37
Hard Disk Drives
The last chapter introduced the general concept of an I/O device and
showed you how the OS might interact with such a beast. In this chapter,
we dive into more detail about one device in particular: the hard disk
drive. These drives have been the main formof persistent data storage in
computer systems for decades and much of the development of file system
technology (coming soon) is predicated on their behavior. Thus, it
is worth understanding the details of a disk’s operation before building
the file system software that manages it. Many of these details are available
in excellent papers by Ruemmler and Wilkes [RW92] and Anderson,
Dykes, and Riedel [ADR03].
CRUX: HOW TO STORE AND ACCESS DATA ON DISK
How do modern hard-disk drives store data? What is the interface?
How is the data actually laid out and accessed? How does disk scheduling
improve performance?
37.1 The Interface
Let’s start by understanding the interface to a modern disk drive. The
basic interface for allmodern drives is straightforward. The drive consists
of a large number of sectors (512-byte blocks), each of which can be read
or written. The sectors are numbered from 0 to n − 1 on a disk with n
sectors. Thus, we can view the disk as an array of sectors; 0 to n − 1 is
thus the address space of the drive.
Multi-sector operations are possible; indeed, many file systems will
read or write 4KB at a time (or more). However, when updating the
disk, the only guarantee drive manufactures make is that a single 512-
byte write is atomic (i.e., it will either complete in its entirety or it won’t
complete at all); thus, if an untimely power loss occurs, only a portion of
a larger write may complete (sometimes called a torn write).
1
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0
11
8 9 10
7
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4 3 2
1
Spindle
Figure 37.1: A DiskWith Just A Single Track
There are some assumptions most clients of disk drives make, but
that are not specified directly in the interface; Schlosser and Ganger have
called this the “unwritten contract” of disk drives [SG04]. Specifically,
one can usually assume that accessing two blocks that are near one-another
within the drive’s address space will be faster than accessing two blocks
that are far apart. One can also usually assume that accessing blocks in
a contiguous chunk (i.e., a sequential read or write) is the fastest access
mode, and usually much faster than any more random access pattern.
37.2 Basic Geometry
Let’s start to understand some of the components of a modern disk.
We start with a platter, a circular hard surface on which data is stored
persistently by inducing magnetic changes to it. A disk may have one
or more platters; each platter has 2 sides, each of which is called a surface.
These platters are usually made of some hard material (such as
aluminum), and then coated with a thin magnetic layer that enables the
drive to persistently store bits even when the drive is powered off.
The platters are all bound together around the spindle, which is connected
to a motor that spins the platters around (while the drive is powered
on) at a constant (fixed) rate. The rate of rotation is oftenmeasured in
rotations per minute (RPM), and typical modern values are in the 7,200
RPM to 15,000 RPM range. Note that we will often be interested in the
time of a single rotation, e.g., a drive that rotates at 10,000 RPM means
that a single rotation takes about 6 milliseconds (6 ms).
Data is encoded on each surface in concentric circles of sectors; we call
one such concentric circle a track. A single surface contains many thousands
and thousands of tracks, tightly packed together, with hundreds of
tracks fitting into the width of a human hair.
To read and write from the surface, we need a mechanism that allows
us to either sense (i.e., read) the magnetic patterns on the disk or to induce
a change in (i.e., write) them. This process of reading and writing is
accomplished by the disk head; there is one such head per surface of the
drive. The disk head is attached to a single disk arm, which moves across
the surface to position the head over the desired track.
37.3 A Simple Disk Drive
Let’s understand how disks work by building up a model one track at
a time. Assume we have a simple disk with a single track (Figure 37.1).
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Head
Arm
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Figure 37.2: A Single Track Plus A Head
This track has just 12 sectors, each of which is 512 bytes in size (our
typical sector size, recall) and addressed therefore by the numbers 0 through
11. The single platter we have here rotates around the spindle, to which
a motor is attached. Of course, the track by itself isn’t too interesting; we
want to be able to read or write those sectors, and thus we need a disk
head, attached to a disk arm, as we now see (Figure 37.2).
In the figure, the disk head, attached to the end of the arm, is positioned
over sector 6, and the surface is rotating counter-clockwise.
Single-track Latency: The Rotational Delay
To understand how a request would be processed on our simple, onetrack
disk, imaginewe now receive a request to read block 0. How should
the disk service this request?
