# ps1sol - Chapter 2. The Basics Problem 2.1 A four segment...

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Chapter 2. The Basics Problem 2.1 A four segment pipeline implements a function and has the following delays for each segment: ( b = . 2) Segment # Maximum delay 11 7 21 5 31 9 41 4 Where c =2ns , 1. What is the cycle time that maximizes performance without allocating multiple cycles to a segment? Since the maximum stage delay is 19 ns, this is the shortest possible delay time that does not require multiple cycles for any given stage. Thus, 19 + 2 = 21 ns is the minimum cycle time for this pipeline. 2. What is the total time to execute the function through all stages? There are four stages, each of which is clocked at 21 ns. 4 × 21 = 84 ns is thus the total delay (latency) through the pipeline. 3. S opt = q (1 - 0 . 2)(17+15+19+14) (0 . 2)(2) 11 . Let T seg =7ns S =11 G = 1 1+10 × . 2 1 7+2 = 37.0 MIPS Let T seg =7 . 5ns S =10 G = 1 1+9 × . 2 1 7 . 5+2 = 37.6 MIPS Let T seg =8 . S =9 G = 1 1+8 × . 2 1 8 . 5+2 = 36.6 MIPS Let T seg . S G = 1 1+7 × . 2 1 9 . 5+2 = 36.2 MIPS The cycle time that maximizes performance = 7 . 5+2=9 . Problem 2.2 Repeat problem 1 if there is a 1 ns clock skew (uncertainty of ± 1ns) in the arrival of each clock pulse. 1

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1. What is the minimum cycle time without allocating multiple cycles to a segment?
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## ps1sol - Chapter 2. The Basics Problem 2.1 A four segment...

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