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Lecture 9 – OOO execution © Avi Mendelson, 5/2005 1 MAMAS – Computer Architecture 234367 Lecture 9 – Out Of Order (OOO) Dr. Avi Mendelson Some of the slides.

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Presentation on theme: "Lecture 9 – OOO execution © Avi Mendelson, 5/2005 1 MAMAS – Computer Architecture 234367 Lecture 9 – Out Of Order (OOO) Dr. Avi Mendelson Some of the slides."— Presentation transcript:

1 Lecture 9 – OOO execution © Avi Mendelson, 5/ MAMAS – Computer Architecture Lecture 9 – Out Of Order (OOO) Dr. Avi Mendelson Some of the slides were taken from: (1) Lihu Rapoport (2) Randi Katz and (3) Petterson

2 Lecture 9 – OOO execution © Avi Mendelson, 5/ Agenda  Introduction  Data flow execution  Out of order execution – principles  Pentium-(II,III and 6) example

3 Lecture 9 – OOO execution © Avi Mendelson, 5/ Introduction  The goal is to increase the performance of a system.  In order to achieve that, so far we discussed – Adding pipe stages in order to increase the clock rate – Increase the potential parallelism within the pipe by  Fetching decoding and executing several instructions in parallel  Are there any other options?

4 Lecture 9 – OOO execution © Avi Mendelson, 5/ Data flow vs. conventional computer ?  In theory “data flow machines” has the best performance – View the program as a parallel operations wait to be executed (will be demonstrate next slide) – Execute instruction as soon as its inputs are ready  So, why computers are Van-Neumann based and not Data-Flow based – Hard to debug – Hard to write Data-Flow programs (need special programming language in order to be efficient)

5 Lecture 9 – OOO execution © Avi Mendelson, 5/ Data flow execution – a different approach for high performance computers  Data flow execution is an alternative for Van-Neumann execution. Here, the instructions are executed in the order of their input dependencies and not in the order they appears in the program  Example : assume that we have as many execution units as we need: (1) r1  r4 / r7 (2) r8  r1 + r2 (3) r5  r5 + 1 (4) r6  r6 - r3 (5) r4  r5 + r6 (6) r7  r8 * r Data Flow Graph We could execute it in 3 cycles

6 Lecture 9 – OOO execution © Avi Mendelson, 5/ Data flow execution - cont  Can we build a machine that will execute the “data flow graph”?  In the early 70 th several machines were built to work according to the data-flow graph. They were called “data flow machines”. They were vanished due to the reasons we mentioned before.  Solution: Let the user think he/she are using Van-Neumann machine, and let the system work in “data-flow mode”

7 Lecture 9 – OOO execution © Avi Mendelson, 5/ OOOE - General Scheme Fetch & Decode Instruction pool Retire (commit) In-order Execute Out-of-order Most of the modern computers are using OOO execution. Most of them are doing the fetching and the retirement IN- ORDER, but it executes in OUT_OF_ORDER

8 Lecture 9 – OOO execution © Avi Mendelson, 5/ Out Of Order Execution Basic idea: – The fetch is done in the program order (in-order) and fast enough in order to “fill-out” a window of instructions. – Out of the instruction window, the system forms a data flow graph and looks for instructions which are ready to be executed:  All the data the instructions are depended on, are ready  Resources are available. – As soon as the instruction is execution it needs to signal to all the instructions which are depend on it that it generate new input. – The instructions are commit in “program’s order” to preserve the “user view”  Advantages: – Help exploit Instruction Level Parallelism (ILP) – Help cover latencies (e.g., cache miss, divide)

9 Lecture 9 – OOO execution © Avi Mendelson, 5/ How to convert “In-order” instruction flow into “data flow”  The problems: 1. Data Flow has only RAW dependencies, while OOOE has also WAR and WAW dependencies 2. How to guarantee the in-order complition.  The Solutions: 1. Register Renaming (based on “Tomuselo algorithm”) solves the WAR and WAW dependencies 2. We need to “enumerate” the instructions at decode time (in order) so we know in what order to retire them

10 Lecture 9 – OOO execution © Avi Mendelson, 5/ Register Renaming  Hold a pool of physical registers.  Architectural registers are mapped into physical registers – When an instruction writes to an architectural register  A free physical register is allocated from the pool  The physical register points to the architectural register  The instruction writes the value to the physical register – When an instruction reads from an architectural register  reads the data from the latest instruction which writes to the same architectural register, and precedes the current instruction.  If no such instruction exists, read directly from the architectural register. – When an instruction commits  Moves the value from the physical register to the architectural register it points.

