Test: File Systems - 1 - Computer Science Engineering (CSE) MCQ

Seek time is time taken by the head to travel to the track of the disk where the data to be accessed is stored.

Circular scanning works just like the elevator to some extent. It begins its scan toward the nearest end and works its way all the way to the end of the system. Once it hits the bottom or top it jumps to the other end and moves in the same direction. Circular SCAN has more head movement than SCAN (elevator) because Circular SCAN has circular jump and it does count as a head movement. SCAN (elevator) is the best choice here.

Total number of possible addresses stored in a disk block = 128/8 = 16

Maximum number of addressable bytes due to direct address block = 8*128
Maximum number of addressable bytes due to 1 single indirect address block = 16*128
Maximum number of addressable bytes due to 1 double indirect address block = 16*16*128

The maximum possible file size = 8*128 + 16*128 + 16*16*128 = 35KB

Higher priority interrupt levels are assigned to requests which, if delayed or interrupted, could have serious consequences. Devices with high speed transfer such as magnetic disks are given high priority, and slow devices such as keyboard receive low priority. Interrupt from CPU temperature sensor would have serious consequences if ignored.

Hardware detects interrupt immediately, but CPU acts only after its current instruction. This is followed to ensure integrity of instructions.

The data on a disk is ordered in the following way. It is first stored on the first sector of the first surface of the first cylinder. Then in the next sector, and next, until all the sectors on the first track are exhausted. Then it moves on to the first sector of the second surface (remains at the same cylinder), then next sector and so on. It exhausts all available surfaces for the first cylinder in this way. After that, it moves on to repeat the process for the next cylinder

The data in hard disk is arranged in the shown manner. The smallest division is sector. Sectors are then combined to make a track. Cylinder is formed by combining the tracks which lie on same dimension of the platters. Read write head access the disk. Head has to reach at a particular track and then wait for the rotation of the platter so that the required sector comes under it. Here, each platter has two surfaces, which is the r/w head can access the platter from the two sides, upper and lower. So,<400,16,29> will represent 400 cylinders are passed(0-399) and thus, for each cylinder 20 surfaces (10 platters * 2 surface each) and each cylinder has 63 sectors per surface. Hence we have passed 0-399 =  400 * 20 * 63 sectors + In 400th cylinder we have passed 16 surfaces(0-15) each of which again contains 63 sectors per cylinder so 16 * 63 sectors. + Now on the 16th surface we are on 29th sector. So, sector no = 400x20x63 + 16×63 + 29 = 505037. 

Whenever head moves from one track to other then its speed and direction changes, which is noting but change in motion or the case of inertia. So answer B

The Unix file system uses an extension of indexed allocation. It uses direct blocks, single indirect blocks, double indirect blocks and triple indirect blocks. Following diagram shows implementation of Unix file system. 

An interrupt service routine will be invoked after the completion of I/O operation and it will place process from block state to ready state, because process performing I/O operation was placed in blocked state till the I/O operation was completed in Synchronous I/O. However, process performing I/O will not be placed in the block state and process continues to execute the remaining instructions in Asynchronous I/O, because handler function will be registered while performing the I/O operation, when the I/O operation completed signal mechanism is used to notify the process that data is available. So, option (B) is false.

In programmed I/O, CPU does continuous polling, To transfer 1B CPU polls for 10^-4 sec = 10^2 micro-sec of processing In interrupt mode CPU is interrupted on completion of i\o, To transfer 1B CPU does 4 micro-sec of processing(since transfer time between other components is negligible). Gain = 10^2 / 4 = 25

In Shortest-Seek-First algorithm, request closest to the current position of the disk arm and head is handled first. In this question, the arm is currently at cylinder number 100. Now the requests come in the queue order for cylinder numbers 30, 85, 90, 100, 105, 110, 135 and 145. The disk will service that request first whose cylinder number is closest to its arm. Hence 1st serviced request is for cylinder no 100 ( as the arm is itself pointing to it ), then 105, then 110, and then the arm comes to service request for cylinder 90. Hence before servicing request for cylinder 90, the disk would had serviced 3 requests. Hence option C.

Since Operating System can execute a single sequential user process at a time, the disk is accessed in FCFS manner always. The OS never has a choice to pick an IO from multiple IOs as there is always one IO at a time

Time takes for 1 rotation = 60/3000 It reads 512*1024 Bytes in one rotation. Time taken to read 4 bytes = 153 ns 153 is approximately 4 cycles (160ns) Percentage of time CPU gets blocked = 40*100/160 = 25

Using larger block size makes disk utilization poorer as more space would be wasted for small data in a block. It may make throughput better as the number of blocks would decrease. A larger block size guarantees that more data from a single file can be written or read at a time into a single block without having to move the disk ́s head to another spot on the disk. The less time you spend moving your heads across the disk, the more continuous reads/writes per second. The smaller the block size, the more frequent it is required to move before a read/write can occur. Larger block size means less number of blocks to fetch and hence better throughput. But larger block size also means space is wasted when only small size is required and hence poor utilization.