HW9-sol

# HW9-sol - 2 5.3.4 a Using equation on page 463 n = 14 bits...

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1 CSE 341: Computer Organization Spring 2011 Solution to HW #9 & Recitation 9 The last 3 problems are HW#9 problems. 5.3.2 a. Binary address: 00000001, 10000110, 11010100, 10000111, 11010101, 10100010 10100001, 00000010, 00101100, 00101001, 11011101 Tag: Binary address >> 4 bits (meaning, the leftmost 28 bits) Index: (Binary address >> 1 bit) mod 8 (meaning the 1 st , 2 nd and 3 rd bits from the right. The 0 th bit is used as word offset.) Hit/Miss: M, M, M, H, H, H, M, M, M, M, M, M b. Binary address: 00000110, 11010110, 10101111, 11010110, 00000110, 01010100, 01000001, 10110000, 01000000, 01101001, 01010101, 11010111 Tag: Binary address >> 4 bits (meaning, the leftmost 28 bits) Index: (Binary address >> 1 bit) mod 8 (meaning the 1 st , 2 nd and 3 rd bits from the right. The 0 th bit is used as word offset.) Hit/Miss: M, M, M, H, M, M, M, M, M, M, H, M

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Unformatted text preview: 2 5.3.4 a. Using equation on page 463, n = 14 bits, m = 0 (1 word per block) 2 14 × (2 × 32 + (32 – 14 – 0 – 2) + 1) = 802 Kbits Calculating for 16 word blocks, m = 4, if n = 10 then the cache is 541 Kbits, and if n = 11 then the cache is 1 Mbit. Thus the cache has 128 KB of data. The larger cache may have a longer access time, leading to lower performance. b. Using equation total cache size = 2 n × (2 m × 32 + (32 – n – m – 2) + 1), n = 13 bits, m = 1 (2 words per block) 2 13 × (2 1 × 32 + (32 – 13 – 1 – 2) + 1) = 2 13 × (64 + 17) = 663 Kbits total cache size For m = 4 (16 word blocks), if n = 10 then the cache is 541 Kbits and if n = 11 then cache is 1 Mbits. Thus the cache has 64 KB of data. The larger cache may have a longer access time, leading to lower performance....
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HW9-sol - 2 5.3.4 a Using equation on page 463 n = 14 bits...

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