Question 5: (2 points) How can you use disk striping to optimize a RAID system f

Question

Question 5: (2 points) How can you use disk striping to optimize a RAID system for high bandwidth (on an individual read operation) versus high transaction throughput (supporting a large number of simultaneous read operations)? Question 6: (2 points) RAID level 5 uses block-level parity, with the parity block distributed among the disks so that no single disk becomes a bottleneck. When a user wants to write a file to the disk, how many block- level reads and writes are required? Assume the disk striping uses large strips so that the file fits in one entire block. Explain your answer.

Explanation / Answer

Question 5:

The RAID is Redundant Array of Independent Disks. The RAID 0 supports the disk striping. The input/output operations related to a data which is stored on multiple disks are in the overlaaped manner.

The performance , bandwidth, or stability ca be increased by using scattered sectors of the same disk.

Disk striping is a feature of RAID systems which is used to increase the bandwidth and performance of the system. The strorage space is divided into the number of sectors in the form of stripes. These stripes can be accessed in an order but these stripes are placed as interleaved.

The stripes should be small in the range of 64 – 512 bytes in case of large number of read operations. If the stripes are in the range of 64 – 512 bytes, then these stripes can be the single piece of information which are placed at interleaved position of the disk.

These stripes containing piece of information or data can be accessed by reading all the disks at same time.

If a block of information is large, then multiuser systems can be used to access these blocks such that it needs less hoping to access next block.

For example:

If the data is divided into 5 hard disks storage in the form of stripes, then the performance and bandwidth will be increased by the ratio 5:1 i.e., 5 times the previous one.

If the rate of input/output operations by all the hard disk is 150 IOPS(I/O operations per second), then the rate of input/output operations will be increased by 5 times i.e., 750 IOPS. The throughput will not cause any problem until and unless there is a disk failure.

Question 6:

The number of operations performed while writing to a file to the disk are as follows:

·       Read the block containing old data.

·       Read the block having old parity.

·       The data of old and new block will be compared against each other and if a bit is changed in the new block, then the corresponding bit in the parity should also be changed.

·       Write the block containing new data.

·       Write the block containing new parity.

The total number of read and write operations needed to write to file to a disk is 2 reads and writes.

The size of block should be equal to the size of disk to make sure that the file fits in one whole block. There is no need of read operations because the size of the block is same and it is not needed to read the old block and observe changes between old and new block.

Since there is no need of read operations if the size of block is equals to the size of the file. Thus, the total number of operations required to write to a file in RAID level 5 will be 2 writes.

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.