Starting from:
$20
Storage - take your times Solved
1 Introduction
In this experiment we will look at secondary storage i.e. drives that we have connected to our computer to provide persistent storage. We will do some performance measurements to try to estimate how long time read and write operations actually take. You can do these experiments using your own computer and preferably should since you then have a more controlled environment.
There is a di erence in performance between a hard-disk drive and a solid-state drive. If possible run these experiments on di erent machines to see the di erence. Do the experiments in a group with di erent machines to see the di erence.
We will use the term drive to mean the physical device, hard-disk or solid-state, not to confuse with driver that is the operating system module that is responsible for talking to the drive.
2 File operations
In the benchmarks that we will implement, we will use the le operations that operate on le descriptors. These are low level operations that the operating system provide and will give us more control over what we are doing. The alternative is using the stdio library but then our benchmark would be running on top of a layer that bu ers our operations to be more e cient. The e ciency is gained by doing write operations in the background, something that often is ne but it will not tell us when the data is actually on the drive.
open, read and write
We will only use three primitives: open, read and write. Look up these function in the man pages; you have state that you want to look at the system calls since there are also shell commands with the same names and these will show up by default.
man 2 open
: int open(const char *pathname, int flags, mode_t mode)
To open a le we specify the path name i.e. the name of the le, some ags that specify how the le should be opened and the mode of the le if we should create it. In our benchmarks we will open the le for reading and writing and specify that it, if it is created it should have user read and write permissions.
int flag = O_RDWR | O_CREAT | O_DSYNC; int mode = S_IRUSR | S_IWUSR | S_IRGRP | S_IROTH; int fd = open(name , flag , mode );
There is one important ag here, the O_DSYNC ag. This ag specify that we do not want a write operation to return unless the thing that we have written is actually pushed to the drive. We will not accept any cheating, the operating system should have pushed the data to the drive and received an acknowledgment.
read, write and lseek
We will only read and write dummy values in our les using the procedures read() and write(). Both of these procedures work the same, they take a le descriptor, the address of a bu er and the number of bytes that they should read or write.
int buffer = 42; write ( fd , &buffer , sizeof ( int ) ) ;
int buffer ; read ( fd , &buffer , sizeof ( int ) ) ;
Both procedures will operate on the le given the current position i.e. an index that tells the operating system where in the le we want to read or write. This position is mostly handled automatically, if you read from the beginning of a le you just read and the position will move forward as you read. We however want to read and write at random locations so we will set this position explicitly. This is done using the procedure lseek() as shown in the examples below.
int pos = 512;
lseek ( fd , pos , SEEK_SET);
:
int step 64; lseek ( fd , step , SEEK_CUR);
:
The third argument to lseek() decides if the value, given as the second argument, should be interpreted as an absolute value (512 in the example above) or as an o set to the current position (64 in the example).
3 Before we go
We will start by generating a large le so that we have something to work with. You could of course take any le that you happen to have but it will have to be large enough for our experiments. We will generate a le that is 512 MiBytes so if you have a movie on your drive you could use it but the code that you write will be reused in writing the benchmark so you might as well generate a new le.
Create a le called generate.c and include some good to have include les. Also de ne the macros BLOCKS and SIZE that describes that we will generate a million blocks of size 512 bytes each.
#include <stdlib .h
#include <stdio .h
#include <unistd .h
#include <sys/types .h
#include <fcntl .h
#include <assert .h
#define BLOCKS 1024∗1024
#define SIZE 512
Start by nding the le name from the command line. Open the le and make sure that we actually managed to open it. When we generate the le we do not have to give the O_DSYNC ag since we’re not interested in how long time each write operation took.
int main( int argc , char ∗argv [ ] ) {
if ( argc < 2) { printf ("usage : generate <f i l e name/n" );
return −1;
}
char ∗name = argv [ 1 ] ;
int mode = S_IRUSR | S_IWUSR | S_IRGRP | S_IROTH; int flag = O_RDWR | O_CREAT; int fd = open(name , flag , mode ); assert ( fd != −1);
Then we loop over the blocks and make sure that we write in each block. We are a bit paranoid here as it would be su cient to just write at the last position, but we don’t want the operating system nor the drive to do clever things behind our backs. int buffer ;
for ( int b=0; b < BLOCKS; b++) {
lseek ( fd , SIZE∗b , SEEK_SET); buffer = b ;
write ( fd , &buffer , sizeof ( int ) ) ;
} close ( fd ); printf ("done\n" );
return 0;
}
Save the le, compile and generate a le foo.txt. Check if the le is actually 512 MiByte in size using the ls -l command. If everything looks OK, we’re ready to go.
