Program Optimization CENG331: Introduction to Computer Systems 11 th Lecture
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Transcript of Program Optimization CENG331: Introduction to Computer Systems 11 th Lecture
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Instructor:
Erol Sahin
Program OptimizationCENG331: Introduction to Computer Systems11th Lecture
Acknowledgement: Most of the slides are adapted from the ones prepared by R.E. Bryant, D.R. O’Hallaron of Carnegie-Mellon Univ.
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OverviewProgram optimization
Removing unnecessary procedure calls Code motion/precomputation Strength reduction Sharing of common subexpressions Optimization blocker: Procedure calls
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Example Matrix Multiplication
Standard desktop computer, compiler, using optimization flags
Both implementations have exactly the same operations count (2n3)
What is going on?
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0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000matrix size
Matrix-Matrix Multiplication (MMM) on 2 x Core 2 Duo 3 GHz Gflop/ s (giga floating point operations per second)
160x
Triple loop
Best code
This code is not obviously stupid
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MMM Plot: Analysis
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Matrix-Matrix Multiplication (MMM) on 2 x Core 2 Duo 3 GHzGflop/ s
Memory hierarchy and other optimizations: 20xVector instructions: 4x
Multiple threads: 4x
Reason for 20x: Blocking or tiling, loop unrolling, array scalarization, instruction scheduling, search to find best choice
Effect: more instruction level parallelism, better register use, less L1/L2 cache misses, less TLB misses
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Harsh RealityThere’s more to runtime performance than asymptotic complexity
One can easily loose 10x, 100x in runtime or even more
What matters: Constants (100n and 5n is both O(n), but ….) Coding style (unnecessary procedure calls, unrolling, reordering, …) Algorithm structure (locality, instruction level parallelism, …) Data representation (complicated structs or simple arrays)
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Harsh RealityMust optimize at multiple levels:
Algorithm Data representations Procedures Loops
Must understand system to optimize performance How programs are compiled and executed
Execution units, memory hierarchy How to measure program performance and identify bottlenecks How to improve performance without destroying code modularity and
generality
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Optimizing Compilers
Use optimization flags, default is no optimization (-O0)!
Good choices for gcc: -O2, -O3, -march=xxx, -m64
Try different flags and maybe different compilers
-O
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Example
Compiled without flags: ~1300 cycles
Compiled with –O3 –m64 -march=… –fno-tree-vectorize~150 cycles
Core 2 Duo, 2.66 GHz
double a[4][4];double b[4][4];
double c[4][4]; # set to zero
/* Multiply 4 x 4 matrices a and b */void mmm(double *a, double *b, double *c, int n) { int i, j, k; for (i = 0; i < 4; i++)
for (j = 0; j < 4; j++) for (k = 0; k < 4; k++)
c[i*4+j] += a[i*4 + k]*b[k*4 + j];}
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Optimizing CompilersCompilers are good at: mapping program to machine
register allocation code selection and ordering (scheduling) dead code elimination eliminating minor inefficiencies
Compilers are not good at: improving asymptotic efficiency up to programmer to select best overall algorithm big-O savings are (often) more important than constant factors
but constant factors also matter
Compilers are not good at: overcoming “optimization blockers” potential memory aliasing potential procedure side-effects
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Limitations of Optimizing CompilersIf in doubt, the compiler is conservative
Operate under fundamental constraints Must not change program behavior under any possible condition Often prevents it from making optimizations when would only affect
behavior under pathological conditions.
Behavior that may be obvious to the programmer can be obfuscated by languages and coding styles e.g., data ranges may be more limited than variable types suggest
Most analysis is performed only within procedures Whole-program analysis is too expensive in most cases
Most analysis is based only on static information Compiler has difficulty anticipating run-time inputs
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Example: Data Type for Vectors
/* data structure for vectors */typedef struct{
int len;double *data;
} vec;
/* retrieve vector element and store at val */double get_vec_element(*vec, idx, double *val){
if (idx < 0 || idx >= v->len)return 0;
*val = v->data[idx];return 1;
}
lendata
0 1 len-1
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Performance
Time quadruples when double string length
Quadratic performance25
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512 1k 2k 4k 8k 16k
32k
64k
128k
256k
0.00010.001
0.010.1
110
1001000CPU Seconds
String Length
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Why is That?
