Transcript In class notes

Pipelining
• Two forms of pipelining
– Instruction unit
• overlap fetch-execute cycle so that multiple instructions are being
processed at the same time, each instruction in a different portion of the
fetch-execute cycle
– Operation unit
• overlap execution of ALU operations
• only useful if execution takes > 1 cycle
– e.g., floating point operations
• We will concentrate mostly on instruction unit-level
• Terms
– Stage – a portion of the pipeline that can accommodate one
instruction, the length of the pipeline is in stages
– Throughput – how often the pipeline delivers a completed
instruction, our goal is 1.0 (or less!), that is, one instruction
leaves the pipeline at the end of each clock cycle (this would
give us an ideal CPI of 1.0)
– Stall – the need to postpone instructions from moving down the
pipeline, stalls are caused by hazards
From Nonpipelined to
Pipelined
Latches are registers,
Denoted as IF/ID.IR
And ID/EX.A for example
We add latches between
stages to control when
the instruction can move
into the new pipe stage
Latches will also contain
logic which will be used to
handle forwarding and insert
stalls
MIPS Pipeline
• The MIPS pipeline is a 5-stage pipeline
– Performance = (n + k – 1 + s) * overhead
•
•
•
•
n = number of instructions
k = 5 (number of stages)
s = stalls, number of stalls inserted is based on the code
overhead = the pipeline latency, which primarily is the time it
takes for the logic in the latches to compute as well as extra
time to open latches, etc
Problems with the Pipeline
• Two instructions might try to alter the PC at the same time
– PC incremented in IF
– A branch instruction in EX could alter the PC
• Two instructions could attempt to access memory at the
same time
– Instruction fetch in IF and data access in MEM
– We will use two separate caches to avoid this problem
(instruction cache accessed by IF stage, data cache accessed by
MEM stage)
• The stages differ in the time it takes to perform their
operation
– IF and MEM are the longest due to cache access time, so we
have to slow the clock speed down to this rate
• Hazards
– Covered in a bit, these result in stalls which lengthens the CPI
from an ideal CPI of 1 to something larger (1 +
stalls/instruction)
Pipelined vs Non-pipelined MIPS
• Assume 1 ns (1 GHz) clock speed but the pipeline
accrues an added .2 ns overhead
• Assume benchmark of
– 40% ALU, 20% branches, 40% for loads and stores
– CPI for unpipelined machine is 4 cycles for ALU and
branches and 5 for loads and stores
• How much faster is the pipelined machine
assuming no stalls?
– Non-pipelined machine has average CPI = .40 * 5 +
.60 * 4 = 4.4
– Pipelined machine has CPI = 1
– Non-pipelined CPU time = 1 ns * 4.4 * IC
– Pipelined CPU time = 1.2 ns * 1 * IC
• Pipelined machine is faster by 4.4 / 1.2 = 3.7
Another Example
• For a non-pipelined version of MIPS, there is
no reason to tune each stage to the same time
– Assume IF & MEM take 1 ns each, ID, WB take .7
ns and EX takes .8 ns
– For the pipelined version, we set the system clock
speed at the longest stage, 1 ns, and assume an
additional .2 ns overhead
• What is the speedup of our pipelined machine?
– Non-pipelined machine executes 1 instruction n 1
+ .7 + .8 + 1 + .7 = 4.2 ns
– Pipelined machine averages 1.2 ns per instruction
– Speedup = 4.2 / 1.2 = 3.5
Another Example
• Assume MIPS unpipelined has a CPI = 3.85 for a given benchmark
and the ideal CPI for pipelined MIPS = 1
• Assume MIPS pipelined has clock cycle time 1.1 times greater than
MIPS unpipelined due to overhead
• Plot the speedup of the MIPS pipelined over the MIPS unpipelined
machine
– for stalls that range from 0 per instruction up to 2 per instruction by units
of .1
• How many stalls / instruction must occur for the two machines
performances to become equal?
The pipelines have equal
Performance when
4
3.5
Speedup
3
2.5
3.85 = 1.1 * (1 + stalls)
2
1.5
Or stalls = 3.85 / 1.1 – 1 =
2.5 stalls per instruction
1
0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
Stalls / Instruction
1.4
1.6
1.8
2
Structural Hazards
• We have already resolved one structural hazard: two
possible cache accesses in one cycle
– This would arise any time we have a load/store instruction
• As it moves down the pipeline and reaches the MEM stage, it would
conflict with the next instruction fetch
• Assuming 35% loads and 15% stores in a program, half of the instructions
would cause this hazard requiring a stall, this would introduce .5 stalls per
instruction or an overall CPI of 1.5!