4 Read performance
Our rst benchmark will be to examine the read performance. Make a copy of your le and call it read.c and make the following updates. We now only list the changes so if you’ve used an existing le you will have to go through the previous section.
We will use clock_gettime() to do time measurement so we need the header le time.h. We will also do a number of read operations separated by STEP bytes and we will read from COUNT number of blocks.
#include <time .h
#define STEP 64
#define COUNT 1000
The idea is that we will choose a block by random and then read the rst bytes of this block, then the bytes at position 64, 128 etc. The question is if there is any di erence in reading the rst entry in a block and reading the second or third entry.
We will gather statistics and present the di erent quartiles in execution time. The easiest way to do this is to collect the execution times and sort them and to do this we need a function that compare to entries.
int compare(const void ∗a , const void ∗b) {
return ( int )(∗( long ∗)a − ∗(long∗)b );
}
We will record the execution time in nano-seconds so we also provide a function that takes two timespec entries and returns the di erence as long.
long n_sec( struct timespec ∗ start , struct timespec ∗stop ) {
long diff_sec = stop−tv_sec − start−tv_sec ;
long diff_nsec stop−tv_nsec − start−tv_nsec ; long wall_nsec = ( diff_sec ∗1000000000) + diff_nsec ;
return wall_nsec ;
}
We have what we need to implement the benchmark. We start as before by opening the le that is given to us on the command line. We then allocate some space to hold a table where we will record the execution time. The table holds an array of time-stamps for each of the indexes that we will access i.e. 0, 64, 128, ... Each of these arrays are COUNT long. int entries = SIZE / STEP; long ∗∗ table = malloc ( sizeof (long∗) ∗ entries );
for ( int i = 0; i < entries ; i++) {
table [ i ] = malloc ( sizeof (long) ∗ COUNT);
}
Now for the actual benchmark, we will access COUNT blocks and generate a radium number from 0 to BLOCKS minus one. We set the current position to the start of this block and then perform a number of read operations. We clock each read operation, store the result and forward the position by SIZE steps.
for ( int c=0; c < COUNT; c++) { int b = rand () % BLOCKS; lseek ( fd , b∗SIZE , SEEK_SET);
for ( int e = 0; e < entries ; e++) {
struct timespec t_start , t_stop ; int buffer ; clock_gettime (CLOCK_MONOTONIC, &t_start ); read ( fd , &buffer , sizeof ( int ) ) ; clock_gettime (CLOCK_MONOTONIC, &t_stop ); lseek ( fd , STEP − sizeof ( int ) , SEEK_CUR);
table [ e ] [ c ] = n_sec(&t_start , &t_stop );
}
}
The only thing that is left is to collect the statistics and we do this one row at a time. Note that we also print the 90 percentile since the maximum value can di er very much.
printf ("#N\tMin\tQ1\tMed\tQ3\tD9\tMax\n" ); for ( int e = 0; e < entries ; e++) {
qsort ( table [ e ] , COUNT, sizeof (long ) , compare );
long min = table [ e ] [ 0 ] ; long q1 = table [ e ] [COUNT/4]; long med = table [ e ] [COUNT/2]; long q3 = table [ e ][3∗(COUNT/4)]; long d9 = table [ e ][9∗(COUNT/10)]; long max = table [ e ] [COUNT−1];
printf ("%d\t%ld\t%ld\t%ld\t%ld\t%ld\t%ld\n" , e∗STEP, min , . . . etc );
}
That’s it, run you rst benchmark as nd out how fast you hard drive is. Hmm, not very bad is it? Just for fun, change the COUNT value to one, run the benchmark and write down the execution time. Now turn your computer o , turn it back on again and run the same test again - hmm, run it again. That was probably only some random operating system thing - repeat, is it reproducible?
the page cache
When a le is read from disk, the operating system will place the read blocks in what is called the page cache. The idea is of course that if we read form the le once it is very likely that we will read from it again so we keep a copy of the blocks that we read in a cache.
