Question (la) A cache system is to be designed to store data from a 256 MB memory space. If each block of main memory contains 16 words, determine the number of blocks that are needed and draw the logical organization of the full address identifying the block ID portion and the word (offset) portion. The IEEE Standard 754 representation of a floating-point number is given as: 01011011001100110101000000000000 base on your knowledge on floating point numbers determine the binary value represented by this number. A digital weighing scale measures the weight of a real-world object, processes the data and stores it in binary form. Using the principles of digital signal processing, practically explain how this phenomenon occurs.

Q: (10 pts) The bus-based dual-core multiprocessor shown in Figure 1 represents a symmetric shared…

A: This graphic is an introduction to the essential elements of using Scrum for agile software…

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A: This type of mapping has an improved and  enhanced form of direct mapping which helps us to remove…

Q: Problem: A computer system uses 32-bit memory addresses and it has a main memory consisting of 1G…

A: Solution: Consecutive addresses refer to bytes.(a) A block has 64 bytes; hence the Word field is 6…

Q: 4. A computer has byte addressable memory with a cache that can hold eight 32-bit data (4 bytes).…

A: Answer: I have given answered in the handwritten format.

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Q: Assume that we have a computer with a cache memory of 512 blocks with a total size of 128K bits.…

A: Assume that we have a computer with a cache memory of 512 blocks with a total size of 128K bits.…

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A: Introduction: Cache mapping is a mechanism for transferring data from the main memory.  There are…

Q: bytes and block size Consider a machine with a byte addressable main memory of 2 of 8 byles. Assume…

A: As per the policy, only first three parts are answered/

Q: Please do a,b,c,d, and e  Consider a machine with a byte addressable main memory of Bytes and block…

A: Answer in step2

Q: 6. For a direct mapped cache comprising 16 single word blocks answer the following questions. Assume…

A: Introduction :

Q: A cache system is to be designed to store data from a 256 MB memory space. If each block of main…

A: According to the asked question, the solution is given below with a proper explanation. According to…

Q: We shall compare different memory organizations based on the following assumptions: • all transfers…

A: Introduction :Given , Clock to send the address = 1clock for access time = 10 bus transfer = 1 clock…

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A: A)List the two main principles used in the design of memory hierarchies and explain them briefly.…

Q: Assume that we are having a computer with the following characteristics: 1MB of main memory; word…

A: Introduction Given,  cache size =64 KB,  4-way cache  main memory size= 1 MB  word size= 1Byte…

Q: 2. This problem concerns the cache shown below. Assume the following: • The memory is…

A: Here we have to find the data in address 01110101 , but as given in the question Addresses are 13…

Q: The 16-bit memory addresses are used in computer systems which has a 4K-byte cache that is…

A: A) Block size is 128 Bytes.  So block offset bits is 7 bits Total number of block inside cache =…

Q: Q4: With the reference of a suppose 93% of the memory accesses found in cache then what is average…

A:

Q: Problem 3: Instructions: Memory Hierarchy For a cache of 64k blocks, a sixteen-word block size, and…

A: Below the solution:

Q: 3. Assume that we have a byte-addressed processor (i.e., addresses specify bytes) with 30- bit…

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A: Let's understand step by step :   Set associative mapping : In this design , lines are grouped to…

Q: Problem 3: Instructions: Memory Hierarchy For a cache of 64k blocks, a sixteen-word block size, and…

A: The memory hierarchy separates computer storage into a hierarchy based on response time.