• We avoid this with 2 caches
• The other source of structural hazard occurs in the EX stage
if an operation takes more than 1 cycle to complete
– We cannot have the next instruction move into EX if the current
instruction is still there
– This happens with longer ALU operations: multiplication,
division, floating point operations
– We will resolve this problem later when we add FP to our
pipeline, for now, assume all ALU operations take 1 cycle
Data Hazards
• The data hazard arises when a value is needed in a later
instruction earlier in the pipeline
– For instance, if we have a LD R1, 0(R2) followed by
DADDI R3, R1, #1, the LD reaches the MEM stage after
the DADDI reaches the ID stage (where it retrieves R1
from the register file)
– We need to stall the DADDI by 3 cycles!
• LD:
• DADDI:
IF ID EX MEM WB
IF stall stall stall ID …
– Another source of data hazard is when two consecutive
ALU operations access the same register, the first
producing the result for the second
• DADD R1, R2, R3:
• DSUB R4, R5, R1:
IF ID EX MEM WB
IF stall stall stall ID …
– Yet another source is an ALU operation which produces a
result used in a branch
• DSUBI R1, R1, #1
• BNEZ R1, top
Data Hazards
Solutions
• We will implement 3 solutions to data hazards
– First, we will only access registers in the first half of
the cycle in WB and the second half of the cycle in ID
• this permits an instruction to place a result in the register file
and in the same cycle another instruction can read the same
register to get the new value
– Second, we will implement forwarding (covered in the
next slide)
• this will shunt a value directly from the ALU as output
directly into the ALU as input
• this will shunt a value received from memory directly into
the ALU as input or directly back to memory
– Third, we will let the compiler fill any remaining stalls
with neutral instructions, this is called compiler
scheduling
Forwarding
• Forwarding can handle ALU to ALU data
dependencies, MEM to ALU data dependences
and MEM to MEM data dependencies
Logic in the ID/EX stage
determines if forwarding
is needed as follows
If source register in ID/EX
= destination register in
EX/MEM or MEM/WB
then forward
See figure C.26 on page C-40
for the full list of forwarding
situations
Forwarding Examples
Notice that
The DADD
And OR do
Not require
Forwarding
Since the
WB write
happens
before the
ID read
Forwarding is Not Enough
• Forwarding will resolve the following situations:
– DADDI R1, R1, #4
– DSUBI R2, R1, R3
• the value of R1 is passed from ALU output to ALU input
– LD R1, 0(R3)
– SD R1, 0(R4)
• the value of R1 is passed from MEM output to MEM input
– DSUBI R1, R1, #1
– BNEZ R1, foo
• the value of R1 is passed from ALU output to ALU input
• It does not resolve these problems
– LD R1, 0(R2)
– DADDI R1, R1, #1
IF ID EX MEM WB
IF ID ….
EX
• the value of R1 is available at the end of MEM but needed in DADDI at the
beginning of EX
– LD R1, 0(R2)
– BNEZ R1, foo
• same
Stalling or Scheduling
• To resolve the last two forms of data hazard, the pipeline has to
either stall the latter instruction or the compiler needs to perform
scheduling
– For a stall, the ID/EX latches look to see if one of the source registers
in this instruction (entering EX) is the same as an instruction entering
MEM, if so, then a 1 cycle stall is inserted, causing the latches in
ID/EX to remain closed
• The compiler can be written to resolve as many of these hazards
as possible by finding an independent instruction (one that does
not use this source/destination register) to place in between the
two dependent instructions
– Consider for example the following code which loads two data from
arrays and adds them together, the code on the right removes all stalls
•
•
•
•
•
•
LD R1, 0(R2)
DADDI R1, R1, #1
LD R3, 0(R4)
DADDI R3, R3, #1
DADD R5, R1, R3
SD R5, 0(R6)
LD R1, 0(R2)
LD R3, 0(R4)
DADDI R1, R1, #1
DADDI R3, R3, #1
DADD R5, R1, R3
SD R5, 0(R6)
Impact of Stalls
• Assume a benchmark of 35% loads, 15% stores, 10% branches,
40% ALU operations
– Of the loads, 50% of the loaded values are used immediately
afterward
– Of the ALU operations, 25% are used immediately afterward either in
other ALU operations, stores or branches
• Without coordinating the ID/WB stages, forwarding or
scheduling, all stalls result in 3 cycle penalties
– Number of stalls per instruction = .35 * .50 * 3 + .40 * .25 * 3 = .825,
or a CPI of 1.825
• With coordinating the ID/WB stages and forwarding, stalls are
reduced to 1 cycle for LD  ALU and LD  Branch operations
– Number of stalls per instruction = .35 * .50 * 1 = .175, or a CPI of
1.175
• Assuming an optimizing compiler can schedule half of these
situations, number of stalls per instruction = .0875 or a CPI of
1.0875
Branch Hazards
• The last form of stall occurs with any branch that is
taken
– Unconditional branches are always taken
– Conditional branches are taken when the condition is true
• Why is the branch a problem?