If you use the following command you will see how the memory is used by the operating system. There is one entry called buff/cache and this is the area used for cached les.
free
total used free shared buff/cache available Mem: 8089288 1321116 5221168 943912 1547004 5476420 Swap: 8301564 6644 8294920
We can direct the operating system to drop les from the cache by running the command below. This will run a shell in supervisor mode and let this shell write a 1 to the le drop_caches. This is an example of how we can interact with the kernel by using the regular le operations. All of the les under /proc directory are special les that allow us to interact with the kernel.
sudo sh -c ’echo 1 /proc/sys/vm/drop_caches’
Check if the size of the cache has been decreased. Run the benchmarks again and see if there is a di erence when we clear the cache.
evaluation
What do the numbers mean? Is there a di erence in access time, why? Are there more caches involved, does the hard-drive have a cache of its own?
Many questions and you will probably not be able top answer them all but you should be able to do a nice graph to go with your ndings. You can quickly generate a nice graph using gnuplot, everything is prepared so we only need to save the data in a le and run some gnuplot commands.
./read foo.txt foo.dat
Now start gnuplot and give the following commands:
gnuplot :
gnuplot set xrange[-30,500] : gnuplot set xtics 64 :
gnuplot set boxwidth 10 :
gnuplot plot ’read.dat’ u 1:3:2:6:5 with candlesticks :
Create a le read.p where you write a gnuplot script that includes the commands above. In the beginning of the le you also specify that you would like to generate a .png le and that we don’t really need the legend. Set the xlabel, ylabel and adjust the yrange to start from 0 ([0:*]). Now you have a nice graph to show you friends when you explain the di erence in read execution time.
set terminal png set output "read.png"
unset key :
Run the script from the command line by giving it as an argument to gnuplot. Now you have a nice graph to show you friends when you explain the di erence in read execution time.
5 Write performance
So now that we know something about the read performance it’s time to test the write performance. Make copy of your benchmark and call it write.c; there are only some small changes that we need to x. Make sure that you have the O_DSYNC ag set, this will prevent the operating system driver from pulling your leg.
int flag = O_RDWR | O_CREAT | O_DSYNC;
Then we simply replace the read operation with a write operation. int buffer = e ; write ( fd , &buffer , sizeof ( int ) ) ;
That’s it, test it and see if there are any di erences compared to the read benchmark.
living on the edge
You might have discovered that writing takes a whole lot longer time than reading, but we can improve things dramatically by cheating. Remember that we set the O_DSYNC ag to make sure that the thing that we wrote was actually pushed out to the disk and not just written to a bu er waiting for better times. Do the same experiment but now without the O_DSYNC ag any improvements?
The operating system will now allow write back i.e. it will acknowledge that the write operation has been performed but the data is still in a bu er allocated by the operating system. The operating system will push the data to drive as soon as it nds a suitable moment, it might have done so already but it is not hanging around waiting for a con rmation. If you trust your power supply this is probably very good strategy but if your computer crashes, the data might not be on the drive.
Note - using O_DSYNC only means that the operating system has pushed the data to the drive. The drive might still have the data in it’s own bu er but it is the responsibility of the drive to push it to the physical disk or ash memory.
6 Memory speed
While we’re at it, we might as well do two more benchmarks. The rst thing we should check is the read and write performance of the regular memory. To do this there are only a few changes we have to do to our benchmark program. Make a copy of read.c and call it mread.c and do the following changes.
We will need to do more than one read operation to get an accurate answer so lets de ne a macro called LOOP.
#define LOOP 64 // can not be 0
Then we allocate a large array on the heap that of course has the size determined by the number of BLOCKS and their SIZE. We of course remove everything that is related to the opening of a le.
int ∗heap = malloc ( sizeof ( int )∗BLOCKS∗SIZE );
Now we replace the read operation in the benchmark with a loop that reads from our array.
for ( int l = 0; l < LOOP; l++) {
buffer = buffer + heap [ b∗SIZE+e+l ] ;
}
The nal trick is to divide the calculated execution time by LOOP to obtain the execution time of one read operation. This is why LOOP can not be set to 0.
table [ e ] [ c ] = (n_sec(&t_start , &t_stop ) / LOOP);
What is the read performance of main memory? Can you trust the number? Do a small change to the program, set LOOP to 0 and remove the division, how long time does it take to do nothing (or rather, measure the time)? Should we compensate for this? What if we add the following to our program:
#define RES ?? // the time i t takes to do nothing
:
table [ e ] [ c ] = (( n_sec(&t_start , &t_stop ) − RES) / LOOP);
This gives us something more accurate but now we have hard coded the benchmark for this particular machine. You could let the benchmark rst determine what the time is to do nothing when it starts but let’s not over do things.
