A processor outputs the hexadecimal address: AA0893BF, with the following partitioning: AA0 (Tag) 890 (Index) BF...
A processor outputs the hexadecimal address: AA0893BF, with the following partitioning:
| AA0 (Tag) | 890 (Index) | BF (Offset) |
Assume that the memory space is Byte addressable. Note that the
parts below are not related to each
other, i.e treat them separately.
a. If Direct Mapped Cache is used, how many bytes are in the
cache?
b. If Two-Way Set Associative Cache is used, how many sets are in
the entire cache?
c. If Two-Way Set Associative Cache is used, how many bytes are in
the entire cache?
d. If Fully Associative Cache is used how many bytes are in one
block?
e. How many blocks are there in memory?
Homework Answers
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A processor outputs the hexadecimal address: AA0893BF, with the
following partitioning:
AA0 (Tag)
890 (Index)
BF...
Text: Explain how a 32-bit byte memory address should be divided into Tag/Index/Offset fields for each...
Text:
Explain how a 32-bit byte memory address should be divided into
Tag/Index/Offset fields for each of the cache configurations below.
Note: 1KB = 210 bytes. You must explain how many bits to assign to
each field and the ordering of the three fields. You get at most
50% of the credit if you give the length of each field without an
explanation.
1) A fully associative cache with cache block size = 2 words and
cache size = 512KB....
Q2. Consider a four-way set associative cache with a data size of 64 KB. The CPU...
Q2. Consider a four-way set associative cache with a data size of 64 KB. The CPU generates a 32-bit byte addressable memory address. Each memory word contains 4 bytes. The block size is 16 bytes. Show the logical partitioning of the memory address into byte offset, cache index, and tag components.
Consider a processor that has a 20-bit address and a 1K Byte Cache. The cache and...
Consider a processor that has a 20-bit address and a 1K Byte Cache. The cache and main memory are divided into blocks where each block is 256 Bytes. If direct mapping is used, what is the tag size of each block in cache and how many tag comparisons are made for a one-cache access? Repeat part (1) for fully associative mapping. Repeat part (1) for 2 way set-associative cache. For the direct map find out which of the following accesses...
Problem 6. Suppose we have a computer with 32 megabytes of main memory, 256 bytes of...
Problem 6. Suppose we have a computer with 32 megabytes of main memory, 256 bytes of cache, and a block size of 16 bytes. For each configuration below, determine the memory address format, indicating the number of bits needed for each appropriate field (i.e. tag, block, set, offset). Show any relevant calculations. Direct cache mapping and memory is byte-addressable a) Direct cache mapping and memory is word-addressable with a word size of 16 bits b) c) 2-way set associative cache...
Suppose a computer using direct mapped cache has 232 byte of byte-addressable main memory, and a...
Suppose a computer using direct mapped cache has 232 byte of byte-addressable main memory, and a cache of 1024 blocks, where each cache block contains 32 bytes. a. How many blocks of main memory are there? b. What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag, block, and offset fields? c.To which cache block will the memory address 0x000063FA map?
Consider a 2-way set associative cache consisting of 8 blocks total of byte-addressable memory with 4...
Consider a 2-way set associative cache consisting of 8 blocks
total of byte-addressable memory with 4 bytes per block. Assume
that the cache is initially empty. Given the following address
sequence, fill in the table below.
Time Access Tag Set Offset 3 10010001 11001001 10110110 10101011 10110010
Given the following 16 bit numbers: A.) 0x8FFF B.) 0x1000 C.) Ox00FO D.) 0x0888 E.) 0xC000 F) 0x9000 . If the numbers are unsigned integers rank them from smallest to larest )rank them from small...
