55:035 Computer Architecture and Organization Lecture 7 155:035 Computer Architecture and Organization

Outline  Cache Memory Introduction  Memory Hierarchy  Direct-Mapped Cache  Set-Associative Cache  Cache Sizes  Cache Performance 255:035 Computer Architecture and Organization

 Memory access time is important to performance!  Users want large memories with fast access times  ideally unlimited fast memory  To use an analogy, think of a bookshelf containing many books: Suppose you are writing a paper on birds. You go to the bookshelf, pull out some of the books on birds and place them on the desk. As you start to look through them you realize that you need more references. So you go back to the bookshelf and get more books on birds and put them on the desk. Now as you begin to write your paper, you have many of the references you need on the desk in front of you.  This is an example of the principle of locality: This principle states that programs access a relatively small portion of their address space at any instant of time. Introduction 355:035 Computer Architecture and Organization

Levels of the Memory Hierarchy 55:035 Computer Architecture and Organization4 Part of The On-chip CPU Datapath ISA Registers One or more levels (Static RAM): Level 1: On-chip 16-64K Level 2: On-chip 256K-2M Level 3: On or Off-chip 1M-16M Registers Cache Level(s) Main Memory Magnetic Disc Optical Disk or Magnetic Tape Farther away from the CPU: Lower Cost/Bit Higher Capacity Increased Access Time/Latency Lower Throughput/ Bandwidth Dynamic RAM (DRAM) 256M-16G Interface: SCSI, RAID, IDE, G-300G CPU

Memory Hierarchy Comparisons 55:035 Computer Architecture and Organization5 CPU Registers 100s Bytes <10s ns Cache K Bytes ns cents/bit Main Memory M Bytes 200ns- 500ns $ cents /bit Disk G Bytes, 10 ms (10,000,000 ns) cents/bit -5-6 Capacity Access Time Cost Tape infinite sec-min Registers Cache Memory Disk Tape Instr. Operands Blocks Pages Files Staging Xfer Unit prog./compiler 1-8 bytes cache cntl bytes OS 4K-16K bytes user/operator Mbytes faster Larger

 We can exploit the natural locality in programs by implementing the memory of a computer as a memory hierarchy. Multiple levels of memory with different speeds and sizes. The fastest memories are more expensive, and usually much smaller in size (see figure).  The user has the illusion of a memory that is both large and fast. Accomplished by using efficient methods for memory structure and organization. Memory Hierarchy 655:035 Computer Architecture and Organization

7 Inventor of Cache M. V. Wilkes, “Slave Memories and Dynamic Storage Allocation,” IEEE Transactions on Electronic Computers, vol. EC-14, no. 2, pp , April 1965.

55:035 Computer Architecture and Organization8 Cache Processor does all memory operations with cache. Miss – If requested word is not in cache, a block of words containing the requested word is brought to cache, and then the processor request is completed. Hit – If the requested word is in cache, read or write operation is performed directly in cache, without accessing main memory. Block – minimum amount of data transferred between cache and main memory. Processor Cache small, fast memory Main memory large, inexpensive (slow) words blocks

55:035 Computer Architecture and Organization9 The Locality Principle A program tends to access data that form a physical cluster in the memory – multiple accesses may be made within the same block. Physical localities are temporal and may shift over longer periods of time – data not used for some time is less likely to be used in the future. Upon miss, the least recently used (LRU) block can be overwritten by a new block. P. J. Denning, “The Locality Principle,” Communications of the ACM, vol. 48, no. 7, pp , July 2005.

