instruction level parallelism and dynamic execution #3:
DESCRIPTION
Instruction Level Parallelism and Dynamic Execution #3:. Based on Prof. David A. Patterson. Example: Evaluating Branch Alternatives. Assume: Conditional & Unconditional = 14%, 65% change PC SchedulingBranchCPIspeedup v. scheme penaltystall Stall pipeline31.421.0 - PowerPoint PPT PresentationTRANSCRIPT
CPSC614Lec 4.1
Instruction Level Parallelism and Dynamic Execution #3:
Based on
Prof. David A. Patterson
CPSC614Lec 4.2
Example: Evaluating Branch Alternatives
Assume: Conditional & Unconditional = 14%, 65% change PC
Scheduling Branch CPIspeedup v. scheme penalty stall
Stall pipeline 3 1.42 1.0Predict taken 1 1.14 1.26Predict not taken 1 1.09 1.29Delayed branch 0.5 1.07 1.31
Pipeline speedup = Pipeline depth1 +Branch frequencyBranch penalty
CPSC614Lec 4.3
Three Stages of Tomasulo Algorithm
1. Issue—get instruction from FP Op Queue If reservation station free (no structural hazard),
control issues instr & sends operands (renames registers).
2. Execute—operate on operands (EX) When both operands ready then execute;
if not ready, watch Common Data Bus for result
3. Write result—finish execution (WB) Write on Common Data Bus to all awaiting units;
mark reservation station available
• Normal data bus: data + destination (“go to” bus)• Common data bus: data + source (“come from” bus)
– 64 bits of data + 4 bits of Functional Unit source address– Write if matches expected Functional Unit (produces result)– Does the broadcast
• Example speed: 3 clocks for Fl .pt. +,-; 10 for * ; 40 clks for /
CPSC614Lec 4.4
Tomasulo ExampleInstruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 Load1 NoLD F2 45+ R3 Load2 NoMULTD F0 F2 F4 Load3 NoSUBD F8 F6 F2DIVD F10 F0 F6ADDD F6 F8 F2
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 NoMult1 NoMult2 No
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F300 FU
Clock cycle counter
FU countdown
Instruction stream
3 Load/Buffers
3 FP Adder R.S.2 FP Mult R.S.
CPSC614Lec 4.5
Tomasulo Example Cycle 1Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 Load1 Yes 34+R2LD F2 45+ R3 Load2 NoMULTD F0 F2 F4 Load3 NoSUBD F8 F6 F2DIVD F10 F0 F6ADDD F6 F8 F2
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 NoMult1 NoMult2 No
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F301 FU Load1
CPSC614Lec 4.6
Tomasulo Example Cycle 2Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 Load1 Yes 34+R2LD F2 45+ R3 2 Load2 Yes 45+R3MULTD F0 F2 F4 Load3 NoSUBD F8 F6 F2DIVD F10 F0 F6ADDD F6 F8 F2
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 NoMult1 NoMult2 No
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F302 FU Load2 Load1
Note: Can have multiple loads outstanding
CPSC614Lec 4.7
Tomasulo Example Cycle 3Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 Load1 Yes 34+R2LD F2 45+ R3 2 Load2 Yes 45+R3MULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2DIVD F10 F0 F6ADDD F6 F8 F2
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 NoMult1 Yes MULTD R(F4) Load2Mult2 No
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F303 FU Mult1 Load2 Load1
• Note: registers names are removed (“renamed”) in Reservation Stations; MULT issued
• Load1 completing; what is waiting for Load1?
CPSC614Lec 4.8
Tomasulo Example Cycle 4Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 Load2 Yes 45+R3MULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4DIVD F10 F0 F6ADDD F6 F8 F2
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 Yes SUBD M(A1) Load2Add2 NoAdd3 NoMult1 Yes MULTD R(F4) Load2Mult2 No
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F304 FU Mult1 Load2 M(A1) Add1
• Load2 completing; what is waiting for Load2?