Given the following 16 bit numbers: A.) 0x8FFF B.) 0x1000 C.) Ox00FO D.) 0x0888 E.) 0xC000 F) 0x9000 . If the numbers are unsigned integers rank them from smallest to larest )rank them from smallest to largest 10. If the above number are signed integers (2's complement 11Acomputer has a 16 bit address field,is byte addressable, the word length is also 16 bits, 32 lines of direct mapped cache and each line of cache holds 8 bytes. A.) How many...
question 2 and 3 2. Determine how many sets of cache blocks will be there for...
question 2 and 3
2. Determine how many sets of cache blocks will be there for the following Cache memory size (in bytes) Direct Mapped Blocks Size (in bits) 32 64 218 2-way Set Associative Block Size (in bits) 32 64 A 2A6 [0.5 * 16 = 8] 4-way Set Associative Block Size (in bits) 32 64 SK 64K 256K 3. The physical memory address generated by a CPU is converted into cache memory addressing scheme using the following mapping...
please answer $5 UXIF map in the computer uses direct mapping Question 18 5 pts Suppose...
please answer
$5 UXIF map in the computer uses direct mapping Question 18 5 pts Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main memory addresses and 32 blocks of cache. Suppose also that each block contains 8 bytes. The size of the block offset field is bits, the bits. size of the set field is bits, and the size of the tag field is 5 pts Question 19 Suppose we have a byte-addressable computer...
Assume the cache can hold 64 kB. Data are transferred between main memory and the cache...
Assume the cache can hold 64 kB. Data are transferred between main memory and the cache in blocks of 4 bytes each. This means that the cache is organized as 16K=2^14 lines of 4 bytes each. The main memory consists of 16 MB, with each byte directly addressable by a 24-bit address (2^24 =16M). Thus, for mapping purposes, we can consider main memory to consist of 4M blocks of 4 bytes each. Please show illustrations too for all work. Part...
Text:
Explain how a 32-bit byte memory address should be divided into
Tag/Index/Offset fields for each of the cache configurations below.
Note: 1KB = 210 bytes. You must explain how many bits to assign to
each field and the ordering of the three fields. You get at most
50% of the credit if you give the length of each field without an
explanation.
1) A fully associative cache with cache block size = 2 words and
cache size = 512KB....
Problem 6. Suppose we have a computer with 32 megabytes of main memory, 256 bytes of cache, and a block size of 16 bytes. For each configuration below, determine the memory address format, indicating the number of bits needed for each appropriate field (i.e. tag, block, set, offset). Show any relevant calculations. Direct cache mapping and memory is byte-addressable a) Direct cache mapping and memory is word-addressable with a word size of 16 bits b) c) 2-way set associative cache...
Consider a 2-way set associative cache consisting of 8 blocks
total of byte-addressable memory with 4 bytes per block. Assume
that the cache is initially empty. Given the following address
sequence, fill in the table below.
Time Access Tag Set Offset 3 10010001 11001001 10110110 10101011 10110010
Given the following 16 bit numbers: A.) 0x8FFF B.) 0x1000 C.) Ox00FO D.) 0x0888 E.) 0xC000 F) 0x9000 . If the numbers are unsigned integers rank them from smallest to larest )rank them from smallest to largest 10. If the above number are signed integers (2's complement 11Acomputer has a 16 bit address field,is byte addressable, the word length is also 16 bits, 32 lines of direct mapped cache and each line of cache holds 8 bytes. A.) How many...
question 2 and 3
2. Determine how many sets of cache blocks will be there for the following Cache memory size (in bytes) Direct Mapped Blocks Size (in bits) 32 64 218 2-way Set Associative Block Size (in bits) 32 64 A 2A6 [0.5 * 16 = 8] 4-way Set Associative Block Size (in bits) 32 64 SK 64K 256K 3. The physical memory address generated by a CPU is converted into cache memory addressing scheme using the following mapping...
please answer
$5 UXIF map in the computer uses direct mapping Question 18 5 pts Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main memory addresses and 32 blocks of cache. Suppose also that each block contains 8 bytes. The size of the block offset field is bits, the bits. size of the set field is bits, and the size of the tag field is 5 pts Question 19 Suppose we have a byte-addressable computer...
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