Below is a list of 64-bit memory address references given as word address.Ox03, Oxb4, Ox2b, Ox01, Oxb7, Ox58, Oxbe, Ox02, Oxb5, Ox2e, Oxb6, Ox5b000040100 810001100100015010191001d11012001060110a1010e11103001170111b1011f1111Given a direct-mapped cache with 16 word blocks, what is the hit ratio?O 0.5O 1O 0.75O 0.25

The solution for the above-given question is given below:

Below is a list of 64-bit memory address references given as word address. Ox03, Oxb4, Ox2b, Ox01, Oxb7, Ox58, Oxbe, Ox02, Oxb5, Ox2e, Oxb6, Ox5b 0000 4 0100 8 1000 1100 5 0110 a 1 0001 0101 9 1001 d 1101 0010 6 1010 e 1110 3 0011 0111 b 1011 f 1111 Given a direct-mapped cache with 16 word blocks, what is the hit ratio? 0.5 O 1 O 0.75 0.25

Below is a list of 64-bit memory address references given as word address. Ox03, Oxb4, Ox2b, Ox01, Oxb7, Ox58, Oxbe, Ox02, Oxb5, Ox2e, Oxb6, Ox5b 0000 4 0100 8 1000 1100 5 0110 a 1 0001 0101 9 1001 d 1101 0010 6 1010 e 1110 3 0011 0111 b 1011 f 1111 Given a direct-mapped cache with 16 word blocks, what is the hit ratio? 0.5 O 1 O 0.75 0.25

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

See similar textbooks

Related questions

Q: Assume that a memory is initialised as follows: Memory Address Cell Contents 0x00C336 0x38 0x000337…

A: The Sequence of Operations Executed in the main memory =>

Q: The 32-bit number 52AB43FC (in hexadecimal) is stored in abyte-addressable memory starting at…

A: Little-Endian notation is a method of storing multi-byte data types (such as integers, floats, and…

Q: EXPLAIN ALL STEPS: Determine the output of the following Pep/9 machine language program if the input…

A: Address Instruction(in hex)0000 D1FC15Convert the instructions into binary…

Q: 1. Given the following information in data memory: Address data 1 4 7 8. 9. A E F data 0x0200 49 20…

A: From the ascii table, we can see that Memory Location Hex Binary Ascii 0x0200 49 01001001 I…

Q: 4.5 The following table represents a small memory. Refer to this table for the following questions…

A: The addresses provided in the above question are in binary to solve this question we need to change…

Q: Determine the main memory block number for the given address and block size. Address: Ox2AE6 Block…

Q: For the Assembler directive 0900:0010 DW 112A, 412B, 33AC, the value AC will be stored at Byte…

A: Explanation: DW: define word 2 byte word Intial location is 10 11 2A : 10 11 41 2B: 12 13 33 AC:…

Q: Assume a direct-mapped cache with 16 one-byte blocks. For each reference, list the binary address,…

A: Block size = 1 B So block offset = 0 bits Total number of block inside cache = 16 So index bits = 4…

Q: Assume a computer that has 32-bit integers. Use a table to clearly show how each of the following…

A: Given that, Size of integers is 32 bit integers Starting address= 0x100 The values are stored…

Q: Assume the following values are stored at the indicated memory addresses and registers: Address OxFF…

A: Fill in the following blank showing the value for the indicated operands: OxFD(%rcx,4)

Q: Assume the following values are stored at the indicated memory addresses and registers: Address…

A: eax - value in %eax = 0x3 $204 - value at memory 204 = 0x56 (%ebx) - value at memory address in %ebx…

Q: maximum number of 16-bit words

A: Here we have to find maximum number of 16-bit words that can be stored in a 2^32 byte physical…

Q: A memory unit with capacity 210 x 10 means:

A: 10 bit word consists of 10 bits in each word and each bit can store either 0 or 1

Q: The statement ( AR – M[AR]) in basic computer means: O A. Fetch effective address from memory O B.…

A: AR (Address register) - Holds address for memory M[AR] - content of the memory word specified by the…

Q: Code on sim 8085

A: Following is the code for the reversed sequence:

Q: Below shows the following hexidecimal values: Address 1000: 13 Address 1001: 03 Address 1002: C5…

A: interprettig the hexadecimal values into 8086 assembly language instructions for all the provided…

