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CSE 331 Computer Organization and Design Fall 2006 Week 15. Section 1: Mary Jane Irwin ( www.cse.psu.edu/~mji ) Section 2: Feihui Li ( www.cse.psu.edu/~feli ) Course material on ANGEL: cms.psu.edu [ adapted from D. Patterson slides ]. Head’s Up. Last week’s material
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CSE 331Computer Organization and DesignFall 2006Week 15 Section 1: Mary Jane Irwin (www.cse.psu.edu/~mji) Section 2: Feihui Li (www.cse.psu.edu/~feli ) Course material on ANGEL: cms.psu.edu [adapted from D. Patterson slides]
Head’s Up • Last week’s material • Intro to pipelined datapath design; memory design • This week’s material • Memory hierarchies • Reading assignment – PH: 7.1-7.2 • Final exam week • Final Exam is Tuesday, Dec 19, 4:40 to 6:30, 110 Business • Reminders • HW 8 (last) is due, Dec. 14 (by 11:55pm) • Quiz 7 (last) is due, Dec. 15 (by 11:55pm) • Dec. 13thdeadline for filing grade corrections/updates
To say that the P6 project had its share of people issues would be an understatement. Any effort that combines extreme creativity, mindbending stress, and colorful personalities has to account for the “people factor,” … The people factor is probably the least understood aspect of hardware design, and nothing you learn in graduate school compares with what a day-to-day, long-term, high-visibility, pressure-cooker project can teach you about your coworkers, or yourself. The Pentium Chronicles, Colwell, pg. 137
Cache Main Memory Secondary Memory (Disk) Review: Major Components of a Computer Processor Devices Control Input Memory Datapath Output
Inclusive– what is in L1$ is a subset of what is in L2$ is a subset of what is in MM that is a subset of is in SM 4-8 bytes (word) 8-32 bytes (block) Increasing distance from the processor in access time 1 to 4 blocks 1,024+ bytes (disk sector = page) (Relative) size of the memory at each level The Memory Hierarchy • Take advantage of the principle of locality to present the user with as much memory as is available in the cheapest technology at the speed offered by the fastest technology Processor L1$ L2$ Main Memory Secondary Memory
The Memory Hierarchy: Why Does it Work? • Temporal Locality (Locality in Time): Keep most recently accessed instructions/datum closer to the processor • Spatial Locality (Locality in Space): Move blocks consisting of contiguous words to the upper levels Lower Level Memory Upper Level Memory To Processor Blk X From Processor Blk Y
Lower Level Memory Upper Level Memory To Processor Blk X From Processor Blk Y The Memory Hierarchy: Terminology • Hit: data is in some block in the upper level (Blk X) • Hit Rate: the fraction of memory accesses found in the upper level • Hit Time: Time to access the upper level which consists of • RAM access time + Time to determine hit/miss • Miss: data is not in the upper level so needs to be retrieve from a block in the lower level (Blk Y) • Miss Rate = 1 - (Hit Rate) • Miss Penalty: Time to replace a block in the upper level + Time to deliver the block the processor • Hit Time << Miss Penalty
