당신은 주제를 찾고 있습니까 “consider two different implementations of the same instruction set architecture – Instruction Set Architectures“? 다음 카테고리의 웹사이트 https://ro.taphoamini.com 에서 귀하의 모든 질문에 답변해 드립니다: https://ro.taphoamini.com/wiki/. 바로 아래에서 답을 찾을 수 있습니다. 작성자 0612 TV w/ NERDfirst 이(가) 작성한 기사에는 조회수 91,701회 및 좋아요 2,124개 개의 좋아요가 있습니다.
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consider two different implementations of the same instruction set architecture 주제에 대한 동영상 보기
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d여기에서 Instruction Set Architectures – consider two different implementations of the same instruction set architecture 주제에 대한 세부정보를 참조하세요
We’ve seen logic components in action in an earlier series, but how do we work with them when they are all packed together in a CPU? In this episode, we discuss instruction sets to better understand how we can control a CPU without in-depth understanding of its construction!
= CONTENTS PAGE =
0:00 Introduction
0:24 Video Overview
1:04 Why even have \”Instructions\”?
1:53 What are instructions?
2:47 Implementation-Independent Instructions
3:57 Representation of Data \u0026 Instructions
5:16 Examples of ISAs – The MIPS ISA
6:21 MIPS – J-Type Instructions
6:45 MIPS – I-Type Instructions
8:30 MIPS – R-Type Instructions
9:35 End of MIPS example
9:53 Assembly Languages
10:31 Example: MIPS Assembly
11:25 Higher Level Languages
12:18 Conclusion
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consider two different implementations of the same instruction set architecture 주제에 대한 자세한 내용은 여기를 참조하세요.
Consider two different implementations of the same instructi
Conser two different implementations of the same instruction set architecture. The instructions can be dived into four es according to their CPI …
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Consider two different implementations of the same instruction
Conser two different implementations of the same instruction set architecture. There are four es of instructions, A, B, C, and D. The clock rate and …
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Consider two different implementations of the same instruction …
Conser two different implementations of the same instruction set architecture. Th e instructions can be dived into four es according to their CPI …
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How to calculate global CPI with dynamic instruction counts …
1.6 [20] <§1.6> Conser two different implementations of the same instruction set architecture. The instructions can be dived into four …
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Tutorial 02 – Canvas – The University of Auckland
Q2 Conser two different implementations of the same instruction set architecture. The instructions can be dived into four es according to their CPI …
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1.4.2 [5] <1.4> What is the global CPI for each implementation?
1 1.4.2 [5] <1.4> What is the global CPI for each implementation? Conser two different implementations of the same instruction set architecture. There …
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Answered: Consider two different implementations… | bartleby
Conser two different implementations of the same instruction set architecture. The instructions can be dived into four es according to their CPI …
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주제와 관련된 이미지 consider two different implementations of the same instruction set architecture
주제와 관련된 더 많은 사진을 참조하십시오 Instruction Set Architectures. 댓글에서 더 많은 관련 이미지를 보거나 필요한 경우 더 많은 관련 기사를 볼 수 있습니다.
주제에 대한 기사 평가 consider two different implementations of the same instruction set architecture
- Author: 0612 TV w/ NERDfirst
- Views: 조회수 91,701회
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- Date Published: 2015. 8. 4.
- Video Url link: https://www.youtube.com/watch?v=1KTW32xSs_k
Consider two different implementations of the same instruction
Consider two different implementations of the same instruction set architecture. There are four classes of instructions, A, B, C, and D. The clock rate and CPI of each implementation are given in the following table. Processor Clock Rate CPI (class A) CPI (class B) CPI (class C) CPI (class D) P 1 2. 5 GHz 1 2 3 3 P 2 3 GHz 2 2 1. 4. 1 [10] <1. 4> Given a program with 106 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D, which implementation is faster? 1. 4. 2 [5] <1. 4> What is the global CPI for each implementation? 1. 4. 3 [5] <1. 4> Find the clock cycles required in both cases. BK TP. HCM
SOLVED: Consider two different implementations of the same instruction set architecture. Th e instructions can be divided into four classes according to their CPI (class A, B, C, and D). P1 with a clo
Video Transcript
for the first part we need. We know that’s a pure time equals instructions into C. P. I divided by oh great. The same place that instructions endured by CPU time is equal to clock rate by cpt I PS We know that it is equal to instructions per CPU time. This will be equal to this as well. These three terms are equal why PS is equal to clock radio, A CPA. So I PS one equals three GHz, divide by 1.5. That is 2 29. I PS two equals 2.5 divided by one. That is 2.5 and 22.9. I PS three equals Clock rate three CP 3. That is four divided by 2.2, That is 1.82 and returned to the part nine. Therefore Professor has de highest performance in instructions per second for the support. The CPU time. There’s 10 seconds. So clock cycles it will be called to CPU time and two o’clock. So instructions one equals I PS one into CPU time for the first but the first processor, so this is equal to Two into 24, 9 into 10. That is 220 10. For the second processor, the instructions, so for the first processor with the clock cycles, the clock cycles Z equals should turn into three GHz. And that is 3-20. And Instructions for the second processor is 2.5 into 10- nine into 10. That is going to save you time too Is equal to 2.5 to 10 to the 10. Clock cycles two is equal to 10 and two 2.5 gigahertz. That is equal to that is equal to 2.5. And to tend to the party. Now, instructions three for the third processor is equal to 1.82 into 10 to 9 into 10. There is 1.8-1 to 2010 Clock cycles for the 3rd professor is equal to turned into four GHz. That is four and two tend to the party. So for the series of part execution dame. You divide by grow great. New will be equal to 0.7 multiplied to instructions or into cp or divided by local. The old clock rate. We know that instructions equals instructions or so we can write cp. I knew cp I knew divided by create new Equals 0.72 cp or nearby clock rate. Hold the c p. I knew Equals 1.2. Cp Ir Simply is that 1.2? Great. New Will be equal to 0.7. Okay, great. All to claw greed. New Equals 1.7. 1 1.71 times though clock rate. So the claw create must increase By 71%. So these are the answers
How to calculate global CPI with dynamic instruction counts and determine which computer is faster?