There are two types of locality: TEMPORAL LOCALITY (locality in time) If an item is referenced, it will likely be referenced again soon. Data is reused. SPATIAL LOCALITY (locality in space) If an item is referenced, items in neighboring addresses will likely be referenced soon Most programs contain natural locality in structure. For example, most programs contain loops in which the instructions and data need to be accessed repeatedly. This is an example of temporal locality. Instructions are usually accessed sequentially, so they contain a high amount of spatial locality. Also, data access to elements in an array is another example of spatial locality. Temporal & Spatial Locality 1055:035 Computer Architecture and Organization

11 Data Locality, Cache, Blocks Increase block size to match locality size Increase cache size to include most blocks Data needed by a program Block 1 Block 2 Memory Cache

 Memory system is organized as a hierarchy with the level closest to the processor being a subset of any level further away, and all of the data is stored at the lowest level (see figure).  Data is copied between only two adjacent levels at any given time. We call the minimum unit of information contained in a two-level hierarchy a block or line. See the highlighted square shown in the figure.  If data requested by the user appears in some block in the upper level it is known as a hit. If data is not found in the upper levels, it is known as a miss. 55:035 Computer Architecture and Organization12 Basic Caching Concepts

Basic Cache Organization TagsData Array Full byte address: Decode & Row Select ? Compare Tags Hit Tag Idx Off Data Word Mux select Block address 1355:035 Computer Architecture and Organization

14 Direct-Mapped Cache Data needed by a program Memory Cache Data needed Swap-out Swap-in LRU Block 1 Block 2

55:035 Computer Architecture and Organization15 Set-Associative Cache Data needed by a program Memory Cache Data needed Swap-in Swap-out LRU Block 1 Block 2 Swap-in

Three Major Placement Schemes 1655:035 Computer Architecture and Organization

Direct-Mapped Placement A block can only go into one place in the cache  Determined by the block’s address (in memory space)  The index number for block placement is usually given by some low- order bits of block’s address. This can also be expressed as: (Index) = (Block address) mod (Number of blocks in cache) Note that in a direct-mapped cache,  Block placement & replacement choices are both completely determined by the address of the new block that is to be accessed. 1755:035 Computer Architecture and Organization

18 Direct-Mapped Cache Main memory Cache of 8 blocks → memory address cache address: tag index 32-word word-addressable memory Block size = 1 word index (local address) tag

55:035 Computer Architecture and Organization19 Direct-Mapped Cache Main memory Cache of 4 blocks → memory address cache address: tag index block offset 32-word word-addressable memory Block size = 2 word index (local address) tag block offset 01

55:035 Computer Architecture and Organization20 Direct-Mapped Cache (Byte Address) Main memory Cache of 8 blocks → memory address cache address: tag index 32-word byte-addressable memory Block size = 1 word index tag byte offset

55:035 Computer Architecture and Organization21 Finding a Word in Cache Valid 2-bit Indexbit TagData byte offset b6 b5 b4 b3 b2 b1 b0 = Data 1 = hit 0 = miss Tag Index Memory address Cache size 8 words Block size = 1 word 32 words byte-address

55:035 Computer Architecture and Organization22 Miss Rate of Direct-Mapped Cache Main memory Cache of 8 blocks → memory address cache address: tag index 32-word word-addressable memory Block size = 1 word index tag byte offset Least recently used (LRU) block This block is needed

55:035 Computer Architecture and Organization23 Miss Rate of Direct-Mapped Cache Main memory Cache of 8 blocks → memory address cache address: tag index 32-word word-addressable memory Block size = 1 word index tag 00 / 01 / 00 / 10 xx 00 xx byte offset Memory references to addresses: 0, 8, 0, 6, 8, miss 2. miss 4. miss 3. miss 5. miss 6. miss

55:035 Computer Architecture and Organization24 Fully-Associative Cache (8-Way Set Associative) Main memory Cache of 8 blocks → memory address cache address: tag 32-word word-addressable memory Block size = 1 word tag byte offset LRU block This block is needed

55:035 Computer Architecture and Organization25 Miss Rate: Fully-Associative Cache xxxxx Main memory Cache of 8 blocks → memory address cache address: tag 32-word word-addressable memory Block size = 1 word tag byte offset Memory references to addresses: 0, 8, 0, 6, 8, miss 2. miss 3. hit 4. miss 5. hit 6. miss

