InClassActivity14-sol - 3/14/03 2:52 PM Activity #14...

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3/14/03 2:52 PM ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2003 Page 1 of 2 Activity #14 (Solution) 3/14/2003 2:52 PM Activity 14.1 Cache organization We consider a cache for a computer with 24-bit byte addresses. The cache holds 8 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 11 11 13 Number of bits in cache index 11 8 6 Number of blocks cache can hold 2K 256 256 Number of sets cache can hold 2K 256 64 Total size of the cache (in bytes) 11,264 8,576 8,640 Overhead of cache organization (%) 37.5 4.7 5.5 1. (15 points) Cache A is direct mapped, and has a block size of 4 bytes. Solution: The cache hold 2K blocks = 2K words of data. In Memory: Byte address = 24 bits Word address = 22 bits Block address = 22 bits In Cache: Cache index for 2K blocks = 11 bits. Tag = 22 – 11 = 11 bits Size = 8K x 8 + 2K x (11 tag bits + 1 valid bit) = 90,112 bits = 11,264 bytes. Overhead % = 100 * (11,264 – 8*1024) / 8*1024 = 37.5 %
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This note was uploaded on 04/25/2010 for the course ECSE ecse-2340 taught by Professor Wozny during the Spring '09 term at Rensselaer Polytechnic Institute.

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InClassActivity14-sol - 3/14/03 2:52 PM Activity #14...

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