Computer Organization and Design /fj^ The Hardware/Software Interface ^hi
instruction memory, fetch instruction Register numbers -> register file, read registers Depending on instruction class
■ Use ALU to calculate
Arithmetic result
Memory address for load/store
Branch target address
■ Access data memory for load/store
■ PC <- target address or PC + 4
M 14
Chapter 4 — The Processor — 3
CPU Overview
Add
PC pH Address Instruction
Instruction memory
Add.
Data
Register #
Registers
Register #
W Register #
Address
Data memory
Data
M 14
Chapter 4 — The Processor — 4
Multiplexers
Add
PC
Address Instruction
Instruction memory
Add.
Can't just join wires together
■ Use multiplexers
Data
Register #
Registers
Register #
W Register #
Address
Data memory
Data
Chapter 4 — The Processor — 5
Control
M
u x
Branch
Add
>
pc pH Address Instruction
Instruction memory
Add,
Data
Register #
Registers
Register #
Data
Zero ALU ALU
result
RegWrite
b. ALU
Chapter 4 — The Processor — 14
Load/Store Instructions
Read register operands
Calculate address using 16-bit offset
■ Use ALU, but sign-extend offset
Load: Read memory and update register Store: Write reaister value to memorv
a. Data memory unit
b. Sign extension unit
M 14
®
Chapter 4 — The Processor — 15
Branch Instructions
Read register operands Compare operands
■ Use ALU, subtract and check Zero output
Calculate target address
■ Sign-extend displacement
■ Shift left 2 places (word displacement) - Add to PC + 4
Already calculated by instruction fetch
M 14
Chapter 4 — The Processor — 16
Branch Instructions
Just re-routes wires
PC+4 from instruction datapath
Instruction
Read
register 1 Read
Read data 1
register 2
Write Registers
register Read
Write data 2
data
<
ALU operation
Branch target
To branch control logic
Sign-bit wire replicated
Chapter 4 — The Processor
Composing the Elements
First-cut data path does an instruction in one clock cycle
■ Each datapath element can only do one function at a time
■ Hence, we need separate instruction and data memories
Use multiplexers where alternate data sources are used for different instructions
M 14
Chapter 4 — The Processor — 18
R-Type/Load/Store Datapath
Instruction
Read
register 1 Read
Read data 1
register 2
Registers Write register Read data 2
Write
data
ALU operation
Address
MemWrite
MemtoReg
Read data
Write data
Data memory
1
M u
x
MemRead
M 14
Chapter 4 — The Processor — 19
Full Datapath
pc
\Add
PCSrc _^
■ , , ALU
Add result
M u
x
Read address
Instruction
Instruction memory
11—**
Read register 1
Read data 1
Read register 2
Write Re9jsters Read
ALUSrc 4 J ALU operation
Zero ALU ALU
result
MemWrite
Address Read data
Write Data
data memory
MemRead
MemtoReg
M u
x
W
Chapter 4 — The Processor — 20
ALU Control
ALU used for
■ Load/Store: F = add
■ Branch: F = subtract
■ R-type: F depends on funct field
ALU control Function
0000 AND
0001 OR
0010 add
0110 subtract
0111 set-on-less-than
1100 NOR
M 14
Chapter 4 — The Processor — 21
ALU Control
Assume 2-bit ALUOp derived from opcode
■ Combinational logic derives ALU control
opcode ALUOp Operation funct ALU function ALU control
Iw 00 load word xxxxxx add 0010
sw 00 store word xxxxxx add 0010
beq 01 branch equal xxxxxx subtract 0110
R-type 10 add 100000 add 0010
subtract 100010 subtract 0110
AND 100100 AND 0000
OR 100101 OR 0001
set-on-less-than 101010 set-on-less-than 0111
M 14
Chapter 4 — The Processor — 22
The Main Control Unit
Control signals derived from instruction
R-type
Load/ Store
Branch
opcode
read, except for load
0 rs rt rd shamt funct
31:26 25:21 20:16 U 5:11 10:6 5:0
35 or 43 rs rt address
31:26 25:21 20:16 \ \ 15:0
4 rs rt \ address
31:26 v. j 25:21 20:16 \ V J 15:0 \
V V
write for R-type and load
sign-extend and add
M 14
Chapter 4 — The Processor — 23
Datapath With Control
>
Add
PC
Read address
Instruction [31-0]
Instruction memory
Instruction [31-26]
RegDst Branch
Mjeft2/^
Add
ALU
result
MemRead
MemtoReg
ALUOp
MemWrite
ALUSrc
Instruction [25-21]
Instruction [20-16]
Instruction [15-11]
M u
X
Instruction [15-0]
<
RegWrite
Read
register 1 Read
Read data 1 register 2
Write register
Read data 2
Write
data Registers
Instruction [5-0]
M u x
>
Zero I—I ALU ALU
Address
Read data
Write
Data
data memorV
M u
X
0
Chapter 4 — The Processor — 24
R-Type Instruction
>
Add
PC
Read address
Instruction [31-0]
Instruction memory
>
Instruction [31-26]
RegDst Branch
* ALU /'Add result
^/shiftV
Hjeft^T
MemRead
MemtoReg
ALUOp
MemWrite
ALUSrc
Instruction [25-21]
Instruction [20-16]
Instruction [15-11]
M u
X
Instruction [15-0]
<
RegWrite
ZD_
Read
register 1 Read
Read data 1 register 2
Write register
Read data 2
Write
data Registers
Instruction [5-0]
M u x
>
Zero I—I ALU ALU
Address
Read data
Write
Data
data memory
M u
X
0
Chapter 4 — The Processor — 25
Load Instruction
>
Add
PC
Read address
Instruction [31-0]
Instruction memory
Instruction [31-26]
RegDst Branch
^left^T
>
Add
MemRead
MemtoReg
ALUOp
MemWrite
ALUSrc
Instruction [25-21]
Instruction [20-16]
Instruction [15-11]
M u
X
Instruction [15-0]
<
RegWrite
Read
register 1 Read
Read data 1 register 2
Write register
Read data 2
Write
data Registers
Instruction [5-0]
ALU result
(0*
M u x
>
Zero I—I ALU ALU
Address
Read data
Write
Data
data memory
M u
X
Chapter 4 — The Processor — 26
Branch-on-Equal Instruction
>
Add
PC
Read address
Instruction [31-0]
Instruction memory
Instruction [31-26]
RegDst Branch
Mjeft2/^
\wd
ALU result
MemRead
MemtoReg
ALUOp
MemWrite
ALUSrc
