1. 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

2. From Non-pipelined 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

3. 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

4. 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)

5. 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 * * 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

6. 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 = 4.2 ns
Pipelined machine averages 1.2 ns per instruction
Speedup = 4.2 / 1.2 = 3.5

7. 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
3.85 = 1.1 * (1 + stalls)
Or stalls = 3.85 / 1.1 – 1 =
2.5 stalls per instruction

8. 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

9. 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: IF ID EX MEM WB
DADDI: 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: IF ID EX MEM WB
DSUB R4, R5, R1: 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

10. Data Hazards

11. 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

12. 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

13. Forwarding Examples Notice that The DADD And OR do Not require
Since the
WB write
happens
before the
ID read

14. 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
It does not resolve these problems
LD R1, 0(R2) IF ID EX MEM WB
DADDI R1, R1, # 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)
same

15. 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) LD R1, 0(R2)
DADDI R1, R1, #1 LD R3, 0(R4)
LD R3, 0(R4) DADDI R1, R1, #1
DADDI R3, R3, #1 DADDI R3, R3, #1
DADD R5, R1, R3 DADD R5, R1, R3
SD R5, 0(R6) SD R5, 0(R6)

16. 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 * * .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 = or a CPI of

17. 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

18. 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

19. 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

20. Continued The new IF and ID stages are shown to the right
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

21. 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

22. 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 % * 2 = 1.2
new pipeline has a penalty of 1 cycle per branch, our CPI goes from 1.0 to % * 1 = 1.1
new pipeline plus scheduling, our CPI goes from 1.0 to % * 40% * 1 = 1.04

23. Scheduling Examples
Loop: LD
R1, 0(R2) IF ID EX
MEM WB
DADDI
R1, R1, #1 s SD
R2, R2, #4
DSUB
R4, R3, R2
BNEZ
R4, Loop
branch delay (LD or next instruction sequential) …
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) IF ID EX
MEM WB
DADDI
R2, R2, #4
DSUB
R4, R3, R2
R1, R1, #1
BNEZ
R4, Loop SD
R1, -4(R2)

24. 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

25. 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
Assume a benchmark with 4% unconditional branches, 6% conditional branches not taken and 70% conditional branches taken
Predict taken penalty = .04 * * * .70 * 2 = .344
Predict not taken penalty = .04 * * * .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
unconditional branch
conditional branch not taken
conditional branch taken
Predict taken 2 3
Predict not taken

26. 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

27. 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

28. Pipelining FP Adder and Multiplier

29. 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?

30. The Need for Stalls 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

31. 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

32. 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

33. 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

34. 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

35. Multiple Out of Order Exceptions
Consider the following two instructions
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

36. 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 out-of-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