In our simple disk, the disk doesn’t have to do much. In particular, it
must just wait for the desired sector to rotate under the disk head. This
wait happens often enough inmodern drives, and is an important enough
component of I/O service time, that it has a special name: rotational delay
(sometimes rotation delay, though that sounds weird). In the example,
if the full rotational delay is R, the disk has to incur a rotational delay
of about R
2 to wait for 0 to come under the read/write head (if we start at
6). A worst-case request on this single track would be to sector 5, causing
nearly a full rotational delay in order to service such a request.
Multiple Tracks: Seek Time
So far our disk just has a single track, which is not too realistic; modern
disks of course have many millions. Let’s thus look at ever-so-slightly
more realistic disk surface, this one with three tracks (Figure 37.3, left).
In the figure, the head is currently positioned over the innermost track
(which contains sectors 24 through 35); the next track over contains the
next set of sectors (12 through 23), and the outermost track contains the
first sectors (0 through 11).

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Remaining rotation
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Figure 37.3: Three Tracks Plus A Head (Right: With Seek)
To understand howthe drivemight access a given sector,we nowtrace
what would happen on a request to a distant sector, e.g., a read to sector
11. To service this read, the drive has to firstmove the disk armto the correct
track (in this case, the outermost one), in a process known as a seek.
Seeks, along with rotations, are one of the most costly disk operations.
The seek, it should be noted, has many phases: first an acceleration
phase as the disk arm gets moving; then coasting as the arm is moving
at full speed, then deceleration as the arm slows down; finally settling as
the head is carefully positioned over the correct track. The settling time
is often quite significant, e.g., 0.5 to 2 ms, as the drive must be certain to
find the right track (imagine if it just got close instead!).
After the seek, the disk arm has positioned the head over the right
track. A depiction of the seek is found in Figure 37.3 (right).
As we can see, during the seek, the armhas been moved to the desired
track, and the platter of course has rotated, in this case about 3 sectors.
Thus, sector 9 is just about to pass under the disk head, and we must
only endure a short rotational delay to complete the transfer.
When sector 11 passes under the disk head, the final phase of I/O
will take place, known as the transfer, where data is either read from or
written to the surface. And thus, we have a complete picture of I/O time:
first a seek, then waiting for the rotational delay, and finally the transfer.
Some Other Details
Though we won’t spend too much time on it, there are some other interesting
details about how hard drives operate. Many drives employ some
kind of track skew to make sure that sequential reads can be properly
serviced even when crossing track boundaries. In our simple example
disk, this might appear as seen in Figure 37.4.
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Track skew: 2 blocks
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Figure 37.4: Three Tracks: Track Skew Of 2
Sectors are often skewed like this because when switching from one
track to another, the disk needs time to reposition the head (even to neighboring
tracks). Without such skew, the head would be moved to the next
track but the desired next block would have already rotated under the
head, and thus the drive would have to wait almost the entire rotational
delay to access the next block.
Another reality is that outer tracks tend to have more sectors than
inner tracks, which is a result of geometry; there is simply more room
out there. These tracks are often referred to as multi-zoned disk drives,
where the disk is organized into multiple zones, and where a zone is consecutive
set of tracks on a surface. Each zone has the same number of
sectors per track, and outer zones have more sectors than inner zones.
Finally, an important part of any modern disk drive is its cache, for
historical reasons sometimes called a track buffer. This cache is just some
small amount of memory (usually around 8 or 16 MB) which the drive
can use to hold data read from or written to the disk. For example, when
reading a sector from the disk, the drive might decide to read in all of the
sectors on that track and cache them in its memory; doing so allows the
drive to quickly respond to any subsequent requests to the same track.
On writes, the drive has a choice: should it acknowledge the write has
completed when it has put the data in its memory, or after the write has
actually been written to disk? The former is called write back caching
(or sometimes immediate reporting), and the latter write through. Write
back caching sometimesmakes the drive appear “faster”, but can be dangerous;
if the file system or applications require that data be written to
disk in a certain order for correctness, write-back caching can lead to
problems (read the chapter on file-system journaling for details).

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ASIDE: DIMENSIONAL ANALYSIS
Remember in Chemistry class, how you solved virtually every problemby
simply setting up the units such that they canceled out, and somehow
the answers popped out as a result? That chemical magic is known
by the highfalutin name of dimensional analysis and it turns out it is
useful in computer systems analysis too.
