Given 256 GB of physical memory, a 2-way set associative cache that is 128 KB in size with a block offset of 4 bits, answer the following: 1.How many bits in the address space? \Hint: The size of the physical memory defines the address space since no other information is given.
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![Given 256 GB of physical memory, a 2-way set associative cache that is 128 KB in size with a block offset of 4 bits, answer the following:
1. How many bits in the address space? \Hint: The size of the physical memory defines the address space since no other information is given.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd117526d-6760-43aa-a178-f23c53b32786%2F20b69415-1b2b-4464-b36a-103f4fda084c%2Fd5qw1v2_processed.jpeg&w=3840&q=75)
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- Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Page Table Physical Memory Physical Address (starting) Oxppppdddd Page Frame Frame Size (hex) Size (dec) Ox10000 Ox10000 2 Охс000 65536 PPpp: page number dddd: page offset 1 1 Оxd000 65536 3 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x00011119 What is the physical address for 0x00000001 What is the logical address for Oxd0000001 ? What is the logical address for Oxc0000002 ?Q: A digital computer has a memory unit of 64k * 16 and a cache memory of 1k words. The cache uses direct mapping with a block size of 4 words. i) How many bits are there in the tag, index, block & words fields of the address formats. ii) How many bits are there in each word of cache? iii) How many blocks can the cache accommodate?Exercise 2: A byte addressable memory has a size of 1024 MBytes. This memory is attached to a direct mapping cache of 32KBytes that contains 1024 lines. a. What is the memory address length? b. What is the block size? c. What is the number of blocks in main memory? d. What is the length in bit of: tag (T), line number (L) and byte number (W)? e. Determine in Hexadecimal the tag (T), line number (L) and byte number (W) of the following Hexadecimal memory address: 000008AE f. What is the block that contains the address 000000DE? g. Which line of the cache can hold the block containing 000000DE?
- A computer of 32 bits has a cache memory of 64 KB with a cache line size of 64 bytes. The cache access time is 20 ns, and the miss penalty is 120 ns. The cache is 2-way associative. a) What is the number of cache lines? b) What is the number of cache sets? c) What is the number of lines per set? d) Draw a scheme of this cache. e) Calculate the time to read a word in case of miss.Part A For each byte sequence listed, determine the Y86 instruction sequence it encodes. If there is some invalid byte in the sequence, show the instruction sequence up to that point and indicate where the invalid value occurs. For each sequence, the starting address, then a colon, and then the byte sequence are shown. 0x100: 30f3fcfffff40630008000000000000 0x100: 30f3fcfffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rrmovq %rsi,0×80A(%rcx) O0x115: 00 halt Ox100: 30f3fcffffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox114: 00 halt 0x100: 30f3fcfffffffff rrmovq $-8,%rbx Ox109: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) 0x200: a06f800c020000000000000030f30a00000000000000 0x113: 00 halt 0x100: 30f3fcffffffffff irmovq $-4,%rbx 0x200: a06f pushq %rsi 0x10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox202: 800c02000000000000 call proc Ox116: 00 halt 0x20b: 00 halt 0x20c: proc: Submit Request Answer 0x20c: 30f30a00000000000000 | irmovq $10,%rbx…Below is a list of 32-bit memory address references, given as memory addresses. 12, 720, 172, 8, 764, 352, 760, 56, 724, 176, 744You would like to access a cache with the given memory addresses. The size of cache is 23 = 8-blocks. Your task is to: (1) find out the binary address, (2) fill out the tag and index for each memory address and (3) indicate whether the access is hit or miss in the following table:
- Suppose a computer using direct mapped cache has 224 bytes of byte- addressable main memory and a cache size of 64K bytes, and each cache block contains 32 bytes. (Note: 64K = 26 * 210) a) How many blocks of main memory are there? b) What is the format of a memory address as seen by cache, i.e., what are the sizes of the tag, block, and offset fields?Memory address Data According to the memory view given below, if RO = Ox20008002 then LDRSB r1, [r0, #-4] is executed as a result of r1 = ?(data overlay big endian)? Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 Øx73 ØX20007FFE ØXD4 ØX20007FFE Lütfen birini seçin: O A. R1 = 0X7F O B. R1 = Oxffffffd4 O C. R1 = Oxffffff7F O D. R1=0XD4000000 O E. R1 = 0XD4Suppose a computer using direct-mapped cache has 232 (that's 232)232) bytes of byte-addressable main memory, and a cache size of 512 bytes, and each cache block contains 64 bytes. How many blocks of main memory are there? What is the format of a memory address as seen by cache, i.e. what are the sizes of the tag, block, and offset fields? To which cache block will the memory address 0x13A4498A map?
- A 2-way set associative cache consists of four sets. Main memory contains 2K blocks of eight words each. Show the main memory address format that allows us to map addresses from main memory to cache. Be sure to include the fields as well as their sizes. Compute the hit ratio for a program that loops 6 times from locations 8 to 51 in main memory. You may leave the hit ratio in terms of a fraction. Please show details how you obtain the result.Suppose a computer using direct-mapped cache has 232 bytes of byte-addressable main memory and a cache size of 512 bytes, and each cache block contains 64 bytes.Q.) What is the format of a memory address as seen by cache; that is, what are the sizes of the tag, block, and offset fields?Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Physical Memory Physical Address (starting) Page Table Oxppppdddd Page | Frame Frame Size (hex) Size (dec) Ox10000 Ox10000 2 Охс000 65536 pppp: page number dddd: page offset 1 1 Oxd000 65536 2 2 Oxe000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x0000ee00 ? What is the physical address for Ox00020001 ? What is the logical address for Oxe0001234 ? What is the logical address for Oxc0004268 ?
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