Instruction [25-21]
Instruction [20-16] i-•-
Instruction [15-11]
M u
X
Instruction [15-0]
<
RegWrite
ZD_
Read
register 1 Read
Read data 1 register 2
Write register
Read data 2
Write
data Registers
Instruction [5-0]
(0*
M u x
>
Zero I—I ALU ALU
Address
Read data
Write
Data
data memory
M u
X
Chapter 4 — The Processor — 27
Implementing Jumps
Jump
address
31:26 25:0
Jump uses word address Update PC with concatenation of
- Top 4 bits of old PC ■ 26-bit jump address
- 00
Need an extra control signal decoded from opcode
Chapter 4 — The Processor — 28
Datapath With Jumps Added
Instruction [25-0] /gh-A Jump address [31-0] "Mien'2^
>
Add
PC
Read address
Instruction [31-0]
Instruction memory
26
28
PC+ 4 [31-28]
Instruction [31-26]
RegDst Jump
Branch
Control
MemRead
MemtoReg
ALUOp
MemWrite
ALUSrc
Reg Write
Instruction [25-21]
Instruction [20-16]
Instruction [15-11]
Instruction [15-0]
<
Read
register 1 Read
Read data 1 register 2
^ Write
reg i
Read ster data2
U Write
data Registers
16
Instruction [5-0]
M u
X
(0^
>
ZeroM
lualu
result
M u x
Address
Read data
Write Data data memory
M
u
X
Chapter 4 — The Processor — 29
Performance Issues
Longest delay determines clock period
■ Critical path: load instruction
■ Instruction memory -» register file -» ALU -> data memory ->• register file
Not feasible to vary period for different instructions
Violates design principle
■ Making the common case fast
We will improve performance by pipelining
M 14
Chapter 4 — The Processor — 30
Pipelining Analogy
■ Pipelined laundry: overlapping execution
■ Parallelism improves performance
Time
Task order
A B C D
Time -
Task order
A B C D
6PM 7 8 9 10 11 12 1 2AM
Q
Q
□
ffl
□
6PM 7 8 9 10 11 12 1 2AM
□
5
□
1
o
Four loads:
Speedup = 8/3.5 = 2.3
Non-stop:
■ Speedup = 2n/0.5n + 1.5*4 = number of stages
M 14
Chapter 4 — The Processor — 31
MIPS Pipeline
Five stages, one step per stage
1. IF: Instruction fetch from memory
2. ID: Instruction decode & register read
3. EX: Execute operation or calculate address
4. MEM: Access memory operand
5. WB: Write result back to register
M 14
Chapter 4 — The Processor — 32
Pipeline Performance
Assume time for stages is
■ 10Ops for register read or write
■ 200ps for other stages
Compare pipelined datapath with single-cycle datapath
Instr Instr fetch Register read ALU op Memory access Register write Total time
Iw 200ps 100 ps 200ps 200ps 100 ps 800ps
sw 200ps 100 ps 200ps 200ps 700ps
R-format 200ps 100 ps 200ps 100 ps 600ps
beq 200ps 100 ps 200ps 500ps
M 14
Chapter 4 — The Processor — 33
Pipeline Performance
Program execution order (in instructions)
Single-cycle (Tc= 800ps)
Time
200
—i—
400
—i—
600
800
—i—
1000 1200 1400 1600 1800
lw $1, 100($0) lw $2, 200($0) lw $3, 300($0)
Instruction fetch
Reg
ALU
Data access
Reg
800 ps
Instruction fetch
Reg
ALU
Data access
Reg
800 ps
Instruction fetch
Program execution order (in instructions)
Pipelined (Tc= 200ps)
800 ps
Time
200
400
600
800
1000 1200 1400
lw $1, 100($0) Instruction fetch Reg ALU Data access Reg
lw $2, 200($0) 200 ps Instruction fetch Reg ALU Data access Reg
lw $3, 300($0) 200 ps Instruction fetch Reg ALU Data access Reg
<
200 ps 200 ps 200 ps 200 ps 200 ps
Chapter 4 — The Processor
— 34
Pipeline Speedup
If all stages are balanced
■ i.e., all take the same time
■ Time between instructionspipe|ined
= Time between instructionsnonpipe|ined Number of stages
If not balanced, speedup is less Speedup due to increased throughput
■ Latency (time for each instruction) does not decrease
M 14
Chapter 4 — The Processor — 35
Pipelining and ISA Design
MIPS ISA designed for pipelining
■ All instructions are 32-bits
Easier to fetch and decode in one cycle c.f. x86:1- to 17-byte instructions
■ Few and regular instruction formats
Can decode and read registers in one step
■ Load/store addressing
Can calculate address in 3rd stage, access memory in 4th stage
■ Alignment of memory operands
Memory access takes only one cycle
M 14
Chapter 4 — The Processor — 36
Hazards
Situations that prevent starting the next instruction in the next cycle
Structure hazards
■ A required resource is busy
Data hazard
■ Need to wait for previous instruction to complete its data read/write
Control hazard
■ Deciding on control action depends on previous instruction
Chapter 4 — The Processor — 37
Structure Hazards
Conflict for use of a resource
In MIPS pipeline with a single memory
■ Load/store requires data access
■ Instruction fetch would have to stall'for that cycle
Would cause a pipeline "bubble"
Hence, pipelined datapaths require separate instruction/data memories
■ Or separate instruction/data caches
M 14
Chapter 4 — The Processor — 38
Data Hazards
An instruction depends on completion of data access by a previous instruction
- add $s0, $t0, $tl sub $t2, $s0, $t3
Time
add $s0, $t0, $t1
sub $t2, $s0, $t3
200
—i—
400
10
>EX
600 800 1000 1200 1400 1600
i i I i i i *
MEM WB ! LJ..J
bubble/ C bubble/ C bubble/ ( bubble/ ( bubble
bubble/ C bubble/ ( bubble/ ( bubble/ ( bubble
ffl
>EX
MEM
WB
Chapter 4 — The Processor — 39
Forwarding (aka Bypassing)
■ Use result when it is computed
■ Don't wait for it to be stored in a register
■ Requires extra connections in the datapath
Program execution
order Time (in instructions)
add $s0, $t0, $t1
sub $t2, $s0, $t3
M 14
Chapter 4 — The Processor — 40
Load-Use Data Hazard
Can't always avoid stalls by forwarding
■ If value not computed when needed
■ Can't forward backward in time!
