lec5_datapath_control

# Reg ir 26 31 to control fsm causewrite1 out pcsource

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Unformatted text preview: struction takes one cycle Each one Spring 2014, Feb 14 . . . Spring ELEC 5200-001/6200-001 Lecture 5 6ns 8ns 7ns 5ns 2ns 8ns 8ns 65 Performance Parameters Performance Average cycles per instruction (CPI) Cycle time (clock period, T) Execution time of a program For single-cycle datapath: CPI = 1 T = hardware time for lw instruction Execution time of a program = T × CPI × (Total instructions executed) Spring 2014, Feb 14 . . . ELEC 5200-001/6200-001 Lecture 5 66 Multicycle Datapath Clock cycle time is determined by the Clock longest operation, ALU or memory: longest Clock cycle time = 2ns Cycles per instruction: lw lw sw sw R-type R-type beq beq j Spring 2014, Feb 14 . . . Spring 5 4 4 3 3 ELEC 5200-001/6200-001 Lecture 5 (10ns) (8ns) (8ns) (6ns) (6ns) 67 CPI of a Multicycle Computer = ∑k (Instructions of type k) × CPIk —————————————— ∑k (instructions of type k) CPIk = Cycles for instruction of type k Note: CPI is dependent on the instruction mix of the program being run. Standard benchmark programs are used for specifying the performance of CPUs. CPI where Spring 2014, Feb 14 . . . Spring ELEC 5200-001/6200-001 Lecture 5 68 Example Consider a program containing: loads loads stores stores branches branches jumps jumps Arithmetic CPI CPI Spring 2014, Feb 14 . . . Spring 25% 10% 11% 2% 52% = 0.25×5 + 0.10×4 + 0.11×3 + ×5 0.02×3 + 0.52×4 0.02×3 = 4.12 for multicycle datapath = 1.00 for single-cycle datapath ELEC 5200-001/6200-001 Lecture 5 69 Multicycle vs. Single-Cycle Performance ratio = Single cycle time / Multicycle time = (CPI × cycle time) for single-cycle ——————————————— (CPI × cycle time) for multicycle = 1.00 × 8ns ————— 4.12 × 2ns = 0.97 Single cycle is faster in this case, but is it always so? Spring 2014, Feb 14 . . . Spring ELEC 5200-001/6200-001 Lecture 5 70 Consider Another Example For this program: loads loads stores stores branches branches jumps jumps Arithmetic CPI CPI Spring 2014, Feb 14 . . . Spring 5% 5% 30% 10% 50% = 0.05×5 + 0.05×4 + 0.30×3 + ×5 0.10×3 + 0.50×4 0.10×3 = 3.65 for multicycle datapath = 1.00 for single-cycle datapath ELEC 5200-001/6200-001 Lecture 5 71 Multicycle vs. Single-Cycle Performance ratio = Single cycle time / Multicycle time = (CPI × cycle time) for single-cycle ——————————————— (CPI × cycle time) for multicycle = 1.00 × 8ns ————— 3.65 × 2ns = 1.096 Multicycle is faster in this case, so the performance ratio depends on the instruction mix. Can we do better? Spring 2014, Feb 14 . . . Spring ELEC 5200-001/6200-001 Lecture 5 72 Next: Pipelining Spring 2014, Feb 14 . . . Spring ELEC 5200-001/6200-001 Lecture 5 73...
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