Activity14 Soln - SOLUTION In-Class Activity #14 3/29/2001...

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ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2001 Page 1 of 2 SOLUTION In-Class Activity #14 3/29/2001 2:07 PM Activity 14.1 Cache organization We consider a cache for a computer with 32-bit byte addresses. The cache holds 256 Kbytes of data. Please calculate the information needed to fill the following table. Cache A Cache B Cache C Number of bits in a tag 14 14 16 Number of bits in cache index 16 13 11 Number of blocks cache can hold 64K 8K 8K Number of sets cache can hold 64K 8K 2K Total number of bytes for cache (bytes) 385,024 277,504 279,552 1 (15) Cache A is direct mapped, and has a block size of 4 bytes. Solution: The cache hold 64K blocks = 64K words of data. In Memory: Byte address = 32 bits Word address = 30 bits Block address = 30 bits In Cache: Cache index for 64K blocks = 16 bits. Tag = 30 – 16 = 14 bits Size = 256K x 8 + 64K x (14 tag bits + 1 valid bit) = 3080192 bits = 385,024 bytes. 2 (15) Cache B is direct mapped, and has a block size of 32 bytes.
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Activity14 Soln - SOLUTION In-Class Activity #14 3/29/2001...

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