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Computer Architecture. Part III-B: Cache Memory. Access Time. If every memory reference to cache required transfer of one word between MM and cache, no increase in speed is achieved. In fact, speed will drop because apart from MM access, there is additional access to cache
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Computer Architecture Part III-B: Cache Memory
Access Time • If every memory reference to cache required transfer of one word between MM and cache, no increase in speed is achieved. In fact, speed will drop because apart from MM access, there is additional access to cache • Suppose reference is repeated n times, and after the first reference, location is always found in the cache • Average access time: tc= cache access time tm = main memory access time n = number of accesses/references
Cache Hit Ratio • The probability that a word will be found in the cache • Depends upon the program and the size and organization of the cache h = Number of times required word found in cache Total number of references h: hit ratio
Access Time ta = Average access time tc = Cache access time (1-h) = miss ratio tm = Memory access time
Fetch Mechanisms • Demand Fetch • Fetch a block from memory when it is needed and is not in the cache • Prefetch • Fetch block/s from memory before they are requested • Selective Fetch • Not always fetching blocks, dependent on some defined criterion; blocks are stored in MM rather than the cache
Write Mechanisms • When words are read from the cache, contents are not modified • However, in general, cache data can be modified and it is possible that data in cache is different from data in MM • Two mechanisms to keep cache and MM in sync • Write-through mechanism • Write-back mechanism
Synchronization Mechanisms • Write-through • Every write operation to the cache is simultaneously repeated for MM • Write-back • The write operation to MM is only done during block-replacement time (i.e. a block displaced by incoming block might be written back to MM, regardless whether the block was altered or not)
Replacement Algorithms • When the word being requested by the CPU is not in the cache, it needs to be transferred from MM. (or it can also be from secondary memory to MM) • A page fault occurs when a page or a block is not in the cache (or MM in the case of secondary memory) • Replacement algorithms determine which page/block to remove or overwrite
Characteristics • Usage based or Non-usage based • Usage based : the choice of page/block to replace is dependent on the how many times each page/block has been referenced • Non-usage based : Use some other criteria for replacement
Assumptions • For a given page size, we only need to consider the page/block number. • If we have a reference (hit) to a page p, then any immediately succeeding references to p does not cause a page fault • The size of memory/cache is represented as the number of pages it is capable of holding (page frame )
Example Consider the following address sequence calls: 0110 0432 0101 0612 0102 0103 0104 0101 0611 0102 0103 0302 which, at 100 bytes per page, can be reduced to the following access string: 1 4 1 6 1 6 1 3 This sequence of page requests is called a reference string.
Replacement Policies • Random replacement algorithm • First-in first-out replacement • Optimal Algorithm • Least recently used algorithm • Least Frequently Used • Most Frequently Used
Random Replacement • A page is chosen randomly at page fault time • There is no relationship between the pages or their use. • Choice is done by a random number generator.
FIFO • Memory treated as a queue • When a page comes in, it is inserted at the tail • When a page is removed, the entry at the head of the queue gets deleted • Easy to understand and program • Performance is not consistently good; dependent on reference string
FIFO Example Consider the following reference string: 7 0 1 2 0 3 0 4 2 With a page frame of 3 * * * * * * * * 7 0 1 2 2 3 0 4 2 7 0 1 1 2 3 0 4 7 0 0 1 2 3 0 An * indicates a miss (the page requested by the CPU is not in the cache or in MM)
FIFO Example #2 Consider the following reference string: 1 2 3 4 1 2 5 1 2 3 4 5 With a page frame of 3 * * * * * * * * * 1 2 3 4 1 2 5 5 5 3 4 4 1 2 3 4 1 2 2 2 5 3 3 1 2 3 4 1 1 1 2 5 5 We have 9 page faults Try performing this FIFO with a page frame of 4
Belady’s Anomaly • An increase in page frame does not necessarily mean a decrease in page faults • More formally, Belady’s anomalyreflects the fact that, for some page-replacement algorithms, the page fault rate may increase as the number of allocated frames increases
Optimal Algorithm • The page that will not be used for the longest period of time is replaced • Guarantees the lowest page fault rate for a fixed number of frames • Difficult to implement because it requires future knowledge of the reference string
Optimal Algorithm Example Consider the following reference string: 7 0 1 2 0 3 0 4 2 With a page frame of 3 We look ahead and see that 7 is the page which will not be used again, so we replace 7; we also note that after our first hit we should not replace 0 immediately, but rather 1 because 1 will not be referenced any more (2 will be referenced last.) * * * * * * 7 0 1 2 2 3 3 4 4 7 0 1 1 2 2 3 3 7 0 0 0 0 2 2
Least Recently Used • Approximates the optimal algorithm • Replaces the page that has not been used for the longest period of time • When all page frames have been used up and every time there is a page hit, the referenced page is placed at the tail to indicate it has been recently accessed
LRU Example Consider the following reference string: 7 0 1 2 0 3 0 4 0 3 0 2 With a page frame of 3 * * * * * * * 7 0 1 2 0 3 0 4 0 3 0 2 7 0 1 2 0 3 0 4 0 3 0 7 0 1 2 2 3 3 4 4 3 We have 7 page faults Try performing this LRU with a page frame of 4
Least Frequently Used • Counts the number of references made to each page; when page is accessed, counter is incremented by one • Page with smallest count is replaced • FIFO is used to resolve a tie • Rationale: Page with the bigger counter is an actively used page • Problem • Page initially actively may never be used again • Solved by using a decaying counter
LFU Example Consider the following reference string: 7 0 1 2 0 3 0 4 0 3 0 2 With a page frame of 3 * * * * * * * 71 01 11 21 21 31 31 41 41 41 41 21 71 01 11 11 21 21 31 31 32 32 32 71 01 02 02 03 03 04 04 05 05 We have 7 page faults Try performing this LFU with a page frame of 4
Most Frequently Used • Opposite of LFU • Replace page with the highest count • Tie is resolved using FIFO • Based on the argument that the page with smallest count has just been probably brought in and is yet to be used • Both LFU and MFU are not common and implementation is expensive.