performance - five minutes rules - the price of one access of disc I/O -
this interesting topic, use following formula compute access interval time:
breakevenintervalinseconds = (pagespermbofram / accessespersecondperdisk) × (priceperdiskdrive / pricepermbofram).
it derived using formulas cost of ram hold page in buffer pool , the cost of (fractional) disk perform i/o every time page needed, equating these 2 costs, , solving equation interval between accesses.
so cost of disc i/o per access priceperdiskdrive / accessespersecondperdisk, question why disc i/o cost per access computed this?
the underlying assumption limit life of disk how many disk seeks there are, while ram has fixed cost size, , fixed lifetime regardless of how accessed. reasonable because seeking disk causes physical wear , tear, , when disk goes, lose whole disk. contrast ram has no physical moving parts, , not wear out use.
with assumption, cost of keeping data on disk depends on frequency of access , cost of disk. cost of keeping data in ram depends on how ram you're using. trying find break point between cheaper keep data on disk or in ram.
however equation given incomplete. while equation identifies relevant factors, there important constant of proportionality missing. how many accesses can average hard drive sustain? how long ram last on average? enter costs keeping data on hard drives , ram, , without them comparing apples , oranges.
this indicative of impression of whole paper. says lot @ great length, important topic, analysis sloppy. slopping , leave critical things out, , don't enough people understand thinking , when analysis appropriate doing. instance if trying maintain low latency system, have keep of data in ram. period. if you're processing large data sets , don't want pay keep in ram, streaming data to/from disk. if you're keeping data in redundant format, instance raid, doing more seeks per read admit.
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