What is automatic storage tiering? Why is it useful? What is object-based storag

Question

What is automatic storage tiering? Why is it useful? What is object-based storage? How is it superior to conventional storage systems? Consider a disk with the following characteristics (these are not parameters of any particular disk unit): block size B = 512 bytes; interblock gap size G = 128 bytes; number of blocks per track = 20; number of tracks per surface = 400. A disk pack consists of 15 double-sided disks. What is the total capacity of a track, and what is its useful capacity (excluding interblock gaps)? How many cylinders are there? What are the total capacity and the useful capacity of a cylinder? What are the total capacity and the useful capacity of a disk pack? Suppose that the disk drive rotates the disk pack at a speed of 2,400 rpm (resolutions per minute); what are the transfer rate (tr) in bytes/msec and the block transfer time (btt) in msec? What is the average rotational delay (rd) in msec? What is the bulk transfer rate? (See Appendix B.) Suppose that the average seek time is 30 msec. How much time does it take (on the average) in msec to locate and transfer a single block, given its block address? Calculate the average time it would take to transfer 20 random blocks, and compare this with the time it would take to transfer 20 consecutive blocks using double buffering to save seek time and rotational delay. A file has r = 20,000 STUDENT records of fixed length. Each record has the following fields: Name (30 bytes), Ssn (9 bytes), Address (40 bytes), PHONE (10 bytes), Birth_date (8 bytes), Sex (1 byte), Major_dept_code (4 bytes), Minor_dept_code (4 byt+F6es), Class_code (4 bytes, integer), and Degree_program (3 bytes). An additional byte is used as a deletion marker. The file is stored on the disk whose parameters are given in Exercise 16.27. Calculate the record size R in bytes. Calculate the blocking factor bfr and the number of file blocks b, assuming an unspanned organization.

Explanation / Answer

R = (30 + 9 + 40 + 10 + 8 + 1 + 4 + 4 + 4 + 3) + 1 = 114 bytes

(b) Calculate the blocking factor bfr and the number of file blocks b assuming an unspanned organization.

bfr = floor(B / R) = floor(512 / 114) = 4 records per block

b = ceil(r / bfr) = ceil(20000 / 4) = 5000 blocks

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