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Cache Memory for Computer Architecture.ppt
- 1. William Stallings Computer Organization andArchitecture 7th Edition Chapter 4 Cache Memory
- 2. Characteristics • Location • Capacity •Unit of transfer • Access method • Performance • Physical type • Physical characteristics • Organisation
- 3. Location • CPU (registers) •Internal (main memory, cache ) • External (disk, tape accessed by I/O controller)
- 4. Capacity • Word size —Thenatural unit of organisation • Number of words —or Bytes
- 5. Unit of Transfer •Internal —Usually governed by data bus width • External —Usually a block which is much larger than a word • Addressable unit —Smallest location which can be uniquely addressed —Word internally —Cluster on M$ disks
- 6. Access Methods (1) •Sequential —Start at the beginning and read through in order —Access time depends on location of data and previous location —e.g. tape • Direct —Individual blocks have unique address —Access is by jumping to vicinity plus sequential search —Access time depends on location and previous location —e.g. disk
- 7. Access Methods (2) •Random —Individual addresses identify locations exactly —Access time is independent of location or previous access —e.g. RAM • Associative —Data is located by a comparison with contents of a portion of the store —Access time is independent of location or previous access —e.g. cache
- 8. Memory Hierarchy • Registers —InCPU • Internal or Main memory —May include one or more levels of cache —“RAM” • External memory —Backing store
- 9. Memory Hierarchy -Diagram
- 10. Physical Types • Semiconductor —RAM •Magnetic —Disk & Tape • Optical —CD & DVD • Others —Bubble —Hologram
- 11. The Bottom Line •How much? —Capacity • How fast? —Time is money • How expensive?
- 12. Hierarchy List • Registers •L1 Cache • L2 Cache • Main memory • Disk cache • Disk • Optical • Tape
- 13. So you wantfast? • It is possible to build a computer which uses only static RAM (see later) • This would be very fast • This would need no cache —How can you cache cache? • This would cost a very large amount
- 14. Locality of Reference •During the course of the execution of a program, memory references tend to cluster • e.g. loops
- 15. Cache • Small amountof fast memory • Sits between normal main memory and CPU • May be located on CPU chip or module
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- 17. Cache operation –overview • CPU requests contents of memory location • Check cache for this data • If present, get from cache (fast) • If not present, read required block from main memory to cache • Then deliver from cache to CPU • Cache includes tags to identify which block of main memory is in each cache slot
- 18. Cache Read Operation- Flowchart
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- 20. Cache Design • Size •Mapping Function • Replacement Algorithm • Write Policy • Block Size • Number of Caches
- 21. Comparison of CacheSizes a Two values seperated by a slash refer to instruction and data caches Processor Type Year of Introduction L1 cache L2 cache L3 cache IBM 360/85 Mainframe 1968 16 to 32 KB — — PDP-11/70 Minicomputer 1975 1 KB — — VAX 11/780 Minicomputer 1978 16 KB — — IBM 3033 Mainframe 1978 64 KB — — IBM 3090 Mainframe 1985 128 to 256 KB — — Intel 80486 PC 1989 8 KB — — Pentium PC 1993 8 KB/8 KB 256 to 512 KB — PowerPC 601 PC 1993 32 KB — — PowerPC 620 PC 1996 32 KB/32 KB — — PowerPC G4 PC/server 1999 32 KB/32 KB 256 KB to 1 MB 2 MB IBM S/390 G4 Mainframe 1997 32 KB 256 KB 2 MB IBM S/390 G6 Mainframe 1999 256 KB 8 MB — Pentium 4 PC/server 2000 8 KB/8 KB 256 KB — IBM SP High-end server/ supercomputer 2000 64 KB/32 KB 8 MB — CRAY MTAb Supercomputer 2000 8 KB 2 MB — Itanium PC/server 2001 16 KB/16 KB 96 KB 4 MB SGI Origin 2001 High-end server 2001 32 KB/32 KB 4 MB — Itanium 2 PC/server 2002 32 KB 256 KB 6 MB IBM POWER5 High-end server 2003 64 KB 1.9 MB 36 MB CRAY XD-1 Supercomputer 2004 64 KB/64 KB 1MB —
- 22. Mapping Function • Cacheof 64kByte • Cache block of 4 bytes —i.e. cache is 16k (214) lines/blocks of 4 bytes • 16MBytes main memory • 24 bit address —(224=16M) —2^22 blocks RAM
- 23. Direct Mapping • Eachblock of main memory maps to only one cache line —i.e. if a block is in cache, it must be in one specific place • Address is in two parts • Least Significant w bits identify unique word • Most Significant s bits specify one memory block • The MSBs are split into a cache line field r and a tag of s-r (most significant)
- 24. Direct Mapping Address Structure Tags-r Line or Slot r Word w 8 14 2 • 24 bit address • 2 bit word identifier (4 byte block) • 22 bit block identifier — 8 bit tag (=22-14) — 14 bit slot or line • No two blocks in the same line have the same Tag field • Check contents of cache by finding line and checking Tag
- 25. Direct Mapping Cache LineTable i=j modulo m • i= cache line number • j= main memory block number • m= number of lines in cache • Cache line Main Memory blocks held • 0 0, m, 2m, 3m…2s-m • 1 1,m+1, 2m+1…2s-m+1 • m-1 m-1, 2m-1,3m-1…2s-1
- 27. Direct Mapping CacheOrganization
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- 29. Direct Mapping Summary •Address length = (s + w) bits • Number of addressable units = 2^(s+w) words or bytes • Block size = line size = 2^w words or bytes • Number of blocks in main memory = 2^(s+ w)/2^w = 2^s • Number of lines in cache = m = 2^r • Size of tag = (s – r) bits
- 30. Direct Mapping pros& cons • Simple • Inexpensive • Fixed location for given block —If a program accesses 2 blocks that map to the same line repeatedly, cache misses are very high