– Branch conditions (conditional branches) and branch target
locations (PC + offset) are both computed in the EX stage
(we do not reset the PC until the MEM stage, but let’s
move that MUX into the EX stage to further reduce the
penalty by 1)
• We have a 2 cycle penalty because we fetched two instructions in
the meantime (one is in IF, one is in ID)
– If the branch is taken, those 2 instructions need to be
flushed from the pipeline, thus taken branches cause a
penalty of 2 cycles
– There are several ways to handle the 2 cycle penalty, both
through hardware and software
Branch Penalty
If the branch is taken, instructions i+1 and i+2 should not have been fetched,
but we do not know this until instruction i completes its EX stage
If the branch is not taken, i+1 and i+2 would need to be fetched anyway,
no penalty
MIPS Solutions to the Branch Penalty
• Hardware solution
– There is no particular reason why the PC + offset and
condition evaluation have to wait until the EX stage
– Let’s add an ADDER to the ID stage to do PC + offset
– We can also move the zero tester into the ID stage so that
the comparison takes place after registers are read
• recall that the ID stage is one of the two shortest (time-wise) in
the pipeline, we should have enough time in this stage to read
from registers and do the zero test
– If the branches are now being determined in the ID stage,
it reduces the branch penalty to 1
• Software solution
– The compiler can try to move a neutral instruction into
that penalty location, known as the branch delay slot
• The new IF and ID stages are
shown to the right
Continued
– The PC + Offset is computed
automatically
– A MUX is used to select which
PC value should be used in the
next fetch, PC + 4 or PC +
Offset, this is based on two
decisions
• is the instruction in ID a branch
and if the instruction is a
conditional branch, is the
condition true? if so, use PC +
Offset
– We simplified our MIPS
instruction set so that the only
two branches are BEQZ and
BNEZ, that is, an integer
register is simply tested against
0, this can be done quickly (in
essence, all bits are NORed
together)
One consequence of this new
architecture is a new source of stall
LW R1, 0(R2)
BEQZ R1, foo // 2 cycles
DSUBI R1, R1, #4
BNEZ R1, foo // 1 cycle stall
Filling the Branch Delay Slot
• The compiler will look
for a neutral instruction
to move down into the
branch delay
– A neutral instruction is
one that does not impact
the branch condition, nor
produces a value that is
used by an instruction
between it and the branch
• If a neutral instruction
can not be found, there
are two other possible
instructions that could be
sought, neither of which
are safe in that, if the
branch is mispredicted,
the instruction would
have to be flushed
Above, (a) is always safe, (b) and
(c) are not, depending on how aggressively
the compiler is set up, it may try to
schedule (b) and (c) type instructions or
not
Impact of Branch Hazards
• Assume a benchmark of 35% loads, 15% stores, 40%
ALU operations, 8% conditional branches and 2%
unconditional branches
– What is the impact on branch hazards if
• we use the original MIPS pipeline with no compiler scheduling
• we use the new MIPS pipeline with no compiler scheduling
• we use the new MIPS pipeline where compiler scheduling can
successfully move a neutral instruction (type a) into the branch
delay slot 60% of the time
– 10% of instructions are branches
• original pipeline has a penalty of 2 cycles per branch, our CPI
goes from 1.0 to 1.0 + 10% * 2 = 1.2
• new pipeline has a penalty of 1 cycle per branch, our CPI goes
from 1.0 to 1.0 + 10% * 1 = 1.1
• new pipeline plus scheduling, our CPI goes from 1.0 to 1.0 +
10% * 40% * 1 = 1.04
Scheduling Examples
Loop: LD
R1, 0(R2) IF ID EX MEM
DADDI R1, R1, #1
IF ID s
SD
R1, 0(R2)
IF
s
DADDI R2, R2, #4
DSUB R4, R3, R2
BNEZ
R4, Loop
branch delay (LD or next instruction sequential)
WB
EX
ID
IF
MEM
EX
ID
IF
WB
MEM
EX
ID
IF
WB
MEM WB
EX
MEM
s
ID
s
WB
EX MEM
IF
…
WB
• Stalls arise after the LD (data hazard), after the DSUB (data hazard
caused by moving the branch computation to ID) and after the BEQZ
(branch hazard)
• Below, the code has been scheduled by the compiler to remove all stalls
with the SD filling the branch delay slot
Loop: LD
R1, 0(R2)
DADDI
R2, R2, #4
DSUB
R4, R3, R2
DADDI
R1, R1, #1
BNEZ
R4, Loop
SD
R1, -4(R2)
IF
ID
EX
MEM WB
IF
ID
EX
MEM WB
IF
ID
EX
MEM WB
IF
ID
EX
MEM WB
IF
ID
EX
MEM WB
IF
ID
EX
MEM WB
Branches in Other Pipelines
• In some pipelines, the stage where the target PC value is
computed occurs earlier than the stage in which the
condition is determined
– This is in part due to the computation of PC + offset being
available earlier
– The condition is usually a test that requires one or more
registers be read first, whereas PC and offset are already
available, so the PC + offset occurs earlier than say R1 == 0 or
R2 != R3
– Thus, in some pipelines, we might implement “assume taken”,
immediately changing the PC as soon as possible, and then
canceling the incorrectly fetched instruction if the branch is not
taken
• Why assume taken for conditional branches?