55:035 Computer Architecture and Organization26 Finding a Word in Associative Cache IndexValid 5-bit Data bit Tag byte offset b6 b5 b4 b3 b2 b1 b0 = Data 1 = hit 0 = miss 5 bit Tag no index Memory address Cache size 8 words Block size = 1 word 32 words byte-address Must compare with all tags in the cache

55:035 Computer Architecture and Organization27 Eight-Way Set-Associative Cache byte offset b31 b30 b29 b28 b27 index b1 b0 Data1 = hit 0 = miss 5 bit Tag Memory address Cache size 8 words Block size = 1 word 32 words byte-address = V | tag | data = = = = = = = 8 to 1 multiplexer

55:035 Computer Architecture and Organization28 Two-Way Set-Associative Cache Main memory Cache of 8 blocks → memory address cache address: tag index 32-word word-addressable memory Block size = 1 word index tags 000 | | | | 111 byte offset LRU block This block is needed

55:035 Computer Architecture and Organization29 Miss Rate: Two-Way Set-Associative Cache Main memory Cache of 8 blocks → memory address cache address: tag index 32-word word-addressable memory Block size = 1 word index tags 000 | 010 xxx | xxx 001 | xxx xxx | xxx byte offset Memory references to addresses: 0, 8, 0, 6, 8, miss 2. miss 4. miss 3. hit 5. hit 6. miss

55:035 Computer Architecture and Organization30 Two-Way Set-Associative Cache byte offset b6 b5 b4 b3 b2 b1 b0 Data 1 = hit 0 = miss 3 bit tag Memory address Cache size 8 words Block size = 1 word 32 words byte-address V | tag | data == to 1 MUX 2 bit index

55:035 Computer Architecture and Organization31 Using Larger Cache Block (4 Words) Val. 16-bit Data Indexbit Tag (4 words=128 bits) byte offset b31… b16 b15… b4 b3 b2 b1 b0 = Data 1 = hit 0 = miss 16 bit Tag 12 bit Index Memory address Cache size 16K words Block size = 4 word 4GB = 1G words byte-address 4K Indexes M U X 2 bit block offset

55:035 Computer Architecture and Organization32 Main memory Size=W words Number of Tag and Index Bits Cache Size = w words Each word in cache has unique index (local addr.) Number of index bits = log 2 w Index bits are shared with block offset when a block contains more words than 1 Assume partitions of w words each in the main memory. W/w such partitions, each identified by a tag Number of tag bits = log 2 (W/w)

55:035 Computer Architecture and Organization33 How Many Bits Does Cache Have?  Consider a main memory:  32 words; byte address is 7 bits wide: b6 b5 b4 b3 b2 b1 b0  Each word is 32 bits wide  Assume that cache block size is 1 word (32 bits data) and it contains 8 blocks.  Cache requires, for each word:  2 bit tag, and one valid bit  Total storage needed in cache = #blocks in cache × (data bits/block + tag bits + valid bit) = 8 (32+2+1) = 280 bits Physical storage/Data storage = 280/256 = 1.094

55:035 Computer Architecture and Organization34 A More Realistic Cache Consider 4 GB, byte-addressable main memory:  1Gwords; byte address is 32 bits wide: b31…b16 b15…b2 b1 b0  Each word is 32 bits wide Assume that cache block size is 1 word (32 bits data) and it contains 64 KB data, or 16K words, i.e., 16K blocks. Number of cache index bits = 14, because 16K = 2 14  Tag size = 32 – byte offset – #index bits = 32 – 2 – 14 = 16 bits Cache requires, for each word:  16 bit tag, and one valid bit  Total storage needed in cache = #blocks in cache × (data bits/block + tag size + valid bits) = 2 14 ( ) = 16×2 10 ×49 = 784×2 10 bits = 784 Kb = 98 KB Physical storage/Data storage = 98/64 = 1.53 But, need to increase the block size to match the size of locality.