String length is called in every iteration! And strlen is O(n), so lower is O(n2)
void lower(char *s){ int i; for (i = 0; i < strlen(s); i++) if (s[i] >= 'A' && s[i] <= 'Z') s[i] -= ('A' - 'a');}
/* My version of strlen */size_t strlen(const char *s){ size_t length = 0; while (*s != '\0') {
s++; length++;
} return length;}
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Improving Performance
Move call to strlen outside of loop
Since result does not change from one iteration to another
Form of code motion/precomputation
void lower(char *s){ int i; int len = strlen(s); for (i = 0; i < len; i++) if (s[i] >= 'A' && s[i] <= 'Z') s[i] -= ('A' - 'a');}
void lower(char *s){ int i; for (i = 0; i < strlen(s); i++) if (s[i] >= 'A' && s[i] <= 'Z') s[i] -= ('A' - 'a');}
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PerformanceLower2: Time doubles when double string length
Linear performance25
6
512 1k 2k 4k 8k 16k
32k
64k
128k
256k
0.0000010.00001
0.00010.001
0.010.1
110
1001000
lower1 lower2
CPU Seconds
String Length
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Optimization Blocker: Procedure CallsWhy couldn’t compiler move strlen out of inner loop?
Procedure may have side effects Function may not return same value for given arguments
Could depend on other parts of global state Procedure lower could interact with strlen
Compiler usually treats procedure call as a black box that cannot be analyzed Consequence: conservative in optimizations
Remedies: Inline the function if possible Do your own code motion
int lencnt = 0;size_t strlen(const char *s){ size_t length = 0; while (*s != '\0') {
s++; length++; } lencnt += length; return length;}
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Today
Program optimization Optimization blocker: Memory aliasing Out of order processing: Instruction level parallelism Understanding branch prediction
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Optimization Blocker: Memory Aliasing
Code updates b[i] (= memory access) on every iteration
Why couldn’t compiler optimize this away?
# sum_rows1 inner loop.L53:
addsd (%rcx), %xmm0 # FP addaddq $8, %rcxdecq %raxmovsd %xmm0, (%rsi,%r8,8) # FP storejne .L53
/* Sums rows of n x n matrix a and stores in vector b */void sum_rows1(double *a, double *b, long n) { long i, j; for (i = 0; i < n; i++) {
b[i] = 0;for (j = 0; j < n; j++) b[i] += a[i*n + j];
}}
ab
Σ
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Reason
If memory is accessed, compiler assumes the possibility of side effects
Example:
double A[9] = { 0, 1, 2, 4, 8, 16}, 32, 64, 128};
double B[3] = A+3;
sum_rows1(A, B, 3);
i = 0: [3, 8, 16]
init: [4, 8, 16]
i = 1: [3, 22, 16]
i = 2: [3, 22, 224]
Value of B:
/* Sums rows of n x n matrix a and stores in vector b */void sum_rows1(double *a, double *b, long n) { long i, j; for (i = 0; i < n; i++) {
b[i] = 0;for (j = 0; j < n; j++) b[i] += a[i*n + j];
}}
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Removing Aliasing
Scalar replacement: Copy array elements that are reused into temporary variables Assumes no memory aliasing (otherwise possibly incorrect)
# sum_rows2 inner loop.L66:
addsd (%rcx), %xmm0 # FP Addaddq $8, %rcxdecq %raxjne .L66
/* Sums rows of n x n matrix a and stores in vector b */void sum_rows2(double *a, double *b, long n) { long i, j; for (i = 0; i < n; i++) {
double val = 0;for (j = 0; j < n; j++) val += a[i*n + j];
b[i] = val; }}
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Unaliased Version When Aliasing Happens