Program execution order Time (in instructions)
Iw $s0, 20($t1)
200 400 600 800 1000 1200 1400
—i-1-1-1-1-1-1—
ID
>EX
MEM
WB
Cbubble) Cbubble) Cbubble)\ (/bubble) (/bubble)
sub $t2, $s0, $t3
MEM
WB
Chapter 4 — The Processor — 41
Code Scheduling to Avoid Stalls
Reorder code to avoid use of load result in the next instruction
stall
stall
e for A = B + e; C B + F
lw $tl, 0($t0) lw $tl, 0($t0)
lw 4($t0) lw 4($t0)
add $t3, $tl, rlw 8($t0)
sw $t3, 12($t0) add $t3, $tl,
lw 8($to;r sw $t3, 12($t0)
add $t5, $ti, add $t5, $tl,
sw $t5, 16($t0) sw $t5, 16($t0)
13 cycles
11 cycles
Chapter 4 — The Processor — 42
Control Hazards
Branch determines flow of control
■ Fetching next instruction depends on branch outcome
■ Pipeline can't always fetch correct instruction
Still working on ID stage of branch
In MIPS pipeline
■ Need to compare registers and compute target early in the pipeline
■ Add hardware to do it in ID stage
M 14
Chapter 4 — The Processor — 43
Stall on Branch
Wait until branch outcome determined before fetching next instruction
Program execution order (in instructions)
Time
200
—I—
400
600
800
1000 1200 1400
add $4, $5, $6 beq $1, $2, 40
or $7, $8, $9
Instruction fetch Reg ALU Data access Reg
Instruction fetch Reg ALU Data access Reg
-► 200 ps
400 ps
T
T
Reg ALU Data access Reg
Chapter 4 — The Processor
Branch Prediction
Longer pipelines can't readily determine branch outcome early
■ Stall penalty becomes unacceptable
Predict outcome of branch
■ Only stall if prediction is wrong
In MIPS pipeline
■ Can predict branches not taken
■ Fetch instruction after branch, with no delay
M 14
Chapter 4 — The Processor — 45
MIPS with Predict Not Taken
Prediction correct
Program execution order (in instructions)
Time
200
400
600 800 1000 1200 1400
add $4, $5, $6 Instruction fetch Reg ALU Data access Reg
beq $1, $2, 40 Instruction Reg ALU Data Reg
200 ps fetch access
Instruction fetch Data access
Iw $3, 300{$0) 200 ps Reg ALU Reg
Prediction incorrect
Program execution order (in instructions)
Time
200
400
600
800
1000 1200 1400
add $4, $5, $6 beq $1, $2, 40
Instruction fetch Reg ALU Data access Reg
Instruction Reg ALU Data Reg
200 ps fetch access
or $7, $8, $9
400 ps
Instruction fetch
Reg ALU Data access Reg
Chapter 4 — The Processor — 46
More-Realistic Branch Prediction
Static branch prediction
■ Based on typical branch behavior
■ Example: loop and if-statement branches
Predict backward branches taken Predict forward branches not taken
Dynamic branch prediction
■ Hardware measures actual branch behavior
e.g., record recent history of each branch
■ Assume future behavior will continue the trend
When wrong, stall while re-fetching, and update history
M 14
Chapter 4 — The Processor — 47
Pipeline Summary
The BIG Picture
Pipelining improves performance by increasing instruction throughput
■ Executes multiple instructions in parallel
■ Each instruction has the same latency
Subject to hazards
■ Structure, data, control
Instruction set design affects complexity of pipeline implementation
Chapter 4 — The Processor — 48
MIPS Pipelined Datapath
WB: Write back
Right-to-left flow leads to hazards
Chapter 4 — The Processor — 49
Pipeline registers
Need registers between stages
■ To hold information produced in previous cycle
0 i M
u PC
x
Address
Instruction memory
Read
register 1 Read data 1
Read
register 2
Registers Read
Write data 2
register
Write
data
I Shift \ I left 2 I-*~
result
16 ^ f Sign- \ 3f \ * í extend ^~
M u
x
Salu
ALU result
Read
Address data
Data
memory
Write
data
(A
M
M 14
Chapter 4 — The Processor — 50
Pipeline Operation
Cycle-by-cycle flow of instructions through the pipelined datapath
■ "Single-clock-cycle" pipeline diagram
Shows pipeline usage in a single cycle Highlight resources used
■ c.f. "multi-clock-cycle" diagram
Graph of operation over time
We'll look at "single-clock-cycle" diagrams for load & store
M 14
Chapter 4 — The Processor — 51
IF for Load, Store,
lw
Instruction fetch
M 14
Chapter 4 — The Processor — 52
ID for Load, Store, ...
