1. What is the difference between primary storage and secondary storage?

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1 1. What is the difference between primary storage and secondary storage? Primary Storage is - Limited - Volatile - Expensive - Fast (May be accessed directly from the CPU) - Retrieving a single character from RAM takes about 150 nanoseconds (150 billionths of a second) Secondary Storage is - Extendible - Persistent - Cheap - slow (must be copied to main memory before being accessed by the CPU) - Retrieving the same character from disk takes about 75 milliseconds (thousandths of a second) - 75 msec is 500,000 times longer than Primary Storage. 2. What are the two major types of secondary storage devices? a. Direct Access Storage Devices (DASDs) - Offer high storage capacity and low cost - Data stored as magnetized areas on magnetic platters - surfaces - Each disk has one or more platters - A disk pack contains several magnetic platters connected to a rotating spindle Examples: - Magnetic Disks Hard Disks (high capacity, low cost per bit) - Optical Disks CD-ROM, DVD-ROM (Read-only/write-once, holds a lot of data, Cheap) b. Serial Devices - Magnetic Tapes 3. Describe the structure of the magnetic disk? - Magnetic disks support direct access to a desired location and consisted of : - Disk blocks - Tracks - Platters - Cylinder - Sectors - Disk heads - Disk Controller - Seek Time - Rotational delay

2 - Some disks have fixed-heads - As many read/write heads as there are tracks on the platter - Track is selected electronically and is therefore much faster - Cost of additional read/write heads is the limiting factor to production - Disks with an actuator are called moveable-head disks - Actuator moves the (single) read/write head per platter to the appropriate track - Disks are divided into concentric circular tracks - On each platter surface - Track capacities vary typically from 4 to 50 Kbytes - The division of the disk into sectors is hard-coded and may not be changed - Subtended at fixed angle at the center of a platter are sectors - Not all disks have their tracks divided into sectors 4. How to calculate disk capacity? - Number of cylinders = number of tracks in a surface - Track capacity = number of sector per track bytes per sector

3 - Cylinder capacity = number of surfaces track capacity - Drive capacity = number of cylinders cylinder capacity Knowing these relationships allows us to compute the amount of disk space a file is likely to require 5. How many cylinders to store a file with 20,000 fixed length records of 256 bytes each on a disk with 512 bytes per sector, 40 sectors per disk and 11 tracks per cylinder? Answer - The file is 20,000 * 256 = 5,120,000 bytes - 5,120,000 / 512 = 10,000 sectors - 10,000 / 40 = 250 tracks / 11 = cylinders - If physically contiguous cylinders are not available then the file will have to be spread out over the disk (fragmentation) 6. How can data organized on disks? Data can be organized on disk in two different ways: 1. Sectors - Physically placement of sector - Adjacent sectors - Interleaving Sectors - Clusters - Extents 2. User defined blocks. 7. How can data organized on disks using Physical placement of sectors? a. Adjacent sectors: fixed size segments of a track used to hold a file Disadvantage: physically it is not possible to read a series of sectors on the same track, after reading one sector, the disk controller take a certain amount of time to transfer data, the start of the next sector will be lost ( able to read only one sector / revolution) b. Interleaving sectors: leave an interval of several fixed sectors between adjacent logical sectors. The following figure indicates interleaving factor =

4 8. How can data organized on disks using Clusters? - The operating system file manager maps using FAT (File Allocation Table) the logical parts of the file to their corresponding physical locations to improve performance. - The file is viewed as a set of a fixed number of contiguous sectors that can be read without requiring additional seeks called clusters. - FAT: contains a list of all the clusters in a file. Each cluster entry in the FAT giving the physical location of the cluster. 9. How can data organized on disks using extents? An extent is a file stored in contiguous sectors, track and cylinders Its clusters are contiguous. The whole file can be accessed with minimum amount of seeking. An extent is possible if a disk has a lot of free space. If no contiguous free space for a file the file can be stored in two or more extents.