Let’s do an example to see how dimensional analysis works and why
it is useful. In this case, assume you have to figure out how long, in milliseconds,
a single rotation of a disk takes. Unfortunately, you are given
only the RPM of the disk, or rotations per minute. Let’s assume we’re
talking about a 10K RPM disk (i.e., it rotates 10,000 times per minute).
How do we set up the dimensional analysis so that we get time per rotation
in milliseconds?
To do so, we start by putting the desired units on the left; in this case,
we wish to obtain the time (in milliseconds) per rotation, so that is exactly
what we write down: Time (ms)
1 Rotation. We then write down everything
we know, making sure to cancel units where possible. First, we obtain
1 minute
10,000 Rotations (keeping rotation on the bottom, as that’s where it is on
the left), then transform minutes into seconds with 60 seconds
1 minute , and then
finally transform seconds in millisecondswith 1000 ms
1 second . The final result is
this equation, with units nicely canceled, is:
Time (ms)
1 Rot. = 1
minute
10,000 Rot. · 60
seconds
1
minute · 1000 ms
1
second = 60,000 ms
10,000 Rot. = 6 ms
Rotation
As you can see from this example, dimensional analysis makes what
seems obvious into a simple and repeatable process. Beyond the RPM
calculation above, it comes in handy with I/O analysis regularly. For
example, you will often be given the transfer rate of a disk, e.g.,
100 MB/second, and then asked: how long does it take to transfer a
512 KB block (in milliseconds)? With dimensional analysis, it’s easy:
Time (ms)
1 Request = 512 KB
1 Request · 1
 MB
1024 KB
· 1
second
100
 MB
· 1000 ms
1
second = 5 ms
Request
37.4 I/O Time: Doing The Math
Now that we have an abstract model of the disk, we can use a little
analysis to better understand disk performance. In particular, we can
now represent I/O time as the sum of three major components:
TI/O = Tseek + Trotation + Ttransfer (37.1)
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Cheetah 15K.5 Barracuda
Capacity 300 GB 1 TB
RPM 15,000 7,200
Average Seek 4 ms 9 ms
Max Transfer 125 MB/s 105 MB/s
Platters 4 4
Cache 16 MB 16/32 MB
Connects via SCSI SATA
Table 37.1: Disk Drive Specs: SCSI Versus SATA
Note that the rate of I/O (RI/O), which is often more easily used for
comparison between drives (as we will do below), is easily computed
from the time. Simply divide the size of the transfer by the time it took:
RI/O =
SizeTransfer
TI/O
(37.2)
To get a better feel for I/O time, let us perform the following calculation.
Assume there are two workloads we are interested in. The first,
known as the randomworkload, issues small (e.g., 4KB) reads to random
locations on the disk. Random workloads are common in many important
applications, including database management systems. The second,
known as the sequential workload, simply reads a large number of sectors
consecutively from the disk, without jumping around. Sequential
access patterns are quite common and thus important as well.
To understand the difference in performance between random and sequential
workloads, we need to make a few assumptions about the disk
drive first. Let’s look at a couple of modern disks from Seagate. The first,
known as the Cheetah 15K.5 [S09b], is a high-performance SCSI drive.
The second, the Barracuda [S09a], is a drive built for capacity. Details on
both are found in Table 37.1.
As you can see, the drives have quite different characteristics, and
in many ways nicely summarize two important components of the disk
drive market. The first is the “high performance” drive market, where
drives are engineered to spin as fast as possible, deliver low seek times,
and transfer data quickly. The second is the “capacity” market, where
cost per byte is the most important aspect; thus, the drives are slower but
pack as many bits as possible into the space available.
From these numbers, we can start to calculate how well the drives
would do under our two workloads outlined above. Let’s start by looking
at the random workload. Assuming each 4 KB read occurs at a random
location on disk, we can calculate how long each such read would take.
On the Cheetah:
Tseek = 4 ms, Trotation = 2 ms, Ttransfer = 30 microsecs (37.3)

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TIP: USE DISKS SEQUENTIALLY
When at all possible, transfer data to and from disks in a sequential manner.
If sequential is not possible, at least think about transferring data
in large chunks: the bigger, the better. If I/O is done in little random
pieces, I/O performance will suffer dramatically. Also, users will suffer.
Also, you will suffer, knowing what suffering you have wrought with
your careless random I/Os.
The average seek time (4milliseconds) is just taken as the average time
reported by the manufacturer; note that a full seek (from one end of the
surface to the other) would likely take two or three times longer. The
average rotational delay is calculated from the RPM directly. 15000 RPM
is equal to 250 RPS (rotations per second); thus, each rotation takes 4 ms.