CPSC614Lec 4.9
Tomasulo Example Cycle 5Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4DIVD F10 F0 F6 5ADDD F6 F8 F2
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
2 Add1 Yes SUBD M(A1) M(A2)Add2 NoAdd3 No
10 Mult1 Yes MULTD M(A2) R(F4)Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F305 FU Mult1 M(A2) M(A1) Add1 Mult2
• Timer starts down for Add1, Mult1
CPSC614Lec 4.10
Tomasulo Example Cycle 6Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4DIVD F10 F0 F6 5ADDD F6 F8 F2 6
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
1 Add1 Yes SUBD M(A1) M(A2)Add2 Yes ADDD M(A2) Add1Add3 No
9 Mult1 Yes MULTD M(A2) R(F4)Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F306 FU Mult1 M(A2) Add2 Add1 Mult2
• Issue ADDD here despite name dependency on F6?
CPSC614Lec 4.11
Tomasulo Example Cycle 7Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4 7DIVD F10 F0 F6 5ADDD F6 F8 F2 6
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
0 Add1 Yes SUBD M(A1) M(A2)Add2 Yes ADDD M(A2) Add1Add3 No
8 Mult1 Yes MULTD M(A2) R(F4)Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F307 FU Mult1 M(A2) Add2 Add1 Mult2
• Add1 (SUBD) completing; what is waiting for it?
CPSC614Lec 4.12
Tomasulo Example Cycle 8Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 No2 Add2 Yes ADDD (M-M) M(A2)
Add3 No7 Mult1 Yes MULTD M(A2) R(F4)
Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F308 FU Mult1 M(A2) Add2 (M-M) Mult2
CPSC614Lec 4.13
Tomasulo Example Cycle 9Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 No1 Add2 Yes ADDD (M-M) M(A2)
Add3 No6 Mult1 Yes MULTD M(A2) R(F4)
Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F309 FU Mult1 M(A2) Add2 (M-M) Mult2
CPSC614Lec 4.14
Tomasulo Example Cycle 10Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6 10
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 No0 Add2 Yes ADDD (M-M) M(A2)
Add3 No5 Mult1 Yes MULTD M(A2) R(F4)
Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3010 FU Mult1 M(A2) Add2 (M-M) Mult2
• Add2 (ADDD) completing; what is waiting for it?
CPSC614Lec 4.15
Tomasulo Example Cycle 11Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 No
4 Mult1 Yes MULTD M(A2) R(F4)Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3011 FU Mult1 M(A2) (M-M+M)(M-M) Mult2
• Write result of ADDD here?• All quick instructions complete in this cycle!
CPSC614Lec 4.16
Tomasulo Example Cycle 12Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 No
3 Mult1 Yes MULTD M(A2) R(F4)Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3012 FU Mult1 M(A2) (M-M+M)(M-M) Mult2
CPSC614Lec 4.17
Tomasulo Example Cycle 13Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 No
2 Mult1 Yes MULTD M(A2) R(F4)Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3013 FU Mult1 M(A2) (M-M+M)(M-M) Mult2
CPSC614Lec 4.18
Tomasulo Example Cycle 14Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 No
1 Mult1 Yes MULTD M(A2) R(F4)Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3014 FU Mult1 M(A2) (M-M+M)(M-M) Mult2
CPSC614Lec 4.19
Tomasulo Example Cycle 15Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 15 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 No
0 Mult1 Yes MULTD M(A2) R(F4)Mult2 Yes DIVD M(A1) Mult1
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3015 FU Mult1 M(A2) (M-M+M)(M-M) Mult2
• Mult1 (MULTD) completing; what is waiting for it?
CPSC614Lec 4.20
Tomasulo Example Cycle 16Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 15 16 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 NoMult1 No
40 Mult2 Yes DIVD M*F4 M(A1)
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3016 FU M*F4 M(A2) (M-M+M)(M-M) Mult2
• Just waiting for Mult2 (DIVD) to complete
CPSC614Lec 4.21
Faster than light computation(skip a couple of cycles)
CPSC614Lec 4.22
Tomasulo Example Cycle 55Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 15 16 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 NoMult1 No
1 Mult2 Yes DIVD M*F4 M(A1)
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3055 FU M*F4 M(A2) (M-M+M)(M-M) Mult2
CPSC614Lec 4.23
Tomasulo Example Cycle 56Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 15 16 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5 56ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 NoMult1 No
0 Mult2 Yes DIVD M*F4 M(A1)
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3056 FU M*F4 M(A2) (M-M+M)(M-M) Mult2
• Mult2 (DIVD) is completing; what is waiting for it?
