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On occasion I've seen comments online along the lines of "make sure you set 'bs=' because the default value will take too long," and my own extremely-unscientific experiences of, "well that seemed to take longer than that other time last week" seem to bear that out. So whenever I use 'dd' (typically in the 1-2GB range) I make sure to specify the bytes parameter. About half the time I use the value specified in whatever online guide I'm copying from; the rest of the time I'll pick some number that makes sense from the 'fdisk -l' listing for what I assume is the slower media (e.g. the SD card I'm writing to).

For a given situation (media type, bus sizes, or whatever else matters), is there a way to determine a "best" value? Is it easy to determine? If not, is there an easy way to get 90-95% of the way there? Or is "just pick something bigger than 512" even the correct answer?

I've thought of trying the experiment myself, but (in addition to being a lot of work) I'm not sure what factors impact the answer, so I don't know how to design a good experiment.

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6 Answers 6

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There's but one way to determine the optimal block size, and that's a benchmark. I've just made a quick benchmark. The test machine is a PC running Debian GNU/Linux, with kernel 2.6.32 and coreutils 8.5. Both filesystems involved are ext3 on LVM volumes on a hard disk partition. The source file is 2GB (2040000kB to be precise). Caching and buffering are enabled. Before each run, I emptied the cache with sync; echo 1 >|/proc/sys/vm/drop_caches. The run times do not include a final sync to flush the buffers; the final sync takes on the order of 1 second.

The same runs were copies on the same filesystem; the diff runs were copies to a filesystem on a different hard disk. For consistency, the times reported are the wall clock times obtained with the time utility, in seconds. I only ran each command once, so I don't know how much variance there is in the timing.

             same   diff
             t (s)  t (s)
dd bs=64M    71.1   51.3
dd bs=1M     73.9   41.8
dd bs=4k     79.6   48.5
dd bs=512    85.3   48.9
cat          76.2   41.7
cp           77.8   45.3

Conclusion: A large block size (several megabytes) helps, but not dramatically (a lot less than I expected for same-drive copies). And cat and cp don't perform so badly. With these numbers, I don't find dd worth bothering with. Go with cat!

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  • 8
    @Nikhil >| is the same as > except that under set -o noclobber, the shell will complain that the file exists if you use >.
    – Gilles 'SO- stop being evil'
    Dec 23, 2011 at 21:29
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    @Masi Yes, if I want to clone a whole disk, I'll use cat. Why are you looking for a better way? What's wrong with cat?
    – Gilles 'SO- stop being evil'
    May 11, 2016 at 11:05
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    @Masi cat just copies its input to its output. If you want to copy from unreliable media, and skip over unreadable parts or retry multiple times, that's a different problem, for which ddrescue works pretty nicely.
    – Gilles 'SO- stop being evil'
    May 11, 2016 at 12:52
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    @sudo You can get the amount of data copied with lsof. Instant speed isn't very relevant with a disk copy because it's uniform so you can divide bytes transferred by elapsed time; if you want something better, you can use pv.
    – Gilles 'SO- stop being evil'
    May 6, 2017 at 18:36
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    @Silv unix.stackexchange.com/questions/32988/…
    – Gilles 'SO- stop being evil'
    May 9, 2020 at 11:27
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dd dates from back when it was needed to translate old IBM mainframe tapes, and the block size had to match the one used to write the tape or data blocks would be skipped or truncated. (9-track tapes were finicky. Be glad they're long dead.) These days, the block size should be a multiple of the device sector size (usually 4KB, but on very recent disks may be much larger and on very small thumb drives may be smaller, but 4KB is a reasonable middle ground regardless) and the larger the better for performance. I often use 1MB block sizes with hard drives. (We have a lot more memory to throw around these days too.)