M 14
Chapter 4 — The Processor — 53
EX for Load
lw
Execution
M 14
Chapter 4 — The Processor — 54
MEM for Load
lw
Memory
M 14
Chapter 4 — The Processor — 55
WB for Load
lw
Write back
0 1
M
u PC
X
Instruction memory
Wrong register number
Read
register 1 Read
data 1
Head
r°qister 2
Registers Read
Write data 2
regis+jr
Write
data
16 / Sign- I
\ ^ 1 extend
32
-V-
Read
Address data
Data
memory
Write
data
M 14
Chapter 4 — The Processor — 56
Corrected Datapath for Load
M
u ^
X
Address
Instruction memory
Head
register 1 Read
data 1
Read
register 2
Registers Read
Write data 2
register
Write
data
1*/ Sign- \ 32 \ extend
/~\ >Add
(am) f ,esu~
I left 2 J-
MEM/WB
Read
Address data
Data
memory
Write
data
M 14
Chapter 4 — The Processor — 57
EX for Store
sw
Execution
MEM/WB
Chapter 4 — The Processor — 58
MEM for Store
sw
Memory
r0 1
M
u PC
X
Instruction memory
Read
register 1 Read
data 1
Read
register 2
Registers Read
Write data 2
register
Write
data
'6 / Sign- 1 -X * I extend
Read
Address data
Data
memory
Write
data
Chapter 4 — The Processor — 59
WB for Store
sw
Write-back
IF/ID
ID/EX
EX/MEM
1^ M
U H
Instruction memory
Read
register 1 Read
data 1
Read
register 2
Registers Head
Write data 2
register
Write
data
16 Sign-
extend
MEM/WB
Read
Address data
□ate
memory
Write
data
M
M 14
Chapter 4 — The Processor — 60
Multi-Cycle Pipeline Diagram
Form showing resource usage
Time (in clock cycles) ->■
CC1 CC2 CC3 CC4 CCS CC 6 CC 7 CC 8 CC 9
Program
execution
order
(in instructions)
lw$10, 20($1)
sub $11, $2, $3
add $12, $3, $4
lw$13, 24($1)
add $14, $5, $6
IM
<
w
-■nReg
_>ALU^ -jJDM —
im I--^Regp
>ALU^
IM [—I pReg_
-Regi
DM
1m]—4
Regp
-J- UM "
-■fieg
—^egi
r DM
r
n
■DM
^egj
Chapter 4 — The Processor — 61
Cycle Pipeline Diagram
Traditional form
Time (in clock cycles) -»
CC1 CC2 CC3 CC4 CC5 CC6 CC7 CC8 CC9
Program
execution
order
(in instructions)
lw$10, 20($1) Instruction fetch Instruction decode Execution Data access Write back
sub $11, $2, $3 Instruction fetch Instruction decode Execution Data access Write back
add $12, $3, $4 Instruction fetch Instruction decode Execution Data access Write back
lw$13, 24($1) Instruction fetch Instruction decode Execution Data access Write back
add $14, $5, $6 Instruction fetch Instruction decode Execution Data access Write back
Chapter 4 — The Processor — 62
90
Single-Cycle Pipeline Diagram
State of pipeline in a given cycle
add $14, $5, $6 Iw $13, 24 ($1) add $12, $3, $4 sub $11, $2, $3 lw$10, 20($1)
Instruction fetch Instruction decode Execution Memory Write-back
ro 1 M
u PC
X
KU
Instruction memory
IF/ID
Read
register 1 Read datal
Read
register 2
Registers Head
Write data 2
register
Write
data
id/ex
I Shift \ I left 2 I-*■
>Add A*
' result
■~^U
EX/MEM
. Zero >LU ALU
mem/wb
Read
Address data
Data
memory
Write
data
\U
M 14
Chapter 4 — The Processor — 63
Pipelined Control (Simplified)
MEM/WB
M 14
Chapter 4 — The Processor — 64
Pipelined Control
■ Control signals derived from instruction
■ As in single-cycle implementation
Instruction
WB
IF/ID
ID/EX
EX/MEM
MEIWWB
M 14
Chapter 4 — The Processor — 65
Pipelined Control
PCSrc
(0s
M
u PC
X
Address
Instruction memory
ID/EX
EX/MEM
Instruction [15-0] 1
Instruction [20-16]
Instruction [15-11]
Read
reqister 1 Read data 1
Read register 2 w.te Registers
register aa,a d
Write data
Branch
MEM/WB
Read
Address data
Data
memory
Write
data
Mem Read
/0\
Chapter 4 — The Processor — 66
Data Hazards in ALU Instructions
Consider this sequence: sub $2, $1,$3
and $12,$2,$5
or $13,$6,$2
add $14,$2,$2
sw $15,100($2)
We can resolve hazards with forwarding
■ How do we detect when to forward?
M 14
Chapter 4 — The Processor —
Dependencies & Forwarding
Time (in clock cycles) -
Value of cc 1 cc 2 cc 3 cc 4 cc 5 cc 6 cc 7 cc 8 cc 9 register $2: 10 10 10 10 10/-20 -20 -20 -20 -20
Program
execution
order
Detecting the Need to Forward
Pass register numbers along pipeline
■ e.g., ID/EX.RegisterRs = register number for Rs sitting in ID/EX pipeline register
ALU operand register numbers in EX stage are given by
- ID/EX.RegisterRs, ID/EX.RegisterRt
Data hazards when
1a. EX/MEM.RegisterRd = ID/EX.RegisterRs 1b. EX/MEM.RegisterRd = ID/EX.RegisterRt 2a. MEM/WB.RegisterRd = ID/EX.RegisterRs 2b. MEM/WB.RegisterRd = ID/EX.RegisterRt
Fwd from EX/MEM pipeline reg
Fwd from MEM/WB pipeline reg
<
Chapter 4 — The Processor — 69
Detecting the Need to Forward
But only if forwarding instruction will write to a register!