CPSC614Lec 4.24
Tomasulo Example Cycle 57Instruction status: Exec Write
Instruction j k Issue Comp Result Busy AddressLD F6 34+ R2 1 3 4 Load1 NoLD F2 45+ R3 2 4 5 Load2 NoMULTD F0 F2 F4 3 15 16 Load3 NoSUBD F8 F6 F2 4 7 8DIVD F10 F0 F6 5 56 57ADDD F6 F8 F2 6 10 11
Reservation Stations: S1 S2 RS RSTime Name Busy Op Vj Vk Qj Qk
Add1 NoAdd2 NoAdd3 NoMult1 NoMult2 Yes DIVD M*F4 M(A1)
Register result status:Clock F0 F2 F4 F6 F8 F10 F12 ... F3056 FU M*F4 M(A2) (M-M+M)(M-M) Result
• Once again: In-order issue, out-of-order execution and out-of-order completion.
CPSC614Lec 4.25
Tomasulo Drawbacks
• Complexity– delays of 360/91, MIPS 10000, Alpha 21264,
IBM PPC 620 in CA:AQA 2/e, but not in silicon!
• Many associative stores (CDB) at high speed
• Performance limited by Common Data Bus– Each CDB must go to multiple functional units
high capacitance, high wiring density– Number of functional units that can complete per
cycle limited to one!» Multiple CDBs more FU logic for parallel assoc
stores
• Non-precise interrupts!– We will address this later
CPSC614Lec 4.26
Tomasulo Loop Example
Loop: LD F0 0 R1
MULTD F4 F0 F2
SD F4 0 R1
SUBI R1 R1 #8
BNEZ R1 Loop
• This time assume Multiply takes 4 clocks• Assume 1st load takes 8 clocks
(L1 cache miss), 2nd load takes 1 clock (hit)• To be clear, will show clocks for SUBI, BNEZ
– Reality: integer instructions ahead of Fl. Pt. Instructions
• Show 2 iterations
CPSC614Lec 4.27
Loop ExampleInstruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 Load1 No1 MULTD F4 F0 F2 Load2 No1 SD F4 0 R1 Load3 No2 LD F0 0 R1 Store1 No2 MULTD F4 F0 F2 Store2 No2 SD F4 0 R1 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 No SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F300 80 Fu
Added Store Buffers
Value of Register used for address, iteration control
Instruction Loop
Iter-ationCount
CPSC614Lec 4.28
Loop Example Cycle 1Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 Load1 Yes 80
Load2 NoLoad3 NoStore1 NoStore2 NoStore3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 No SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F301 80 Fu Load1
CPSC614Lec 4.29
Loop Example Cycle 2Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 Load1 Yes 801 MULTD F4 F0 F2 2 Load2 No
Load3 NoStore1 NoStore2 NoStore3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load1 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F302 80 Fu Load1 Mult1
CPSC614Lec 4.30
Loop Example Cycle 3Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 Load1 Yes 801 MULTD F4 F0 F2 2 Load2 No1 SD F4 0 R1 3 Load3 No
Store1 Yes 80 Mult1Store2 NoStore3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load1 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F303 80 Fu Load1 Mult1
• Implicit renaming sets up data flow graph
CPSC614Lec 4.31
Loop Example Cycle 4Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 Load1 Yes 801 MULTD F4 F0 F2 2 Load2 No1 SD F4 0 R1 3 Load3 No
Store1 Yes 80 Mult1Store2 NoStore3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load1 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F304 80 Fu Load1 Mult1
• Dispatching SUBI Instruction (not in FP queue)
CPSC614Lec 4.32
Loop Example Cycle 5Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 Load1 Yes 801 MULTD F4 F0 F2 2 Load2 No1 SD F4 0 R1 3 Load3 No
Store1 Yes 80 Mult1Store2 NoStore3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load1 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F305 72 Fu Load1 Mult1
• And, BNEZ instruction (not in FP queue)
CPSC614Lec 4.33
