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Consider a HDD with:
- data transfer rate: 240 MB/s
- rotation speed: 10000 RPM
- mean seek time: 20 ms
- overhead controller: 0.3 ms
The mean I/O service time to transfer a sector of 8 KB
T_Over = 0.3 ms
T_Seek = 20 ms
T_Rot = 3 ms
T_Transfer = 0.032 ms
T_I/O = 23.332 ms
Consider a HDD with:
- block size: 3 KB
- mean I/O service time per block (with no locality): 9.0 ms
- transfer time: 0.09 ms
- overhead controller: 0.8 ms
How long does it take to transfer a file of 50 MB if we assume a locality of 70%
T_BlocksWLocality = T_Transfer + T_Over = 0.89 ms
T_BlocksWOLocality = 9 ms
Number of Blocks = 50 MB / 3 KB = 17067
NumBlocksWLocality = 0.7 * 17067 = 11947
NumBlocksWOLocality = 0.3 * 17067 = 5120
T_I/O = 11947 * 0.89 + 5120 * 9 = 56713 ms
A HDD has a rotation speed of 10000 RPM, an average seek time of 4 ms, a negligible controller overhead and a transfer rate of 256 MB/s. Files are stored into blocks whose size is 4 KB
a. The rotational latency of the disk
b. The time required to read a 400 KB file devised into 5 sets of contiguous blocks
c. The time required to read a 400 KB file with a locality of 95%
a. 3 ms
b.
T_I/O = T_Transfer400KB + 5 * (T_Seek + T_Rot) = 36.526 ms
c. 36.526 ms
Consider to have 6 disks, each one with a capacity of 1TB.
What will be to total storage capacity of the system if they are in the following configurations?
a. RAID 0
b. RAID 1
c. RAID 0+1
d. RAID 1+0
e. RAID 5
f. RAID 6
a. RAID 0 - 0.0 Fear => 6 TB
b. RAID 1 - 1.0 Fear => 1 TB
c. RAID 0+1 - (0.0 + 1.0) / 2.0 Fear => 3 TB
d. RAID 1+0 - (0.0 + 1.0) / 2.0 Fear => 3 TB
e. RAID 5 - (N - 1) * Disk_Capacity = 5 TB
f. RAID 6 - (N - 2) * Disk_Capacity = 4 TB
Consider the following RAID 0 setup:
- n = 5 disks
- MTTR = 8 h
- MTTF(one disk) = 1600 days
The MTTDL will be
Failure Rate = 1/MTTF
Failure Rate System = n * 1/MTTF
MTTDL = 1/Failure Rate
MTTDL System = n * 1/Failure Rate System = 5 * 1/1600 = 320 days
Consider the following RAID 1 setup:
- n = 2 disks
- MTTR = 8 days
- MTTF(one disk) = 1800 day
The MTTDL will be
Failure Rate = 1/MTTF
Failure Rate System = N * Failure Rate (chance to loose any of the disks) * (Failure Rate * MTTR) (loosing the other before reparing it) = 2/1800 (8/1800) = 16/1800^2
MTTDL = 1800^2/16 days
Consider 2 groups (RAID 0) of 2 disks each (RAID 1), for a total of 4 disks in configuration RAID 1+0
- MTTR = 3 days
- MTTF(one disk) = 1400 day
The MTTDL will be
In a RAID 1+0, the same copy in both groups has to fail
Failure Rate System = N/MTTF (chance of any disk to fail) * (1/MTTF (chance of the specific replica that contains the same data to fail) * MTTR) = 12 / 1400^2
MTTDL = 1400^2/12 days
Consider 2 groups (RAID 1) of 4 disks each (RAID 0), for a total of 8 disks in configuration RAID 0+1
- MTTR = 4 days
- MTTF(one disk) = 2200 day
The MTTDL will be
In a RAID 0+1 when one disk in a stripe group fails the entire group goes off
Failure Rate System = N/MTTF (chance of any disk to fail) * (N/2 * 1/MTTF (chance of any of the other stripe group fail) * MTTR) = 128/MTTR^2
MTTDL = MTTR^2/128
A system administrator has to decide to use a stock of disks characterized by:
- MTTF = 800 days
- MTTR = 20 days
The target lifetime of the system is 3 years
The maximum number of disks that could be used in a RAID 0+1 to have a MTTDL larger than the system lifetime is
Failure Rate System = N / MTTF *(N / 2 * 1/MTTF * MTTR)
MTTDL = 1/Failure Rate System = 800^2/(N^2/2 * 20) = 7,…
Consider the following RAID 5 setup:
- n = 4
- MTTR = 3 days
- MTTF(one disk) = 1000 day
The MTTDL will be
Failure Rate System = N/MTTF * ((N-1)/MTTF (chance of failure of any other disk) * MTTR) = 36 / MTTF^2
MTTDL = MTTF^2/36
Consider the following RAID 6 setup:
- n = 5
- MTTR = 2 days
- MTTF(one disk) = 1100 day
The MTTDL will be
Failure Rate System = N/MTTF * ((N-1)/MTTF (chance of failure of any other disk) * MTTR) ((N-2)/MTTF (chance of a third failure) * MTTR/2 (average overlapping period between replacements)) = 120/MTTF^3
MTTDL = MTTF^3/120