. EX/MEM.RegWrite, MEM/WB. Reg Write
And only if Rd for that instruction is not $zero
. EX/MEM.RegisterRd t 0, MEM/WB.RegisterRd#0
M 14
Chapter 4 — The Processor — 70
Forwarding Paths
ID/EX
EX/MEM
MEM/WB
Registers
Rs
M u
X
ForwardA
M u
X
Y
ForwardB
M u
X
ALU
Forwarding V«_ unit /*■
Data memory
EX/MEM. RegisterRd
MEM/WB.RegisterRd
M u
X
b. With forwarding
<
Chapter 4 — The Processor — 71
Forwarding Conditions
EX hazard
. if (EX/MEM.RegWrite and (EX/MEM.RegisterRd * 0) and (EX/MEM.RegisterRd = ID/EX.RegisterRs))
ForwardA = 10
. if (EX/MEM.RegWrite and (EX/MEM.RegisterRd * 0) and (EX/MEM.RegisterRd = ID/EX.RegisterRt))
Forward B = 10
MEM hazard
. if (MEM/WB.RegWrite and (MEM/WB.RegisterRd * 0) and (MEM/WB.RegisterRd = ID/EX.RegisterRs))
ForwardA = 01
. if (MEM/WB.RegWrite and (MEM/WB.RegisterRd * 0) and (MEM/WB.RegisterRd = ID/EX.RegisterRt))
ForwardB = 01
M 14
Chapter 4 — The Processor — 72
Double Data Hazard
Consider the sequence:
add $l,$l,$2
add $1,$1,$3 add $1,$1,$4
Both hazards occur
■ Want to use the most recent
Revise MEM hazard condition
■ Only fwd if EX hazard condition isn't true
M 14
Chapter 4 — The Processor — 73
Revised Forwarding Condition
MEM hazard
. if (MEM/WB.RegWrite and (MEM/WB.RegisterRd * 0)
and not (EX/MEM.RegWrite and (EX/MEM.RegisterRd * 0)
and (EX/MEM.RegisterRd = ID/EX.RegisterRs)) and (MEM/WB.RegisterRd = ID/EX.RegisterRs)) Forward A = 01
. if (MEM/WB.RegWrite and (MEM/WB.RegisterRd * 0)
and not (EX/MEM.RegWrite and (EX/MEM.RegisterRd * 0)
and (EX/MEM.RegisterRd = ID/EX.RegisterRt)) and (MEM/WB.RegisterRd = ID/EX.RegisterRt)) ForwardB = 01
M 14
Chapter 4 — The Processor — 74
Datapath with Forwarding
ID/EX
MEM/WB
PC-
Instruction memory
<
Chapter 4 — The Processor — 75
Load-Use Data Hazard
Time (in clock cycles)
CC1 CC2 CC3 CC4 CC 5 CC6 CC 7 CC 8 CC 9
Program
execution
order
(in instructions) Iw $2, 20($1)
and $4, $2, $5
or $8, $2, $6
add $9, $4, $2
sit $1, $6, $7
<
IM
n
^Reg
Need to stall for one cycle
MRegi
Chapter 4 — The Processor — 76
Load-Use Hazard Detection
Check when using instruction is decoded in ID stage
ALU operand register numbers in ID stage are given by
. IF/ID.RegisterRs, IF/ID.RegisterRt
Load-use hazard when
- ID/EX. Mem Read and
((ID/EX.RegisterRt = IF/ID.RegisterRs) or (ID/EX.RegisterRt = IF/ID.RegisterRt))
If detected, stall and insert bubble
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Chapter 4 — The Processor — 77
How to Stall the Pipeline
Force control values in ID/EX register toO
■ EX, MEM and WB do nop (no-operation)
Prevent update of PC and IF/ID register
■ Using instruction is decoded again
■ Following instruction is fetched again
■ 1-cycle stall allows MEM to read data for lw
Can subsequently forward to EX stage
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Chapter 4 — The Processor — 78
Stall/Bubble in the Pipeline
Time {in clock cycles)-+~
CC1 CC2 CC 3 CC4 CC 5 CC 6 CC 7 CC 8 CC 9 CC10
Program
execution
order
(in instructions) lw$2, 20($1)
and becomes nop
and $4, $2, $5
or $8, $2, $6
add $9, $4, $2
IM
^Reg
IM
Stall inserted here
DM
Reg
Chapter 4 — The Processor — 79
Stall/Bubble in the Pipeline
Time (in clock cycles)-
CC 1 CC2 CC3 CC4 CC 5 CC 6 CC 7 CC 8 CC 9 CC 10
Program
execution
order
(in instructions)
Iw $2, 20($1)
IM -
and becomes nop
IM--RRe>
tJRe^
and $4, $2, $5 stalled in ID
or $8, $2, $6 stalled in IF
r-i-|DM|—I
^Reg
IM -
Reg,
n bubble
U
^Reg
L _
add $9, $4, $2
Or, more accurately.
IM -
Chapter 4 — The Processor —
Datapath with Hazard Detection
Q.