Loop Example Cycle 6Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 Load1 Yes 801 MULTD F4 F0 F2 2 Load2 Yes 721 SD F4 0 R1 3 Load3 No2 LD F0 0 R1 6 Store1 Yes 80 Mult1
Store2 NoStore3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load1 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F306 72 Fu Load2 Mult1
• Notice that F0 never sees Load from location 80
CPSC614Lec 4.34
Loop Example Cycle 7Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 Load1 Yes 801 MULTD F4 F0 F2 2 Load2 Yes 721 SD F4 0 R1 3 Load3 No2 LD F0 0 R1 6 Store1 Yes 80 Mult12 MULTD F4 F0 F2 7 Store2 No
Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load1 SUBI R1 R1 #8Mult2 Yes Multd R(F2) Load2 BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F307 72 Fu Load2 Mult2
• Register file completely detached from computation• First and Second iteration completely overlapped
CPSC614Lec 4.35
Loop Example Cycle 8Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 Load1 Yes 801 MULTD F4 F0 F2 2 Load2 Yes 721 SD F4 0 R1 3 Load3 No2 LD F0 0 R1 6 Store1 Yes 80 Mult12 MULTD F4 F0 F2 7 Store2 Yes 72 Mult22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load1 SUBI R1 R1 #8Mult2 Yes Multd R(F2) Load2 BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F308 72 Fu Load2 Mult2
CPSC614Lec 4.36
Loop Example Cycle 9Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 Load1 Yes 801 MULTD F4 F0 F2 2 Load2 Yes 721 SD F4 0 R1 3 Load3 No2 LD F0 0 R1 6 Store1 Yes 80 Mult12 MULTD F4 F0 F2 7 Store2 Yes 72 Mult22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load1 SUBI R1 R1 #8Mult2 Yes Multd R(F2) Load2 BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F309 72 Fu Load2 Mult2
• Load1 completing: who is waiting?• Note: Dispatching SUBI
CPSC614Lec 4.37
Loop Example Cycle 10Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 Load2 Yes 721 SD F4 0 R1 3 Load3 No2 LD F0 0 R1 6 10 Store1 Yes 80 Mult12 MULTD F4 F0 F2 7 Store2 Yes 72 Mult22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1
4 Mult1 Yes Multd M[80] R(F2) SUBI R1 R1 #8Mult2 Yes Multd R(F2) Load2 BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3010 64 Fu Load2 Mult2
• Load2 completing: who is waiting?• Note: Dispatching BNEZ
CPSC614Lec 4.38
Loop Example Cycle 11Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 Load2 No1 SD F4 0 R1 3 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 Yes 80 Mult12 MULTD F4 F0 F2 7 Store2 Yes 72 Mult22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1
3 Mult1 Yes Multd M[80] R(F2) SUBI R1 R1 #84 Mult2 Yes Multd M[72] R(F2) BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3011 64 Fu Load3 Mult2
• Next load in sequence
CPSC614Lec 4.39
Loop Example Cycle 12Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 Load2 No1 SD F4 0 R1 3 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 Yes 80 Mult12 MULTD F4 F0 F2 7 Store2 Yes 72 Mult22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1
2 Mult1 Yes Multd M[80] R(F2) SUBI R1 R1 #83 Mult2 Yes Multd M[72] R(F2) BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3012 64 Fu Load3 Mult2
• Why not issue third multiply?
CPSC614Lec 4.40
Loop Example Cycle 13Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 Load2 No1 SD F4 0 R1 3 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 Yes 80 Mult12 MULTD F4 F0 F2 7 Store2 Yes 72 Mult22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1
1 Mult1 Yes Multd M[80] R(F2) SUBI R1 R1 #82 Mult2 Yes Multd M[72] R(F2) BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3013 64 Fu Load3 Mult2
• Why not issue third store?