Aliasing still creates interference
Result different than before
double A[9] = { 0, 1, 2, 4, 8, 16}, 32, 64, 128};
double B[3] = A+3;
sum_rows1(A, B, 3);
i = 0: [3, 8, 16]
init: [4, 8, 16]
i = 1: [3, 27, 16]
i = 2: [3, 27, 224]
Value of B:
/* Sum rows is of n X n matrix a and store in vector b */void sum_rows2(double *a, double *b, long n) { long i, j; for (i = 0; i < n; i++) {
double val = 0;for (j = 0; j < n; j++) val += a[i*n + j];
b[i] = val; }}
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Optimization Blocker: Memory AliasingMemory aliasing: Two different memory references write
to the same location
Easy to have happen in C Since allowed to do address arithmetic Direct access to storage structures
Hard to analyze = compiler cannot figure it out Hence is conservative
Solution: Scalar replacement in innermost loop Copy memory variables that are reused into local variables Basic scheme:
Load: t1 = a[i], t2 = b[i+1], …. Compute: t4 = t1 * t2; …. Store: a[i] = t12, b[i+1] = t7, …
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TodayProgram optimization
Optimization blocker: Memory aliasing Out of order processing: Instruction level parallelism Understanding branch prediction
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Example: Compute Factorials Machines
Intel Pentium 4 Nocona, 3.2 GHz Intel Core 2, 2.7 GHz
Compiler Versions GCC 3.4.2
int rfact(int n){ if (n <= 1)
return 1; return n * rfact(n-1);}
int fact(int n){ int i; int result = 1;
for (i = n; i > 0; i--)result = result * i;
return result;}
Machine Nocona Core 2
rfact 15.5 6.0
fact 10.0 3.0
Cycles per element (or per mult)
Something changed from Pentium 4 to Core: Details later
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Optimization 1: Loop Unrolling
Compute more values per iteration
Does not help here
Why? Branch prediction – details later
int fact_u3a(int n){ int i; int result = 1;
for (i = n; i >= 3; i-=3) {result = result * i * (i-1) * (i-2);
} for (; i > 0; i--)
result *= i; return result;}
Machine Nocona Core 2
rfact 15.5 6.0
fact 10.0 3.0
fact_u3a 10.0 3.0
Cycles per element (or per mult)
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Optimization 2: Multiple Accumulators
That seems to help. Can one get even faster?
Explanation: instruction level parallelism – details later
int fact_u3b(int n){ int i; int result0 = 1; int result1 = 1; int result2 = 1;
for (i = n; i >= 3; i-=3) {result0 *= i;result1 *= (i-1);result2 *= (i-2);
} for (; i > 0; i--)
result0 *= i; return result0 * result1 * result2;}
Machine Nocona Core 2
rfact 15.5 6.0
fact 10.0 3.0
fact_u3a 10.0 3.0
fact_u3b 3.3 1.0
Cycles per element (or per mult)
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Modern CPU Design
Execution
FunctionalUnits
Instruction Control
Integer/Branch
FPAdd
FPMult/Div Load Store
InstructionCache
DataCache
FetchControl
InstructionDecode
Address
Instructions
Operations
Prediction OK?
DataData
Addr. Addr.
GeneralInteger
Operation Results
RetirementUnit
RegisterFile
Register Updates
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Superscalar ProcessorDefinition: A superscalar processor can issue and execute
multiple instructions in one cycle. The instructions are retrieved from a sequential instruction stream and are usually scheduled dynamically.
Benefit: without programming effort, superscalar processor can take advantage of the instruction level parallelism that most programs have
Most CPUs since about 1998 are superscalar.
