Question one
"chmod" is a well known command line utility, that is used for manually managing the access and permissions to files and directories on MacOS, Linux and other Unix like operating systems. Explain how the file permissions are changed after executing this command:
$ chmod 624 SEUfile.txt
Answer:
624 is made up of 3 numbers, each of on the "octals" represents each of the groups that have access to a file. For example the octal 624 presents a situation where 6 is the octal for setting Owner permissions, 2 represents Group permissions, and 4 represents Other permissions.
Owner - 6
Group - 2
Other - 4
Each one of these numbers allows the Read, Write, and Execute permissions per group. =RWX
Owner Permissions - 6 = 4+2+0 = rw-
Group Permissions - 2 = 0+2+0 = -w-
Other Permissions - 4 = 4+0+0 = r—
It becomes –rw--w-r--, where file type is a file denoted by the leading "-", owner is able to read, write. Group is only able to write, and Other is only able to read.
Question two
Consider a file system on a disk that has both logical and physical block sizes of 512 bytes. Assume that the information about each file is already in memory. For each of the three allocation strategies (contiguous, linked, and indexed), answer these questions:
a. How is the logical-to-physical address mapping accomplished in this system? (For the indexed allocation, assume that a file is always less than 512 blocks long.)
b. If we are currently at logical block 10 (the last block accessed was block 10) and want to access logical block 4, how many physical blocks must be read from the disk?
Answer:
Let Z be the starting file address (block number).
• Contiguous. Divide the logical address by 512 with X and Y the resulting quotient and remainder respectively.
a. Add X to Z to obtain the physical block number. Y is the displacement into that block.
b. 1
• Linked. Divide the logical physical address by 511 with X and Y the resulting quotient and remainder respectively.
a. Chase down the linked list (getting X + 1 blocks). Y + 1 is the displacement into the last physical block.
b. 4 •
Indexed. Divide the logical address by 512 with X and Y the resulting quotient and remainder respectively.
a. Get the index block into memory. Physical block address is contained in the index block at location X. Y is the displacement into the desired physical block.
b. 2
Question three
a) Explain the ways of disk attachments used by computers to access memory disk.
Answer:
Computers access disk storage in two ways. One way is via I/O ports (or host-attached storage). The other way is via a remote host in a distributed file system (or network-attached storage).
1. Host-Attached Storage
Host-attached storage is storage accessed through local I/O ports. The desktop PC uses an I/O bus architecture called IDE or ATA. This architecture supports a maximum of two drives per I/O bus. A newer, similar protocol that has simplified cabling is SATA. A wide variety of storage devices are suitable for use as host-attached storage. Among these are hard disk drives, RAID arrays, and CD, DVD and tape drives.
2. Network-Attached Storage
A network-attached storage (NAS) device is a special-purpose storage system that is accessed remotely over a data network. Clients access network-attached storage via a RPC (remote-procedure-call) interface such as NFS for UNIX systems or CIFS for Windows machines. The RPCs are carried via TCP or UDP over an IP like LAN.
b) Explain C-SCAN disk scheduling algorithm with example?
Answer:
C-SCAN stands for Circular SCAN which is a variant of SCAN disk scheduling which provides more uniform waiting time. C-SCAN moves the disk head from one end of the disk to the other, servicing requests which are on the way. When the head reaches the other end, it immediately returns to the beginning of the disk without servicing any requests during the return trip. Then it starts the onward journey servicing the remaining requests. The C-SCAN scheduling algorithm treats the cylinders as a circular list that wraps around from the final cylinder to the first one.
As an example, suppose following requests arrive in the queue:
Queue = 98, 183, 37, 122, 14, 124, 65, 67 with the disk head at 53.
Question four
Suppose that a disk drive has 5,000 cylinders, numbered 0 to 4,999. The drive is currently serving a request at cylinder 2,150, and the previous request was at cylinder 1,805. The queue of pending requests, in FIFO order is as below:
2069, 1212, 2296, 2800, 544, 1618, 356, 1523, 4965, 3681
Starting from the current head position, what is the total distance (in cylinders) that the disk arm moves to satisfy all the pending requests for each of the following disk-scheduling algorithms?
a) LOOK
Answer:
The order of access of cylinders is as below:
2150, 2296, 2800, 3681, 4965, 2069, 1618, 1523, 1212, 544, 256
The distance moved by the disk arm:
146 + 504 + 881 + 1284 + 2896 + 451 + 95 + 311 + 668 + 188 = 7424 cylinders
b) C-LOOK
Answer:
The order of access of cylinders is as below:
2150, 2296, 2800, 3681, 4965, 356, 544, 1212, 1523, 1618, 2069
The distance moved by the disk arm:
146 + 504 + 881 + 1284 + 4609 + 188 + 668 + 311 + 95 + 451 = 9137 cylinders
واجب 3
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