On average, the disk will encounter a half rotation and thus 2 ms is the
average time. Finally, the transfer time is just the size of the transfer over
the peak transfer rate; here it is vanishingly small (30 microseconds; note
that we need 1000 microseconds just to get 1 millisecond!).
Thus, from our equation above, TI/O for the Cheetah roughly equals
6 ms. To compute the rate of I/O, we just divide the size of the transfer
by the average time, and thus arrive at RI/O for the Cheetah under the
random workload of about 0.66 MB/s. The same calculation for the Barracuda
yields a TI/O of about 13.2 ms, more than twice as slow, and thus
a rate of about 0.31 MB/s.
Now let’s look at the sequential workload. Here we can assume there
is a single seek and rotation before a very long transfer. For simplicity,
assume the size of the transfer is 100 MB. Thus, TI/O for the Barracuda
and Cheetah is about 800 ms and 950 ms, respectively. The rates of I/O
are thus very nearly the peak transfer rates of 125 MB/s and 105 MB/s,
respectively. Table 37.2 summarizes these numbers.
The table shows us a number of important things. First, and most
importantly, there is a huge gap in drive performance between random
and sequential workloads, almost a factor of 200 or so for the Cheetah
and more than a factor 300 difference for the Barracuda. And thus we
arrive at the most obvious design tip in the history of computing.
A second,more subtle point: there is a large difference in performance
between high-end “performance” drives and low-end “capacity” drives.
For this reason (and others), people are often willing to pay top dollar for
the former while trying to get the latter as cheaply as possible.
Cheetah Barracuda
RI/O Random 0.66 MB/s 0.31 MB/s
RI/O Sequential 125 MB/s 105 MB/s
Table 37.2: Disk Drive Performance: SCSI Versus SATA
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ASIDE: COMPUTING THE “AVERAGE” SEEK
In many books and papers, you will see average disk-seek time cited
as being roughly one-third of the full seek time. Where does this come
from?
Turns out it arises from a simple calculation based on average seek
distance, not time. Imagine the disk as a set of tracks, from 0 to N. The
seek distance between any two tracks x and y is thus computed as the
absolute value of the difference between them: |x − y|.
To compute the average seek distance, all you need to do is to first add
up all possible seek distances:
N
Xx
=0
N
Xy
=0
|x − y|. (37.4)
Then, divide this by the number of different possible seeks: N2. To
compute the sum, we’ll just use the integral form:
Z N
x=0 Z N
y=0
|x − y| dy dx. (37.5)
To compute the inner integral, let’s break out the absolute value:
Z x
y=0
(x − y) dy + Z N
y=x
(y − x) dy. (37.6)
Solving this leads to (xy − 1
2 y2)
x
0
+ ( 1
2 y2 − xy)
N
x
which can be simplified
to (x2 −Nx + 1
2N2). Now we have to compute the outer integral:
Z N
x=0
(x2 − Nx +
1
2
N2) dx, (37.7)
which results in:
(
1
3
x3 −
N
2
x2 +
N2
2
x)
N
0
=
N3
3
. (37.8)
Remember that we still have to divide by the total number of seeks
(N2) to compute the average seek distance: (N3
3 )/(N2) = 1
3N. Thus the
average seek distance on a disk, over all possible seeks, is one-third the
full distance. And now when you hear that an average seek is one-third
of a full seek, you’ll know where it came from.

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Figure 37.5: SSTF: Scheduling Requests 21 And 2
37.5 Disk Scheduling
Because of the high cost of I/O, the OS has historically played a role in
deciding the order of I/Os issued to the disk. More specifically, given a
set of I/O requests, the disk scheduler examines the requests and decides
which one to schedule next [SCO90, JW91].
Unlike job scheduling, where the length of each job is usually unknown,
with disk scheduling, we can make a good guess at how long
a “job” (i.e., disk request) will take. By estimating the seek and possible
the rotational delay of a request, the disk scheduler can know how long
each request will take, and thus (greedily) pick the one that will take the
least time to service first. Thus, the disk scheduler will try to follow the
principle of SJF (shortest job first) in its operation.
