1 Chapter Seven

2 Users want large and fast memories! SRAM access times are ns at cost of $100 to $250 per Mbyte. DRAM access times are ns at cost of $5 to $10 per Mbyte. Disk access times are 10 to 20 million ns at cost of $.10 to $.20 per Mbyte. Try and give it to them anyway –build a memory hierarchy Exploiting Memory Hierarchy 1997

3 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. Why does code have locality? Our initial focus: two levels (upper, lower) –block: minimum unit of data moved to cache. –hit: data requested is in the upper level –miss: data requested is not in the upper level

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5 Two issues: –How do we know if a data item is in the cache? –If it is, how do we find it? Our first example: – block size is one word of data – "direct mapped" For each item of data at the lower level, there is exactly one location in the cache where it might be. e.g., lots of items at the lower level share locations in the upper level Cache

6 Cache memory Level of memory hierarchy between main memory and the CPU. Cache is constructed from very fast memory. It would be too expensive to construct main memory from this very fast memory. Cache memory is quite common. Note: For Cache line is often used instead of block. Address is broken into Tag, index and byte offset. The index is used to select the block in cache. The tag is compared with the value of the tag field in the cache blocked indexed. The offset is used to select a particular word in the cache.

7 Direct Mapped cache

8 Mapping: address is modulo the number of blocks in the cache Direct Mapped Cache

9 Taking advantage of spatial locality: Direct Mapped Cache (a). The instruction cache is direct mapped cache, and contains 64 lines. Each line contains 8 instructions. How many address bits are required for the tag? Answer: 64 lines require 6 bits for the index (2 6 = 64). 4 X 8 = 2 5 bytes in a line requires 5 bits for the offset. The number of bits for the tag = 32 – 6 – 5 = 21 bits.

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19 Homework A certain MIPS computer has a separate cache for instructions and data. (a). The instruction cache is direct mapped cache, and contains 64 lines. Each line contains 8 instructions. How many address bits are required for the tag? (b). The data cache is four way set associative and can hold a maximum or 4096 bytes of data. If four address bits are required for the index, how many bits are used for the tag? (c). Would it be possible to build the instruction cache such that the most significant bits of the address are used for the index instead of being used for the tag? (d). Explain either why this cannot be done, or if it can be done why it is not done.

20 Read hits –this is what we want! Read misses –stall the CPU, fetch block from memory, deliver to cache, restart Write hits: –can replace data in cache and memory (write-through) –write the data only into the cache (write-back the cache later) Write misses: –read the entire block into the cache, then write the word Hits vs. Misses

21 Make reading multiple words easier by using banks of memory It can get a lot more complicated... Hardware Issues

22 Increasing the block size tends to decrease miss rate: Use split caches because there is more spatial locality in code: Performance

23 Performance Simplified model: execution time = (execution cycles + stall cycles)  cycle time stall cycles = # of instructions  miss ratio  miss penalty(in cycles) Two ways of improving performance: –decreasing the miss ratio –decreasing the miss penalty What happens if we increase block size?

24 Compared to direct mapped, give a series of references that: –results in a lower miss ratio using a 2-way set associative cache –results in a higher miss ratio using a 2-way set associative cache assuming we use the “least recently used” replacement strategy Decreasing miss ratio with associativity

25 An implementation (b). The data cache is four way set associative and can hold a maximum or 4096 bytes of data. If four address bits are required for the index, how many bits are used for the tag? Answer: Each set contains 4096/4 = 1024 = 2 10 bytes of data. 4 index bits implies 2 4 lines in each set. Thus, there are 2 10 /2 4 = 2 6 bytes in each line requiring 6 bits for the offset. The number of tag bits = 32 – 4 – 6 = 22.

26 Performance

27 Decreasing miss penalty with multilevel caches Add a second level cache: –often primary cache is on the same chip as the processor –use SRAMs to add another cache above primary memory (DRAM) –miss penalty goes down if data is in 2nd level cache Example: –CPI of 1.0 on a 500Mhz machine with a 5% miss rate, 200ns DRAM access –Adding 2nd level cache with 20ns access time decreases miss rate to 2% Using multilevel caches: –try and optimize the hit time on the 1st level cache –try and optimize the miss rate on the 2nd level cache

28 Cache Summary Processor performance has increased faster than memory performance. Cache memory helps resolve this problem. When a word is not found in cache multiple words (called a block or line) are moved form main memory to cache memory. Each cache line contains a tag to identify the main memory address of the line. Set associative memories a use to approximate a fully associative cache memory. Average memory access time = Hit time + Miss rate X Miss penalty –Modern processors may execute other instructions when a miss occurs to reduce the effect of the miss.