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15-447 Computer ArchitectureFall 2008 © October 27th, 2008 Majd F. Sakr CS-447– Computer Architecture.

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Presentation on theme: "15-447 Computer ArchitectureFall 2008 © October 27th, 2008 Majd F. Sakr CS-447– Computer Architecture."— Presentation transcript:

1 15-447 Computer ArchitectureFall 2008 © October 27th, 2008 Majd F. Sakr msakr@qatar.cmu.edu www.qatar.cmu.edu/~msakr/15447-f08/ CS-447– Computer Architecture Lecture 19 Memory Hierarchy

2 15-447 Computer ArchitectureFall 2008 © During This Lecture °Introduction to the Memory Hierarchy Processor Memory Gap Locality Latency Hiding

3 15-447 Computer ArchitectureFall 2008 © The Big Picture Processor (active) Computer Control (“brain”) Datapath (“brawn”) Memory (passive) (where programs, data live when running) Devices Input Output Keyboard, Mouse Display, Printer Disk, Network

4 15-447 Computer ArchitectureFall 2008 © Processor-DRAM Memory Gap (latency) µProc 60%/yr. (2X/1.5yr) DRAM 9%/yr. (2X/10 yrs) 1 10 100 1000 19801981198319841985198619871988198919901991199219931994199519961997199819992000 DRAM CPU 1982 Processor-Memory Performance Gap: (grows 50% / year) Performance Time “Moore’s Law”

5 15-447 Computer ArchitectureFall 2008 © °SRAM: value is stored on a pair of inverting gates very fast but takes up more space than DRAM (4 to 6 transistors) Memories:

6 15-447 Computer ArchitectureFall 2008 © DRAM: value is stored as a charge on capacitor (must be refreshed) very small but slower than SRAM (factor of 5 to 10) Word line Pass Transistor Bit line Capacitor Memories:

7 15-447 Computer ArchitectureFall 2008 © °Users want large and fast memories! °SRAM access times are.5 – 5ns at cost of $4000 to $10,000 per GB. °DRAM access times are 50-70ns at cost of $100 to $200 per GB. °Disk access times are 5 to 20 million ns at cost of $.50 to $2 per GB. Memory 2004

8 15-447 Computer ArchitectureFall 2008 © Storage Trends metric1980198519901995200020052005:1980 $/MB8,0008801003010.2040,000 access (ns)3752001007060508 typical size(MB) 0.0640.256416641,00015,000 DRAM metric1980198519901995200020052005:1980 $/MB19,2002,90032025610075256 access (ns)3001503515121030 SRAM metric1980198519901995200020052005:1980 $/MB50010080.300.050.00110,000 access (ms)877528108422 typical size(MB) 1101601,0009,000400,000400,000 Disk

9 15-447 Computer ArchitectureFall 2008 © CPU Clock Rates 1980198519901995200020052005:1980 processor 8080286386PentiumP-IIIP-4 clock rate(MHz) 16201507503,0003,000 cycle time(ns)1,0001665061.30.33,333

10 15-447 Computer ArchitectureFall 2008 © The CPU-Memory Gap The gap widens between DRAM, disk, and CPU speeds.

11 15-447 Computer ArchitectureFall 2008 © Locality °Principle of Locality: Programs tend to reuse data and instructions near those they have used recently, or that were recently referenced themselves. Temporal locality: Recently referenced items are likely to be referenced in the near future. Spatial locality: Items with nearby addresses tend to be referenced close together in time. Locality Example: Data –Reference array elements in succession (stride-1 reference pattern): –Reference sum each iteration: Instructions –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

12 15-447 Computer ArchitectureFall 2008 © Locality Example °Claim: Being able to look at code and get a qualitative sense of its locality is a key skill for a professional programmer. °Question: Does this function have good locality? int sum_array_rows(int a[M][N]) { int i, j, sum = 0; for (i = 0; i < M; i++) for (j = 0; j < N; j++) sum += a[i][j]; return sum; }

13 15-447 Computer ArchitectureFall 2008 © Locality Example °Question: Does this function have good locality? int sum_array_cols(int a[M][N]) { int i, j, sum = 0; for (j = 0; j < N; j++) for (i = 0; i < M; i++) sum += a[i][j]; return sum; }

14 15-447 Computer ArchitectureFall 2008 © Memory Hierarchy (1/3) °Processor executes instructions on order of nanoseconds to picoseconds holds a small amount of code and data in registers °Memory More capacity than registers, still limited Access time ~50-100 ns °Disk HUGE capacity (virtually limitless) VERY slow: runs ~milliseconds

15 15-447 Computer ArchitectureFall 2008 © Memory Hierarchy (2/3) Processor Size of memory at each level Increasing Distance from Proc., Decreasing speed Level 1 Level 2 Level n Level 3... Higher Lower Levels in memory hierarchy As we move to deeper levels the latency goes up and price per bit goes down.

16 15-447 Computer ArchitectureFall 2008 © Memory Hierarchy (3/3) °If level closer to Processor, it must be: smaller faster subset of lower levels (contains most recently used data) °Lowest Level (usually disk) contains all available data °Other levels?