– In loops, the conditional branch is typically taken (to branch
back to the top of the loop) and perhaps 50% of conditional
branches are taken for if and if-else statements, so we might
assume a conditional branch is taken 60-70% of the time
Example
• The MIPS R4000 pipeline is 8 stages where branch target
locations are known in stage 3 and branch conditions are
evaluated in stage 4
conditional
branch not
taken
unconditional
branch
Predict taken
Predict not taken
2
2
conditional
branch taken
3
0
2
3
– Assume a benchmark with 4% unconditional branches, 6%
conditional branches not taken and 70% conditional branches taken
– Predict taken penalty = .04 * 2 + .06 * 3 + .06 * .70 * 2 = .344
– Predict not taken penalty = .04 * 2 + .06 * 0 + 06 * .70 * 3 = .206
• This argues that, like MIPS, assuming a branch is not taken
makes more sense than assuming branches are taken
– However, this may not be the case in even longer pipelines or for
benchmarks that have more conditional branches and fewer
unconditional branches – we will visit this in some example problems
out of class
Adding Floating Point to MIPS
• FP operations take longer than integer
– Even a FP addition takes more time because we have to
normalize both numbers (line up their decimal point) and then
put them back into FP notation when done with the operation
– We could either lengthen the clock cycle time
• this impacts all operations
– Or alter our EX stage to handle variable lengths
– We choose the latter approach as it has less impact on the
CPU’s performance although it causes new problems with
handling exceptions
• We will replace the current EX stage with a 4-device EX
stage
–
–
–
–
The integer ALU
An FP adder
An FP multiplier (which will also be used for int multiplies)
An FP divider
New Pipeline
• The integer EX unit will
still complete all
instructions in 1 cycle
• The EX adder will take 4
cycles
• The EX/int multiplier
will take 7 cycles
• The EX/int divider will
take 25 cycles
• The adder and multiplier
will be pipelined
– No need to pipeline the
integer unit
– The divider will not be
used often enough to
warrant it being
pipelined
Functional Unit
Latency
Integer ALU
0
Data Memory
1
FP Add
3
FP/Int Multiply
6
FP/Int Divide/Sqrt
24
Initiation Interval
1
1
1
1
25
Pipelining FP Adder and Multiplier
New Complications
• Forwarding is still available from M7/A4/Div/Ex to
Ex/M1/A1/Div but more data hazard stalls may be needed
• What happens if two instructions reach MEM at the same time?
• What happens if a later instruction reaches MEM before an
earlier instruction? (out of order completion)
• What happens if 2 divisions occur within 25 cycles of each
other?
• What happens if an earlier instruction raises an interrupt after a
later instruction leaves the pipeline?