Do some experiments and change the number of COUNTS between 10 and 10000, what is going on? Change also the write benchmark in the same way and include for example the operation below in the loop. Run with a small number of counts - what is happening?
heap [ b∗SIZE+e+l ] = e+l ;
Let’s try to access the whole array before we run the benchmark loop. Add the following to your programs, both to mread.c and mwrite.c. Add it immediately after allocating the array - any di erence? for ( int b =0; b < BLOCKS; b++) {
heap [ b∗SIZE ] = b ;
}
On my computer a write operation seams to be faster than a read operation, this is probably only a problem with the clock - right? (when I say something like this it is to make you think, there is nothing wrong with my clock)
7 A mix of both
This experiment is only possible if you have sudo access to the machine that you’re running on. We will mount a new le system using something called tmpfs. This will create a drive that looks like a regular drive from the outside but is actually implemented as a segment of the physical DRAM memory. Things will be faster than any regular disk drive but once you turn o the power everything will be lost.
Start by creating a directory called for example dram. Then you run the following command in a shell. Your user id is probably 1000 but check this using the command id.
sudo mount -t tmpfs -o mode=744,uid=1000,gid=1000 tmpfs ./dram
So far so good, what you have now done is created a virtual drive and mounted it using the directory dram. Anything we place in this directory will be allocated in physical memory (if you have enough of it). Start by copying the le foo.txt to this directory. Then run the benchmarks read and write using the copy as the target.
./read dram/foo.txt :
Any di erence in read performance? How about write performance?
8 Summary
You should now have a better understanding of the performance of secondary storage i.e. your hard-disk or solid-state drive. We have looked at the time it takes to do a read or write operation to a random block in a large le and made some observations. As an exercise write down the approximate execution time for the following operations:
• Reading or writing to a random location in memory.
• Reading or writing to a random location of a le that has been cached by the drive and/or the operating system.
• Reading or writing to several positions in the same le block.
• Reading or writing to a random location of a le from a cold start i.e. not cached by the operating system nor drive.
The performance of a hard-disk drive di ers from the performance of a solid-state drive - what is the di erence and when does it matter?
In this experiment we will look at secondary storage i.e. drives that we have connected to our computer to provide persistent storage. We will do some performance measurements to try to estimate how long time read and write operations actually take. You can do these experiments using your own computer and preferably should since you then have a more controlled environment.
There is a di erence in performance between a hard-disk drive and a solid-state drive. If possible run these experiments on di erent machines to see the di erence. Do the experiments in a group with di erent machines to see the di erence.
We will use the term drive to mean the physical device, hard-disk or solid-state, not to confuse with driver that is the operating system module that is responsible for talking to the drive.
2 File operations
In the benchmarks that we will implement, we will use the le operations that operate on le descriptors. These are low level operations that the operating system provide and will give us more control over what we are doing. The alternative is using the stdio library but then our benchmark would be running on top of a layer that bu ers our operations to be more e cient. The e ciency is gained by doing write operations in the background, something that often is ne but it will not tell us when the data is actually on the drive.
open, read and write
We will only use three primitives: open, read and write. Look up these function in the man pages; you have state that you want to look at the system calls since there are also shell commands with the same names and these will show up by default.
man 2 open
: int open(const char *pathname, int flags, mode_t mode)
To open a le we specify the path name i.e. the name of the le, some ags that specify how the le should be opened and the mode of the le if we should create it. In our benchmarks we will open the le for reading and writing and specify that it, if it is created it should have user read and write permissions.
int flag = O_RDWR | O_CREAT | O_DSYNC; int mode = S_IRUSR | S_IWUSR | S_IRGRP | S_IROTH; int fd = open(name , flag , mode );
There is one important ag here, the O_DSYNC ag. This ag specify that we do not want a write operation to return unless the thing that we have written is actually pushed to the drive. We will not accept any cheating, the operating system should have pushed the data to the drive and received an acknowledgment.
read, write and lseek
We will only read and write dummy values in our les using the procedures read() and write(). Both of these procedures work the same, they take a le descriptor, the address of a bu er and the number of bytes that they should read or write.
int buffer = 42; write ( fd , &buffer , sizeof ( int ) ) ;
int buffer ; read ( fd , &buffer , sizeof ( int ) ) ;
Both procedures will operate on the le given the current position i.e. an index that tells the operating system where in the le we want to read or write. This position is mostly handled automatically, if you read from the beginning of a le you just read and the position will move forward as you read. We however want to read and write at random locations so we will set this position explicitly. This is done using the procedure lseek() as shown in the examples below.
int pos = 512;
lseek ( fd , pos , SEEK_SET);
:
int step 64; lseek ( fd , step , SEEK_CUR);
:
The third argument to lseek() decides if the value, given as the second argument, should be interpreted as an absolute value (512 in the example above) or as an o set to the current position (64 in the example).