Intel: since Pentium Pro
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Pentium 4 Nocona CPUMultiple instructions can execute in parallel
1 load, with address computation1 store, with address computation2 simple integer (one may be branch)1 complex integer (multiply/divide)1 FP/SSE3 unit1 FP move (does all conversions)
Some instructions take > 1 cycle, but can be pipelinedInstruction Latency Cycles/IssueLoad / Store 5 1Integer Multiply 10 1Integer/Long Divide 36/106 36/106Single/Double FP Multiply 7 2Single/Double FP Add 5 2Single/Double FP Divide 32/46 32/46
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Latency versus Throughput
Last slide: latency cycles/issueInteger Multiply 10 1
Step 11 cycle
Step 21 cycle
Step 101 cycle
Consequence: How fast can 10 independent int mults be executed?
t1 = t2*t3; t4 = t5*t6; … How fast can 10 sequentially dependent int mults be executed?
t1 = t2*t3; t4 = t5*t1; t6 = t7*t4; …
Major problem for fast execution: Keep pipelines filled
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Hard Bounds
How many cycles at least if Function requires n int mults? Function requires n float adds? Function requires n float ops (adds and mults)?
Latency and throughput of instructionsInstruction Latency Cycles/IssueLoad / Store5 1Integer Multiply 10 1Integer/Long Divide 36/106 36/106Single/Double FP Multiply 7 2Single/Double FP Add 5 2Single/Double FP Divide 32/46 32/46
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Performance in Numerical Computing
Numerical computing = computing dominated by floating point operations
Example: Matrix multiplication
Performance measure: Floating point operations per second (flop/s) Counting only floating point adds and mults Higher is better Like inverse runtime
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Nocona vs. Core 2Nocona (3.2 GHz)
Instruction Latency Cycles/IssueLoad / Store5 1Integer Multiply 10 1Integer/Long Divide 36/106 36/106Single/Double FP Multiply 7 2Single/Double FP Add 5 2Single/Double FP Divide 32/46 32/46
Core 2 (2.7 GHz) (Recent Intel microprocessors)Instruction LatencyCycles/IssueLoad / Store5 1Integer Multiply 3 1Integer/Long Divide 18/50 18/50Single/Double FP Multiply 4/5 1Single/Double FP Add 3 1Single/Double FP Divide 18/32 18/32
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Instruction Control
Grabs instruction bytes from memory Based on current PC + predicted targets for predicted branches Hardware dynamically guesses whether branches taken/not taken and
(possibly) branch target
Translates instructions into micro-operations (for CISC style CPUs) Micro-op = primitive step required to perform instruction Typical instruction requires 1–3 operations
Converts register references into tags Abstract identifier linking destination of one operation with sources of later
operations
Instruction Control
InstructionCache
FetchControl
InstructionDecode
Address
Instructions
Operations
RetirementUnit
RegisterFile
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Translating into Micro-Operations
Goal: Each operation utilizes single functional unit Requires: Load, integer arithmetic, store
Exact form and format of operations is trade secret
imulq %rax, 8(%rbx,%rdx,4)
load 8(%rbx,%rdx,4) temp1imulq %rax, temp1 temp2store temp2, 8(%rbx,%rdx,4)
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Traditional View of Instruction Execution