CPSC614Lec 4.41
Loop Example Cycle 14Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 14 Load2 No1 SD F4 0 R1 3 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 Yes 80 Mult12 MULTD F4 F0 F2 7 Store2 Yes 72 Mult22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1
0 Mult1 Yes Multd M[80] R(F2) SUBI R1 R1 #81 Mult2 Yes Multd M[72] R(F2) BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3014 64 Fu Load3 Mult2
• Mult1 completing. Who is waiting?
CPSC614Lec 4.42
Loop Example Cycle 15Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 14 15 Load2 No1 SD F4 0 R1 3 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 Yes 80 [80]*R22 MULTD F4 F0 F2 7 15 Store2 Yes 72 Mult22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 No SUBI R1 R1 #8
0 Mult2 Yes Multd M[72] R(F2) BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3015 64 Fu Load3 Mult2
• Mult2 completing. Who is waiting?
CPSC614Lec 4.43
Loop Example Cycle 16Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 14 15 Load2 No1 SD F4 0 R1 3 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 Yes 80 [80]*R22 MULTD F4 F0 F2 7 15 16 Store2 Yes 72 [72]*R22 SD F4 0 R1 8 Store3 No
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1
4 Mult1 Yes Multd R(F2) Load3 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3016 64 Fu Load3 Mult1
CPSC614Lec 4.44
Loop Example Cycle 17Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 14 15 Load2 No1 SD F4 0 R1 3 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 Yes 80 [80]*R22 MULTD F4 F0 F2 7 15 16 Store2 Yes 72 [72]*R22 SD F4 0 R1 8 Store3 Yes 64 Mult1
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load3 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3017 64 Fu Load3 Mult1
CPSC614Lec 4.45
Loop Example Cycle 18Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 14 15 Load2 No1 SD F4 0 R1 3 18 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 Yes 80 [80]*R22 MULTD F4 F0 F2 7 15 16 Store2 Yes 72 [72]*R22 SD F4 0 R1 8 Store3 Yes 64 Mult1
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load3 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3018 64 Fu Load3 Mult1
CPSC614Lec 4.46
Loop Example Cycle 19Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 No1 MULTD F4 F0 F2 2 14 15 Load2 No1 SD F4 0 R1 3 18 19 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 No2 MULTD F4 F0 F2 7 15 16 Store2 Yes 72 [72]*R22 SD F4 0 R1 8 19 Store3 Yes 64 Mult1
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load3 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3019 56 Fu Load3 Mult1
CPSC614Lec 4.47
Loop Example Cycle 20Instruction status: Exec Write
ITER Instruction j k Issue CompResult Busy Addr Fu1 LD F0 0 R1 1 9 10 Load1 Yes 561 MULTD F4 F0 F2 2 14 15 Load2 No1 SD F4 0 R1 3 18 19 Load3 Yes 642 LD F0 0 R1 6 10 11 Store1 No2 MULTD F4 F0 F2 7 15 16 Store2 No2 SD F4 0 R1 8 19 20 Store3 Yes 64 Mult1
Reservation Stations: S1 S2 RS Time Name Busy Op Vj Vk Qj Qk Code:
Add1 No LD F0 0 R1Add2 No MULTD F4 F0 F2Add3 No SD F4 0 R1Mult1 Yes Multd R(F2) Load3 SUBI R1 R1 #8Mult2 No BNEZ R1 Loop
Register result status
Clock R1 F0 F2 F4 F6 F8 F10 F12 ... F3020 56 Fu Load1 Mult1
• Once again: In-order issue, out-of-order execution and out-of-order completion.
CPSC614Lec 4.48
Why can Tomasulo overlap iterations of loops?
• Register renaming– Multiple iterations use different physical destinations
for registers (dynamic loop unrolling).
• Reservation stations – Permit instruction issue to advance past integer control
flow operations– Also buffer old values of registers - totally avoiding the
WAR stall that we saw in the scoreboard.
