Information Technology Reference
In-Depth Information
13 milliseconds = 158 MB/s. The random read bandwidth is thus
158/270 = 59% of the sequential read bandwidth.
Example: Random write workload.
Question: For the solid state disk described in Figure 12.6, consider a work-
load consisting of 500 write requests, each of a randomly chosen
page. How long will servicing these requests take?
Answer: The disk can service random write requests at a rate of 2000 per
second (assuming the disk is nearly full), so 500 requests will
take 500/2000 = 250 ms.
Example: Random v. sequential writes.
Question: How does this random write performance compare to the drive's
sequential write performance?
Answer: The effective bandwidth in this case is 500 requests * 4
KB
request
/ 250 milliseconds = 8.2 MB/s. The random write bandwidth is
thus 8.2/210 = 3.9% of the sequential write bandwidth.
Exercises
12. Suppose that you have a 256 GB solid state drive that the operating sys-
tem and drive both support the TRIM command. To evaluate the drive,
you do an experiment where you time the system's write performance for
random page-sized when the file system is empty compared to its perfor-
mance when the file system holds 255 GB of data, and you find that write
performance is significantly worse in the latter case.
What is the likely reason for this worse performance as the disk fills despite
its support for TRIM?
What can be done to mitigate this slowdown?
13. Suppose you have a flash drive such as the one described in Figure 12.10
on page 383 and you have a workload consisting of 10000 4KB reads to
pages randomly scattered across the drive. Assuming that you wait for
request i to finish before you issue request i+ 1, how long will these 10000
request take (total)?