3 Before we go
We will start by generating a large le so that we have something to work with. You could of course take any le that you happen to have but it will have to be large enough for our experiments. We will generate a le that is 512 MiBytes so if you have a movie on your drive you could use it but the code that you write will be reused in writing the benchmark so you might as well generate a new le.
Create a le called generate.c and include some good to have include les. Also de ne the macros BLOCKS and SIZE that describes that we will generate a million blocks of size 512 bytes each.
#include <stdlib .h
#include <stdio .h
#include <unistd .h
#include <sys/types .h
#include <fcntl .h
#include <assert .h
#define BLOCKS 1024∗1024
#define SIZE 512
Start by nding the le name from the command line. Open the le and make sure that we actually managed to open it. When we generate the le we do not have to give the O_DSYNC ag since we’re not interested in how long time each write operation took.
int main( int argc , char ∗argv [ ] ) {
if ( argc < 2) { printf ("usage : generate <f i l e name/n" );
return −1;
}
char ∗name = argv [ 1 ] ;
int mode = S_IRUSR | S_IWUSR | S_IRGRP | S_IROTH; int flag = O_RDWR | O_CREAT; int fd = open(name , flag , mode ); assert ( fd != −1);
Then we loop over the blocks and make sure that we write in each block. We are a bit paranoid here as it would be su cient to just write at the last position, but we don’t want the operating system nor the drive to do clever things behind our backs. int buffer ;
for ( int b=0; b < BLOCKS; b++) {
lseek ( fd , SIZE∗b , SEEK_SET); buffer = b ;
write ( fd , &buffer , sizeof ( int ) ) ;
} close ( fd ); printf ("done\n" );
return 0;
}
Save the le, compile and generate a le foo.txt. Check if the le is actually 512 MiByte in size using the ls -l command. If everything looks OK, we’re ready to go.
4 Read performance
Our rst benchmark will be to examine the read performance. Make a copy of your le and call it read.c and make the following updates. We now only list the changes so if you’ve used an existing le you will have to go through the previous section.
We will use clock_gettime() to do time measurement so we need the header le time.h. We will also do a number of read operations separated by STEP bytes and we will read from COUNT number of blocks.
#include <time .h
#define STEP 64
#define COUNT 1000
The idea is that we will choose a block by random and then read the rst bytes of this block, then the bytes at position 64, 128 etc. The question is if there is any di erence in reading the rst entry in a block and reading the second or third entry.
We will gather statistics and present the di erent quartiles in execution time. The easiest way to do this is to collect the execution times and sort them and to do this we need a function that compare to entries.
int compare(const void ∗a , const void ∗b) {
return ( int )(∗( long ∗)a − ∗(long∗)b );
}
We will record the execution time in nano-seconds so we also provide a function that takes two timespec entries and returns the di erence as long.
long n_sec( struct timespec ∗ start , struct timespec ∗stop ) {
long diff_sec = stop−tv_sec − start−tv_sec ;
long diff_nsec stop−tv_nsec − start−tv_nsec ; long wall_nsec = ( diff_sec ∗1000000000) + diff_nsec ;
return wall_nsec ;
}
We have what we need to implement the benchmark. We start as before by opening the le that is given to us on the command line. We then allocate some space to hold a table where we will record the execution time. The table holds an array of time-stamps for each of the indexes that we will access i.e. 0, 64, 128, ... Each of these arrays are COUNT long. int entries = SIZE / STEP; long ∗∗ table = malloc ( sizeof (long∗) ∗ entries );
for ( int i = 0; i < entries ; i++) {
table [ i ] = malloc ( sizeof (long) ∗ COUNT);
}
Now for the actual benchmark, we will access COUNT blocks and generate a radium number from 0 to BLOCKS minus one. We set the current position to the start of this block and then perform a number of read operations. We clock each read operation, store the result and forward the position by SIZE steps.