Q: 1.Assume your 32-bit computer (memory address 32-bits) has 16-KB (only L1-data) direct mapped cache.…

A: Here block size = 64B Block offset = log(64)= 6 bits Index bits = log (cache size/block size) =…

Q: ume the following: • The memory is byte addressable. • Memory accesses are to 1-byte words (not to…

A: Block offset bits = 2 bits Set index bits = 3 bits For a hit in set 3, the tag bits should be 0x06…

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Q: Please explain this 1,2, and 3 Consider following cache elements Cache can hold 64 kB Data are…

A: GIVEN: Please explain this 1,2, and 3 Consider following cache elements Cache can hold 64 kB Data…

Q: 3. Given a 32-byte cache (byte-addressable, initially empty) and a sequence of access address (6)10,…

A:

Q: Function NSU-1 For a system, RAM-64KB, Block size-4 bytes, Cache sine- 128 bytes, Direct mapped…

A: RAM -64 KB Block size -4 byte Cache size 128 byte Hit ratio while using direct mapped cache: To…

Q: computer Witn the wing total of TNabyle Content of cach addressable location is I byte; block size…

A: Here the block size is given as 16 bytes and the word size is 1 byte Therefore there are 16 words…

Q: 3) The physical address is the actual location within the RAM. It is pu bus by the CPU to be decoded…

A: 1. Logical address =CS:IP                                    =426:A436  2.offset address = IP…

Q: Problem 3: Instructions: Memory Hierarchy For a cache of 64k blocks, a sixteen-word block size, and…

A: Below is the answer to the above question. I hope this will be helpful for you....

Q: 21. The idea of cache memory is based on a. The property of locality of reference b. The…

A: 1) The Idea of the cache memory is based on the property of Locality of reference 2) Locality of…

Q: The following equation was suggested both for cache memory and disk cache memory Ts = Tc + M * Tp…

A: The Answer is

Q: Given a byte-addressable computer with a cache that holds 4 blocks of 2 bytes each. If each memory…

A: please see the next step for solution

Q: 4. For a system, assume, RAM- 64KB, block size -4 bytes, cache size - 64 bytes, 2-way set…

A: Given : RAM= 64KB, block size = 4 bytes, cache size = 64 bytes, and the mapping is 2-way set…

Q: Data are transferred between main memory and the cache in blocks of 8 bytes each Main memory…

A: Block size = 8B =>  Block offset = log 8 = 3 bit Memory size = 256MB => memory bits = 28 bits…

Q: Assume that we have a computer with a cache memory of 512 blocks with a total size of 128K bits.…

A: Because the bits are 128K in cache size. Thus byte cache size = 128K bits / 8 = 16KB. It is divided…

Q: Assume that we have a computer with a cache memory of 512 blocks with a total size of 128K bits.…

A: Given the cache's capacity of 128K bits. Cache capacity in bytes = 128K bits / 8 = 16KB Because this…

Q: Assume that we have a computer with a cache memory of 512 blocks with a total size of 128K bits.…

A: Given:

Q: Q3 (a) Analyze the differences between Sequential and Direct access methods in Computer storage…

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Q: Answer the following questions. A. A computer system has a main memory access time as 60ns. you as a…

A: ANSWER:-

Q: Assume the following: • The memory is byte addressable. • Memory accesses are to 1-byte words (not…

A:

Q: 1. Consider a computer with the following characteristics: total of 1Mbyte of main memory; Content…

A: GIVEN . Consider a computer with the following characteristics: total of 1Mbyte of main memory;…

Q: Assume that a system’s memory has 128M bytes. Blocks are64 bytes in length, and the cache consists…

A: Ans:)  Given, Memory = 128M bytes, Block size = 64 bytes, cache size = 32K blocks

Q: For a direct-mapped cache design with a 64-bit address, the following bits of the address are used…

A:

Q: Assume A and B are two distinct memory blocks and are mapped to two separate cache lines. Consider…

A: The answer as given below:

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Q: Compare between the following techniques that are used for computer memory cache: (A) Direct-mapped.…

A: Cache mapping is the method of bringing main memory data blocks into cache storage blocks. It is of…

Q: Compare between the following techniques that are used for computer memory cache: (A) Direct-mapped.…

A: Below i explain the answer in my own words and also in brief by which you understand it well.

Q: If memory read cycle takes 100 ns and a cache read cycle takes 20 ns, then for four continuous…

A: Memory Read cycle : 100nsCache Read cycle : 20ns Four continuous reference is done - one reference…