In a system with fixed partitions, there are 4 partitions available to jobs. Their sizes are 16, 32, 48, and 64 KB. Four jobs executed by the system at the moment have the following sizes: 12, 22, 42, and 55 KB. The amount of internal fragmentation expressed in KB is: ܘ ܝ ܘ ܝ ܘ ܩ ܗ a. b. C. d. e. f. 8. h. 0 12 17 25 29 34 41 56

In a system with fixed partitions, there are 4 partitions available to jobs. Their sizes are 16, 32, 48, and 64 KB. Four jobs executed by the system at the moment have the following sizes: 12, 22, 42, and 55 KB. The amount of internal fragmentation expressed in KB is: ܘ ܝ ܘ ܝ ܘ ܩ ܗ a. b. C. d. e. f. 8. h. 0 12 17 25 29 34 41 56

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

A disk has a capacity of two tera-byte size. Say the file system uses a multi-level inode structure for locating the data blocks of afile.The inode stores pointers to data blocks, including a single indirect block, a double indirect block, and several direct blocks in the 64 B of available space.Now, it is given that the disk has a block size of 512 B. The maximum file size that can be stored in such a file system in MB (round off upto 2 decimal places).
In a main memory-disk virtual storage system, the page size is 1KByte and the FIFO algorithm is used for page replacements. A given program has been allocated three page frames in the main memory and it makes the following 16 memory references when it starts executing (the addresses are given in decimal):500, 2000, 2500, 800, 4000, 1000, 5500, 1500, 2800, 400, 5000, 700, 2100, 3500, 900, 2400 Fill in the contents of the three page frames after each memory reference in a table and calculate the hit ratio. Hint: denote by 'a' the page consisting of locations 0 through 1023 in memory. Similarly, b: 1024-2047, c: 2048-3071, d: 3072-4095, e: 4096-5119 and f: 5120-6143. Round to three decimal places.
There are many parameters that could be used to describe disk performance; among them are: number of bits per track disk capacity (in bits) number of disk surfaces rotational speed rotational latency transfer rate tracks per surface sectors per track blocks per track sectors per block seek time speed of disk arm block-read time number of blocks Some of these parameters are independent, and others are (approximately) linearly related. That is, doubling one doubles the other. Decide which of these parameters are linearly related. Then, select from the list below, the relationship that is true, to within a close approximation. Note: none of the statements may be true exactly, but one will always be much closer to the truth than the other three. Also note: you should assume all dimensions and parameters of the disk are unchanged except for the ones mentioned. a) If you double the number of sectors per block, then you double the number of blocks on the disk. b)…
There are many parameters that could be used to describe disk performance; among them are: number of bits per track disk capacity (in bits) number of disk surfaces rotational speed rotational latency transfer rate tracks per surface sectors per track blocks per track sectors per block seek time speed of disk arm block-read time number of blocks Some of these parameters are independent, and others are (approximately) linearly related. That is, doubling one doubles the other. Decide which of these parameters are linearly related. Then, select from the list below, the relationship that is true, to within a close approximation. Note: none of the statements may be true exactly, but one will always be much closer to the truth than the other three. Also note: you should assume all dimensions and parameters of the disk are unchanged except for the ones mentioned. a) If you divide tracks into half as many blocks, then you double the read time for a block.…
Consider a file system that uses inodes to represent files. Disk blocks are 8 KB in size, and a pointer to a disk block requires 8 bytes. This file system has 12 direct disk blocks, as well as 1 single and 1 double indirect disk blocks. What is the maximum size of a file that can be stored in this file system in GBs (round to the nearest integer)?
Consider a disk pack with 32 surfaces, 64 tracks and 512 sectors per pack. 256 bytes of data are stored in a bit serial manner in a sector. The number of bits required to specify a particular sector in the disk is.
In a microcomputer, the addresses of memory locations are binary numbers that identify each memory circuit where a byte is stored. The number of bits that make up an address depends on how many memory locations there are. Since the number of bits can be very large, the addresses are often specified in hex instead of binary. If a microcomputer uses a 24-bit address, how many different memory locations are there? What is the largest number that can be How many hex digits are needed to represent the address of a memory location? What is the hex address of the 172th memory location? (Note: The first address is always ) The computer program is stored in the lowest 3 kbyte block of memory. Give the start and end address of this What is the BCD code of the hex address of the 172th memory location?
The index node (inode) of a hypothetical file system has 20 direct, 256 double-indirect and 2 triple-indirect pointers. If system is word addressable and each word is equal to 4 bytes. The disk block size is 2 KW and the disk block address 8 Word long. The maximum possible file size is (rounded off to 1 decimal place) GB (answer up-to 1 decimal places).
Suppose that there is a file system with blocks of 512 bytes in size. For each inode, there are 20 direct blocks, 2 singly-indirect blocks, and 2 doubly-indirect blocks. The size of a pointer is 4 bytes. Calculate the maximum possible size of a file in bytes. Show your work.
Given rax = 0x0000000200000100, rbx = 0x0000000000000100, and rcx = 0x0000000000000001,and the following values in memoryaddress -> byte at that address0x0000000000000100 -> 0x010x0000000000000101 -> 0x000x0000000000000102 -> 0x000x0000000000000103 -> 0x000x0000000000000104 -> 0x020x0000000000000105 -> 0x000x0000000000000106 -> 0x000x0000000000000107 -> 0x00what is the new value in %rax after the following operation?subq -0x04(%rbx, %rcx, 4), %rax
physcal addresses are 4s ng 4 Ame dat in a cetain compe, te addresses can be translaled without y TLB entries At most how many ditina vid the address translation peh has 12 vld The Translation Look aside Bulfer (TLB)i sine is kB and the word size iby The memory is word addresible. The pe virtual addresses are 64 bea long d th sine is miss?
In a certain computer, the virtual addresses are 32 bits long and the physical addresses are 48 bits long. The memory is word addressable. The page size is 16 kB and the word size is 2 bytes. The Translation Look-aside Buffer (TLB) in the address translation path has 64 valid entries. Hit ratio of TLB is 100% then maximum number of distinct virtual addresses that can be translated is K.