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2122 Tunable Parameters
Database administrators can tune a database system at three levels The lowest level is at the hardware level Options for tuning systems at this level include adding disks or using a RAID system if disk I/O is a bottleneck, adding more memory if the disk buffer size is a bottleneck, or moving to a faster processor if CPU use is a bottleneck The second level consists of the database-system parameters, such as buffer size and checkpointing intervals The exact set of database-system parameters that can be tuned depends on the speci c database system Most database-system manuals provide information on what database-system parameters can be adjusted, and how you should choose values for the parameters Well-designed database systems perform as much tuning as possible automatically, freeing the user or database administrator from the burden For instance, in many database systems the buffer size is xed but tunable If the system automatically adjusts the buffer size by observing indicators such as page-fault rates, then the user will not have to worry about tuning the buffer size The third level is the highest level It includes the schema and transactions The administrator can tune the design of the schema, the indices that are created, and the transactions that are executed, to improve performance Tuning at this level is comparatively system independent The three levels of tuning interact with one another; we must consider them together when tuning a system For example, tuning at a higher level may result in the hardware bottleneck changing from the disk system to the CPU, or vice versa
2123 Tuning of Hardware
Even in a well-designed transaction processing system, each transaction usually has to do at least a few I/O operations, if the data required by the transaction is on disk An important factor in tuning a transaction processing system is to make sure that the disk subsystem can handle the rate at which I/O operations are required For instance, disks today have an access time of about 10 milliseconds, and transfer times of 20 MB per second, which gives about 100 random access I/O operations of 1 KB each If each transaction requires just 2 I/O operations, a single disk would support at most 50 transactions per second The only way to support more transactions per second is to increase the number of disks If the system needs to support n transactions per second, each performing 2 I/O operations, data must be striped (or otherwise partitioned) across n/50 disks (ignoring skew) Notice here that the limiting factor is not the capacity of the disk, but the speed at which random data can be accessed (limited in turn by the speed at which the disk arm can move) The number of I/O operations per transaction can be reduced by storing more data in memory If all data are in memory, there will be no disk I/O except for writes Keeping frequently used data in memory reduces the number of disk I/Os, and is worth the extra cost of memory Keeping very infrequently used data in memory would be a waste, since memory is much more expensive than disk The question is, for a given amount of money available for spending on disks or memory, what is the best way to spend the money to achieve maximum number of
Silberschatz Korth Sudarshan: Database System Concepts, Fourth Edition
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Performance Tuning
transactions per second A reduction of 1 I/O per second saves (price per disk drive) / (access per second per disk) Thus, if a particular page is accessed n times per second, the saving due to keeping it in memory is n times the above value Storing a page in memory costs (price per MB of memory) / (pages per MB of memory) Thus, the break-even point is n price per MB of memory price per disk drive = access per second per disk pages per MB of memory
We can rearrange the equation, and substitute current values for each of the above parameters to get a value for n; if a page is accessed more frequently than this, it is worth buying enough memory to store it Current disk technology and memory and disk prices give a value of n around 1/300 times per second (or equivalently, once in 5 minutes) for pages that are randomly accessed This reasoning is captured by the rule of thumb called the 5-minute rule: If a page is used more frequently than once in 5 minutes, it should be cached in memory In other words, it is worth buying enough memory to cache all pages that are accessed at least once in 5 minutes on an average For data that are accessed less frequently, buy enough disks to support the rate of I/O required for the data The formula for nding the break-even point depends on factors, such as the costs of disks and memory, that have changed by factors of 100 or 1000 over the past decade However, it is interesting to note that the ratios of the changes have been such that the break-even point has remained at roughly 5 minutes; the 5-minute rule has not changed to say, a 1-hour rule or a 1-second rule! For data that are sequentially accessed, signi cantly more pages can be read per second Assuming 1 MB of data is read at a time, we get the 1-minute rule, which says that sequentially accessed data should be cached in memory if they are used at least once in 1 minute The rules of thumb take only number of I/O operations into account, and do not consider factors such as response time Some applications need to keep even infrequently used data in memory, to support response times that are less than or comparable to disk access time Another aspect of tuning is in whether to use RAID 1 or RAID 5 The answer depends on how frequently the data are updated, since RAID 5 is much slower than RAID 1 on random writes: RAID 5 requires 2 reads and 2 writes to execute a single random write request If an application performs r random reads and w random writes per second to support a particular throughput, a RAID 5 implementation would require r + 4w I/O operations per second whereas a RAID 1 implementation would require r + w I/O operations per second We can then calculate the number of disks required to support the required I/O operations per second by dividing the result of the calculation by 100 I/O operations per second (for current generation disks) For many applications, r and w are large enough that the (r + w)/100 disks can easily hold two copies of all the data For such applications, if RAID 1 is used, the required number of disks is actually less than the required number of disks if RAID 5 is used! Thus RAID 5 is useful only when the data storage requirements are very large, but the I/O rates and data transfer requirements are small, that is, for very large and very cold data
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