for ( int c=0; c < COUNT; c++) { int b = rand () % BLOCKS; lseek ( fd , b∗SIZE , SEEK_SET);
for ( int e = 0; e < entries ; e++) {
struct timespec t_start , t_stop ; int buffer ; clock_gettime (CLOCK_MONOTONIC, &t_start ); read ( fd , &buffer , sizeof ( int ) ) ; clock_gettime (CLOCK_MONOTONIC, &t_stop ); lseek ( fd , STEP − sizeof ( int ) , SEEK_CUR);
table [ e ] [ c ] = n_sec(&t_start , &t_stop );
}
}
The only thing that is left is to collect the statistics and we do this one row at a time. Note that we also print the 90 percentile since the maximum value can di er very much.
printf ("#N\tMin\tQ1\tMed\tQ3\tD9\tMax\n" ); for ( int e = 0; e < entries ; e++) {
qsort ( table [ e ] , COUNT, sizeof (long ) , compare );
long min = table [ e ] [ 0 ] ; long q1 = table [ e ] [COUNT/4]; long med = table [ e ] [COUNT/2]; long q3 = table [ e ][3∗(COUNT/4)]; long d9 = table [ e ][9∗(COUNT/10)]; long max = table [ e ] [COUNT−1];
printf ("%d\t%ld\t%ld\t%ld\t%ld\t%ld\t%ld\n" , e∗STEP, min , . . . etc );
}
That’s it, run you rst benchmark as nd out how fast you hard drive is. Hmm, not very bad is it? Just for fun, change the COUNT value to one, run the benchmark and write down the execution time. Now turn your computer o , turn it back on again and run the same test again - hmm, run it again. That was probably only some random operating system thing - repeat, is it reproducible?
the page cache
When a le is read from disk, the operating system will place the read blocks in what is called the page cache. The idea is of course that if we read form the le once it is very likely that we will read from it again so we keep a copy of the blocks that we read in a cache.
If you use the following command you will see how the memory is used by the operating system. There is one entry called buff/cache and this is the area used for cached les.
free
total used free shared buff/cache available Mem: 8089288 1321116 5221168 943912 1547004 5476420 Swap: 8301564 6644 8294920
We can direct the operating system to drop les from the cache by running the command below. This will run a shell in supervisor mode and let this shell write a 1 to the le drop_caches. This is an example of how we can interact with the kernel by using the regular le operations. All of the les under /proc directory are special les that allow us to interact with the kernel.
sudo sh -c ’echo 1 /proc/sys/vm/drop_caches’
Check if the size of the cache has been decreased. Run the benchmarks again and see if there is a di erence when we clear the cache.
evaluation
What do the numbers mean? Is there a di erence in access time, why? Are there more caches involved, does the hard-drive have a cache of its own?
Many questions and you will probably not be able top answer them all but you should be able to do a nice graph to go with your ndings. You can quickly generate a nice graph using gnuplot, everything is prepared so we only need to save the data in a le and run some gnuplot commands.
./read foo.txt foo.dat
Now start gnuplot and give the following commands:
gnuplot :
gnuplot set xrange[-30,500] : gnuplot set xtics 64 :
gnuplot set boxwidth 10 :
gnuplot plot ’read.dat’ u 1:3:2:6:5 with candlesticks :
Create a le read.p where you write a gnuplot script that includes the commands above. In the beginning of the le you also specify that you would like to generate a .png le and that we don’t really need the legend. Set the xlabel, ylabel and adjust the yrange to start from 0 ([0:*]). Now you have a nice graph to show you friends when you explain the di erence in read execution time.
set terminal png set output "read.png"
unset key :
Run the script from the command line by giving it as an argument to gnuplot. Now you have a nice graph to show you friends when you explain the di erence in read execution time.
5 Write performance
So now that we know something about the read performance it’s time to test the write performance. Make copy of your benchmark and call it write.c; there are only some small changes that we need to x. Make sure that you have the O_DSYNC ag set, this will prevent the operating system driver from pulling your leg.
int flag = O_RDWR | O_CREAT | O_DSYNC;
Then we simply replace the read operation with a write operation. int buffer = e ; write ( fd , &buffer , sizeof ( int ) ) ;
That’s it, test it and see if there are any di erences compared to the read benchmark.
living on the edge
You might have discovered that writing takes a whole lot longer time than reading, but we can improve things dramatically by cheating. Remember that we set the O_DSYNC ag to make sure that the thing that we wrote was actually pushed out to the disk and not just written to a bu er waiting for better times. Do the same experiment but now without the O_DSYNC ag any improvements?