55:035 Computer Architecture and Organization35 Cache Bits for 4-Word Block Consider 4 GB, byte-addressable main memory:  1Gwords; byte address is 32 bits wide: b31…b16 b15…b2 b1 b0  Each word is 32 bits wide Assume that cache block size is 4 words (128 bits data) and it contains 64 KB data, or 16K words, i.e., 4K blocks. Number of cache index bits = 12, because 4K = 2 12  Tag size = 32 – byte offset – #block offset bits – #index bits = 32 – 2 – 2 – 12 = 16 bits Cache requires, for each word:  16 bit tag, and one valid bit  Total storage needed in cache = #blocks in cache × (data bits/block + tag size + valid bit) = 2 12 (4× ) = 4×2 10 ×145 = 580×2 10 bits =580 Kb = 72.5 KB Physical storage/Data storage = 72.5/64 = 1.13

Cache size equation Simple equation for the size of a cache: (Cache size) = (Block size) × (Number of sets) × (Set Associativity) Can relate to the size of various address fields: (Block size) = 2 (# of offset bits) (Number of sets) = 2 (# of index bits) (# of tag bits) = (# of memory address bits)  (# of index bits)  (# of offset bits) Memory address 3655:035 Computer Architecture and Organization

37 Interleaved Memory  Reduces miss penalty.  Memory designed to read words of a block simultaneously in one read operation.  Example:  Cache block size = 4 words  Interleaved memory with 4 banks  Suppose memory access ~15 cycles  Miss penalty = 1 cycle to send address + 15 cycles to read a block + 4 cycles to send data to cache = 20 cycles  Without interleaving, Miss penalty = 65 cycles Processor Cache Small, fast memory words blocks Memory bank 0 Memory bank 1 Memory bank 2 Memory bank 3 Main memory

Cache Design  The level’s design is described by four behaviors: Block Placement:  Where could a new block be placed in the given level? Block Identification:  How is a existing block found, if it is in the level? Block Replacement:  Which existing block should be replaced, if necessary? Write Strategy:  How are writes to the block handled? 55:035 Computer Architecture and Organization38

55:035 Computer Architecture and Organization39 Handling a Miss  Miss occurs when data at the required memory address is not found in cache.  Controller actions: Stall pipeline Freeze contents of all registers Activate a separate cache controller  If cache is full  select the least recently used (LRU) block in cache for over-writing  If selected block has inconsistent data, take proper action  Copy the block containing the requested address from memory Restart Instruction

55:035 Computer Architecture and Organization40 Miss During Instruction Fetch  Send original PC value (PC – 4) to the memory.  Instruct main memory to perform a read and wait for the memory to complete the access.  Write cache entry.  Restart the instruction whose fetch failed.

55:035 Computer Architecture and Organization41 Writing to Memory  Cache and memory become inconsistent when data is written into cache, but not to memory – the cache coherence problem.  Strategies to handle inconsistent data: Write-through  Write to memory and cache simultaneously always.  Write to memory is ~100 times slower than to (L1) cache. Write-back  Write to cache and mark block as “dirty”.  Write to memory occurs later, when dirty block is cast-out from the cache to make room for another block

55:035 Computer Architecture and Organization42 Writing to Memory: Write-Back Write-back (or copy back) writes only to cache but sets a “dirty bit” in the block where write is performed. When a block with dirty bit “on” is to be overwritten in the cache, it is first written to the memory. “Unnecessary” writes may occur for both write-through and write-back  write-through has extra writes because each store instruction causes a transaction to memory (e.g. eight 32-bit transactions versus 1 32-byte burst transaction for a cache line)  write-back has extra writes because unmodified words in a cache line get written even if they haven’t been changed  penalty for write-through is much greater, thus write-back is far more popular

55:035 Computer Architecture and Organization43 Cache Hierarchy  Average access time = T1 + (1 – h1) [ T2 + (1 – h2)Tm ]  Where  T1 = L1 cache access time (smallest)  T2 = L2 cache access time (small)  Tm = memory access time (large)  h1, h2 = hit rates (0 ≤ h1, h2 ≤ 1)  Average access time reduces by adding a cache. Processor L1 Cache (SRAM) Main memory large, inexpensive (slow) Access time = T1 Access time = Tm L2 Cache (DRAM) Access time = T2

