Question 7: A computer system has a main memory of 64 Mbytes, 256 Kbytes cache memory and two-way set associative mapping with block size of 16 bytes. Assume that a computer uses a byte-addressable mechanism. Determine: a) Main memory address structure. b) Number of blocks in the main memory. c) Number of lines in each cache memory set.

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A: A direct mapped cache is one of the simplest technique of memory mapping, in which each address of…

Q: c) ( >) If a CPU has 24 address lines and 24 bit data bus. Show the design of L3 (direct mapped…

A: please check the solution below

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A: Physical address size: 36 bits (byte addressable memory)Word size: 4 bytes (32 bits)Cache size: 256…

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A: Computer memory is a device that stores data and allows us to retrieve it when needed. Primary…

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

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.

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

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

Q: A byte-addressable computer uses 32-bit address to access main memory. Suppose the data cache…

A: To determine the number of bits in the tag, index, and offset, as well as the total cache size, we…

Q: Assume a processor has data cache and instruction cache each of 32 KB, with 16-word blocks. System…

A: Given:- Direct Mapped cache Cache size = 32 KB =215 = 15 number of bits required. Block size = 16…

Q: Q1) Consider a machine with a byte addressable main memory of 2 bytes and block size of 4 bytes.…

A: Solution:  Q. How is a 12-bit memory address divided into tags? line number. and byte number? A.…

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A: Since cache uses write-through, write misses will not bring blocks into the cache and the data…

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

A: question 1 Assuming that each memory address has 8 bits, the address format for the cache would be…

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A: A virtual address space or address space is the set of ranges of virtual addresses that an operating…

Q: een-bit wor resent the

A:

Q: 2. 2. Suppose a computer using direct-mapped cache has 2³2 bytes of byte-addressable main memory and…

A: Given Data Main Memory Size = 232 Bytes Number of Cache Blocks = 1024 = 210 Block Size = 32 Bytes =…

Q: 2. A computer system contaias a main memory of 32KB. It also has a 1KB cache divided into…

A:

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Q: Design a machine with a byte addressable main memory of 216 bytes and block size of 8 byte. Assume…

A: Consider a machine with a byte addressable main memory of 216 bytes and block size of 8 bytes.…

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A: 16bit processor,16bit(word-sized), can address 64k of RAM. 220(1048576)bytes, so can construct 20bit…

Q: Set No. Block No. Location Within Block No. of bits (b) For the main memory address 1:1:3, briefly…

A:

Q: We want to build a byte organized main memory of 8 GB for a 32-bit CPU architecture composed of byte…

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Q: mapped cache consisting of 32 lines is used with this machine. (2 baytlık bayt adreslenebilir ana…

A: A)  Block size = 8 bytes So block offset is 3 bits Total number of blocks is 32. So index bits is 5…

Q: A CPU has 32-bit address and an 8-way set associative 4MB cache with the cache block size of 64B, we…

A: Answer is given below-

Q: 01. A block-set associative cache ryסnסרתה consists of 128 blocks divided into four block sets The…

A: GIVEN:

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A: Step 1  Introduction: Buffer A designated province of memory within the program for storing the…

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Q: Question 35 A processor running a program makes 100 accesses to memory addresses. 80% are to L1…

A: A processor running a program makes 100 accesses to memory addresses.80% are to L1 only,10% are to…

Q: If a computer specified as a 32-bit processor and can execute up to 64 instructions. Show the size…

A: Dear Student, As the computer can execute 64 = 26 instructions so, bits required for opcode is 6bits…

Q: (i) to calculate the Cache Entry (2) (ii) as the Cache Tag (2)

A: Cache memory : It the fastest memory of the computer.

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.