Topic: BinaryFill in the boxesWill give you high rating thank you!! A byte is always 8 bits;Each byte has its own addressAddr AddrContents(Bin)000 1 0 0 0 1 10 1001 0 000 o 0 00010 1 1 1 0 1 0 10011 0 1 1 1 1 1 1 1100 1 0 0 00 100101 1 0 10 10 1 11Word size is 4 bytesStarting address 023Word size is 4 bytesStarting address 46110 0 0 11 0 0 10111 0 1 1 0 o 0 1 1Notice that each word starts with an address that is a multiple of 4. Some systems require that addresses be aligned tothese boundaries, i.e. address provided must be a multiple of 4, while other systems allow unaligned memory accesses.It depends on the way the system is designed.Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of thefirst word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides inaddresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. (Recall that the word sizeis 4 bytes.)Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, thesecond word starts atthe third word starts atand the last word starts atCheck4)7.

Here in the given diagram, Memory size is 1KB Word size is 64bit

Answered: 4 bytes.) ow assume the same 1kB of…

4 bytes.) ow assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the cond word starts at the third word starts at and the last word starts at

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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