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COMP25212 Further Pipeline Issues. Cray 1 COMP25212 Designed in 1976 Cost $8,800,000 8MB Main Memory Max performance 160 MFLOPS Weight 5.5 Tons Power.

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Presentation on theme: "COMP25212 Further Pipeline Issues. Cray 1 COMP25212 Designed in 1976 Cost $8,800,000 8MB Main Memory Max performance 160 MFLOPS Weight 5.5 Tons Power."— Presentation transcript:

1 COMP25212 Further Pipeline Issues

2 Cray 1 COMP25212 Designed in 1976 Cost $8,800,000 8MB Main Memory Max performance 160 MFLOPS Weight 5.5 Tons Power 115 KW (250KW inc Storage and cooling)

3 COMP25212 Further Pipeline Issues

4 COMP25212 More Pipeline Detail Register Bank Data Cache PC Instruction Cache MUX ALU IF ID EX MEM WB

5 COMP25212 Data Hazards Pipeline can cause other problems Consider ADD R1,R2,R3 MUL R0,R1,R1 The ADD instruction is producing a value in R1 The following MUL instruction uses R1 as input

6 COMP25212 Instructions in the Pipeline Register Bank Data Cache PC Instruction Cache MUX ALU IF ID EX MEM WB ADD R1,R2,R3MUL R0,R1,R1

7 COMP25212 The Data isn’t Ready At end of ID cycle, MUL instruction should have selected value in R1 to put into buffer at input to EX stage But the correct value for R1 from ADD instruction is being put into the buffer at output of EX stage at this time It won’t get to input of Register Bank until one cycle later – then probably another cycle to write into R1

8 COMP25212 Insert Delays? One solution is to detect such data dependencies in hardware and hold instruction in decode stage until data is ready – ‘bubbles’ & wasted cycles again Another is to use the compiler to try to reorder instructions Only works if we can find something useful to do – otherwise insert NOPs - waste

9 COMP25212 Forwarding Register Bank Data Cache PC Instruction Cache MUX ALU ADD R1,R2,R3MUL R0,R1,R1 We can add extra paths for specific cases Control becomes more complex

10 COMP25212 Why did it Occur? Due to the design of our pipeline In this case, the result we want is ready one stage ahead of where it was needed, why pass it down the pipeline? But what if we have the sequence LDR R1,[R2,R3] MUL R0,R1,R1 LDR instruction means load R1 from memory address R2+R3

11 COMP25212 Pipeline Sequence for LDR Fetch Decode and read registers (R2 & R3) Execute – add R2+R3 to form address Memory access, read from address Now we can write the value into register R1 We have designed the ‘worst case’ pipeline to work for all instructions

12 Forwarding Register Bank Data Cache PC Instruction Cache MUX ALU NOPMUL R0,R1,R1 We can add extra paths for specific cases Control becomes more complex LDR R1,[R2,R3]

13 COMP25212 Longer Pipelines As mentioned previously we can go to longer pipelines –Do less per pipeline stage –Each step takes less time –So can increase clock frequency –But greater penalty for hazards –More complex control Negative returns?

14 COMP25212 Where Next? Despite these difficulties it is possible to build processors which approach 1 cycle per instruction (cpi) Given that the computational model is one of serial instruction execution can we do any better than this?


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