11 Lecture 9 – OOO execution © Avi Mendelson, 5/ OOOE with Register Renaming: Example Before renaming After renaming (1)r1  mem1t1  mem1 (2)r2  r2 + r1t2  r2 + t1 (3)r1  mem2 t3  mem2 (4)r3  r3 + r1 t4  r3 + t3 (5)r1  mem3t5  mem3 (6)r4  r5 + r1t6  r5 + t5 (7)r5  2t7  2 (8)r6  r5 + 2t8  t7 + 2 After renaming all the false dependencies (WAW and WAR) were removed WAW WAR

12 Lecture 9 – OOO execution © Avi Mendelson, 5/ Executing Beyond Branches  The scheme we saw so far does not search beyond a branch  Limited to the parallelism within a basic-block – A basic-block is ~5 instruction long (1) r1  r4 / r7 (2)r2  r2 + r1 (3)r3  r2 - 5 (4)beq r3,0,300 If the beq is predicted NT, (5)r8  r8 + 1 Inst 5 can be spec executed  We would like to look beyond branches – But what if we execute an instruction beyond a branch and then it turns out that we predicted the wrong path ? Solution: Speculative Execution

13 Lecture 9 – OOO execution © Avi Mendelson, 5/ Speculative Execution Execution of instructions from a predicted (yet unsure) path Eventually, path may turn wrong Implementation: – Hold a pool of all “not yet executed” instructions – Fetch instructions into the pool from a predicted path – Instructions for which all operands are “ready” can be executed – An instruction may change the processor state (commit) only when it is safe  An instruction commits only when all previous (in-order) instructions had committed  Instructions commit in-order  Instructions which follow a branch commit only after the branch commits  If a predicted branch is wrong all the instructions which follow it are flushed Register Renaming helps speculative execution – Renamed registers are kept until speculation is verified to be correct

14 Lecture 9 – OOO execution © Avi Mendelson, 5/ The magic of the modern X86 architectures (Intel, AMD, etc.)  The user view of the X86 machine is as a CISC architecture.  The machine supports this view by keeping the in-order parts as close as possible to the X86 view.  While moving from the In-order part (front-end) to the OOO part (execution), the hardware translates each X86 instruction into a set of uop operations, which are the internal machine operations. These operations are RISC like (load-store based).  During this translation, the hardware performs the register renaming. So, during the execution time it uses internal registers and not the X86 ones. The number of these registers can be changed from one generation to another.  While moving back from the OOO part (execution) to the In-Order part (commit), the hardware translates the registers back to X86, in order to keep for the user a coherent picture.

15 Lecture 9 – OOO execution © Avi Mendelson, 5/ OOOE Architecture: based on Pentium-II Write back bus Instruction cache Data cache Bus Interface Unit IFU Instr. Fetch ID Instr. Decode and rename RATROB RS MOB Load/Store Operations Arithmetic Operations Retire (commit) Logic  In Order Front-end 1. Fetch from instruction cache, base on Branch prediction. 2. Decode and rename:  Translate to Uops  Use the RAT table for renaming  Put ALL instructions in ROB  Put all “arithmetic instructions” in the RS queue  Put all Load/Store instructions in MOB  Out-Of-Order execution 3. Do in Parallel:  Load and store operations are executed based on MOB information  Arithmetic operations are executed based on RS information. 4. All results are written back to ROB, while RS and MOB “still” values they need  In Order completion (retirement) 5. The retire logic (commit logic) moves instructions out of the ROB and updates the architectural registers

16 Lecture 9 – OOO execution © Avi Mendelson, 5/ Re-order Buffer (ROB)  Mechanism for keeping the in-order view of the user.  Basic ROB functions – Provide large physical register space for register renaming – Keeps intermediate results, some of them may not be commit if the branch prediction was wrong (we will discuss this mechanism later on) – Keeps information on what is the “Real Register” the commit need to update