5 10. Define Internal Fragmentation of a disk? Internal fragmentation of a disk is the unused disk space which cannot be used by other files. To Store a file of 300-byte records in a disk of sector size 512 bytes. - Store a record in a sector. This will cause the loss of disk space, i.e., internal fragmentation. - Allow records to span in two sectors. This will save disk space. But, it may require the retrieval of two sectors when accessing a record. If the number of bytes in a file is not a multiple of the cluster size, internal fragmentation will occur in the last extent of the file. 11. Describe how can data organized on a disk using blocks? Disk tracks can be divided into integral numbers of user-defined blocks. Block size can have fixed or variable length. In block organization, different amount of data can be transferred in a single I/O operation. A block organization does not have sector-spanning and fragmentation problems. - track organization by sectors - track organization by blocks A block may contain one or several records. Each block is usually accompanied by one or more subblocks containing extra information about the data blocks.

6 - Count subblock: counting the number bytes in the accompanied data block. - Key subblock: containing the key for the last record in the data block. - When key subblocks are used, a track can be searched by the disk controller for a block or record identified by a given key. - This search is more efficient than sector-addressable schemes because it does not load the keys into primary memory. 12. What are Blocks? And why are they important? - A track is divided into blocks or pages - Block size (generally) fixed for each operating system - Typical block sizes range from 512 bytes to 4096 bytes (4 kb) - A disk with hard-coded sectors often has the sectors further subdivided onto blocks - Importance :Whole blocks are transferred between disk and main memory for processing 13. What is the difference between a cluster and FAT? - A cluster is a fixed number of contiguous sectors - All clusters on a disk are the same size - To view a file as a series of clusters and still maintain the sectored view, the file manager ties the logical sectors to the physical clusters using a file allocation table (FAT) - The file manager allocates an integer number of clusters to a file. An example: Sector size: 512 bytes, Cluster size: 2 sectors - If a file contains 10 bytes, a cluster is allocated (1024 bytes). - There may be unused space in the last cluster of a file. - This unused space contributes to internal fragmentation The FAT contains a linked list of all the clusters in a file, ordered according to the logical order of the sectors in a cluster - With each entry in the FAT is an entry giving the physical location of the cluster 14. How can operating system Read a particular byte from a disk file?

7 - The computer operating system find the correct surface, track and sector, reads the entire sector into a special area of memory called buffer - And then find the requested byte in the buffer. 15. How to choose cluster size? - Some OS allow the system administrator to choose the cluster size. - When to use large cluster size? - When disks contain large files likely to be processed sequentially. Example: Updates in a master file of bank accounts (in batch mode) - When to use Small cluster size? - When disks contain small files and/or files likely to be accessed randomly Example : online updates for airline reservation 16. Describe what is mean by Overhead? a. Preformatting overhead for sector-addressable disks - Stored at the beginning of each sector, including information about sector address, track address, and condition (whether the sector is usable or defective). - Preformatting also involves placing gaps and synchronization marks between fields of information. b. Nondata overhead for block-addressable disks - Subblocks and interblock gaps. - Block factors: number of bytes per track/block length. - In general, block factors is the greater the better. - However, larger blocks have higher potential of internal track fragmentation. 17. Calculate the space on the disk used to store a file? Suppose we need to store a file with 50,000 fixed length data records on a disk with the following characteristics: - Number of bytes/sector = Number of sectors / track = 63 - Number of tracks / cylinder = 16 - Number of cylinders = 4096 a. How many cylinders does the file require if each data record requires 256 bytes? Number of sectors for a file = (50000* 256) /512 = sector Cylinder size = 16 track * 63 sector * 512 byte = Cylinders for the file = 50000* 256 / Cylinder size (516096) = 24.8 cylinders Of course it may be that the disk does not have 24.8 physically contiguous cylinders available. So the file has to be spread out over dozens even hundreds of cylinders. 18. A Disk drive have byte / track, the amount of space taken by blocks and inter block gaps = 300 byte/block. We need to store file with 100 byte/ record.

