Homework07-sol

# Homework07-sol - 10/26/07 7:36 PM SOLUTION Homework #07...

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10/26/07 7:36 PM SOLUTION Homework #07 1. [20] Direct-mapped word-level cache. 10/26/07 7:36 PM 1a. A MIPS computer has a 32-bit byte-address space. It has a 32KB direct-mapped cache organized as 8K one-word blocks. Complete the following table (keep all values in hex: it is easier). Ignore valid bits, reference bits and dirty bits. Byte address Word address Cache index Tag 0488C04B 01223012 1012 00911 84CC0488 21330122 0122 10998 7FFFFBC7 1FFFFEF1 1EF1 0FFFF 00003BC5 00000EF1 0EF1 00000 The 30-bit word address is bits 31 - 2 of the byte address. The cache index must have 13 bits to address 8K = 8192 blocks. The (30-13)=17 high-order bits of the word address are the tags. 1b. What is the total size of this cache in bytes? For data: 32KB x 8 bits = 262,144 bits For tags: 8KB x 17 bits = 139,264 bits 401,408 bits = 49.0 KB Grading: Here, and in the next three questions, 2 points for each correct index and tag, and 4 points for correct cache size. 10 points partial grade for correct number and location of tag and index bits, but incorrect word/block address calculations. 2. [20] Direct-mapped block-level cache.

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## This homework help was uploaded on 04/08/2008 for the course ECE 562 taught by Professor Zhou during the Fall '07 term at New Hampshire.

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Homework07-sol - 10/26/07 7:36 PM SOLUTION Homework #07...

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