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In the table below, to have 8 memory addresses where each memory address holds one byte of data (byte-addressable), we need 3 bits to index every address. Memory Address Data Stored 000 A 001 010 011 1 100 2 101 110 M 111 N If you have 64 GiB of RAM memory, then how many bits are needed to have memory addresses for every byte-addressable memory location?

In the table below, to have 8 memory addresses where each memory address holds one byte of data (byte-addressable), we need 3 bits to index every address. Memory Address Data Stored 000 A 001 010 011 1 100 2 101 110 M 111 N If you have 64 GiB of RAM memory, then how many bits are needed to have memory addresses for every byte-addressable memory location?

Database System Concepts
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
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In the table below, to have 8 memory addresses where each memory address holds one 
byte of data (byte-addressable), we need 3 bits to index every address.  (check image) 

If you have 64 GiB of RAM memory, then how many bits are needed to have memory  addresses for every byte-addressable memory location?

In the table below, to have 8 memory addresses where each memory address holds one byte of data (byte-addressable), we need 3 bits to index every address. Memory Address Data Stored 000 A 001 010 011 1 100 101 110 M 111 N If you have 64 GiİB of RAM memory, then how many bits are needed to have memory addresses for every byte-addressable memory location? 2.
Transcribed Image Text:In the table below, to have 8 memory addresses where each memory address holds one byte of data (byte-addressable), we need 3 bits to index every address. Memory Address Data Stored 000 A 001 010 011 1 100 101 110 M 111 N If you have 64 GiİB of RAM memory, then how many bits are needed to have memory addresses for every byte-addressable memory location? 2.
Expert Solution
Step 1

With n bits, we can identify 2n different memory locations.

For example,

1 bit: 21 = 22 bits: 22 = 43 bits: 23 = 8
and so on...

 

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