How is the Hierarchy Managed? • registers memory • by compiler (programmer?) • cache main memory • by the cache controller hardware • main memory disks • by the operating system (virtual memory) • virtual to physical address mapping assisted by the hardware (TLB) • by the programmer (files)
Why? The “Memory Wall” • The Processor vs DRAM speed “gap” continues to grow Clocks per DRAM access Clocks per instruction • Good cache design is increasingly important to overall performance
The Cache • Two questions to answer (in hardware): • Q1: How do we know if a data item is in the cache? • Q2: If it is, how do we find it? • Direct mapped • For each item of data at the lower level, there is exactly one location in the cache where it might be - so lots of items at the lower level must share locations in the upper level • Address mapping: (block address) modulo (# of blocks in the cache) • First consider block sizes of one word
Caching: A Simple First Example Main Memory 0000xx 0001xx 0010xx 0011xx 0100xx 0101xx 0110xx 0111xx 1000xx 1001xx 1010xx 1011xx 1100xx 1101xx 1110xx 1111xx Two low order bits define the byte in the word (32b words) Cache Index Valid Tag Data 00 01 10 11 Q2: How do we find it? Use next 2 low order memory address bits – the index – to determine which cache block (i.e., modulo the number of blocks in the cache) Q1: Is it there? Compare the cache tag to the high order 2 memory address bits to tell if the memory block is in the cache (block address) modulo (# of blocks in the cache)
01 4 11 15 Direct Mapped Cache • Consider the main memory word reference string 0 1 2 3 4 3 4 15 Start with an empty cache - all blocks initially marked as not valid 0 miss 1 miss 2 miss 3 miss 00 Mem(0) 00 Mem(1) 00 Mem(0) 00 Mem(0) 00 Mem(1) 00 Mem(2) 00 Mem(0) 00 Mem(1) 00 Mem(2) 00 Mem(3) miss 3 hit 4 hit 15 miss 4 00 Mem(0) 00 Mem(1) 00 Mem(2) 00 Mem(3) 01 Mem(4) 00 Mem(1) 00 Mem(2) 00 Mem(3) 01 Mem(4) 00 Mem(1) 00 Mem(2) 00 Mem(3) 01 Mem(4) 00 Mem(1) 00 Mem(2) 00 Mem(3) • 8 requests, 6 misses
Byte offset 31 30 . . . 13 12 11 . . . 2 1 0 Tag 20 Data 10 Hit Index Index Valid Tag Data 0 1 2 . . . 1021 1022 1023 20 32 MIPS Direct Mapped Cache Example • One word/block, cache size = 1K words What kind of locality are we taking advantage of?
Handling Cache Hits • Read hits (I$ and D$) • this is what we want! • Write hits (D$ only) • allow cache and memory to be inconsistent • write the data only into the cache (then write-back the cache contents to the memory when that cache block is “evicted”) • need a dirty bit for each cache block to tell if it needs to be written back to memory when it is evicted • require the cache and memory to be consistent • always write the data into both the cache and the memory (write-through) • don’t need a dirty bit • writes run at the speed of the main memory - slow! – or can use a write buffer, so only have to stall if the write buffer is full
Time (clock cycles) Inst 0 I n s t r. O r d e r D$ Reg D$ D$ D$ D$ Reg Reg Reg Reg Reg Reg Reg Reg Reg I$ I$ I$ I$ I$ ALU ALU ALU ALU ALU Inst 1 Inst 2 Inst 3 Inst 4 Review: Why Pipeline? For Throughput! • To avoid a structural hazard need two caches on-chip: one for instructions (I$) and one for data (D$) To keep the pipeline running at its maximum rate both I$ and D$ need to satisfy a request from the datapath every cycle. What happens when they can’t do that?
