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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
Let us now consider a generic components D. Computer, the minimum integer value of MTTF of D in order to have a T equal to five days a reliability, greater or equal to 0.96
Yeah, exponential distribution we know that they reliability of a component is equal to Euler elevated to minus lambda times time.
We also know that the MTTF is equal to one above lambda
MTTF ge 122.48 days => 123 days
What is fault tolerance?
It consists of noticing active faults and component subsystem failures, and doing something helpful in response
What is error containment?
It is a helpful response, derived from the fault tolerance of the system, which is another close relative of modularity and the building of system out of subsystems
The boundary adopted for error containment is usually the boundary of the smallest subsystem inside which the error occurred
Can be of four types:
- Masking
- Fail Fast
- Fail Stop
- Do Nothing
Discuss the main advantages of the server consolidation approach enabled by utilization technology
Server consolidation enabled by virtualization offers several advantages:
- Improved Resource Utilization: Virtualization allows multiple virtual servers to run on a single physical server, leading to better utilization of hardware resources. This reduces the amount of underutilized hardware.
- Cost Savings: By consolidating servers, organizations can reduce the need for physical hardware, leading to savings on hardware purchases, maintenance, and energy consumption.
- Simplified Management: Managing fewer physical servers simplifies the IT infrastructure, making it easier to deploy, update, and maintain servers. Centralized management tools provided by virtualization platforms further streamline these processes.
- Increased Flexibility and Scalability: Virtual environments are highly flexible, allowing for quick deployment and scaling of resources as needed. Virtual machines (VMs) can be easily created, modified, or moved between physical hosts.
- Disaster Recovery and High Availability: Virtualization enhances disaster recovery capabilities through features like snapshots, cloning, and live migration. These features allow for quick recovery of VMs and reduce downtime.
- Isolation and Security: Virtualization provides better isolation of workloads, improving security by containing any issues within individual VMs without affecting others.
- Reduced Physical Space: Fewer physical servers mean less space is required in data centers, which can be a significant benefit in terms of real estate and cooling requirements.
- Better Testing and Development Environments: Virtualization enables the creation of isolated testing and development environments that mimic production systems without the need for dedicated hardware.
- Energy Efficiency: Consolidating servers leads to lower energy consumption due to fewer physical devices needing power and cooling, contributing to greener IT practices.
- Legacy System Support: Virtualization can support legacy operating systems and applications by running them on VMs, reducing the need for outdated hardware.
Overall, server consolidation through virtualization optimizes IT infrastructure, reduces costs, enhances flexibility, and improves disaster recovery and security.