17 Lecture 9 – OOO execution © Avi Mendelson, 5/ Reservation station (RS)  Pool of all “not yet executed” uops – Holds both the uop attributes as well as the values of the input data  For each operand, it keeps indication if it is ready – Operand that need to be retrieved from the RRF is always ready – Operand that waits for another Uop to generate its value, will “lesson” to the WB bus. When the value appears on the bus (the value is always associated with the ROB number it needs to update), all RS entries how need to consume this value, “still” it from the bus and mark the input as ready (this is done in parallel to the ROB update. – Uops whose all operands are ready can be dispatched for execution – Dispatcher chooses which of the ready uops to execute next. If can also do “forwarding”; i.e., schedule the instruction at the same cycle the information is written to the RS entry.  As soon as Uop completes its execution, it is deleted from the RS.  If the RS is full, it stalls the decoder

18 Lecture 9 – OOO execution © Avi Mendelson, 5/ Memory Order Buffer (MOB)  Goal – Manipulates the Load and Store operations. If possible, it allows out-of-order among memory operations  Structure similar in concept to ROB  Every memory uop allocates new entry in-order.  Address need to be updated when known  Problem- Memory dependencies cannot be fully resolved statically (memory disambiguation) – store r1,a; load r2,b  can advance load before store – store r1,[r3]; load r2,b  load should wait till r3 is known  In most of the modern processors, Loads may pass loads/stores but Stores must be execute in order (among stores).  For simplicity, this course assumes that all MOB operations are executed in order.

19 Lecture 9 – OOO execution © Avi Mendelson, 5/ An example of OOO Execution

20 Lecture 9 – OOO execution © Avi Mendelson, 5/ Instruction Q MOB RS ROB Execute Retire RAT R0 R1 R2 R3

21 Lecture 9 – OOO execution © Avi Mendelson, 5/ Instruction Q MOB RS ROB Execute Retire RAT R0 R1 R2 R3 LD R1,X R2 <- R3 R1 <- R1+R0  I4 I5

22 Lecture 9 – OOO execution © Avi Mendelson, 5/ Instruction Q MOB RS ROB Execute Retire RAT R0 R1 R2 R3 LD R1,X R2 <- R3 R1 <- R1+R0  LD R1,X RB0 M0 LD RB0,X I4 I5 Takes 3 cycles

23 Lecture 9 – OOO execution © Avi Mendelson, 5/ Instruction Q MOB RS ROB Execute Retire RAT R0 R1 R2 R3 LD R1,X R2 <- R3 R1 <- R1+R0  LD R1,X RB0 M0 LD RB0,X R2 <- R3 RB1 RB1 <- R3 RS0 I4 I5

24 Lecture 9 – OOO execution © Avi Mendelson, 5/ Instruction Q MOB RS ROB Execute Retire RAT R0 R1 R2 R3 LD R1,X R2 <- R3 R1 <- R1+R0  LD R1,X RB2 M0 LD RB0,X R2 <- R3 RB1 RB1 <- R3 RS0 R1 <- R1+R0 RB1 <- R3 RS1 I4 RB2 <- RB0+R0 I5

25 Lecture 9 – OOO execution © Avi Mendelson, 5/ Instruction Q MOB RS ROB Execute Retire RAT R0 R1 R2 R3 LD R1,X R2 <- R3 R1 <- R1+R0  LD R1,X RB2 M0 LD RB0,X R2 <- R3 RB1 O.K R1 <- R1+R0 RS1 I4 RB2 <- RB0+R0 I5 Got the value now I4 Cannot execute since the data is not ready yet

26 Lecture 9 – OOO execution © Avi Mendelson, 5/ Instruction Q MOB RS ROB Execute Retire RAT R0 R1 R2 R3 LD R1,X R2 <- R3 R1 <- R1+R0  LD R1,X RB2 OK R2 <- R3 RB1 OK R1 <- RB0+R0 RS1 I4 I5 I4 RB2 <- RB0+R0 RS2 I4 I5 RS3 I5

27 Lecture 9 – OOO execution © Avi Mendelson, 5/ Instruction Q MOB RS ROB Execute Retire RAT R0 R1 R2 R3 LD R1,X R2 <- R3 R1 <- R1+R0  LD R1,X RB3 R2 <- R3 R1 <- RB1+R0 OK I4 I5 I4 I5 I6 I4 I5 I4I5 I6 rs2 rs3 rs0

28 Lecture 9 – OOO execution © Avi Mendelson, 5/ Backup


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