• Other perspective: Tomasulo building data flow dependency graph on the fly.
CPSC614Lec 4.49
Tomasulo’s scheme offers 2 major advantages
(1)the distribution of the hazard detection logic– distributed reservation stations and the CDB– If multiple instructions waiting on single result,
& each instruction has other operand, then instructions can be released simultaneously by broadcast on CDB
– If a centralized register file were used, the units would have to read their results from the registers when register buses are available.
(2) the elimination of stalls for WAW and WAR hazards
CPSC614Lec 4.50
Dynamic Hardware Branch Prediction
• As the amount of ILP grows, control dependences rapidly become the limiting factor.
• Basic Schemes (Static)– Prediction Not Taken– Delayed Branch– Action taken does not depend on the dynamic behavior of
the branch.
CPSC614Lec 4.51
Dynamic Hardware Prediction
• Goal– Allow the processor to resolve the
outcome of a branch early, thus preventing control dependences from causing stalls.
CPSC614Lec 4.52
Branch-Prediction Buffer
• Branch history buffer• Small memory indexed by the lower
portion of the address of the branch instruction
• 1 bit says whether the branch was taken or not
• Simplest
CPSC614Lec 4.53
Example
• Consider a loop branch whose behavior is taken nine times in a row, then not taken once. What is the prediction accuracy for this branch?
Even if a branch is almost always taken, we predict incorrectly twice when it is not taken.
CPSC614Lec 4.54
2-bit Prediction Scheme
• A prediction must miss twice before it is changed.
CPSC614Lec 4.55
n-bit Saturating Counter
• Values: 0 ~ 2n-1
• When the counter is greater than or equal to one-half of its maximum value, the branch is predicted as taken. Otherwise, not taken.
• Studies have shown that the 2-bit predictors do almost do well, and thus most systems rely on 2-bit branch predictors.
CPSC614Lec 4.56
4096-entry 2-bitprediction buffer
CPSC614Lec 4.57
CPSC614Lec 4.58
Correlating Branch Predictor
• It may possible to improve the accuracy if we look at the behavior of other branches.
if (aa == 2)aa = 0;
if (bb == 0)bb = 0;
if (aa != bb)
The behavior of b3 is correlated with the behavior of b1 and b2.
CPSC614Lec 4.59
Correlating Predictors
• Two-level predictors
if (d == 0)d = 1;
if (d == 1)
CPSC614Lec 4.60
initial value of d
b1 value of d before b2
b2
0
1
2
CPSC614Lec 4.61
1-bit Predictor (Initialized to NT)
d b1 predic
b1 action
new b1 pr
b2 predic
b2 action
new b2 pr
2
0
2
0
CPSC614Lec 4.62
(1,1) Predictor
• Every branch has two separate prediction bits.– First bit: the prediction if the last branch in the
program is not taken.– Second bit: the prediction if the last branch in
the program is taken.
• Write the pair of prediction bits together.
CPSC614Lec 4.63
Combinations & Meaning
Prediction bits Prediction if not taken
Prediction if taken
CPSC614Lec 4.64
(m,n) Predictor
• Uses the last m branches to choose from 2m branch predictors, each of which is an n-bit predictor.
• Yields higher prediction rates than 2-bit scheme
• Requires a trivial amount of additional hardware
• The global history of the most recent m branches are recorded in an m-bit shift register.
CPSC614Lec 4.65
CPSC614Lec 4.66
(m,n) Predictor
• Total number of bits:
= 2m x n x #prediction entries selected by the branch address
• Examples
CPSC614Lec 4.67
CPSC614Lec 4.68
Tournament Predictors
• Most popular multilevel branch predictors
CPSC614Lec 4.69
Tournament Predictors
• By using multiple predictors (one based on global information, one based on local information, and combining them with a selector), it can select the right predictor for the right branch.
• Alpha 21264– Uses most sophisticated branch predictor as of
2001.
CPSC614Lec 4.70
CPSC614Lec 4.71