8 a. How many records can be stored per track if the blocking factor is 10? Blocking factor = 10 Each block holds = 10 (block factor) * 100 (record size) (block data) = 1300 byte. The number of blocks that can fit on track = / 1300 = = 15 block. Number of records can be stored = 15 block * 10 record = 150 record. b. How many records can be stored per track if the blocking factor is 60? Blocking factor = 60 Each block holds = 60 (block factor) * 100 (record size) (block data) = 6300 byte. The number of blocks that can fit on track = / 6300 = 3 block. Number of records can be stored = 3 block * 60 record = 180 record. c. What do you understand from the above results? The larger blocking factor can lead to more efficient use of storage 19. Consider the following parameters of a disk: Disk capacity = MB Tracks per recording surface = 100 Number of recording surfaces = 10 Bytes per sector = 512 Min. seek time = 2 ms Avg. seek time = 8 ms Spindle speed = rpm (Assume 1 MB=1024 B) a. How many cylinders are there? Number of cylinders = Number of tracks per surface = 100 tracks/surf = 100 cylinders b. What is the capacity of a cylinder? Cylinder capacity = Total disk Capacity / No. of cylinders = MB / 100 cylinders = 200 MB/cylinder c. What is the capacity of each recording surface? Surface Capacity = Total disk Capacity / No. of recording surfaces = MB / 10 surfaces = 2000 MB/surface d. What is the capacity of one track?

9 Track Capacity = Surface Capacity / No. of tracks per surface = 2000(MB/surf) / 100(tracks/surf) = 20 MB/track e. How many sectors are there in one track? Number of sector in one track = 20(MB/track)/512(B/sector) = 20*1024/512 sectors/track = 40 sectors/track f. What is the average rotational delay? Spindle speed = rpm = rotations/60 sec Max rotational delay = 60*1000 ms/10000 rotations = 6 ms Min rotational delay = 0 ms Average rotational delay = expected value of rot. delay = (6ms+0)/2 = 3 ms g. What is the time to find and read one sector? Time find and read one sector = Average seek time + average rot. Delay + time to read (max rotational delay) = /(40) = 11 + to 3/20 = ms 20. Assume that a disk drive uses block addressing. The following subblock organization for nondate is available: - Countdata, where the extra space used by count subblock. - Interblock gaps is equivalent to 95 Bytes. This disk has usable Bytes per track, 30 tracks per cylinder, and 496 cylinders per drive. Suppose you have a file with 350,000 60Byte records that you want to store on this drive. Answer the following questions. Unless otherwise directed, assume that the blocking factor is 10 and that the Countdata subblock organization is used. a. How many blocks can be stored on one track? Block size = Blocking Factor * Record Size + Interblock gap = 10 * Bytes = 695 B Number of blocks per track = Track size / Block size = B / 695 (B/block) = 24 blocks/track b. How many records can be stored on one track? Number of records/ track = (Number of blocks /track) * (Number of records/block) = 24 blocks/track * 10 records/block = 240 records/track c. How many cylinders are required to hold the file (blocking factor 10 and Countdata format)? Number of records/cylinder = 240 rec/track * 30 tracks/cylinder = 7200 records/cylinder.

10 Number of cylinder required to hold a file = Number of records in file / number of records in cylinder = rec in file /7200 rec in cylinder = 49 cylinders for the file d. How much space will go unused totally due to internal track fragmentation and count subblocks? Total file is records = / 10 records per block = blocks. For each block 95 B/block is unused because of count subblock, Total unused space for count subblocks = (95 unused space /block) * Number of blocks = 95 * = B. Number of blocks in one track = (17017 / 695 ) = 24 Block and 337 due to internal fragmentation for one track. Number of tracks in the file = (block/file) /24 (block/track) = 1459 tracks/file. Total unused space for internal fragmentation = Number of tracks in file * unused fragmentation for one track = 1459 * 337 B = B. Therefore the total unused space = Count subblock unused space + fragmentation unused space = B = B In the rest of the questions assume that the file is organized into clusters and extents. Suppose the cluster size = 4 blocks and extent size = 15 clusters. e. How many clusters are needed to store this file? Number of clusters in file = Number of blocks in file / Cluster size = ( records / 10 (rec/block)) /4 (blocks per cluster) = 8750 clusters f. What is the number of extents? Number of Extents in file = Number of Clusters in file / extent size = ceiling (8750 clusters/ (15 clusters/extent)) = 584 extents 21. What are the factors that affect the cost of disk access? - Seek time: the time required to move the access arm to the correct cylinder. - average seek time - Rotational delay: the time required to rotate the disk so the desired sector can be placed under the read/write head. - Maximum rotational delay: time for one resolution - Average rotational delay: half of maximum rotational delay - Transfer time: the time required to read the data from the disk (Number of bytes transferred number of bytes on a track) * Rotation time or (Number of sectors transferred number of sectors in a track) * Rotation time