55:035 Computer Architecture and Organization44 Average Access Time T1+T2+Tm T1 0 h1=1 1 h1=0 miss rate, 1- h1 Access time T1+T2+Tm / 2 T1+T2 T1 < T2 < Tm h2 = 0 h2 = 1 h2 = 0.5 T1 + (1 – h1) [ T2 + (1 – h2)Tm ]

55:035 Computer Architecture and Organization45 Processor Performance Without Cache  5GHz processor, cycle time = 0.2ns  Memory access time = 100ns = 500 cycles  Ignoring memory access, Clocks Per Instruction (CPI) = 1  Assuming no memory data access: CPI= 1 + # stall cycles = = 501

55:035 Computer Architecture and Organization46 Performance with Level 1 Cache  Assume hit rate, h1 = 0.95  L1 access time = 0.2ns = 1 cycle  CPI= 1 + # stall cycles = x 500 = 26  Processor speed increase due to cache = 501/26 = 19.3

55:035 Computer Architecture and Organization47 Performance with L1 and L2 Caches  Assume: L1 hit rate, h1 = 0.95 L2 hit rate, h2 = 0.90 (this is very optimistic!) L2 access time = 5ns = 25 cycles  CPI = 1 + # stall cycles = ( x 500) = = 4.75  Processor speed increase due to both caches = 501/4.75=  Speed increase due to L2 cache = 26/4.75= 5.47

If the tag bits do not match, then a miss occurs. Upon a cache miss:  The CPU is stalled  Desired block of data is fetched from memory and placed in cache.  Execution is restarted at the cycle that caused the cache miss. Recall that we have two different types of memory accesses:  reads (loads) or writes (stores). Thus, overall we can have 4 kinds of cache events:  read hits, read misses, write hits and write misses. 55:035 Computer Architecture and Organization48 Cache Miss Behavior

Fully-Associative Placement  One alternative to direct-mapped is: Allow block to fill any empty place in the cache.  How do we then locate the block later? Can associate each stored block with a tag  Identifies the block’s home address in main memory. When the block is needed, we can use the cache as an associative memory, using the tag to match all locations in parallel, to pull out the appropriate block. 4955:035 Computer Architecture and Organization

Set-Associative Placement The block address determines not a single location, but a set.  A set is several locations, grouped together. (set #) = (Block address) mod (# of sets) The block can be placed associatively anywhere within that set.  Where? This is part of the placement strategy. If there are n locations in each set, the scheme is called “n-way set-associative”.  Direct mapped = 1-way set-associative.  Fully associative = There is only 1 set. 5055:035 Computer Architecture and Organization

Replacement Strategies Which existing block do we replace, when a new block comes in? With a direct-mapped cache:  There’s only one choice! (Same as placement) With a (fully- or set-) associative cache:  If any “way” in the set is empty, pick one of those  Otherwise, there are many possible strategies:  (Pseudo-) Random: Simple, fast, and fairly effective  (Pseudo-) Least-Recently Used (LRU)  Makes little difference in L2 (and higher) caches 5155:035 Computer Architecture and Organization

Write Strategies Most accesses are reads, not writes  Especially if instruction reads are included Optimize for reads!  Direct mapped can return value before valid check Writes are more difficult, because:  We can’t write to cache till we know the right block  Object written may have various sizes (1-8 bytes) When to synchronize cache with memory?  Write through - Write to cache & to memory  Prone to stalls due to high mem. bandwidth requirements  Write back - Write to memory upon replacement  Memory may be left out of date for a long time 5255:035 Computer Architecture and Organization