17 15-447 Computer ArchitectureFall 2008 © Memory Caching °We’ve discussed three levels in the hierarchy: processor, memory, disk °Mismatch between processor and memory speeds leads us to add a new level: a memory cache °Implemented with SRAM technology

18 15-447 Computer ArchitectureFall 2008 © Memory Hierarchy Analogy: Library (1/2) °You’re writing a term paper (Processor) at a table in Library °Library is equivalent to disk essentially limitless capacity very slow to retrieve a book °Table is memory smaller capacity: means you must return book when table fills up easier and faster to find a book there once you’ve already retrieved it

19 15-447 Computer ArchitectureFall 2008 © Memory Hierarchy Analogy: Library (2/2) °Open books on table are cache smaller capacity: can have very few open books fit on table; again, when table fills up, you must close a book much, much faster to retrieve data °Illusion created: whole library open on the tabletop Keep as many recently used books open on table as possible since likely to use again Also keep as many books on table as possible, since faster than going to library

20 15-447 Computer ArchitectureFall 2008 © Memory Hierarchy Basis °Disk contains everything. °When Processor needs something, bring it into to all higher levels of memory. °Cache contains copies of data in memory that are being used. °Memory contains copies of data on disk that are being used. °Entire idea is based on Temporal Locality: if we use it now, we’ll want to use it again soon (a Big Idea)

21 15-447 Computer ArchitectureFall 2008 © Locality °A principle that makes having a memory hierarchy a good idea °If an item is referenced, temporal locality: it will tend to be referenced again soon spatial locality: nearby items will tend to be referenced soon.

22 15-447 Computer ArchitectureFall 2008 © A View of the Memory Hierarchy Regs L2 Cache Memory Disk Tape Instr. Operands Blocks Pages Files Upper Level Lower Level Faster Larger Cache Blocks

23 15-447 Computer ArchitectureFall 2008 © Our initial focus: two levels (upper, lower) block: minimum unit of data hit: data requested is in the upper level miss: data requested is not in the upper level Cache

24 15-447 Computer ArchitectureFall 2008 © Cache Design °How do we organize cache? °Where does each memory address map to? (Remember that cache is subset of memory, so multiple memory addresses map to the same cache location.) °How do we know which elements are in cache? °How do we quickly locate them?

25 15-447 Computer ArchitectureFall 2008 © Direct Mapped Cache °Mapping: address is modulo the number of blocks in the cache

26 15-447 Computer ArchitectureFall 2008 © Direct-Mapped Cache (1/2) °In a direct-mapped cache, each memory address is associated with one possible block within the cache Therefore, we only need to look in a single location in the cache for the data if it exists in the cache Block is the unit of transfer between cache and memory

27 15-447 Computer ArchitectureFall 2008 © Direct-Mapped Cache (2/2) °Cache Location 0 can be occupied by data from: Memory location 0, 4, 8,... 4 blocks => any memory location that is multiple of 4 Memory Memory Address 0 1 2 3 4 5 6 7 8 9 A B C D E F 4 Byte Direct Mapped Cache Cache Index 0 1 2 3

28 15-447 Computer ArchitectureFall 2008 © Issues with Direct-Mapped °Since multiple memory addresses map to same cache index, how do we tell which one is in there? °What if we have a block size > 1 byte? °Answer: divide memory address into three fields ttttttttttttttttt iiiiiiiiii oooo tagindexbyte to checkto offset if have selectwithin correct blockblockblock WIDTHHEIGHT Tag Index Offset

29 15-447 Computer ArchitectureFall 2008 © Direct-Mapped Cache Terminology °All fields are read as unsigned integers. °Index: specifies the cache index (which “row” of the cache we should look in) °Offset: once we’ve found correct block, specifies which byte within the block we want -- I.e., which “column” °Tag: the remaining bits after offset and index are determined; these are used to distinguish between all the memory addresses that map to the same location

30 15-447 Computer ArchitectureFall 2008 © Direct Mapped Cache (for MIPS)

31 15-447 Computer ArchitectureFall 2008 © Direct-Mapped Cache Example (1/3) °Suppose we have a 16KB of data in a direct-mapped cache with 4 word blocks °Determine the size of the tag, index and offset fields if we’re using a 32-bit architecture °Offset need to specify correct byte within a block block contains 4 words = 16 bytes = 2 4 bytes need 4 bits to specify correct byte

32 15-447 Computer ArchitectureFall 2008 © Direct-Mapped Cache Example (2/3) °Index: (~index into an “array of blocks”) need to specify correct row in cache cache contains 16 KB = 2 14 bytes block contains 2 4 bytes (4 words) # blocks/cache =bytes/cache bytes/block = 2 14 bytes/cache 2 4 bytes/block =2 10 blocks/cache need 10 bits to specify this many rows

33 15-447 Computer ArchitectureFall 2008 © Direct-Mapped Cache Example (3/3) °Tag: use remaining bits as tag tag length = addr length - offset - index = 32 - 4 - 10 bits = 18 bits so tag is leftmost 18 bits of memory address °Why not full 32 bit address as tag? All bytes within block need same address (4b) Index must be same for every address within a block, so its redundant in tag check, thus can leave off to save memory (10 bits in this example)

34 15-447 Computer ArchitectureFall 2008 © TIO cache mnemonic AREA (cache size, B) = HEIGHT (# of blocks) * WIDTH (size of one block, B/block) WIDTH (size of one block, B/block) HEIGHT (# of blocks) AREA (cache size, B) 2 (H+W) = 2 H * 2 W Tag Index Offset


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