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  • Hard drives or USB mass storage devices are either 512 or 4096 (newer) bytes. Optical and direct access flash media is 2048 bytes. Can't go wrong with 4096 bytes.
    – LawrenceC
    Mar 17, 2011 at 13:34
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    Why the copying program's block size should have anything to do with the underlying device's characteristics (tapes excepted)? The kernel does its own buffering (and sometimes prefetching) anyway.
    – Gilles 'SO- stop being evil'
    Mar 17, 2011 at 22:43
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    To minimize fractional buffers; things in general go faster when you use aligned buffers because the kernel can start buffer reads/writes at sector (or better, track or cylinder, but I think modern drives lie about those) and kernel buffer boundaries, because the kernel isn't having to skip over stuff or read extra stuff or manage partial buffers. Certainly you can just let the kernel deal with it all, but if you're copying gigabytes of data that extra work can cut the copy time down considerably.
    – geekosaur
    Mar 17, 2011 at 22:53
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    You (generally) need to include @Gilles if you want me to be notified of your comment reply, see How do comment @replies work?. Since I happened to be passing by: the kernel will deal with it all anyway. Your claim that “that extra work can cut the copy time down considerably” doesn't agree with my benchmarks, but different systems may have different behaviors, so please contribute timings too!
    – Gilles 'SO- stop being evil'
    Mar 17, 2011 at 23:07
  • @Gilles: sorry, I had mistaken you for the original asker.
    – geekosaur
    Mar 17, 2011 at 23:09
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I agree with geekosaur's answer that the size should be a multiple of the block size, which is often 4K.

If you want to find the block size stat -c "%o" filename is probably the easiest option.

But say you do dd bs=4K, that means it does read(4096); write(4096); read(4096); write(4096)...

Each system call involves a context switch, which involves some overhead, and depending on the I/O scheduler, reads with interspersed writes could cause the disk to do lots of seeks. (Probably not a major issue with the Linux scheduler, but nonetheless something to think about.)

So if you do bs=8K, you allow the disk to read two blocks at a time, which are probably close together on the disk, before seeking somewhere else to do the write (or to service I/O for another process).

By that logic, bs=16K is even better, etc.

So what I'd like to know is if there is an upper limit where performance starts to get worse, or if it's only bounded by memory.

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  • 5
    Profile, don't speculate!
    – Gilles 'SO- stop being evil'
    Mar 17, 2011 at 23:30
  • 4
    Interestingly, their benchmarks suggest there is little benefit above 4K however.
    – Mikel
    Mar 18, 2011 at 1:24
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    Also, apparently the default file read ahead window is 128 KB, so that value might be beneficial.
    – Mikel
    Mar 18, 2011 at 1:33
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    I have access to a 24 drive RAID50 here, where bs=8K gets me 197MB/s but bs=1M gets me 2.2 GB/s which is close to the theoretical throughput of the RAID. So bs matters ALOT. However using bs=10M I only get 1.7GB/s. So it appears to get worse over some threshold, but not sure why.
    – Joseph Garvin
    Nov 2, 2015 at 18:22
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    @JosephGarvin Yes, I always find it worthwhile to play around a bit with the block size if I'm doing a large transfer. It seems the optimum block size depends on a lot of different things.
    – sudo
    May 6, 2017 at 21:01
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As Gilles says, you can determine the optimal parameter for the bs option to dd by benchmarking. This, though, begs the question: how can you conveniently benchmark this parameter?

My tentative answer to this question is: use dd-opt, the utility I've recently started working on to solve precisely this problem :)

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  • 1
    What is the sensitivity of the output? 90-95% or >95%? I do not find that you can change it.
    – Léo Léopold Hertz 준영
    May 11, 2016 at 8:54
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    @Masi, I'm afraid I haven't worked on dd-opt in a long time. However, it is free software licensed under the AGPLv3. So, feel free to improve it and to evalutate its sensitivity/accuracy!
    – user6860
    May 11, 2016 at 11:17
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If not, is there an easy way to get 90-95% of the way there?

Use bs=1M

It'll give you more than 95% optimal performance over more than 85% your devices, from slow USB2/3 flash drives, SD cards and hard-drives to NVMe SSDs and even RAM-only devices such as /dev/zero.

Source?

Voices in my head.

And some empirical testing over 10+ years combined with pseudo-scientific benchmarking and biased common sense.

Hey, you asked about the easy way!

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I optimized for sdcard reader usb2.0 which seems to run best at bs=10M. I tried 4k, on up to 16M, after 8-10M no improvement. You can see how the transfer rate measurement degrades...most likely due to loading up the buffers on the device then waiting for the device to transfer to the actual medium.

angstrom/sdcard# dd if=/dev/zero of=/dev/sdb bs=10M
123+0 records in
123+0 records out
1289748480 bytes (1.3 GB) copied, 21.4684 s, 60.1 MB/s
341+0 records in
341+0 records out
3575644160 bytes (3.6 GB) copied, 117.636 s, 30.4 MB/s
816+0 records in
816+0 records out
8556380160 bytes (8.6 GB) copied, 326.588 s, 26.2 MB/s
955+0 records in
955+0 records out
10013900800 bytes (10 GB) copied, 387.456 s, 25.8 MB/s
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