SSTF: Shortest Seek Time First
One early disk scheduling approach is known as shortest-seek-time-first
(SSTF) (also called shortest-seek-first or SSF). SSTF orders the queue of
I/O requests by track, picking requests on the nearest track to complete
first. For example, assuming the current position of the head is over the
inner track, and we have requests for sectors 21 (middle track) and 2
(outer track), we would then issue the request to 21 first, wait for it to
complete, and then issue the request to 2 (Figure 37.5).
SSTF works well in this example, seeking to the middle track first and
then the outer track. However, SSTF is not a panacea, for the following
reasons. First, the drive geometry is not available to the host OS; rather,
it sees an array of blocks. Fortunately, this problem is rather easily fixed.
Instead of SSTF, an OS can simply implement nearest-block-first (NBF),
which schedules the request with the nearest block address next.
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The second problem is more fundamental: starvation. Imagine in
our example above if there were a steady stream of requests to the inner
track, where the head currently is positioned. Requests to any other
tracks would then be ignored completely by a pure SSTF approach. And
thus the crux of the problem:
CRUX: HOW TO HANDLE DISK STARVATION
How can we implement SSTF-like scheduling but avoid starvation?
Elevator (a.k.a. SCAN or C-SCAN)
The answer to this query was developed some time ago (see [CKR72]
for example), and is relatively straightforward. The algorithm, originally
called SCAN, simply moves across the disk servicing requests in order
across the tracks. Let us call a single pass across the disk a sweep. Thus, if
a request comes for a block on a track that has already been serviced on
this sweep of the disk, it is not handled immediately, but rather queued
until the next sweep.
SCAN has a number of variants, all of which do about the same thing.
For example, Coffman et al. introduced F-SCAN,which freezes the queue
to be serviced when it is doing a sweep [CKR72]; this action places requests
that come in during the sweep into a queue to be serviced later.
Doing so avoids starvation of far-away requests, by delaying the servicing
of late-arriving (but nearer by) requests.
C-SCAN is another common variant, short for Circular SCAN. Instead
of sweeping in one direction across the disk, the algorithm sweeps
from outer-to-inner, and then inner-to-outer, etc.
For reasons that should now be obvious, this algorithm (and its variants)
is sometimes referred to as the elevator algorithm, because it behaves
like an elevator which is either going up or down and not just servicing
requests to floors based on which floor is closer. Imagine how annoying
it would be if you were going down from floor 10 to 1, and somebody
got on at 3 and pressed 4, and the elevator went up to 4 because it
was “closer” than 1! As you can see, the elevator algorithm, when used
in real life, prevents fights from taking place on elevators. In disks, it just
prevents starvation.
Unfortunately, SCANand its cousins do not represent the best scheduling
technology. In particular, SCAN(or SSTF even) do not actually adhere
as closely to the principle of SJF as they could. In particular, they ignore
rotation. And thus, another crux:
CRUX: HOW TO ACCOUNT FOR DISK ROTATION COSTS
How can we implement an algorithm that more closely approximates SJF
by taking both seek and rotation into account?

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Figure 37.6: SSTF: Sometimes Not Good Enough
SPTF: Shortest Positioning Time First
Before discussing shortest positioning time first or SPTF scheduling (sometimes
also called shortest access time first or SATF), which is the solution
to our problem, let us make sure we understand the problem in more detail.
Figure 37.6 presents an example.
In the example, the head is currently positioned over sector 30 on the
inner track. The scheduler thus has to decide: should it schedule sector 16
(on the middle track) or sector 8 (on the outer track) for its next request.
So which should it service next?
The answer, of course, is “it depends”. In engineering, it turns out
“it depends” is almost always the answer, reflecting that trade-offs are
part of the life of the engineer; such maxims are also good in a pinch,
e.g., when you don’t know an answer to your boss’s question, you might
want to try this gem. However, it is almost always better to know why it
depends, which is what we discuss here.
What it depends on here is the relative time of seeking as compared
to rotation. If, in our example, seek time is much higher than rotational
delay, then SSTF (and variants) are just fine. However, imagine if seek is
quite a bit faster than rotation. Then, in our example, it wouldmakemore
sense to seek further to service request 8 on the outer track than it would
to performthe shorter seek to the middle track to service 16, which has to
rotate all the way around before passing under the disk head.
Onmodern drives, aswe sawabove, both seek and rotation are roughly
equivalent (depending, of course, on the exact requests), and thus SPTF
is useful and improves performance. However, it is even more difficult
to implement in an OS, which generally does not have a good idea where
track boundaries are or where the disk head currently is (in a rotational
sense). Thus, SPTF is usually performed inside a drive, described below.
