power Old V 2 * F 3. Problem 1.9, page 60 part (a): FIT = 10 9 / MTTF and FIT = 100, so MTTF = 10 9 / FIT = 10 9 / 100 = 10,000,000 part (b): Availability = MTTF / [MTTF + MTTR] = 107 / (107 + 24) 100%
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4. Problem 1.12, parts (a) and (b), page 61 part (a): Normalized to
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Unformatted text preview: Pentium D part (b): Arithmetic Means Chip Memory Performance Dhrystone Performance Arithmetic Mean of Metrics Arithmetic Mean of Normalized Athlon 64 X2 4800+ 1.14 1.36 12070.5 1.25 Pentium EE 840 1.08 1.24 11060.5 1.16 Pentium D 820 1.00 1.00 9110 1.00 Athlon 64 X2 3800+ 0.98 1.13 10035 1.05 Pentium 4 0.91 0.50 5176 0.71 Athlon 64 3000+ 0.98 0.50 5290.5 0.74 Pentium 4 570 1.17 0.74 7355.5 0.95 Processor X 2.33 0.33 6000 1.33 4. Problem 1.14, page 62 a. Amdahl’s Law: speedup = 1/[0.6 + (0.4/2)] = 1.25 b. Amdahl’s Law: speedup = 1/[0.01 + (0.99/2)] = 1.98 c. Amdahl’s Law: speedup = 1/[0.2 + 0.8*[0.6 + (0.4/2)] ]= 1.19 d. Amdahl’s Law: speedup = 1/[0.8 + 0.2*[0.01 + (0.99/2)] ]= 1.11...
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