CS计算机代考程序代写 data structure compiler file system cache Optimizing Sequential Programs — Memory Hierarchy

Optimizing Sequential Programs — Memory Hierarchy

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Performance Analysis of
Parallel Programs

Presented by

Shuaiwen Leon Song

USYD Future System Architecture Lab (FSA)

https://shuaiwen-leon-song.github.io/

Slides are modified based on similar
classes offered by UC Berkeley James
Demmel and Tom Doeppner’s cs0330 at
Brown

https://shuaiwen-leon-song.github.io/

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Write a 2d Array

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Write a 2d array via single pointer

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Using an array of pointers

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Using pointer to a pointer

Principle of Locality

 Principle of Locality: Programs tend to use data and instructions with
addresses near or equal to those they have used recently

 Temporal locality:
▪ Recently referenced items are likely

to be referenced again in the near future

 Spatial locality:
▪ Items with nearby addresses tend

to be referenced close together in time

Locality Example

 Data references
▪ Reference array elements in succession

(stride-1 reference pattern).

▪ Reference variable sum each iteration.

 Instruction references
▪ Reference instructions in sequence.

▪ Cycle through loop repeatedly.

sum = 0;

for (i = 0; i < n; i++) sum += a[i]; return sum; Spatial locality Temporal locality Spatial locality Temporal locality Memory Hierarchies  Some fundamental and enduring properties of hardware and software: ▪ Fast storage technologies cost more per byte, have less capacity, and require more power (heat!). ▪ The gap between CPU and main memory speed is widening. ▪ Well-written programs tend to exhibit good locality.  These fundamental properties complement each other beautifully.  They suggest an approach for organizing memory and storage systems known as a memory hierarchy. An Example Memory Hierarchy Registers L1 cache (SRAM) Main memory (DRAM) Local secondary storage (local disks) Larger, slower, cheaper per byte Remote secondary storage (tapes, distributed file systems, Web servers) Local disks hold files retrieved from disks on remote network servers Main memory holds disk blocks retrieved from local disks L2 cache (SRAM) L1 cache holds cache lines retrieved from L2 cache CPU registers hold words retrieved from L1 cache L2 cache holds cache lines retrieved from main memory L0: L1: L2: L3: L4: L5: Smaller, faster, more expensive per byte 10 Memory Wall Problem  The "memory wall" is the growing disparity of speed between CPU and memory outside the CPU chip.  CPU speed improved at an annual rate of 55% while memory speed only improved at 10%.  CPU speed improvements slowed significantly partly due to major physical barriers and partly because current CPU designs have already hit the memory wall in some sense. https://en.wikipedia.org/wiki/Central_processing_unit 11 An Example: Intel Nehalem Core i-7 Register access, ~1 cycle (0.376 ns) L1 Cache access, ~4 cycles L2 Cache access, ~10 cycles L3 Cache access, line unshared, ~40 cycles L3 Cache access, shared line in another core, ~65 cycles L3 Cache access, modified in another core, ~75 cycles DRAM access, ~150 cycles Disk access, ~10,000 cycles CPU is very fast: 4 or 8 operations per cycle 12 An Example: Intel Nehalem Core i-7 Register access, ~1 cycle (0.376 ns) L1 Cache access, ~4 cycles L2 Cache access, ~10 cycles L3 Cache access, line unshared, ~40 cycles L3 Cache access, shared line in another core, ~65 cycles L3 Cache access, modified in another core, ~75 cycles DRAM access, ~150 cycles Disk access, ~10,000 cycles CPU is very fast: 4 or 8 operations per cycle Based on idea of spatial and temporal data locality • Cache hit: data found in cache • Cache miss: data not found in cache and must be copied from a lower level • Compulsory miss: first reference miss • Capacity miss: cache runs out of room for new data • Conflict miss: many data items map to same location in cache • True sharing miss: Occurs when a thread in another processor wants the same data. Cure: Minimize sharing • False sharing miss: Occurs when another processor uses different data in the same cache line. Cure: Pad data. 13 True sharing vs False Sharing True sharing: true sharing, would require programmatic synchronization constructs to ensure ordered data access. False sharing: The frequent coordination required between processors when cache lines are marked ‘Invalid’ requires cache lines to be written to memory and subsequently loaded. False sharing increases this coordination and can significantly degrade application performance. 14 1. C = C + A * B 2. for ( i=0; i

}

 Every array element a[…], b[…] is used twice within
▪ Define 4 registers to replace a[…], 4 registers to replace b[…] within

 Every array element c[…] is used n times in the k-loop
▪ Define 4 registers to replace c[…] before the k-loop begin

c[i*n + j] = a[i*n + k]*b[k*n + j] + a[i*n + k+1]*b[(k+1)*n + j]

+ c[i*n + j]

c[(i+1)*n + j] = a[(i+1)*n + k]*b[k*n + j] + a[(i+1)*n + k+1]*b[(k+1)*n + j]

+ c[(i+1)*n + j]

c[i*n + (j+1)] = a[i*n + k]*b[k*n + (j+1)] + a[i*n + k+1]*b[(k+1)*n + (j+1)]

+ c[i*n + (j+1)]

c[(i+1)*n + (j+1)] = a[(i+1)*n + k]*b[k*n + (j+1)]

+ a[(i+1)*n + k+1]*b[(k+1)*n + (j+1)] + c[(i+1)*n + (j+1)]

