For the following problems assume 1 kilobyte (KB) = 1024 bytes and1 megabyte (MB) = 1024 kilobytes. For this problem, assume you have address translation hardware with the following properties: 1. • Virtual addresses, physical addresses, and page table entries are 32 bits wide. The page size in the system is 4 KB. A virtual address is a page number followed by a byte offset within the page. (a) How many bits of the virtual address must be used for the offset, so that every byte in the page can have a unique address? (b) How many bits are left over in the virtual address to store the page number? (c) How many different page numbers does an address space in this system support? (You can express this as a power of two) (d) If a page table consists of a page table entry for each page number in an address space, how much space in MB would the page table take up if it were stored in physical memory?

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
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For the following problems assume 1 kilobyte (KB) = 1024 bytes and 1 megabyte (MB) =
1024 kilobytes.
%3D
%3D
1. For this problem, assume you have address translation hardware with the following
properties:
Virtual addresses, physical addresses, and page table entries are 32 bits wide.
The page size in the system is 4 KB.
A virtual address is a page number followed by a byte offset within the page.
(a) How many bits of the virtual address must be used for the offset, so that every
byte in the page can have a unique address?
(b) How many bits are left over in the virtual address to store the page number?
(c) How many different page numbers does an address space in this system
support? (You can express this as a power of two)
(d) If a
table consists of a page table entry for each page number in an address
page
space, how much space in MB would the page table take up if it were stored in
physical memory?
Transcribed Image Text:For the following problems assume 1 kilobyte (KB) = 1024 bytes and 1 megabyte (MB) = 1024 kilobytes. %3D %3D 1. For this problem, assume you have address translation hardware with the following properties: Virtual addresses, physical addresses, and page table entries are 32 bits wide. The page size in the system is 4 KB. A virtual address is a page number followed by a byte offset within the page. (a) How many bits of the virtual address must be used for the offset, so that every byte in the page can have a unique address? (b) How many bits are left over in the virtual address to store the page number? (c) How many different page numbers does an address space in this system support? (You can express this as a power of two) (d) If a table consists of a page table entry for each page number in an address page space, how much space in MB would the page table take up if it were stored in physical memory?
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