Imperative View Registers are fixed storage locations
Individual instructions read & write them Instructions must be executed in specified sequence to guarantee
proper program behavior
addq %rax, %rbx # I1 andq %rbx, %rdx # I2 mulq %rcx, %rbx # I3 xorq %rbx, %rdi # I4
rax
+
rbx rdx rcx rdi
I1
I2
I3
I4
&
*
^
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Dataflow View of Instruction Execution
Functional View View each write as creating new instance of value Operations can be performed as soon as operands available No need to execute in original sequence
rax.0
+
rbx.0 rdx.0 rcx.0 rdi.0
I1
I2/I3
I4
*
^
rbx.1
rdx.1rbx.2
rdi.1
addq %rax, %rbx # I1 andq %rbx, %rdx # I2 mulq %rcx, %rbx # I3 xorq %rbx, %rdi # I4
&
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Example Computation
Data Types Use different declarations for data_t
int float double
void combine4(vec_ptr v, data_t *dest){ int i; int length = vec_length(v); data_t *d = get_vec_start(v); data_t t = IDENT; for (i = 0; i < length; i++) t = t OP d[i]; *dest = t;}
Operations Use different definitions of OP
and IDENT + / 0 * / 1
d[1] OP d[2] OP d[3] OP … OP d[length-1]
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Cycles Per Element (CPE)Convenient way to express performance of program that operators on
vectors or lists
Length = n
In our case: CPE = cycles per OP (gives hard lower bound)
T = CPE*n + Overhead
0
100
200
300
400
500
600
700
800
900
1000
0 50 100 150 200
Cyc
les
n = Number of elements
vsum1: Slope = 4.0
vsum2: Slope = 3.5
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x86-64 Compilation of Combine4
Inner Loop (Case: Integer Multiply)L33: # Loop:
movl (%eax,%edx,4), %ebx # temp = d[i]incl %edx # i++imull %ebx, %ecx # x *= tempcmpl %esi, %edx # i:lengthjl L33 # if < goto Loop
void combine4(vec_ptr v, data_t *dest){ int i; int length = vec_length(v); data_t *d = get_vec_start(v); data_t t = IDENT; for (i = 0; i < length; i++) t = t OP d[i]; *dest = t;}
Method Int (add/mult) Float (add/mult)
combine4 2.2 10.0 5.0 7.0
bound 1.0 1.0 2.0 2.0
Cycles per element (or per OP)
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Combine4 = Serial Computation (OP = *)Computation (length=8)
((((((((1 * d[0]) * d[1]) * d[2]) * d[3]) * d[4]) * d[5]) * d[6]) * d[7])
Sequential dependence! Hence, Performance: determined by latency of OP!
*
*
1 d0
d1
*
d2
*
d3
*
d4
*
d5
*
d6
*
d7
Method Int (add/mult) Float (add/mult)
combine4 2.2 10.0 5.0 7.0
bound 1.0 1.0 2.0 2.0
Cycles per element (or per OP)
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Loop Unrolling
Perform 2x more useful work per iteration
void unroll2a_combine(vec_ptr v, data_t *dest){ int length = vec_length(v); int limit = length-1; data_t *d = get_vec_start(v); data_t x = IDENT; int i; /* Combine 2 elements at a time */ for (i = 0; i < limit; i+=2) {
x = (x OP d[i]) OP d[i+1]; } /* Finish any remaining elements */ for (; i < length; i++) {
x = x OP d[i]; } *dest = x;}
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Effect of Loop Unrolling
Helps integer sum
Others don’t improve. Why? Still sequential dependency
x = (x OP d[i]) OP d[i+1];
Method Int (add/mult) Float (add/mult)
combine4 2.2 10.0 5.0 7.0
unroll2 1.5 10.0 5.0 7.0
bound 1.0 1.0 2.0 2.0
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– 47 –
Loop Unrolling with Reassociation
Can this change the result of the computation?
Yes, for FP. Why?