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TIP: IT ALWAYS DEPENDS (LIVNY’S LAW)
Almost any question can be answeredwith “it depends”, as our colleague
Miron Livny always says. However, use with caution, as if you answer
too many questions this way, people will stop asking you questions altogether.
For example, somebody asks: “want to go to lunch?” You reply:
“it depends, are you coming along?”
Other Scheduling Issues
There are many other issues we do not discuss in this brief description
of basic disk operation, scheduling, and related topics. One such issue
is this: where is disk scheduling performed on modern systems? In
older systems, the operating system did all the scheduling; after looking
through the set of pending requests, the OS would pick the best one, and
issue it to the disk. When that request completed, the next one would be
chosen, and so forth. Disks were simpler then, and so was life.
In modern systems, disks can accommodate multiple outstanding requests,
and have sophisticated internal schedulers themselves (which can
implement SPTF accurately; inside the disk controller, all relevant details
are available, including exact head position). Thus, the OS scheduler usually
pickswhat it thinks the best fewrequests are (say 16) and issues them
all to disk; the disk then uses its internal knowledge of head position and
detailed track layout information to service said requests in the best possible
(SPTF) order.
Another important related task performed by disk schedulers is I/O
merging. For example, imagine a series of requests to read blocks 33,
then 8, then 34, as in Figure 37.6. In this case, the scheduler shouldmerge
the requests for blocks 33 and 34 into a single two-block request; any reordering
that the scheduler does is performed upon the merged requests.
Merging is particularly important at the OS level, as it reduces the number
of requests sent to the disk and thus lowers overheads.
One final problem that modern schedulers address is this: how long
should the system wait before issuing an I/O to disk? One might naively
think that the disk, once it has even a single I/O, should immediately
issue the request to the drive; this approach is called work-conserving, as
the diskwill never be idle if there are requests to serve. However, research
on anticipatory disk scheduling has shown that sometimes it is better to
wait for a bit [ID01], in what is called a non-work-conserving approach.
By waiting, a new and “better” request may arrive at the disk, and thus
overall efficiency is increased. Of course, deciding when to wait, and for
how long, can be tricky; see the research paper for details, or check out
the Linux kernel implementation to see how such ideas are transitioned
into practice (if you are the ambitious sort).

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37.6 Summary
We have presented a summary of how disks work. The summary is
actually a detailed functional model; it does not describe the amazing
physics, electronics, and material science that goes into actual drive design.
For those interested in even more details of that nature, we suggest
a different major (or perhaps minor); for those that are happy with this
model, good! We can now proceed to using the model to build more interesting
systems on top of these incredible devices.
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References
[ADR03] “More Than an Interface: SCSI vs. ATA”
Dave Anderson, Jim Dykes, Erik Riedel
FAST ’03, 2003
One of the best recent-ish references on how modern disk drives really work; a must read for anyone
interested in knowing more.
[CKR72] “Analysis of Scanning Policies for Reducing Disk Seek Times”
E.G. Coffman, L.A. Klimko, B. Ryan
SIAM Journal of Computing, September 1972, Vol 1. No 3.
Some of the early work in the field of disk scheduling.
[ID01] “Anticipatory Scheduling: A Disk-scheduling Framework
To Overcome Deceptive Idleness In Synchronous I/O”
Sitaram Iyer, Peter Druschel
SOSP ’01, October 2001
A cool paper showing how waiting can improve disk scheduling: better requests may be on their way!
[JW91] “Disk Scheduling Algorithms Based On Rotational Position”
D. Jacobson, J. Wilkes
Technical Report HPL-CSP-91-7rev1, Hewlett-Packard (February 1991)
A more modern take on disk scheduling. It remains a technical report (and not a published paper)
because the authors were scooped by Seltzer et al. [SCO90].
[RW92] “An Introduction to Disk Drive Modeling”
C. Ruemmler, J.Wilkes
IEEE Computer, 27:3, pp. 17-28, March 1994
A terrific introduction to the basics of disk operation. Some pieces are out of date, but most of the basics
remain.
[SCO90] “Disk Scheduling Revisited”
Margo Seltzer, Peter Chen, John Ousterhout
USENIX 1990
A paper that talks about how rotation matters too in the world of disk scheduling.
[SG04] “MEMS-based storage devices and standard disk interfaces: A square peg in a round
hole?”