PC
Instruction memory
IR ID
J Hazard detection A unit
ID/EX.MemRead
Registers
IF/ID.RegisterRs
IF/ID.RegisterRt
IF/ID.RegisterRt
IF/ID.RegisterRd
ID/EX. RegisterRt
EX/MEM
M u
X
M u
BL
Rs
V
M u
X
WB
M
Rt
Forwarding unit
MEM/WB
Data memory
WB
M u
X
Chapter 4 — The Processor — 81
Stalls and Performance
The BIG Picture
Stalls reduce performance
■ But are required to get correct results
Compiler can arrange code to avoid hazards and stalls
■ Requires knowledge of the pipeline structure
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Chapter 4 — The Processor — 82
Branch Hazards
If branch outcome determined in MEM
Time (in clock cycles)
CC1 CC 2 CC 3 CC 4 CC 5 CC 6 CC 7 CC 8 CC 9
Program
execution
order
(in instructions)
40 beq $1, $3, 28
w
44 and $12, $2, $5
48 orSl3, $6, $2
52 add $14, $2, $2
72 Iw $4, 50($7)
3Reg
r-p
IM
3Reg
|-|-|dm"—i
~Jm]--°^eg^
im —
-Reg'
Dm[— —Reg
Reg__>
im
PC
3 Reg
Jjm|--Reg|
DM —
Flush these instructions (Set control values to 0)
J
Regj
Chapter 4 — The Processor — 83
Reducing Branch Delay
Move hardware to determine outcome to ID stage
■ Target address adder
■ Register comparator
Example: branch taken
36: sub $10, $4, $8
40: beq $1, $3, 7
44: and $12, $2, $5
48: or $13, $2, $6
52: : add $14, $4, $2
56: : sit $15, $6, $7
72: ■ ■ ■ : lw $4, 50($7)
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Chapter 4 — The Processor — 84
Example: Branch Taken
before<2>
Chapter 4 — The Processor — 85
Example: Branch Taken
if.FIush
Iw $4, 50{$7)
Bubble (nop)
M
u PC
X . J 76
IF,
76
72
Instruction ^mer^^l
,ID
IH- Control L-
73 V___>
Hazard detection unit
Clock 4
Hegisters
beq $1, $3, 7
sub $10,... I before<1>
M
ID/EX
wb|-
EX
M
M
EX/^EM
.11
\
)ALU
$3
10
Forwarding unit
MEM/WB
Data memory
Chapter 4 — The Processor — 86
Data Hazards for Branches
If a comparison register is a destination of 2nd or 3rd preceding ALU instruction
add $1, $2, $3 add $4, $5, $6
IF
ID
EX
MEM
WB
IF
ID
EX
rVIEM
WB
IF
ID
beq $1, $4, target
IF
ID>
MEM
WB
EX
MEM
WB
Can resolve using forwarding
Chapter 4 — The Processor — 87
Data Hazards for Branches
If a comparison register is a destination of preceding ALU instruction or 2nd preceding load instruction
■ Need 1 stall cycle
lw L, addr |_f add $4, $5, $6 beq stalled beq $1, $4, target
ID
IF
EX
ID
IF
MEM
EX
ID
WB
MEM npj| WB
\o\ o
pi EX MEM
WB
Chapter 4 — The Processor — 88
Data Hazards for Branches
If a comparison register is a destination of immediately preceding load instruction
■ Need 2 stall cycles
lw $1, addr if
beq stalled beq stalled beq $1, $0, target
id
if
EX
MEM
u
WB
id
id
CP o o
id
EX
MEM
WB
Chapter 4 — The Processor — 89
Dynamic Branch Prediction
In deeper and superscalar pipelines, branch penalty is more significant
Use dynamic prediction
■ Branch prediction buffer (aka branch history table)
■ Indexed by recent branch instruction addresses
■ Stores outcome (taken/not taken)
■ To execute a branch
Check table, expect the same outcome Start fetching from fall-through or target If wrong, flush pipeline and flip prediction
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Chapter 4 — The Processor — 90
1-Bit Predictor: Shortcoming
Inner loop branches mispredicted twice!
outer:
inner:
beq ..., ..., inner
beq ..., ..., outer
Mispredict as taken on last iteration of inner loop
Then mispredict as not taken on first iteration of inner loop next time around
Chapter 4 — The Processor — 91
2-Bit Predictor
Only change prediction on two successive mispredictions
Calculating the Branch Target
Even with predictor, still need to calculate the target address
■ 1 -cycle penalty for a taken branch
' Branch target buffer
■ Cache of target addresses
■ Indexed by PC when instruction fetched
If hit and instruction is branch predicted taken, can fetch target immediately
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Chapter 4 — The Processor — 93
Exceptions and Interrupts
"Unexpected" events requiring change in flow of control
■ Different ISAs use the terms differently
Exception
■ Arises within the CPU
e.g., undefined opcode, overflow, syscall, ...
Interrupt
■ From an external I/O controller
Dealing with them without sacrificing performance is hard
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Chapter 4 — The Processor —
Handling Exceptions
In MIPS, exceptions managed by a System Control Coprocessor (CPO)
Save PC of offending (or interrupted) instruction
■ In MIPS: Exception Program Counter (EPC)
Save indication of the problem
■ In MIPS: Cause register
■ We'll assume 1-bit
0 for undefined opcode, 1 for overflow
Jump to handler at 8000 00180
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Chapter 4 — The Processor — 95
An Alternate Mechanism
Vectored Interrupts
■ Handler address determined by the cause
Example:
- Undefined opcode: C000 0000
. Overflow: C000 0020
■ C000 0040
Instructions either
■ Deal with the interrupt, or
■ Jump to real handler
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Chapter 4 — The Processor — 96
Handler Actions
Read cause, and transfer to relevant handler
Determine action required If restartable
■ Take corrective action
■ use EPC to return to program
Otherwise
■ Terminate program
■ Report error using EPC, cause, ...
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Chapter 4 — The Processor — 97
Exceptions in a Pipeline
Another form of control hazard Consider overflow on add in EX stage
add $1, $2, $1
■ Prevent $1 from being clobbered
■ Complete previous instructions
■ Flush add and subsequent instructions
■ Set Cause and EPC register values
■ Transfer control to handler
Similar to mispredicted branch
■ Use much of the same hardware
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Chapter 4 — The Processor — 98
Pipeline with Exceptions
Chapter 4 — The Processor — 99
Exception Properties
Restartable exceptions
■ Pipeline can flush the instruction
■ Handler executes, then returns to the instruction
Refetched and executed from scratch
PC saved in EPC register
■ Identifies causing instruction
■ Actually PC + 4 is saved
Handler must adjust
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Chapter 4 — The Processor — 100
Exception Example
Exception on add in
40 sub $11, $2, $4
44 and $12, $2, $5
48 or $13, $2, $6
4C add $1, $2, $1
50 sit $15, $6, $7
54 lw $16, 50($7)
Handler
80000180 sw $25, 1000($0) 80000184 sw $26, 1004($0)
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Chapter 4 — The Processor — 101
Exception Example
lw $16, 50($7) IF. Flush
Sit $15, $6, $7
add $1, $2, $1
EX.FIush
Clock 6
Chapter 4 — The Processor — 102
Exception Example
sw$25, 1000($0) .Flush
bubble (nop)
bubble
EX.FIush
bubble
or $13, .