01 4 00 01 0 4 00 0 01 4 00 0 01 4 Another Reference String Mapping • Consider the main memory word reference string 0 4 0 4 0 4 0 4 Start with an empty cache - all blocks initially marked as not valid miss miss miss miss 0 4 0 4 00 Mem(0) 00 Mem(0) 01 Mem(4) 00 Mem(0) 4 0 4 0 miss miss miss miss 01 Mem(4) 00 Mem(0) 01 Mem(4) 00 Mem(0) • 8 requests, 8 misses • Ping pong effect due to conflict misses - two memory locations that map into the same cache block
Sources of Cache Misses • Compulsory (cold start or process migration, first reference): • First access to a block, “cold” fact of life, not a whole lot you can do about it • If you are going to run “millions” of instruction, compulsory misses are insignificant • Conflict (collision) • Multiple memory locations mapped to the same cache location • Solution 1: increase cache size • Solution 2: increase associativity • Capacity • Cache cannot contain all blocks accessed by the program • Solution: increase cache size
Handling Cache Misses • Read misses (I$ and D$) • stall the pipeline, fetch the block from the next level in the memory hierarchy, write the word+tag in the cache and send the requested word to the processor, let the pipeline resume • Write misses (D$ only) • stall the pipeline, fetch the block from next level in the memory hierarchy, install it in the cache (may involve having to evict a dirty block if using a write-back cache), write the word+tag in the cache, let the pipeline resume or (normally used in write-back caches) • Write allocate – just write the word+tag into the cache (may involve having to evict a dirty block), no need to check for cache hit, no need to stall or (normally used in write-through caches with a write buffer) • No-write allocate – skip the cache write (but must invalidate that cache block since it will now hold stale data) and just write the word to the write buffer (and eventually to the next memory level), no need to stall if the write buffer isn’t full
Byte offset Hit 31 30 . . . 13 12 11 . . . 4 3 2 1 0 Data 20 Tag 8 Block offset Index Data Index Valid Tag 0 1 2 . . . 253 254 255 20 32 Multiword Block Direct Mapped Cache • Four words/block, cache size = 1K words What kind of locality are we taking advantage of?
0 1 2 3 4 3 11 01 5 15 14 4 4 15 Taking Advantage of Spatial Locality • Let cache block hold more than one word 0 1 2 3 4 3 4 15 Start with an empty cache - all blocks initially marked as not valid miss hit miss 00 Mem(1) Mem(0) 00 Mem(1) Mem(0) 00 Mem(1) Mem(0) 00 Mem(3) Mem(2) hit miss hit 00 Mem(1) Mem(0) 00 Mem(1) Mem(0) 01 Mem(5) Mem(4) 00 Mem(3) Mem(2) 00 Mem(3) Mem(2) 00 Mem(3) Mem(2) hit miss 01 Mem(5) Mem(4) 01 Mem(5) Mem(4) 00 Mem(3) Mem(2) 00 Mem(3) Mem(2) • 8 requests, 4 misses
We concluded that no single quantifiable metric could reliably predict career success. However, we also felt that fast-trackers have identifiable qualities and show definite trends. One quality is a whatever-it-takes attitude. All high-output engineers have a willingness to do whatever it takes to make a project succeed, especially their particular corner of it. If that means working weekends, doing extracurricular research, or rewriting tools, they do it. The Pentium Chronicles, Colwell, pg. 140
Miss Rate vs Block Size vs Cache Size • Miss rate goes up if the block size becomes a significant fraction of the cache size because the number of blocks that can be held in the same size cache is smaller (increasing capacity misses)
Average Access Time Miss Rate Miss Penalty Exploits Spatial Locality Increased Miss Penalty & Miss Rate Fewer blocks compromises Temporal Locality Block Size Block Size Block Size Block Size Tradeoff • Larger block sizes take advantage of spatial locality but • If the block size is too big relative to the cache size, the miss rate will go up • Larger block size means larger miss penalty • Latency to first word in block + transfer time for remaining words • In general, Average Memory Access Time = Hit Time + Miss Penalty x Miss Rate
Multiword Block Considerations • Read misses (I$ and D$) • Processed the same as for single word blocks – a miss returns the entire block from memory • Miss penalty grows as block size grows • Early restart – datapath resumes execution as soon as the requested word of the block is returned • Requested word first – requested word is transferred from the memory to the cache (and datapath) first • Nonblocking cache – allows the datapath to continue to access the cache while the cache is handling an earlier miss • Write misses (D$) • Can’t use write allocate or will end up with a “garbled” block in the cache (e.g., for 4 word blocks, a new tag, one word of data from the new block, and three words of data from the old block), so must fetch the block from memory first and pay the stall time
