COMP25212 Further Pipeline Issues. Cray 1 COMP25212 Designed in 1976 Cost $8,800,000 8MB Main Memory...
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Transcript of COMP25212 Further Pipeline Issues. Cray 1 COMP25212 Designed in 1976 Cost $8,800,000 8MB Main Memory...
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)
COMP25212
More Pipeline Detail
Register B
ank
Data
Cache
PC
Instruction
Cache
MU
X
ALU
IF ID EX MEM WB
COMP25212
Data Hazards
• Pipeline can cause other problems
• ConsiderADD R1,R2,R3
MUL R0,R1,R1
• The ADD instruction is producing a value in R1
• The following MUL instruction uses R1 as input
COMP25212
Instructions in the Pipeline
Register B
ank
Data
Cache
PC
Instruction
Cache
MU
X
ALU
IF ID EX MEM WB
ADD R1,R2,R3MUL R0,R1,R1
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
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
COMP25212
Forwarding
Register B
ank
Data
Cache
PC
Instruction
Cache
MU
X
ALU
ADD R1,R2,R3MUL R0,R1,R1
• We can add extra paths for specific cases• Control becomes more complex
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 sequenceLDR R1,[R2,R3]MUL R0,R1,R1
• LDR instruction means load R1 from memory address R2+R3
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
Forwarding
Register B
ank
Data
Cache
PC
Instruction
Cache
MU
X
ALU
NOPMUL R0,R1,R1
• We can add extra paths for specific cases• Control becomes more complex
LDR R1,[R2,R3]
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?