Did I do this problem right? My answer is that P2(0.667ms) is faster than P1 (1.04ms). Which number is the global CPI?
1.6 [20] <§1.6> Consider two different implementations of the same instruction set architecture. The instructions can be divided into four classes according to their CPI (class A, B, C, and D). P1 clock rate of 2.5 GHz and CPIs of 1 (10%), 2 (20%), 3(50%), and 3 (20%). P2 clock rate of 3 GHz and CPIs of 2 (10%), 2 (20%), 2 (50%), and 2 (20%).
Given a program with a dynamic instruction count of 1.0E6 (1.0 * 10^6) instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D, which implementation is faster?
1.4.2 [5] <1.4> What is the global CPI for each implementation?
Presentation on theme: “1.4.2 [5] <1.4> What is the global CPI for each implementation?”— Presentation transcript:
1 1.4.2 [5] <1.4> What is the global CPI for each implementation?
Consider two different implementations of the same instruction set architecture. There are four classes of instructions, A, B, C, and D. The clock rate and CPI of each implementation are given in the following table. 1.4.1 [10] <1.4> Given a program with 106 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D, which implementation is faster? 1.4.2 [5] <1.4> What is the global CPI for each implementation? 1.4.3 [5] <1.4> Find the clock cycles required in both cases. Processor Clock Rate CPI (class A) CPI (class B) CPI (class C) CPI (class D) P1 2.5 GHz 1 2 3 P2 3 GHz
2 Class A (10% of 106 instr.): 105 instr.
Solution 1.4.1 Class A (10% of 106 instr.): 105 instr. Class B (20% of 106 instr.): 2 × 105 instr. Class C(50% of 106 instr.): 5 × 105 instr. Class D(20% of 106 instr.): 2 × 105 instr. CPU-Time = AƩD(ICi x CPIi) x Clock Cycle Time = (IC x CPI)/Clock Rate Used for: 1. CPU-Time(P1) = ( × 105 × × 105 × × 105 × 3)/(2.5 × 109) = × 10−4 s CPU-Time(P2) = (105 × × 105 × × 105 × × 105 × 2)/(3 × 109) = 6.66 × 10−4 s Return
3 Global CPI = (CPU-Time x Clock Rate)/IC Therefore:
Solution 1.4.2 Global CPI = (CPU-Time x Clock Rate)/IC Therefore: CPI (P1) = 10.4 × 10−4 × 2.5 × 109/106 = 2.6 CPI (P2) = 6.66 × 10−4 × 3 × 109/106 = 2.0 Return
4 Class A (10% of 106 instr.): 105 instr.
Solution 1.4.3 Class A (10% of 106 instr.): 105 instr. Class B (20% of 106 instr.): 2 × 105 instr. Class C(50% of 106 instr.): 5 × 105 instr. Class D(20% of 106 instr.): 2 × 105 instr. CPU-Time = AƩD(ICi x CPIi) x Clock Cycle Time Clock Cycles = (AƩD(ICi x CPIi) Therefore: clock cycles (P1) = 105 × × 105 × × 105 × × 105 × 3 = × 105 clock cycles (P2) = 105 × × 105 × × 105 × × 105 × 2 = 20 × 105 Return
Answered: Consider two different implementations…
Consider two different implementations of the same instruction set architecture. The instructions can be divided into four classes according to their CPI (class A, B, C and D). P1 with a clock rate of 3GHz and CPIs of 3, 2, 1, 4, and P2 with a clock rate of 2.5GHz and CPIs of 2, 2, 2, 2.
Processor Class A Class B Class C Class D P1 3 2 1 4 P2 2 2 2 2
Given a program with a dynamic instruction count of 1.0E5 instructions divided into classes as follows: 10% class A, 30% class B, 40% class C, and 20% class D.
Which implementation is faster?
What is the global CPI for each implementation?
Find the clock cycles required in both cases?
키워드에 대한 정보 consider two different implementations of the same instruction set architecture
다음은 Bing에서 consider two different implementations of the same instruction set architecture 주제에 대한 검색 결과입니다. 필요한 경우 더 읽을 수 있습니다.
이 기사는 인터넷의 다양한 출처에서 편집되었습니다. 이 기사가 유용했기를 바랍니다. 이 기사가 유용하다고 생각되면 공유하십시오. 매우 감사합니다!
사람들이 주제에 대해 자주 검색하는 키워드 Instruction Set Architectures
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