void unroll2aa_combine(vec_ptr v, data_t *dest){ int length = vec_length(v); int limit = length-1; data_t *d = get_vec_start(v); data_t x = IDENT; int i; /* Combine 2 elements at a time */ for (i = 0; i < limit; i+=2) {
x = x OP (d[i] OP d[i+1]); } /* Finish any remaining elements */ for (; i < length; i++) {
x = x OP d[i]; } *dest = x;}
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– 48 –
Effect of Reassociation
Nearly 2x speedup for Int *, FP +, FP * Reason: Breaks sequential dependency
Why is that? (next slide)
x = x OP (d[i] OP d[i+1]);
Method Int (add/mult) Float (add/mult)
combine4 2.2 10.0 5.0 7.0
unroll2 1.5 10.0 5.0 7.0
unroll2-ra 1.56 5.0 2.75 3.62
bound 1.0 1.0 2.0 2.0
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– 49 –
Reassociated Computation
What changed: Ops in the next iteration can be
started early (no dependency)
Overall Performance N elements, D cycles latency/op Should be (N/2+1)*D cycles:
CPE = D/2 Measured CPE slightly worse for
FP
*
*
1
*
*
*
d1d0
*
d3d2
*
d5d4
*
d7d6
x = x OP (d[i] OP d[i+1]);
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– 50 –
Loop Unrolling with Separate Accumulators
Different form of reassociation
void unroll2a_combine(vec_ptr v, data_t *dest){ int length = vec_length(v); int limit = length-1; data_t *d = get_vec_start(v); data_t x0 = IDENT; data_t x1 = IDENT; int i; /* Combine 2 elements at a time */ for (i = 0; i < limit; i+=2) { x0 = x0 OP d[i]; x1 = x1 OP d[i+1]; } /* Finish any remaining elements */ for (; i < length; i++) {
x0 = x0 OP d[i]; } *dest = x0 OP x1;}
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– 51 –
Effect of Separate Accumulators
Amost exact 2x speedup (over unroll2) for Int *, FP +, FP * Breaks sequential dependency in a “cleaner,” more obvious way
x0 = x0 OP d[i]; x1 = x1 OP d[i+1];
Method Int (add/mult) Float (add/mult)
combine4 2.2 10.0 5.0 7.0
unroll2 1.5 10.0 5.0 7.0
unroll2-ra 1.56 5.0 2.75 3.62
unroll2-sa 1.50 5.0 2.5 3.5
bound 1.0 1.0 2.0 2.0
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Separate Accumulators
*
*
1 d1
d3
*
d5
*
d7
*
*
*
1 d0
d2
*
d4
*
d6
x0 = x0 OP d[i]; x1 = x1 OP d[i+1];
What changed: Two independent “streams” of
operations
Overall Performance N elements, D cycles latency/op Should be (N/2+1)*D cycles:
CPE = D/2 CPE matches prediction!
What Now?
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Unrolling & AccumulatingIdea
Can unroll to any degree L Can accumulate K results in parallel L must be multiple of K
Limitations Diminishing returns
Cannot go beyond throughput limitations of execution units Large overhead for short lengths
Finish off iterations sequentially
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– 54 –
Unrolling & Accumulating: Intel FP *Case
Intel Nocona (Saltwater fish machines) FP Multiplication Theoretical Limit: 2.00
FP * Unrolling Factor LK 1 2 3 4 6 8 10 121 7.00 7.00 7.01 7.002 3.50 3.50 3.503 2.344 2.01 2.006 2.00 2.018 2.01
10 2.0012 2.00
Accu
mul
ator
s
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Unrolling & Accumulating: Intel FP +Case
Intel Nocona (Saltwater fish machines) FP Addition Theoretical Limit: 2.00
FP + Unrolling Factor LK 1 2 3 4 6 8 10 121 5.00 5.00 5.02 5.002 2.50 2.51 2.513 2.004 2.01 2.006 2.00 1.998 2.01
10 2.0012 2.00
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– 56 –
Unrolling & Accumulating: Intel Int *Case
Intel Nocona (Saltwater fish machines) Integer Multiplication Theoretical Limit: 1.00
Int * Unrolling Factor LK 1 2 3 4 6 8 10 121 10.00 10.00 10.00 10.012 5.00 5.01 5.003 3.334 2.50 2.516 1.67 1.678 1.25
10 1.0912 1.14
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– 57 –
Unrolling & Accumulating: Intel Int +Case
Intel Nocona (Saltwater fish machines) Integer addition Theoretical Limit: 1.00 (unrolling enough)
Int + Unrolling Factor LK 1 2 3 4 6 8 10 121 2.20 1.50 1.10 1.032 1.50 1.10 1.033 1.344 1.09 1.036 1.01 1.018 1.03
10 1.0412 1.11
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– 58 –
FP *: Nocona versus Core 2
Machines Intel Nocona
3.2 GHz Intel Core 2
2.7 GHz
Performance Core 2 lower latency
& fully pipelined(1 cycle/issue)
FP * Unrolling Factor LK 1 2 3 4 6 8 10 121 4.00 4.00 4.00 4.012 2.00 2.00 2.003 1.344 1.00 1.006 1.00 1.008 1.00
10 1.0012 1.00
FP * Unrolling Factor LK 1 2 3 4 6 8 10 121 7.00 7.00 7.01 7.002 3.50 3.50 3.50
3 2.344 2.01 2.006 2.00 2.018 2.01
10 2.0012 2.00
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Can We Go Faster?Yes, SSE!