Clock 7
Chapter 4 — The Processor — 103
Multiple Exceptions
Pipelining overlaps multiple instructions
■ Could have multiple exceptions at once
Simple approach: deal with exception from earliest instruction
■ Flush subsequent instructions
■ "Precise" exceptions
In complex pipelines
■ Multiple instructions issued per cycle
■ Out-of-order completion
■ Maintaining precise exceptions is difficult!
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Chapter 4 — The Processor — 104
Imprecise Exceptions
Just stop pipeline and save state
■ Including exception cause(s)
Let the handler work out
■ Which instruction(s) had exceptions
■ Which to complete or flush
May require "manual" completion
Simplifies hardware, but more complex handler software
Not feasible for complex multiple-issue out-of-order pipelines
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Chapter 4 — The Processor — 105
Instruction-Level Parallelism (ILP)
Pipelining: executing multiple instructions in parallel
To increase ILP
■ Deeper pipeline
Less work per stage =^> shorter clock cycle
■ Multiple issue
Replicate pipeline stages =^> multiple pipelines Start multiple instructions per clock cycle CPI < 1, so use Instructions Per Cycle (IPC) E.g., 4GHz 4-way multiple-issue
16 BIPS, peak CPI = 0.25, peak IPC = 4 But dependencies reduce this in practice
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Chapter 4 — The Processor
Multiple Issue
Static multiple issue
■ Compiler groups instructions to be issued together
■ Packages them into "issue slots"
■ Compiler detects and avoids hazards
Dynamic multiple issue
■ CPU examines instruction stream and chooses instructions to issue each cycle
■ Compiler can help by reordering instructions
■ CPU resolves hazards using advanced techniques at runtime
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Chapter 4 — The Processor — 107
Speculation
"Guess" what to do with an instruction
■ Start operation as soon as possible
■ Check whether guess was right
If so, complete the operation
If not, roll-back and do the right thing
Common to static and dynamic multiple issue Examples
■ Speculate on branch outcome
Roll back if path taken is different
■ Speculate on load
Roll back if location is updated
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Chapter 4 — The Processor — 108
Compiler/Hardware Speculation
Compiler can reorder instructions
■ e.g., move load before branch
■ Can include "fix-up" instructions to recover from incorrect guess
Hardware can look ahead for instructions to execute
■ Buffer results until it determines they are actually needed
■ Flush buffers on incorrect speculation
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Chapter 4 — The Processor — 109
Speculation and Exceptions
What if exception occurs on a speculatively executed instruction?
■ e.g., speculative load before null-pointer check
Static speculation
■ Can add ISA support for deferring exceptions
Dynamic speculation
■ Can buffer exceptions until instruction completion (which may not occur)
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Chapter 4 — The Processor — 110
Static Multiple Issue
Compiler groups instructions into "issue packets"
■ Group of instructions that can be issued on a single cycle
■ Determined by pipeline resources required
Think of an issue packet as a very long instruction
■ Specifies multiple concurrent operations
■ => Very Long Instruction Word (VLIW)
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Chapter 4 — The Processor — 111
Scheduling Static Multiple Issue
Compiler must remove some/all hazards
■ Reorder instructions into issue packets
■ No dependencies with a packet
■ Possibly some dependencies between packets
Varies between ISAs; compiler must know!
■ Pad with nop if necessary
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Chapter 4 — The Processor — 112
MIPS with Static Dual Issue
Two-issue packets
■ One ALU/branch instruction
■ One load/store instruction
■ 64-bit aligned
ALU/branch, then load/store
Pad an unused instruction with nop
Address Instruction type Pipeline Stages
n ALU/branch IF ID EX MEM WB
n + 4 Load/store IF ID EX MEM WB
n + 8 ALU/branch IF ID EX MEM WB
n + 12 Load/store IF ID EX MEM WB
n + 16 ALU/branch IF ID EX MEM WB
n + 20 Load/store IF ID EX MEM WB
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Chapter 4 — The Processor — 113
MIPS with Static Dual Issue
80000180
M
u PC
X
Chapter 4 — The Processor — 114
Hazards in the Dual-Issue MIPS
More instructions executing in parallel EX data hazard
■ Forwarding avoided stalls with single-issue
■ Now can't use ALU result in load/store in same packet
add $t0, $s0, $sl load $s2, 0($t0)
Split into two packets, effectively a stall
Load-use hazard
■ Still one cycle use latency, but now two instructions
More aggressive scheduling required
Chapter 4 — The Processor — 115
Scheduling Example
Schedule this for dual-issue MIPS
Loop:
lw $t0, 0($sl)
addu $t0, $t0, $s2
sw $t0, 0($sl)
addi $sl, $sl,-4
# $tO=array element
# add scalar in $s2
# store result
# decrement pointer
bne L, $zero, Loop # branch $sl!=0
ALU/branch Load/store cycle
Loop: noD lw $t0, 0($sl) 1
addi $sl, $sl,-4 nop 2
addu $t0, $t0, $s2 nop 3
bne L, $zero, Loop sw $t0, 4($sl) 4
IPC = 5/4 = 1.25 (c.f. peak IPC = 2)
<
Chapter 4 — The Processor — 116
Loop Unrolling
Replicate loop body to expose more parallelism
■ Reduces loop-control overhead
Use different registers per replication
■ Called "register renaming"
■ Avoid loop-carried "anti-dependencies"
Store followed by a load of the same register
Aka "name dependence"
Reuse of a register name
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Chapter 4 — The Processor — 117
Loop Unrolling Example
ALU/branch Load/store cycle
Loop: addi $sl, $sl,-16 lw $t0, 0($sl) 1
nop lw $tl, 12($sl) 2
addu $t0, $t0, $s2 lw $t2, 8($sl) 3
addu $tl, $tl, $s2 lw $t3, 4($sl) 4
addu $t2, $t2, $s2 sw $t0, 16($sl) 5
addu $t3, $t4, $s2 sw $tl, 12($sl) 6
sw $t2, 8($sl) 7
bne L, $zero, Loop sw $t3, 4($sl) 8
IPC = 14/8 = 1.75
■ Closer to 2, but at cost of registers and code size
