Memory and Storage for SI

Memory and Storage for SI, Part 7: The Five-Minute Rule in Microseconds

Gray and Putzolu’s five-minute rule redone for flash that answers in microseconds, 512-byte reads, and GPU hosts, with every input sourced or labeled.

On this page 7 sections

In 1987, Jim Gray and Franco Putzolu turned a pricing question into a rule. If a page of data is read again within about five minutes, it’s cheaper to keep it in memory than to fetch it from disk. Zhang and colleagues rebuilt the rule from first principles in a model that uses normalized costs, where the boundary between DRAM and flash falls to a few seconds for GPU hosts paired with SSDs built for small random reads.

Memory and Storage for SI, Part 7: a balance with a DRAM module on one pan and a flash SSD on the other, and a clock dial whose sweep shrinksA level balance beam on a central post. On the left pan sits a DRAM module, on the right pan a flash SSD. The pivot of the beam is a clock dial: a long dashed sweep stands for the old interval, and a short bright sweep inside it for the new one.GUIDE / MEMORY AND STORAGE FOR SI, PART 7Five minutes to secondsRENT · READ · BALANCE

The Memory and Storage Hierarchy tells the rule’s history through 2017 and leaves the sum for today undone. Part 4 covered the drives that changed one of its inputs, and Part 6 the drives at the other end of the cluster. This part redoes the sum for flash that answers in microseconds, with each input marked as sourced, modeled, or forecast.

The rule prices one access

Gray’s rule weighs two costs. Keeping a page in DRAM means paying rent on that memory for as long as the page stays there. Reading it from storage instead uses up a slice of the drive each time, because a drive can only serve so many reads per second. Zhang’s team writes the classical rule as one ratio. It sets the break-even interval as the cost of one storage access divided by the cost of holding one page in DRAM. If a page comes back sooner than that, it’s cheaper to keep in memory. If it comes back later, it’s cheaper to leave on the drive.

The rule prices one access

The rule weighs DRAM rent against the price of one read

The balance point is the break-even interval

A balance beam held level on a pivot labeled break-even interval. The left pan holds the rent for one page held in DRAM. The right pan holds one read from the drive. Under the beam, the two-ratio formula is drawn as two boxes joined by a times sign. The first box, marked hardware, reads pages per MB of DRAM divided by reads per second per drive. The second box, marked market, reads price of a drive divided by price of a MB of DRAM. A note at the bottom gives the 1987 case: 15 random accesses per second per disc, one access every 400 seconds for a 1KB record.break-even intervalRent: one pageheld in DRAMOne read fromthe drivehardware(pages per MB of DRAM /reads per second per drive)market(price of a drive /price of a MB of DRAM)×1987: 15 random accesses per second per disc,one access every 400 seconds for a 1KB recordDRAMFlash driveBalance
Figure 7.1 · The interval is a hardware ratio times a market ratio. Source: Gray and Putzolu 1987; Appuswamy and colleagues; Zhang and colleagues.

In Gray and Putzolu’s paper, a disc and half a controller deliver 15 random accesses per second. At Tandem’s prices that put the break-even at one access every 400 seconds for a 1KB record, and they called that five minutes.

For the sums below I use the two-ratio form from Appuswamy and colleagues, and the hierarchy page explains where it comes from. It multiplies one ratio set by the hardware by one set by the market.

interval = (pages per MB of DRAM / reads per second per drive) × (price of a drive / price of a MB of DRAM)

From milliseconds to microseconds

Reads per second is the input that moved the most. A disc doing 15 random accesses a second spends about 67ms on each one. The hierarchy page follows hard drives and flash through 2017, and Figure 7.2 adds today’s drives to that history. Goetz Graefe’s 2007 flash disk served 6,200 reads per second, and Appuswamy’s 2017 SATA SSD served 67,000.

