4. The figure below shows an 8-way interleaved, byte-addressable memory. Thetotal size of the memory is 4 KB.The elements A[i]G) of a 2-dimensional array A are 4-bytes (one-word) in lengthand can be stored in the memory as shown, where 0 < ij <7.The width of the bus between CPU and memory is 32 bits, that is, it can carryonly one word at a time.Bank 1Bank 7Bank 0317A[7][0]A[I][0]A[0]|0]32A[7]1]64A[7]|2]A[I][2]A[0||2]RANKO255224AI기기A[I][7]A[0||7)RANKN......Since the address space of the memory is 4 KB, 12 bits are needed to uniquelyidentify each memory location, i.e., Addr[11:0]. Find out and explain which bitsof the address will be used for:• Byte on bus: Addr […... :...]Bank index bits within a bank: Addr [.... .]• Chip select address bits within a rank: Addr […... .]• Rank bits: Addr […... :....]

Dear Student, As 12 bits are needed to uniquely identify each memory location the Address bits start…

figure below shows an 8-way interleaved, byte-addressable memory. The total size of the memory is 4 KB. The elements A[i]G) of a 2-dimensional array A are 4-bytes (one-word) in length and can be stored in the memory as shown, where 0

figure below shows an 8-way interleaved, byte-addressable memory. The total size of the memory is 4 KB. The elements A[i]G) of a 2-dimensional array A are 4-bytes (one-word) in length and can be stored in the memory as shown, where 0

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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Concept explainers

Hash Table

A hash table also called a hash map, is a data structure that maps key-value pairs. The hash table can be used in complex systems to provide fast access and compact storage. A hash table is a structure that stores data in an abstract data type kind of associative array. It keeps data in an array format, with a distinct index value for every data value. If the index of the required data is known, then the data can be accessed very easily.

Question

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4. The figure below shows an 8-way interleaved, byte-addressable memory. The
total size of the memory is 4 KB.
The elements A[i]G) of a 2-dimensional array A are 4-bytes (one-word) in length
and can be stored in the memory as shown, where 0 < ij <7.
The width of the bus between CPU and memory is 32 bits, that is, it can carry
only one word at a time.
Bank 1
Bank 7
Bank 0
31
7
A[7][0]
A[I][0]
A[0]|0]
32
A[7]1]
64
A[7]|2]
A[I][2]
A[0||2]
RANKO
255
224
AI기기
A[I][7]
A[0||7)
RANKN
......
Since the address space of the memory is 4 KB, 12 bits are needed to uniquely
identify each memory location, i.e., Addr[11:0]. Find out and explain which bits
of the address will be used for:
• Byte on bus: Addr […... :...]
Bank index bits within a bank: Addr [.... .]
• Chip select address bits within a rank: Addr […... .]
• Rank bits: Addr […... :....]

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