StevenW. Schlosser, Gregory R. Ganger
FAST ’04, pp. 87-100, 2004
While the MEMS aspect of this paper hasn’t yet made an impact, the discussion of the contract between
file systems and disks is wonderful and a lasting contribution.
[S09a] “Barracuda ES.2 data sheet”
http://www.seagate.com/docs/pdf/datasheet/disc/ds cheetah 15k 5.pdfA data
sheet; read at your own risk. Risk of what? Boredom.
[S09b] “Cheetah 15K.5”
http://www.seagate.com/docs/pdf/datasheet/disc/ds barracuda es.pdf See above
commentary on data sheets.

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Homework
This homework uses disk.py to familiarize you with how a modern
hard drive works. It has a lot of different options, and unlike most of
the other simulations, has a graphical animator to show you exactly what
happens when the disk is in action. See the README for details.
1. Compute the seek, rotation, and transfer times for the following
sets of requests: -a 0, -a 6, -a 30, -a 7,30,8, and finally -a
10,11,12,13.
2. Do the same requests above, but change the seek rate to different
values: -S 2, -S 4, -S 8, -S 10, -S 40, -S 0.1. How do the
times change?
3. Do the same requests above, but change the rotation rate: -R 0.1,
-R 0.5, -R 0.01. How do the times change?
4. You might have noticed that some request streams would be better
served with a policy better than FIFO. For example, with the
request stream-a 7,30,8, what order should the requests be processed
in? Now run the shortest seek-time first (SSTF) scheduler
(-p SSTF) on the same workload; how long should it take (seek,
rotation, transfer) for each request to be served?
5. Now do the same thing, but using the shortest access-time first
(SATF) scheduler (-p SATF). Does it make any difference for the
set of requests as specified by -a 7,30,8? Find a set of requests
where SATF does noticeably better than SSTF; what are the conditions
for a noticeable difference to arise?
6. You might have noticed that the request stream -a 10,11,12,13
wasn’t particularly well handled by the disk. Why is that? Can you
introduce a track skew to address this problem (-o skew, where
skew is a non-negative integer)? Given the default seek rate, what
should the skew be to minimize the total time for this set of requests?
What about for different seek rates (e.g., -S 2, -S 4)? In
general, could you write a formula to figure out the skew, given the
seek rate and sector layout information?
7. Multi-zone disks pack more sectors into the outer tracks. To configure
this disk in such a way, run with the -z flag. Specifically, try
running some requests against a disk run with -z 10,20,30 (the
numbers specify the angular space occupied by a sector, per track;
in this example, the outer track will be packed with a sector every
10 degrees, the middle track every 20 degrees, and the inner track
with a sector every 30 degrees). Run some random requests (e.g.,
-a -1 -A 5,-1,0, which specifies that random requests should
be used via the -a -1 flag and that five requests ranging from 0 to
the max be generated), and see if you can compute the seek, rotation,
and transfer times. Use different random seeds (-s 1, -s 2,
etc.). What is the bandwidth (in sectors per unit time) on the outer,
middle, and inner tracks?
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8. Scheduling windows determine how many sector requests a disk
can examine at once in order to determine which sector to serve
next. Generate some random workloads of a lot of requests (e.g.,
-A 1000,-1,0, with different seeds perhaps) and see how long
the SATF scheduler takes when the scheduling window is changed
from 1 up to the number of requests (e.g., -w 1 up to -w 1000,
and some values in between). How big of scheduling window is
needed to approach the best possible performance? Make a graph
and see. Hint: use the -c flag and don’t turn on graphics with -G
to run these more quickly. When the scheduling window is set to 1,
does it matter which policy you are using?
9. Avoiding starvation is important in a scheduler. Can you think of a
series of requests such that a particular sector is delayed for a very
long time given a policy such as SATF? Given that sequence, how
does it perform if you use a bounded SATF or BSATF scheduling
approach? In this approach, you specify the scheduling window
(e.g., -w 4) as well as the BSATF policy (-p BSATF); the scheduler
then will only move onto the next window of requests when all of
the requests in the current window have been serviced. Does this
solve the starvation problem? How does it perform, as compared
to SATF? In general, how should a diskmake this trade-off between
performance and starvation avoidance?
10. All the scheduling policies we have looked at thus far are greedy,
in that they simply pick the next best option instead of looking for
the optimal schedule over a set of requests. Can you find a set of
requests in which this greedy approach is not optimal?

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