From milliseconds to microseconds

Random reads per second per drive rose from 15 to 10 million

About 67ms per access in 1987; under 5µs per XL-FLASH read

A horizontal bar chart on a logarithmic axis of random reads per second per drive, from 1 to 10 million. Seven drives, oldest first. A disc in 1987 does 15. A hard drive in 2007 does 83 and one in 2017 does 200. A flash disk in 2007 does 6,200. A SATA SSD in 2017 does 67,000. The Micron 9650 is rated at 5.4 million. The Kioxia GP1 is rated at 10 million at 512 bytes, drawn dashed because it is a maker rating at that read size. Disks are neutral and flash is teal. Each gridline is ten times the one before, so the bars climb about six orders of magnitude from the disc to the GP1.Drive and year101,000100,00010,000,000Random reads per second per drive, log scaleDisc, 198715Hard drive, 200783Hard drive, 2017200Flash disk, 20076,200SATA SSD, 201767,000Micron 96505.4 millionKioxia GP1, 512 bytes10 millionDisksFlashMaker rating at 512 bytes
Figure 7.2 · Each bar is one drive; the axis is logarithmic, so each gridline is ten times the one before. Source: Gray and Putzolu; Appuswamy and colleagues; Graefe; Micron 9650 product brief; Kioxia GP1 release; Kioxia XL-FLASH page.

Today’s drives are rated far higher. Micron rates its 9650 at 5.4 million random reads per second, and Kioxia rates its GP1 at up to 10 million per second at 512 bytes. Each read is faster too. Kioxia gives its XL-FLASH a read latency under 5µs, where a 1987 disc spent about 67ms on each access. That sets a die’s read latency against a whole disc’s service time, which are different measures.

Reads per second sit in the first ratio, so ten times the reads cuts the interval by ten, as long as the drive’s price doesn’t climb with them.

Redoing the sum

Appuswamy’s 2017 SATA SSD makes a clean worked example, because every input sits in one table. It serves 67,000 random reads per second and costs 560 USD, and DRAM costs 0.005 USD per MB. A megabyte holds 256 pages of 4KB, so the sum is:

interval = (256 / 67,000) × (560 / 0.005) = about 428s

That’s about 7 minutes, and the paper gives the same seven minutes for 4KB reads. The paper gets 41 seconds for the Intel 750, which was its NVMe drive of the time.

Redoing the sum

Page size, reads per second and prices each scale the interval

One input changed at a time, read top to bottom

A ladder of four boxes joined by three downward arrows, each arrow carrying a factor. The top box is the start: a 2017 SATA SSD with 4KB pages, about 428 seconds, about 7 minutes. The first arrow, times 8, leads to 512-byte pages at about 3,424 seconds, 57 minutes. The second arrow, times 1.6 net, leads to five times the reads at 512 bytes, a net 1.6 times the 4KB value. The third arrow, about times 0.8, leads to the first-quarter 2026 price forecasts, DRAM up 90% to 95% and enterprise SSD up 53% to 58%. The first two boxes are solid because the page gives their seconds; the last two are dashed because they carry factors only. This is a model, not a measurement.2017 SATA SSD, 4KB pages: about 428s (about 7 minutes)512-byte pages: x8, about 3,424s (57 minutes)5 times the reads at 512 bytes: net x1.6 of the 4KB value1Q26 price forecasts (DRAM up 90% to 95%,enterprise SSD up 53% to 58%): about x0.8x8x1.6x0.8Seconds on the pageFactor onlyOne input changed
Figure 7.3 · Model: Appuswamy’s 2017 inputs with one input changed at a time; the price step is a sensitivity on TrendForce forecasts. Source: Appuswamy and colleagues; Zhang and colleagues; TrendForce.

The page shrinks

Changing one input at a time shows what each one does. A megabyte holds 2,048 pages of 512 bytes, which is 8 times as many as at 4KB. At the same price, the same drive then breaks even at 8 times the interval, which works out to about 3,424s, or 57 minutes. If the drive keeps its read rate with the smaller pages, each read still takes the same share of it, while each page now holds an eighth as much DRAM, so a small page is cheap to keep and costly to fetch.

Drives built for small reads work against that. Zhang’s team says makers are developing drives with up to 10x more IOPS per dollar that scale as reads shrink. Their example is 50 million reads per second at 512 bytes against 10 million at 4KB. At the same drive price, five times the reads divides the eightfold rise by five, so in this thought experiment the 512-byte interval ends up 1.6 times the 4KB value.

The prices move

Because the formula divides the drive’s price by DRAM’s, the 2026 shortage moves the answer too. In February 2026, TrendForce forecast conventional DRAM contract prices up 90% to 95% for 1Q26 and enterprise SSD prices up 53% to 58%. That quarter has since ended. If nothing else changed, those two forecasts would cut the interval to about 0.8 of its earlier value in one quarter. That’s a sensitivity on forecast inputs, not a result. A proper 2026 number needs a drive’s rated small-read IOPS and its price from the same date, and none of the sources saved for Part 4 gives a price.