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Chapter 4 — The Processor — 118
Dynamic Multiple Issue
"Superscalar" processors
CPU decides whether to issue 0, 1,2, ... each cycle
■ Avoiding structural and data hazards
Avoids the need for compiler scheduling
■ Though it may still help
■ Code semantics ensured by the CPU
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Chapter 4 — The Processor — 119
Dynamic Pipeline Scheduling
Allow the CPU to execute instructions out of order to avoid stalls
■ But commit result to registers in order
Example
lw $t0, 20($s2)
addu $tl, $t0, $t2
sub $s4, $s4, $t3
slti $t5, $s4, 20
■ Can start sub while addu is waiting for lw
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Chapter 4 — The Processor — 120
Dynamically Scheduled CPU
Instruction fetch and decode unit Ii
1
Reservation station Reservation station Reservation station Reservation station
In-order issue
Functional units
Integer
Integer
J
Floating point
Out-of-order execute
Reorders buffer for register writes
Commit unit
In-order commit
Can supply operands for issued instructions
Preserves dependencies
Hold pending operands
Results also sent to any waiting reservation stations
Chapter 4 — The Processor — 121
Register Renaming
Reservation stations and reorder buffer effectively provide register renaming
On instruction issue to reservation station
■ If operand is available in register file or reorder buffer
Copied to reservation station
No longer required in the register; can be overwritten
■ If operand is not yet available
It will be provided to the reservation station by a function unit
Register update may not be required
Chapter 4 — The Processor — 122
Speculation
Predict branch and continue issuing
■ Don't commit until branch outcome determined
Load speculation
■ Avoid load and cache miss delay
Predict the effective address Predict loaded value
Load before completing outstanding stores Bypass stored values to load unit
■ Don't commit load until speculation cleared
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Chapter 4 — The Processor — 123
Why Do Dynamic Scheduling?
Why not just let the compiler schedule code?
Not all stalls are predicable
■ e.g., cache misses
Can't always schedule around branches
■ Branch outcome is dynamically determined
Different implementations of an ISA have different latencies and hazards
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Chapter 4 — The Processor — 124
Does Multiple Issue Work?
The BIG Picture
Yes, but not as much as we'd like
Programs have real dependencies that limit ILP
Some dependencies are hard to eliminate
■ e.g., pointer aliasing
Some parallelism is hard to expose
■ Limited window size during instruction issue
Memory delays and limited bandwidth
■ Hard to keep pipelines full
Speculation can help if done well
Chapter 4 — The Processor — 125
Power Efficiency
Complexity of dynamic scheduling and speculations requires power
Multiple simpler cores may be better
Microprocessor Year Clock Rate Pipeline Stages Issue width Out-of-order/ Speculation Cores Power
i486 1989 25MHz 5 1 No 1 5W
Pentium 1993 66MHz 5 2 No 1 10W
Pentium Pro 1997 200MHz 10 3 Yes 1 29W
P4 Willamette 2001 2000MHz 22 3 Yes 1 75W
P4 Prescott 2004 3600MHz 31 3 Yes 1 103W
Core 2006 2930MHz 14 4 Yes 2 75W
UltraSparc III 2003 1950MHz 14 4 No 1 90W
UltraSparcTI 2005 1200MHz 6 1 No 8 70W
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Chapter 4 — The Processor — 126
The Opteron X4 Microarchitecture
Instruction cache
Branch prediction
Instruction prefetch and decode
T
RISC-operation queue
I
Dispatch and register remaining
Integer and floating-point operation queue
Integer ALU. Multiplier
Integer ALU
Integer ALU
Floating point Adder /SSE
Floating point
Multiplier /SSE
Load/Store queue
Data cache
<
Commit unit
72 physical registers
Chapter 4 — The Processor — 127
The Opteron X4 Pipeline Flow
For integer operations
Instruction Fetch
II
Number of clock cycles
Decode
and translate
queue Reorder f buffer allocation + register renaming 1 buffer
—
—► 1-
Scheduling + dispatch unit
Execution
Data Cache/I Commit
■ FP is 5 stages longer
■ Up to 106 RISC-ops in progress
Bottlenecks
■ Complex instructions with long dependencies
■ Branch mispredictions
■ Memory access delays
Chapter 4 — The Processor — 128
Fallacies
Pipelining is easy (!)
■ The basic idea is easy
■ The devil is in the details
e.g., detecting data hazards
Pipelining is independent of technology
■ So why haven't we always done pipelining?
■ More transistors make more advanced techniques feasible
■ Pipeline-related ISA design needs to take account of technology trends
e.g., predicated instructions
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Chapter 4 — The Processor — 1
Pitfalls
Poor ISA design can make pipelining harder
■ e.g., complex instruction sets (VAX, IA-32)
Significant overhead to make pipelining work IA-32 micro-op approach
■ e.g., complex addressing modes
Register update side effects, memory indirection
■ e.g., delayed branches
Advanced pipelines have long delay slots
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Chapter 4 — The Processor — 130
Concluding Remarks
ISA influences design of datapath and control
Datapath and control influence design of ISA
Pipelining improves instruction throughput using parallelism
■ More instructions completed per second
■ Latency for each instruction not reduced
Hazards: structural, data, control
Multiple issue and dynamic scheduling (ILP)
■ Dependencies limit achievable parallelism
■ Complexity leads to the power wall
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Chapter 4 — The Processor —
Exercises
Answer the following exercises, and send your answers as a PDF attachment to the email address listed below
xamiri@fi.muni.cz I Leave body of the email blank Deadline is April 28th Exercises
4.1.1(b), 4.2.4(a), 4.7.3(a), 4.9.2(a)~4.9.4(a), 4.10.4(a), 4.11.1(a)~4.11.4(a), 4.13.1(a), 4.13.5(a), 4.16.4(b), 4.18.1(b), 4.21.2(a), 4.21.4(a), 4.22.1(b), 4.23.2(a), 4.24.2(b)
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Chapter 1 — Computer Abstractions and Technology — 132