Other Ways to Reduce Cache Miss Rates • Allow more flexible block placement • In a direct mappedcache a memory block maps to exactly one cache block • At the other extreme, could allow a memory block to be mapped to any cache block – fully associative cache • A compromise is to divide the cache into sets each of which consists of n “ways” (n-way set associative) • Use multiple levels of caches • Add a second level of caches on chip – normally a unified L2 cache (i.e., it holds both instructions and data) • L1 caches focuses on minimizing hit time in support of a shorter clock cycle (smaller with smaller block sizes) • L2 cache focuses on reducing miss rate to reduce the penalty of long main memory access times (larger with larger block sizes)
Cache Summary • The Principle of Locality: • Program likely to access a relatively small portion of the address space at any instant of time • Temporal Locality: Locality in Time • Spatial Locality: Locality in Space • Three major categories of cache misses: • Compulsory misses: sad facts of life, e.g., cold start misses • Conflict misses: increase cache size and/or associativity Nightmare Scenario: ping pong effect! • Capacity misses: increase cache size • Cache design space • total size, block size, associativity (replacement policy) • write-hit policy (write-through, write-back) • write-miss policy (write allocate, write buffers)
Improving Cache Performance Reduce the miss rate • bigger cache • associative cache • larger blocks (16 to 64 bytes) • use a victim cache – a small buffer that holds the most recently discarded blocks Reduce the miss penalty • smaller blocks • for large blocks fetch critical word first • use a write buffer • check write buffer (and/or victim cache) on read miss – may get lucky • use multiple cache levels – L2 cache not tied to CPU clock rate • faster backing store/improved memorybandwidth • wider buses • SDRAMs Reduce the hit time • smaller cache • direct mapped cache • smaller blocks • for writes • no write allocate – just write to write buffer • write allocate – write to a delayed write buffer that then writes to the cache
N x M SRAM M bit planes Cycle Time Review: (DDR) SDRAM Operation Column Address +1 • After a row is read into the SRAM register • Input CAS as the starting “burst” address along with a burst length • Transfers a burst of data (ideally a cache block) from a series of sequential addr’s within that row • A clock controls transfer of successive words in the burst N cols DRAM Row Address N rows M-bit Output 1st M-bit Access 2nd M-bit 3rd M-bit 4th M-bit RAS CAS Col Address Row Add Row Address
Memory Systems that Support Caches • The off-chip interconnect and memory architecture affects overall system performance dramatically on-chip • Assume • 1 clock cycle (1 ns) to send the address from the cache to the Main Memory • 50 ns (50 processor clock cycles) for DRAM first word access time, 10 ns (10 clock cycles) cycle time (remaining words in burst for SDRAM) • 1 clock cycle (1 ns) to return a word of data from the Main Memory to the cache • Memory-Bus to Cache bandwidth • number of bytes accessed from Main Memory and transferred to cache/CPU per clock cycle CPU Cache bus 32-bit data & 32-bit addr per cycle Main Memory
One Word Wide Memory Organization • If the block size is one word, then for a memory access due to a cache miss, the pipeline will have to stall the number of cycles required to return one data word from memory cycle(s) to send address cycle(s) to read DRAM cycle(s) to return data total clock cycles miss penalty • Number of bytes transferred per clock cycle (bandwidth) for a miss is bytes per clock on-chip CPU Cache 1 50 1 52 bus Main Memory 4/52 = 0.077
50 cycles 10 cycles 10 cycles 10 cycles Burst Memory Organization • What if the block size is four words and a (DDR) SDRAM is used? cycle(s) to send 1st address cycle(s) to read DRAM cycle(s) to return last data word total clock cycles miss penalty • Number of bytes transferred per clock cycle (bandwidth) for a single miss is bytes per clock on-chip CPU 1 50 + 3*10 = 80 1 82 Cache bus Main Memory (4 x 4)/82 = 0.183
DRAM Memory System Summary • Its important to match the cache characteristics • caches access one block at a time (usually more than one word) with the DRAM characteristics • use DRAMs that support fast multiple word accesses, preferably ones that match the block size of the cache with the memory-bus characteristics • make sure the memory-bus can support the DRAM access rates and patterns • with the goal of increasing the Memory-Bus to Cache bandwidth