1. Assume that P2 issues the logical address 0xD98A. What is the logical page number contained in the given logical address, in hexadecimal?Answer is: 0x 6 But I do not knot well how to calculate For the following questions, consider a system that has a physical main memory sizeof 64KB (216) and a page frame size (and hence page size) of 8KB (213). Assume thatthe system uses pure demand paging and that the system supports up to 17-bitlogical/virtual addresses. Assume that a process P2 with a logical address space of128KB (217) executes on this system. Assume that the current state of the page tablefor P2 in as shown below. The index of each entry in the page table is given for yourconvenience. Note that both the page and page frame number are in decimal. If alogical page is not currently in the physical main memory, no frame number isspecified for that logical page.Page Frame012345сл678910111213141517025I346

A logical address is a reference to a memory location that is expressed in terms of an abstract…

Answered: logical page is not currently in the…

logical page is not currently in the physical main memory, no frame number is specified for that logical page. Page Frame 0 1 2 ♡2 3 4 5 6 7 8 9 10 11 ~ 3 12 N 13 14 15 1 7 0 2 5 32 4 6

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

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5. Let the content of register $t1 represent a 2's complement number N. Write a MIPS code fragment such that the register $t2 stores the absolute value of N. In other words, register $t2 has a copy of register $t1 if N is positive, and $t2 should hold the negative of N, if N is negative.
write a program to count the number of data elements equal (FF)H in a set of data starting at address (0100)H and ended with the value of (00)H.
Question 10 Match the data types on the left with range of values they can represent. The range depends on the number of bytes (8-bits per byte). Assume that a modern 64-bit operating system is being used. NOTE: You can verify the size of each data type using the sizeof operator. unsigned long long float double address of char address of long long address of double -9,223,372,036,854,775,808 [Choose] -32,768 to 32,767 0 to 65,535 0 to 4,294,967,295 -128 to 127 (+/-)1.2E-38 to (+/-)3.4E+38 0 to 18,446,744,073,709,551,615 (+/-)2.3E-308 to (+/-)1.7E+308 -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 -2,147,483,648 to 2,147,483,647 0 to 255 [Choose] [Choose]
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The address returned to the caller of z on line 9 is 33792 (beginning of free memory block) + 1k (header) = 34816.this doesn't follow the line 5 and line 7. Could you please explain why 33792 returned to in line 9 and 32k in line 5 and 7?thank you
Provide the correct answers to each question
please don't copy the answer cuz its not right and clrear
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