The Need for Stalls
L.D
MUL.D
ADD.D
S.D
F3, 0(R2) IF
F0, F3, F6
F2, F0, F8
F2, 0(R2)
ID
IF
EX
ID
IF
MEM WB
stall M1
stall ID
IF
M2 M3 M4 M5 M6 M7 MEM WB
stall stall stall stall stall stall A1
A2
stall stall stall stall stall stall ID
EX
A3
A4
MEM WB
stall stall MEM WB
• Above is a timing diagram for four FP operations
– MUL.D stalled 1 cycle because of the data hazard with L.D
– MUL.D produces a result for the ADD.D, the MUL.D takes 7 EX
cycles, so the ADD.D is stalled for 6 additional cycles
– ADD.D produces a result for the S.D, the ADD.D takes 4 EX cycles
(although the S.D doesn’t need the result until the beginning of its
MEM stage), so 2 stalls (instead of 3)
– But now notice how the ADD.D and S.D collide in the MEM stage
– We could permit this if we realize the ADD.D does nothing in MEM
and S.D does nothing in WB, otherwise we would have to stall the
S.D 1 additional cycle
• Both structural hazards (colliding in the MEM stage) and need
for forwarding and stalls is handled in the ID stage as before,
but now the possible situations is more complex, there are more
possibilities to check so this requires a greater amount of logic
in the ID/EX latches
Another Problem – WAW Hazards
• We classify data hazards into three categories
– RAW (read after write) – this is the type we have seen
earlier, an instruction needs to read from a register but
the write takes place later, so we have to use
forwarding and stalls to enforce the read after the write
– WAW (write after write) – the same register is written
to by 2 instructions, but the order of the writes happen
in the wrong order
– WAR (write after read) – this does not happen in MIPS
and we will hold off on discussing it for now
• The WAW hazard could not arise in the 5-stage
MIPS pipeline but could happen in the floating
point pipeline
MUL.D
ADD.D
F0, F1, F2 IF
F0, F3, F4
ID
IF
M1
ID
M2
A1
M3
A2
M4
A3
M5
A4
M6 M7
MEM WB
MEM WB
Handling WAW Hazards
• The WAW hazard should not happen because from a coding
perspective, its like doing this:
– x = y * 5;
– x = z + 1;
– With no instructions in between the two, the first instruction
makes no sense
• However, WAW hazards can arise because of the
optimizing compiler performing scheduling, dealing with
branch delay slots, and especially branch delays filled with
instructions that might not be safe
• The solution in MIPS – when discovered, shut off the
earlier instruction
– Even though the earlier instruction will take longer to execute,
we do not let it write its results to the register because the later
instruction’s result is the only one that matters
Another Issue: Exceptions (Interrupts)
• In a non-pipelined machine, interrupts are handled at the
end of each fetch-execute cycle
– But where do we handle them in a pipelined machine?
• In a non-pipelined machine, to handle an interrupt, the
current register values are saved (e.g., PC, IR, etc)
– But in a pipelined machine, there are multiple instructions and so
multiple register values (which PC do we save? Recall that a
branch instruction midway through the pipeline might have
altered the PC already!)
• Exceptions are somewhat simplified in MIPS due to the
division of functions performed in each stage
–
–
–
–
–
IF: page fault, memory violation, misaligned memory access
ID: undefined or illegal op code
EX: arithmetic exception
MEM: same as IF
WB: none
• This list does not include break points or hardware
interrupts
Simple Solution
• At the stage an interrupt arises
– Shut down all register writes and memory writes for
instructions from that point back to the beginning of the
pipeline
• Instructions further down the pipeline can complete (MEM and
WB stages)
– Insert a TRAP instruction in the next IF stage rather than
an instruction fetch
– Save the PC of the faulting instruction when the TRAP is
executed
• Problem: what if the faulting instruction is in the
branch delay slot? In such a case, if the branch is
taken, the PC value is already replaced with the
branch target location
– To resolve this problem, we can pass the old PC value
down the pipeline in the latches
Multiple Out of Order Exceptions
• Consider the following two instructions
LD
DADD
R1, O(R5) IF
R2, R3, R4
ID
IF
EX MEM WB
ID EX
MEM WB
• What if the LD raises an exception in the MEM stage
and the DADD raises an exception earlier in time in
the IF stage or at the same time in the EX stage?
– We should handle all exceptions in the order they arise in
terms of the sequential order of the instructions, not
temporally
– Thus, we should only handle an exception when the
instruction reaches the WB stage to force exceptions to be
handled in order
– Similar to passing down the PC value of each instruction
as it moves down the pipeline, we also pass down a status
vector regarding possible interrupts
Other Concerns
• Handling exceptions is trickier in longer pipelines
– There might be multiple stages where registers can be written to
or where memory can be written to at different times of
instruction execution
– Pipelines that have variable-length execution units can have outof-order instruction completion as with the MIPS FP pipeline
• what happens if an instruction which takes longer to execute raises an
interrupt after a later instruction completes?
• we will visit solutions to this problem later in the semester
– If condition codes are used, they also have to be passed down
the pipeline
• A precise exception is one that can be handled as if the
machine were not pipelined, but pipelined machines may
not be able to easily handle precise exceptions
– Some pipelines use two modes, imprecise modes in which
exceptions can be handled out of order (which may lead to
errors) or precise mode which may cause a slower performance