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EECE 476: Computer Architecture Slide Set #3: Instruction Set Architecture Design

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Computer Systems Research

Computer Systems Research. Faculty: Krishna Kavi Phil Sweany Hubert Bahr Saraju Mohanty Hao Li Recent PhD Graduate: Mehran Rezaei (at UTA) Ph.D. Students: Wentong Li, Afrin Naz, Paul Lin Henry Liu, Peng Chen, Wenming Li MS Students: Craig Webb

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Loop Unrolling & Predication

Loop Unrolling & Predication. CSE 820. Software Pipelining. With software pipelining a reorganized loop contains instructions from different iterations of the original loop. Sometimes called symbolic loop unrolling. Software Pipelined Loop.

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Pipelining and Hazards

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Pipelining and Hazards

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Midterm 2 review

Midterm 2 review. Chapter 2 - 4. Instruction Set Architecture. Interface between the hardware and software Easy to program with and efficient to implement Regularity, simple tradeoff, constant operands Constant size, small instruction set What operations to include?

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A Multithreaded Architecture

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Comp4611 Tutorial 5

Comp4611 Tutorial 5. Pipelining and Control Hazards Oct. 15 2014. Outline . Branch Hazard Modified Pipelined MIPS Datapath 4 Static Ways to Reduce Branch Penalty Dynamic Branch Prediction Exercise on Pipeline Time Chart. Outline . Branch Hazard Modified Pipelined MIPS Datapath

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Example

Example. Loop: L.D F0, 0(R1) Add.D F0, F0, F2 S.D 0(R1), F0 L.D F0, 0(R2) Mult.D F0, F0, F2 S.D 0(R2), F0 SUBI R1, R1, 8 SUBI R2, R2, 8 BNEZ R2, Loop. Speculative Dynamic Machine specification Issue rate of 1 One broadcast per cycle for CDB branch takes 1 cycle,

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(Recap) Pipeline Hazards

(Recap) Pipeline Hazards. Reg. Reg. Reg. Reg. Reg. Reg. Reg. Reg. Reg. Reg. ALU. ALU. ALU. ALU. ALU. Ifetch. Ifetch. Ifetch. Ifetch. Ifetch. DMem. DMem. DMem. DMem. DMem. EX. WB. MEM. IF. ID/RF. I n s t r. O r d e r. add r1 ,r2,r3. sub r4, r1 ,r3.

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Chapter 4 Predication

Chapter 4 Predication. CSE 820. Go over midterm exam. Review. Predicated instructions have a predicate which determines if the instruction will complete execution or become a NOP. Example. if (A==0) A=B; else A= A+4;

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