The operating system will now allow write back i.e. it will acknowledge that the write operation has been performed but the data is still in a bu er allocated by the operating system. The operating system will push the data to drive as soon as it nds a suitable moment, it might have done so already but it is not hanging around waiting for a con rmation. If you trust your power supply this is probably very good strategy but if your computer crashes, the data might not be on the drive.
Note - using O_DSYNC only means that the operating system has pushed the data to the drive. The drive might still have the data in it’s own bu er but it is the responsibility of the drive to push it to the physical disk or ash memory.
6 Memory speed
While we’re at it, we might as well do two more benchmarks. The rst thing we should check is the read and write performance of the regular memory. To do this there are only a few changes we have to do to our benchmark program. Make a copy of read.c and call it mread.c and do the following changes.
We will need to do more than one read operation to get an accurate answer so lets de ne a macro called LOOP.
#define LOOP 64 // can not be 0
Then we allocate a large array on the heap that of course has the size determined by the number of BLOCKS and their SIZE. We of course remove everything that is related to the opening of a le.
int ∗heap = malloc ( sizeof ( int )∗BLOCKS∗SIZE );
Now we replace the read operation in the benchmark with a loop that reads from our array.
for ( int l = 0; l < LOOP; l++) {
buffer = buffer + heap [ b∗SIZE+e+l ] ;
}
The nal trick is to divide the calculated execution time by LOOP to obtain the execution time of one read operation. This is why LOOP can not be set to 0.
table [ e ] [ c ] = (n_sec(&t_start , &t_stop ) / LOOP);
What is the read performance of main memory? Can you trust the number? Do a small change to the program, set LOOP to 0 and remove the division, how long time does it take to do nothing (or rather, measure the time)? Should we compensate for this? What if we add the following to our program:
#define RES ?? // the time i t takes to do nothing
:
table [ e ] [ c ] = (( n_sec(&t_start , &t_stop ) − RES) / LOOP);
This gives us something more accurate but now we have hard coded the benchmark for this particular machine. You could let the benchmark rst determine what the time is to do nothing when it starts but let’s not over do things.
Do some experiments and change the number of COUNTS between 10 and 10000, what is going on? Change also the write benchmark in the same way and include for example the operation below in the loop. Run with a small number of counts - what is happening?
heap [ b∗SIZE+e+l ] = e+l ;
Let’s try to access the whole array before we run the benchmark loop. Add the following to your programs, both to mread.c and mwrite.c. Add it immediately after allocating the array - any di erence? for ( int b =0; b < BLOCKS; b++) {
heap [ b∗SIZE ] = b ;
}
On my computer a write operation seams to be faster than a read operation, this is probably only a problem with the clock - right? (when I say something like this it is to make you think, there is nothing wrong with my clock)
7 A mix of both
This experiment is only possible if you have sudo access to the machine that you’re running on. We will mount a new le system using something called tmpfs. This will create a drive that looks like a regular drive from the outside but is actually implemented as a segment of the physical DRAM memory. Things will be faster than any regular disk drive but once you turn o the power everything will be lost.
Start by creating a directory called for example dram. Then you run the following command in a shell. Your user id is probably 1000 but check this using the command id.
sudo mount -t tmpfs -o mode=744,uid=1000,gid=1000 tmpfs ./dram
So far so good, what you have now done is created a virtual drive and mounted it using the directory dram. Anything we place in this directory will be allocated in physical memory (if you have enough of it). Start by copying the le foo.txt to this directory. Then run the benchmarks read and write using the copy as the target.
./read dram/foo.txt :
Any di erence in read performance? How about write performance?
8 Summary
You should now have a better understanding of the performance of secondary storage i.e. your hard-disk or solid-state drive. We have looked at the time it takes to do a read or write operation to a random block in a large le and made some observations. As an exercise write down the approximate execution time for the following operations:
• Reading or writing to a random location in memory.
• Reading or writing to a random location of a le that has been cached by the drive and/or the operating system.
• Reading or writing to several positions in the same le block.
• Reading or writing to a random location of a le from a cold start i.e. not cached by the operating system nor drive.
The performance of a hard-disk drive di ers from the performance of a solid-state drive - what is the di erence and when does it matter?
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