• ### 电路与模拟电子技术基础_查丽斌_课后答案[1-9章]

这是电路与模拟电子技术基础的课后答案。大学阶段课程的答案。

2019-04-24
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• ### 计算机组成与设计（第四版）答案

Solution 1.4 1.4.1 P2 Class A: 105 instr. Class B: 2 × 105 instr. Class C: 5 × 105 instr. Class D: 2 × 105 instr. Time = No. instr. × CPI/clock rate P1: Time class A = 0.66 × 10−4 Time class B = 2.66 × 10−4 Time class C = 10 × 10−4 Time class D = 5.33 × 10−4 Total time P1 = 18.65 × 10−4 P2: Time class A = 10−4 Time class B = 2 × 10−4 Time class C = 5 × 10−4 Time class D = 3 × 10−4 Total time P2 = 11 × 10−4 1.4.2 CPI = time × clock rate/No. instr. CPI(P1) = 18.65 × 10−4 × 1.5 × 109/106 = 2.79 CPI(P2) = 11 × 10−4 × 2 × 109/106 = 2.2 1.4.3 clock cycles(P1) = 105 × 1 + 2 × 105 × 2 + 5 × 105 × 3 + 2 × 105 × 4 = 28 × 105 clock cycles(P2) = 105 × 2 + 2 × 105 × 2 + 5 × 105 × 2 + 2 × 105 × 3 = 22 × 105 1.4.4 (500 × 1 + 50 × 5 + 100 × 5 + 50 × 2) × 0.5 × 10–9 = 675 ns 1.4.5 CPI = time × clock rate/No. instr. CPI = 675 × 10–9 × 2 × 109/700 = 1.92 1.4.6 Time = (500 × 1 + 50 × 5 + 50 × 5 + 50 × 2) × 0.5 × 10–9 = 550 ns Speed-up = 675 ns/550 ns = 1.22 CPI = 550 × 10–9 × 2 × 109/700 = 1.57

2019-04-24
30
• ### 数字逻辑基础与verilog硬件描述语言答案

大学，数字逻辑基础与verilog硬件描述语言课后答案。通过verilog语言

2018-11-15
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