whether the flash memory extends the buffer pool or the disk. The central question is, therefore, not what to keep in cache but how to manage flash memory contents and its lifetime.

In database systems, flash memory can also be used for recovery logs, because its short access times permit very fast transaction commit. Limitations in write bandwidth discourage such use, however. Perhaps systems with dual logs can combine low latency and high bandwidth, one log on a traditional disk and one log on an array of flash chips.

 

O THER HARD WARE

In all cases, RAM is assumed to be a substantial size, although probably less than flash memory or disk. The relative sizes should be governed by the five-minute rule. ¹¹ Note that despite similar transfer bandwidth, the short access latency of flash memory compared with disk results in surprising retention times for data in RAM.

Finally, we assume sufficient processing bandwidth as provided by modern many-core processors. Moreover, we believe that forthcoming transactional memory (in hardware and in the software runtime system) permits highly concurrent maintenance of complex data structures. For example, page replacement heuristics might use priority queues rather than bitmaps or linked lists. Similarly, advanced lock management might benefit from more complex data structures. Nonetheless, we do not assume or require data structures more complex than those already in common use for page replacement and location tracking.

THE FIVE-MINUTE RULE

If flash memory is introduced as an intermediate level in the memory hierarchy, relative sizing of memory levels requires renewed consideration. Tuning can be based on purchasing cost, total cost of ownership, power, mean time to failure, mean time to data loss, or a combination of metrics. Following Gray and Putzolu, ¹² this article focuses on purchasing cost. Other metrics and appropriate formulas to determine relative sizes can be derived similarly (e.g., by replacing dollar costs with energy use for caching and moving data).

Gray and Putzolu introduced the following formula: ^{13, 14}

BreakEvenIntervalinSeconds = (PagesPerMBofRAM / AccessesPerSecondPerDisk) × (PricePerDiskDrive / PricePerMBofRAM)

It is derived using formulas for the costs of RAM to hold a page in the buffer pool and of a (fractional) disk to perform I/O every time a page is needed, equating these two costs, and solving the equation for the interval between accesses.

Assuming modern RAM, a disk drive using 4-KB pages, and the values from tables 1 and 2, this produces:

 

(256 / 83) × ($80 / $0.047) = 5,248 seconds = 90 minutes = 1½ hours

 

(The “=” sign indicates rounding in this article.) This compares with two minutes (for 4-KB pages) 20 years ago.

If there is a surprise in this change, it is that the break-even interval has grown by less than two orders of magnitude. Recall that RAM was estimated in 1987 at about $5,000 per megabyte, whereas the 2007 cost is about $0.05 per megabyte, a difference of five orders of magnitude. On the other hand, disk prices have also tumbled ($15,000 per disk in 1987), and disk latency and bandwidth have improved considerably (from 15 accesses per second to about 100 on SATA and about 200 on high-performance SCSI disks).

For RAM and flash disks of 32 GB, the break-even interval is

(256 / 6,200) × ($999 / $0.047) = 876 seconds = 15 minutes

 

If the 2007 price for flash disks includes a “novelty premium” and comes down closer to the price of raw flash memory—say, to $400 (a price also anticipated by Gray and Fitzgerald¹⁵), then the break-even interval is 351 seconds = 6 minutes.

An important consequence is that in systems tuned using economic considerations, turn-over in RAM is about 15 times faster (90 minutes / 6 minutes) if flash memory rather than a traditional disk is the next level in the storage hierarchy. Much less RAM is required, resulting in lower costs for purchase, power, and cooling.

Perhaps most interesting, applying the same formula to flash and disk results in the following:

(256 / 83) × ($80 / $0.03) = 8,070 seconds = 2¼ hours

Thus, all active data will remain in RAM and flash memory.

Without a doubt, two hours is longer than any common checkpoint interval, which implies that dirty pages in flash are forced to disk not by page replacement but by checkpoints. Pages that are updated frequently must be