- 31. Associative Mapping • Amain memory block can load into any line of cache • Memory address is interpreted as tag and word • Tag uniquely identifies block of memory • Every line’s tag is examined for a match • Cache searching gets expensive
- 32. Fully Associative CacheOrganization
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- 34. Tag 22 bit Word 2bit Associative Mapping Address Structure • 22 bit tag stored with each 32 bit block of data • Compare tag field with tag entry in cache to check for hit • Least significant 2 bits of address identify which 16 bit word is required from 32 bit data block • e.g. —Address Tag Data Cache line —16339C 058CE7 24682468 3FFF — 0001 0110 0011 0011 1001 1100 =16339C —0000 0101 1000 1100 1110 0111 =058BE7 tag
- 35. Associative Mapping Summary •Advantage of associative mapping: —1. There is flexibility when mapping a block to any line of the cache • Disadvantages of associative mapping: —1. A replacement algorithm must be used to determine which line of cache to swap out —2. More space is needed for the tag field —3. The most important disadvantage is the complex circuitry needed to examine all of the tags in parallel in the cache
- 36. Set Associative Mapping •Cache is divided into a number of sets • Each set contains a number of lines • A given block maps to any line in a given set —e.g. Block B can be in any line of set i • e.g. 2 lines per set —2 way associative mapping —A given block can be in one of 2 lines in only one set • The divided into v sets, each of which consists of k lines (k- way set associative memory) m= v×k i=j modulo v • i=cache set number • j= main memory block number • m=number of lines in cache
- 37. Set Associative Mapping Example •13 bit set number • Block number in main memory is modulo 213 • 000000, 00A000, 00B000, 00C000 … map to same set 0
- 38. Two Way SetAssociative Cache Organization
- 39. Set Associative Mapping AddressStructure • Use set field to determine cache set to look in • Compare tag field to see if we have a hit • e.g —Address Tag Data Set number —1FF 7FFC 1FF 12345678 1FFF —001 7FFC 001 11223344 1FFF Tag 9 bit Set 13 bit Word 2 bit
- 40.
- 41. Set Associative MappingSummary • Address length = (s + w) bits • Number of addressable units = 2^(s+w) words or bytes • Block size = line size = 2^w words or bytes • Number of blocks in main memory = 2^(s+w) / 2^w= 2^s • Number of lines in set = k • Number of sets = v = 2^d • Number of lines in cache = kv = k * 2^d • Size of tag = (s – d) bits
- 42. • The dividedinto v sets, each of which consists of k lines m= v×k i=j modulo v • i=cache set number • j= main memory block number • m=number of lines in cache • V=m k=1 direct mapping • V=1 k=m associative mapping
- 43. Replacement Algorithms (1) Directmapping • No choice • Each block only maps to one line • Replace that line/block
- 44. Replacement Algorithms (2) Associative& Set Associative • Hardware implemented algorithm (speed) • Least Recently used (LRU) • e.g. in 2 way set associative —Which of the 2 block is lru? • First in first out (FIFO) —replace block that has been in cache longest • Least frequently used —replace block which has had fewest hits (counter ) • Random
- 45. • In 2-wayset associative scheme we can attach a USE bit with both the lines in a set • Whenever a line is read its USE bit is set and that of other line reset • For replacement, the line with USE bit 0 is selected • In a fully associative cache, an index of line use is maintained and least used line is chosen for replacement. • Some other algorithms are FIFO, LFU (Least Frequently Used), or random. • All algorithms are implemented in hardware
- 46. Write Policy • Mustnot overwrite a cache block unless main memory is up to date • Multiple CPUs may have individual caches • I/O may address main memory directly
- 47. Write through • Allwrites go to main memory as well as cache • Multiple CPUs can monitor main memory traffic to keep local (to CPU) cache up to date • Disadvantage —Lots of traffic —Slows down writes
- 48. Write back • Updatesinitially made in cache only • Update bit for cache slot/line is set when update occurs • If block is to be replaced, write to main memory only if update bit is set • Other caches get out of sync • I/O must access main memory through cache
- 50. Tag will storewith each word. problem15
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- 55. • Cache accesstime Tc = 100 ns Memory access time Tm = 500 ns If the effective access time is 10% greater than the cache access time, what is the hit ratio H? effective access time =cache hit ratio*cache access time+ cache miss ratio *(cache access time +main memory access time) effective access time =10% greater the cache access time ==>110 let cache hit ratio is h 110 = h*100ns +(1-h) (100+500) 110= 100h+600-600h 500h= 490 h= 490/500= .98 = 98%
- 56. • Two levelmemory contains cache and main memory. Cache access time is 20ns and main memory access time is 120ns/word. The size of block is 4 words. Main memory is referred 20% of times. What is the average access time: Here memory hierarchy is followed , so for each main memory reference block need to be copied back to cache So 1 word memory access time =120 ns so 4 word memory access time=Tm=4*120=480 ns Hc=cache hit ratio= 80%=0.8 (1-Hc)=memory hit ratio=cache miss ratio=20%=0.2 Tc=20ns Tm=480ns Average Access time=Hc*TC + (1- Hc)*(Tm+Tc)=(0.8*20)+0.2*(480+20)=116ns