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Exploit more aggressive register reuse

for(i = 0; i < n; i += 2) for(j = 0; j < n; j += 2) { register int t = i*n+j; register int tt = t+n; register double c00 = c2[t]; register double c01 = c2[t+1]; register double c10 = c2[tt]; register double c11 = c2[tt+1]; for(k = 0; k < n; k += 2) { /* 2 by 2 mini matrix multiplication using registers*/ register int ta = i*n+k; register int tta = ta+n; register int tb = k*n+j; register int ttb = tb+n; register double a00 = a[ta]; register double a01 = a[ta+1]; register double a10 = a[tta]; register double a11 = a[tta+1]; register double b00 = b[tb]; register double b01 = b[tb+1]; register double b10 = b[ttb]; register double b11 = b[ttb+1]; c00 += a00*b00 + a01*b10; c01 += a00*b01 + a01*b11; c10 += a10*b00 + a11*b10; c11 += a10*b01 + a11*b11; } c2[t] = c00; c2[t+1] = c01; c2[tt] = c10; c2[tt+1] = c11; }  Every array element a[…], b[…] is used twice ▪ Define 4 registers to replace a[…], 4 registers to replace b[…] within

 Every array element c[…] is used n times in the k-loop
▪ Define 4 registers to replace c[…] before the k-loop begin

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Can We Use Less Than 12 Registers for 2X2 MM?

for(i = 0; i < n; i += 2) for(j = 0; j < n; j += 2) { register int t = i*n+j; register int tt = t+n; register double c00 = c2[t]; register double c01 = c2[t+1]; register double c10 = c2[tt]; register double c11 = c2[tt+1]; for(k = 0; k < n; k += 2) { /* 2 by 2 mini matrix multiplication using registers*/ register int ta = i*n+k; register int tta = ta+n; register int tb = k*n+j; register int ttb = tb+n; register double a00 = a[ta]; register double a10 = a[tta]; register double b00 = b[tb]; register double b01 = b[tb+1]; c00 += a00*b00 ; c01 += a00*b01 ; c10 += a10*b00 ; c11 += a10*b01 ; a00 = a[ta+1]; a10 = a[tta+1]; b00 = b[ttb]; b01 = b[ttb+1]; c00 += a00*b00 ; c01 += a00*b01 ; c10 += a10*b00 ; c11 += a10*b01 ; } c2[t] = c00; c2[t+1] = c01; c2[tt] = c10; c2[tt+1] = c11; }  Exercise: Assume you only have 16 floating point registers ➢ How to reorganize the loops to maximize the data reuse in registers? 20 How to calculate time wasted in accessing operands that are not in register  First, cycle time: 1 cycle = 1/frequency e.g., 5GHz = 5* 10^9 Hz, so 1 cycle time is 1/(5*10^9) sec.  Total time wasted in accessing operands that are not in registers? (Total time – Total Floating point operations time) / Total Time Caches  Cache: A smaller, faster storage device that acts as a staging area for a subset of the data in a larger, slower device.  Fundamental idea of a memory hierarchy: ▪ For each k, the faster, smaller device at level k serves as a cache for the larger, slower device at level k+1.  Why do memory hierarchies work? ▪ Because of locality, programs tend to access the data at level k more often than they access the data at level k+1. ▪ Thus, the storage at level k+1 can be slower, and thus larger and cheaper per bit.  Big Idea: The memory hierarchy creates a large pool of storage that costs as much as the cheap storage near the bottom, but that serves data to programs at the rate of the fast storage near the top. Cache Performance Metrics  Miss Rate ▪ Fraction of memory references not found in cache (misses / accesses) = 1 – hit rate ▪ Typical numbers (in percentages): ▪ 3-10% for L1 ▪ can be quite small (e.g., < 1%) for L2, depending on size, etc.  Hit Time ▪ Time to deliver a line in the cache to the processor ▪ includes time to determine whether the line is in the cache ▪ Typical numbers: ▪ 1-2 clock cycle for L1 ▪ 5-20 clock cycles for L2  Miss Penalty ▪ Additional time required because of a miss ▪ typically 50-200 cycles for main memory (Trend: increasing!) Lets think about those numbers  Huge difference between a hit and a miss ▪ Could be 100x, if just L1 and main memory  Would you believe 99% hits is twice as good as 97%? ▪ Consider: cache hit time of 1 cycle miss penalty of 100 cycles ▪ Average access time: 97% hits: 1 cycle + 0.03 * 100 cycles = 4 cycles 99% hits: 1 cycle + 0.01 * 100 cycles = 2 cycles  This is why “miss rate” is used instead of “hit rate” effective-access-time = cache-access-time + miss-rate * miss-penalty Writing Cache Friendly Code  Make the common case go fast ▪ Focus on the inner loops of the core functions  Minimize the misses in the inner loops ▪ Repeated references to variables are good (temporal locality) ▪ Stride-1 reference patterns are good (spatial locality) Key idea: Our qualitative notion of locality is quantified through our understanding of cache memories. Miss Rate Analysis for Matrix Multiply  Assume: ▪ Line size = 32Bytes (big enough for 4 64-bit double precision floating point numbers) ▪ Matrix dimension (N) is very large ▪ Approximate 1/N as 0.0 ▪ Cache is not even big enough to hold multiple rows  Analysis Method: ▪ Look at access pattern of inner loop A k i B k j C i j Matrix Multiplication Example  Description: ▪ Multiply N x N matrices ▪ O(N3) total operations ▪ N reads per source element ▪ N values summed per destination ▪ but may be able to hold in register /* ijk */ for (i=0; i 8 bytes (i.e., larger than one array element) , then exploit spatial locality

▪ compulsory miss rate = 8 bytes / B

 Stepping through rows in one column:
▪ for (i = 0; i < n; i++) sum += a[i][0]; ▪ accesses distant elements ▪ no spatial locality! ▪ compulsory miss rate = 1 (i.e. 100%) Matrix Multiplication (ijk) /* ijk */ for (i=0; i