What the revisit adds

Zhang’s team keeps Gray’s balance and adds two costs that only show up with a fast drive. Their calibrated rule also charges for queue handling, interrupts and the rest of the host processor work on each I/O, and for the host DRAM bandwidth each transfer uses. At a few reads per second those costs are too small to notice, but at millions per second they add up to a real share of the total.

What the revisit adds

Zhang’s model adds host costs and lands on seconds

Left, a fraction. The top is three cost boxes added together: host processor per I/O, where a CPU core costs 4 and serves 1 million IOPS and a GPU streaming multiprocessor costs 3 and serves 4 million IOPS; host DRAM bandwidth per transfer; and SSD cost per I/O. A bracket under them is labeled saved by caching. A fraction bar divides that sum by DRAM rent per page. An arrow carries the result to three chips on the right, all model output: a CPU host at 512 bytes, about 34 seconds; a CPU host at 4KB, about 10 seconds; and a GPU host at 512 bytes, about 5 seconds, 7 times shorter. A note says the costs are normalized from manufacturing parameters.Host processor per I/OCPU core: cost 4, 1 million IOPSGPU SM: cost 3, 4 million IOPSHost DRAM bandwidth per transferSSD cost per I/O++saved by cachingDRAM rent per pageCPU host, 512 bytes:about 34sCPU host, 4KB:about 10sGPU host, 512 bytes:about 5s (7x shorter)Normalized costs from manufacturing parametersCost termModel resultCarried forward
Figure 7.4 · Model output. Source: Zhang and colleagues.

Their model takes its costs from die area and process node rather than market prices, which the authors say avoids buyer bias. In those units a server CPU core costs 4 and serves 1 million IOPS. A GPU streaming multiprocessor costs 3 and serves 4 million, and the authors take that GPU figure from NVIDIA’s SCADA work covered in Part 4.

These results come out of a model built on those normalized costs, so the seconds aren’t a price list for any drive you can buy. In the model, an SLC Storage-Next SSD on a CPU host breaks even at about 34s for 512-byte pages and about 10s for 4KB. Moving the same reads to a GPU host cuts the 512-byte figure from about 34s to about 5s, because the GPU serves more I/O for its cost.

Slower flash pushes the other way. As sensing latency grows from 5µs for SLC to 40µs for TLC, the drive’s IOPS per dollar falls and its share of the total cost rises. Relaxing the latency target from 7µs to 85µs shortens the 512-byte GPU interval by only about 1.5s. The paper finds that host processor IOPS matter most.

What to watch

Once one of the Storage-Next drives in Part 4 gets a published price next to its rated small-read IOPS, I can redo the sum on this page with real numbers instead of a sensitivity. To check and vary the model, Zhang’s team built a calibrated SSD simulator called MQSim-Next.

What to watch

The break-even interval went from minutes to seconds

Break-even interval between DRAM and the next tier, by date

A vertical timeline of the break-even interval between DRAM and the next tier, oldest at the top. In 1987, on disk, 400 seconds, five minutes. In 2007, on a flash disk, about 15 minutes, 876 seconds. In 2017, about 7 minutes on a SATA SSD and 41 seconds on NVMe. In the model, a GPU host with a Storage-Next SSD, about 5 seconds. The last mark is open and dashed: next is a priced drive with a rated small-read IOPS, which the page is waiting for. Disk is neutral, flash is teal, the model result is a ringed teal mark, and the line of time is copper.1987, disk: 400 seconds (five minutes)2007, flash disk: about 15 minutes (876 seconds)2017, SATA SSD: about 7 minutes; NVMe: 41 secondsModel, GPU host with a Storage-Next SSD: about 5 secondsNext: a priced drive with a rated small-read IOPSDiskFlashModelOpen
Figure 7.5 · The last figure is model output; the open mark waits on a priced drive with a rated small-read IOPS. Source: Gray and Putzolu; Graefe; Appuswamy and colleagues; Zhang and colleagues (model).

I think an interval measured in seconds makes the economic case for flash tiers like the ones in Part 5. In this model, data that’s read again within seconds is worth keeping in memory, and data that waits longer is cheaper on flash. KV cache blocks aren’t the paper’s pages, though, so their cutoff stays unknown until someone runs the model on them.

In the last part of the series, I step back from the arithmetic and place each memory family by the job it could do.

KEEP GOING / NEXT CHAPTERContinue to Part 8

Sources

Primary is the original paper, standard or datasheet. Secondary is a review, a vendor summary or a news report. Measured is my own test data.

Show all 8 sources, 7 primary