But not in this class
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TodayProgram optimization
Optimization blocker: Memory aliasing Out of order processing: Instruction level parallelism Understanding branch prediction
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– 65 –
What About Branches?Challenge
Instruction Control Unit must work well ahead of Execution Unitto generate enough operations to keep EU busy
When encounters conditional branch, cannot reliably determine where to continue fetching
80489f3: movl $0x1,%ecx 80489f8: xorl %edx,%edx 80489fa: cmpl %esi,%edx 80489fc: jnl 8048a25 80489fe: movl %esi,%esi
8048a00: imull (%eax,%edx,4),%ecx
Executing
How to continue?
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– 66 –
Branch OutcomesWhen encounter conditional branch, cannot determine where to continue
fetchingBranch Taken: Transfer control to branch targetBranch Not-Taken: Continue with next instruction in sequence
Cannot resolve until outcome determined by branch/integer unit
80489f3: movl $0x1,%ecx 80489f8: xorl %edx,%edx 80489fa: cmpl %esi,%edx 80489fc: jnl 8048a25 80489fe: movl %esi,%esi
8048a00: imull (%eax,%edx,4),%ecx
8048a25: cmpl %edi,%edx 8048a27: jl 8048a20
8048a29: movl 0xc(%ebp),%eax 8048a2c: leal 0xffffffe8(%ebp),%esp
8048a2f: movl %ecx,(%eax)
Branch Taken
Branch Not-Taken
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– 67 –
Branch PredictionIdea
Guess which way branch will go Begin executing instructions at predicted position
But don’t actually modify register or memory data
80489f3: movl $0x1,%ecx 80489f8: xorl %edx,%edx 80489fa: cmpl %esi,%edx 80489fc: jnl 8048a25
. . .
8048a25: cmpl %edi,%edx 8048a27: jl 8048a20
8048a29: movl 0xc(%ebp),%eax 8048a2c: leal 0xffffffe8(%ebp),%esp
8048a2f: movl %ecx,(%eax)
Predict Taken
BeginExecution
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– 68 –
Branch Prediction Through Loop 80488b1: movl (%ecx,%edx,4),%eax
80488b4: addl %eax,(%edi) 80488b6: incl %edx
80488b7: cmpl %esi,%edx 80488b9: jl 80488b1
80488b1: movl (%ecx,%edx,4),%eax 80488b4: addl %eax,(%edi)
80488b6: incl %edx 80488b7: cmpl %esi,%edx 80488b9: jl 80488b1
80488b1: movl (%ecx,%edx,4),%eax 80488b4: addl %eax,(%edi)
80488b6: incl %edx 80488b7: cmpl %esi,%edx 80488b9: jl 80488b1
i = 98
i = 99
i = 100
Predict Taken (OK)
Predict Taken(Oops)
80488b1: movl (%ecx,%edx,4),%eax 80488b4: addl %eax,(%edi)
80488b6: incl %edx 80488b7: cmpl %esi,%edx 80488b9: jl 80488b1
i = 101
Assume vector length = 100
Read invalid
location
Executed
Fetched
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– 69 –
Branch Misprediction Invalidation 80488b1: movl (%ecx,%edx,4),%eax
80488b4: addl %eax,(%edi) 80488b6: incl %edx
80488b7: cmpl %esi,%edx 80488b9: jl 80488b1
80488b1: movl (%ecx,%edx,4),%eax 80488b4: addl %eax,(%edi)
80488b6: incl %edx 80488b7: cmpl %esi,%edx 80488b9: jl 80488b1
80488b1: movl (%ecx,%edx,4),%eax 80488b4: addl %eax,(%edi)
80488b6: incl %edx 80488b7: cmpl %esi,%edx 80488b9: jl 80488b1
i = 98
i = 99
i = 100
Predict Taken (OK)
Predict Taken (Oops)