Question

Below is a list of 64-bit memory address references given as word address.
Ox03, Oxb4, Ox2b, Ox01, Oxb7, Ox58, Oxbe, Ox02, Oxb5, Ox2e, Oxb6, Ox5b
0000
4
0100 8
1000
1100
1
0001
5
0101
9
1001
d
1101
2
0010
6
0110
a
1010
e
1110
3
0011
7
0111
b
1011
f
1111
Given a direct-mapped cache with 16 word blocks, what is the hit ratio?
O 0.5
O 1
O 0.75
O 0.25

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Similar questions

Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes (like LC-3): ADDRESS OxFFFF OXOCOE OXOCOD Ox0C0C OXOCOB OXOCOA 0x0C09 0x0000 CONTENTS 1111 1111 1111 1111 1111 1110 1101 1100 0001 1011 1100 0101 0110 0101 1000 0111 1100 0000 0100 0000 0011 0001 0101 0010 0000 1100 0000 1101 0000 0000 0000 0000 The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) What are the contents in hex of the memory location at following address in binary: 0000 1100 0000 1110? (Enter hex like the following example: Ox2A3F)
A memory system has 4 KB byte-addressable main memory and a direct-mapped cache that consists of 8 blocks with 16 bytes per block. The following shows the main memory address format that allows us to map addresses from main memory to cache. Note: 12 bit address, 4 bit offset, 3 bit $block, and 5 bit tag Assume the cache directory shown below:
Assume that a memory is initialised as follows: Memory Address Cell Contents 0x000336 0x38 0x000337 0x26 0x000338 0x48 0x000339 0x38 0x00C33A 0x00C33B 0x77 0x38 The following sequence of operations is executed in the given memory: Write(0x00C337, 0x0205) Write (0x00C339, 0x0106) Write(0x00C338, 0x0109) Assume that the read and write operations are done in groups of two bytes, and the system is a big endian system. What is the result of read(0x00C337) +read (0x00C338)? Write your answer in Base 10 (decimal)
Please explain on a paper Thank you for your help
The memory location at address 00002001 contains the memory variable in binary form. What is the data memory variable in hexadecimal form? MEMORY 1110 1011 00002001 1110 1010 00002000 1110 1001 00001999 1110 1000 00001998 1110 0111 00001997 1110 0110 00001996 DATA ADDRESS The data memory variable in hexadecimal form is E7. a. b. The data memory variable in hexadecimal form is EA. The data memory variable in hexadecimal form is EB. C. The data memory variable in hexadecimal form is E9. Od.
Please help me with this computer architecture homework
In 8Gi x 32 RAM chip, 8Gi represents the total number of unique addressable memory locations. True False
(1) Convert the following binary machine code to Pep/9 assembly language. (2) Describe what the code does. (3) What is the output? Address Machine Language 0000 1101 0001 1111 1100 0001 0101 0003 1111 0001 0000 0000 0001 1001 0006 1101 0001 1111 1100 0001 0101 0009 1111 0001 0000 0000 0001 1010 000C 1101 0001 0000 0000 0001 1010 000F 1111 0001 1111 1100 0001 0110 0012 1101 0001 0000 0000 0001 1001 0015 1111 0001 1111 1100 0001 0110 0018 0000 0000
Draw a picture illustrating the contents of memory, given the following data declarations: (You need to draw a diagram showing the memory addresses and their contents.) Assume that your data segment starts at 0x1000 in memory. Your answer should fill the table below and explain why. Addres Hexadecimal Hexadecimal S Value Value 0x1000 Ox |0x Ox Ox 0x Name: asciiz "Jones" Age: byte 48 Numbers: word 11, 20 Letter1: asciiz "Car" Letter2: .byte 25 Hexadecimal Value Hexadecimal Value
It is common practise to divide computer memory into many categories using bytes, each of which consists of 8 bits. How many dist values can a tell? How many distinct valueytes may be represented by two separate bytes?
The memory size might be specified as follows: Memory Size = Number of Words x The number of bits in a word What is the number of address and data lines necessary to build a 256 X 32 memory circuit utilising a decoder and a multiplexer?
Show how the decimal value 2019111310 would be stored by byte-addressable machines with 32-bit words, using little endian format and then big endian format. Assume the value starts at address 201916. Draw a diagram of memory and place the appropriate values in the correct (and labeled) memory locations.