185, 17, 124, 83, 40, 29, 76, 112 Identify the disk scheduling algorithms based on the following descriptions: i Selects the disk I'O request that requires the least amount of travel from the current position of the disc arm, regardless of direction. 11. The disc arm moves back and forth over the disk, satisfying requests in its path.

185, 17, 124, 83, 40, 29, 76, 112 Identify the disk scheduling algorithms based on the following descriptions: i Selects the disk I'O request that requires the least amount of travel from the current position of the disc arm, regardless of direction. 11. The disc arm moves back and forth over the disk, satisfying requests in its path.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

os first come first serve shortest seek time first LOOK which one is more efficient? Explain
Both disks and main memory support direct access to any desired location (page). On average, main memory accesses are faster, of course. What is the other important difference between the two (from the perspective of the time required to access a desired page)?
Different algorithms are used for scheduling the disk head so that seek time may be reduced. Calculate a total number of heads movement for the following algorithms for the given numbers of disk cylinders. Shortest Seek Time First (SSTF) b) Circular Look (C-Look) Head Start: 71 Queue: 82 , 153 , 14 , 149 , 6 , 189 , 62 , 99 , 121
Suppose that we want to schedule disk requests on a disk containing 200 cylinders numbered from 0 to 199. The Disk requests to be addressed are 82,170,43,140,24,16,189. And the Read/Write arm is at 48, and it is also given that the disk arm should move "towards the larger value". Q₁- Illustrate on a graph the total read/write head movement for the SCAN Disk scheduling algorithm. Q2- Calculate the total overhead movement (total distance covered by the disk arm).
Disk requests come to disk driver for cylinders 10, 153, 34, 90, 120, 205, 319, 256 and 400 in that order at a time when the disk drive is reading from cylinder 99. The cylinders are numbered from 0 to 499. Compute the total head movement incurred while servicing these requests if the disk arm scheduling algorithm is first-come-first-serve.
8. Consider one disk with 400 cylinders, numbered 0 to 399. Assume the current position of head is at cylinder 130. The request queue is given as follows: 150, 40, 350, 100, 280 For each of the disk-scheduling algorithms given below, show (by plotting in the graphs below) the current position of head based on the disk arm movement to satisfy the above requests, and count the total distance (in cylinders) of the disk arm movement. a. First Come First Served (FCFS) 0 40 100 1.30 150 Total distance (disk am movement) = b. Shortest Seek Time First (SSTF) 40 100 130 150 0 Total distance (disk arm movement) = c. LOOK - in the direction of increasing track number 0 40 100 130 150 Total distance (disk arm movement) = 280 280 280 350 350 350 399 399 399
Serial access memories are useful in applications whereA. Data consists of numbersB. Short access time is requiredC. Each stored word is processed differentlyD. Data naturally needs to flow in and out in the serial form.
Question 4 1 pts locality means that an instruction or data item that has been recently accessed will likely be re- accessed soon in the future. O Associative O Temporal O Context O Spatial Question 5 1 pts Consider the typical memory hierarchy in modern computer systems. Which of the following components has the fastest access time in general? O Main memory O Registers O Cache O Solid-state disk Question 6 1 pts is a fault resulting from a reference to a non-memory-resident page of memory. O Disk fault O CPU fault O /0 fault O Page fault
Both disks and main memory support direct access to any desired location (page). On average, main memory accesses are faster, of course. What is the other important difference between the two (from the perspective of the time required to access a desired page)?
Alert dont submit AI generated answer.
It is given that on a system, on average, an instruction executes in 1 nano second. Suppose that it takes 20 micro seconds of processor time for page fault. While, it takes 300 micro seconds of disk time for reading and writing a single page. Suppose that on average 1/3 of the pages that need to be paged out are modified. What is the average number of instructions between page faults that would cause the disk to be busy doing page transfers all the time?