80488b1: movl (%ecx,%edx,4),%eax 80488b4: addl %eax,(%edi)
80488b6: incl %edx i = 101
Invalidate
Assume vector length = 100
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– 70 –
Branch Misprediction Recovery
Performance Cost Multiple clock cycles on modern processor Can be a major performance limiter
80488b1: movl (%ecx,%edx,4),%eax 80488b4: addl %eax,(%edi)
80488b6: incl %edx 80488b7: cmpl %esi,%edx 80488b9: jl 80488b1
80488bb: leal 0xffffffe8(%ebp),%esp 80488be: popl %ebx 80488bf: popl %esi 80488c0: popl %edi
i = 99
Definitely not taken
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– 71 –
Determining Misprediction PenaltyGCC/x86-64 tries to minimize
use of Branches Generates conditional
moves when possible/sensible
int cnt_gt = 0;int cnt_le = 0;int cnt_all = 0;
int choose_cmov(int x, int y){ int result; if (x > y) { result = cnt_gt; } else { result = cnt_le; } ++cnt_all; return result;} choose_cmov:
cmpl %esi, %edi # x:y movl cnt_le(%rip), %eax # r = cnt_le cmovg cnt_gt(%rip), %eax # if >= r=cnt_gt incl cnt_all(%rip) # cnt_all++ ret # return r
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Forcing ConditionalCannot use conditional move
when either outcome has side effect
int cnt_gt = 0;int cnt_le = 0;
int choose_cond(int x, int y){ int result; if (x > y) { result = ++cnt_gt; } else { result = ++cnt_le; } return result;}
choose_cond: cmpl %esi, %edi jle .L8 movl cnt_gt(%rip), %eax incl %eax movl %eax, cnt_gt(%rip) ret.L8: movl cnt_le(%rip), %eax incl %eax movl %eax, cnt_le(%rip) ret
If
Then
Else
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– 73 –
Testing Methodology
Idea Measure procedure under two different prediction probabilities
P = 1.0: Perfect predictionP = 0.5: Random data
Test Data x = 0, y = 1Case +1: y = [+1, +1, +1, …, +1, +1]Case −1: y = [−1, −1, −1, …, −1, −1]Case A: y = [+1, −1, +1, …, +1, −1] (alternate)Case R: y = [+1, −1, −1, …, −1, +1] (random)
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– 74 –
Testing Outcomes
Observations: Conditional move insensitive to data Perfect prediction for regular patterns
Else case requires 6 (Nocona), 2 (AMD), or 1 (Core 2) extra cycles Averages to 15.2
Branch penalties: (for R, processor will get it right half of the time) Nocona: 2 * (31.2-15.2) = 32 cycles AMD: 2 * (15.7-9.2) = 13 cycles Core 2: 2 * (17.7-8.7) = 18 cycles
Case cmov cond
+1 12.3 18.2
−1 12.3 12.2
A 12.3 15.2
R 12.3 31.2
Case cmov cond
+1 8.05 10.1
−1 8.05 8.1
A 8.05 9.2
R 8.05 15.7
Intel Nocona AMD OpteronCase cmov cond
+1 7.17 9.2
−1 7.17 8.2
A 7.17 8.7
R 7.17 17.7
Intel Core 2
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Getting High Performance So FarGood compiler and flags
Don’t do anything stupid Watch out for hidden algorithmic inefficiencies Write compiler-friendly code
Watch out for optimization blockers: procedure calls & memory references
Careful with implemented abstract data types Look carefully at innermost loops (where most work is done)
Tune code for machine Exploit instruction-level parallelism Avoid unpredictable branches Make code cache friendly (Covered later in course)