Action on Cache Hits vs. Misses Read hits:  Desirable Read misses:  Stall the CPU, fetch block from memory, deliver to cache, restart Write hits:  Write-through: replace data in cache and memory at same time  Write-back: write the data only into the cache. It is written to main memory only when it is replaced Write misses:  No write-allocate: write the data to memory only.  Write-allocate: read the entire block into the cache, then write the word 55:035 Computer Architecture and Organization53

Cache Hits vs. Cache Misses  Consider the write-through strategy: every block written to cache is automatically written to memory.  Pro: Simple; memory is always up-to-date with the cache  No write-back required on block replacement.  Con: Creates lots of extra traffic on the memory bus.  Write hit time may be increased if CPU must wait for bus.  One solution to write time problem is to use a write buffer to store the data while it is waiting to be written to memory.  After storing data in cache and write buffer, processor can continue execution.  Alternately, a write-back strategy writes data to main memory only a block is replaced.  Pros: Reduces memory bandwidth used by writes.  Cons: Complicates multi-processor systems 55:035 Computer Architecture and Organization54

Hit/Miss Rate, Hit Time, Miss Penalty The hit rate or hit ratio is  fraction of memory accesses found in upper level. The miss rate (= 1 – hit rate) is  fraction of memory accesses not found in upper levels. The hit time is  the time to access the upper level of the memory hierarchy, which includes the time needed to determine whether the access is a hit or miss. The miss penalty is  the time needed to replace a block in the upper level with a corresponding block from the lower level.  may include the time to write back an evicted block. 55:035 Computer Architecture and Organization55

Cache Performance Analysis Performance is always a key issue for caches. We consider improving cache performance by:  (1) reducing the miss rate, and  (2) reducing the miss penalty. For (1) we can reduce the probability that different memory blocks will contend for the same cache location. For (2), we can add additional levels to the hierarchy, which is called multilevel caching. We can determine the CPU time as 55:035 Computer Architecture and Organization56

Cache Performance The memory-stall clock cycles come from cache misses. It can be defined as the sum of the stall cycles coming from writes + those coming from reads:  Memory-Stall CC = Read-stall cycles + Write-stall cycles, where 55:035 Computer Architecture and Organization57

Cache Performance Formulas Useful formulas for analyzing ISA/cache interactions :  (CPU time) = [(CPU cycles) + (Memory stall cycles)] × (Clock cycle time)  (Memory stall cycles) = (Instruction count) × (Accesses per instruction) × (Miss rate) × (Miss penalty) But, are not the best measure for cache design by itself:  Focus on time per-program, not per-access  But accesses-per-program isn’t up to the cache design  We can limit our attention to individual accesses  Neglects hit penalty  Cache design may affect #cycles taken even by a cache hit  Neglects cycle length  May be impacted by a poor cache design 55:035 Computer Architecture and Organization58

More Cache Performance Metrics Can split access time into instructions & data:  Avg. mem. acc. time = (% instruction accesses) × (inst. mem. access time) + (% data accesses) × (data mem. access time) Another simple formula:  CPU time = (CPU execution clock cycles + Memory stall clock cycles) × cycle time  Useful for exploring ISA changes Can break stalls into reads and writes:  Memory stall cycles = (Reads × read miss rate × read miss penalty) + (Writes × write miss rate × write miss penalty) 55:035 Computer Architecture and Organization59

Factoring out Instruction Count Gives (lumping together reads & writes): May replace:  So that miss rates aren’t affected by redundant accesses to same location within an instruction. 6055:035 Computer Architecture and Organization

Improving Cache Performance Consider the cache performance equation: It obviously follows that there are three basic ways to improve cache performance:  A. Reducing miss rate  B. Reducing miss penalty  C. Reducing hit time Note that by Amdahl’s Law, there will be diminishing returns from reducing only hit time or amortized miss penalty by itself, instead of both together. (Average memory access time) = (Hit time) + (Miss rate)×(Miss penalty) “Amortized miss penalty” Reducing amortized miss penalty 6155:035 Computer Architecture and Organization

62 AMD Opteron Microprocessor L1 (split 64KB each) Block 64B Write-back L2 1MB Block 64B Write-back