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24 September, 03:13

2. Similar to problem 1, assume your computer system has a 32-bit byte-addressable architecture where addresses and data are each 32 bits. It has a 16K-byte (16,384 bytes) direct-mapped cache, but now the block size is 32 bytes. Answer the following question to observe how the design change impacts the cache size. [10pts]

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  1. 24 September, 03:30
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    Question:

    The question is not complete. The question to answer was not added. See below the possible question and the answer.

    a. How many blocks are in the cache with this new arrangement?

    b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

    C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

    Answer:

    (a) Number of blocks = 512 blocks

    (b) Tag is 18

    (c) Total size of the cache = 8388608 bytes

    Explanation:

    a.

    block size = 32 bytes

    cache size = 16384 bytes

    No. of blocks = 16384 / 32

    No. pf blocks = 512 blocks

    b.

    Total address size = 32 bits

    Address bits = Tag + Line index + block offset

    Block Size = 32 bytes.

    So block size = 25 bytes.

    Hence Offset is 5

    No. of Cache blocks = 512 blocks = 29 blocks

    Hence line offset is 9

    We know that Address bits = Tag + Line index + block offset

    So, 32 = tag+9+5

    tag = 32 - (9+5)

    So Tag is 18

    c.

    Data bits = 32 bits

    Tag=18 bits

    Valid bit is 1 bit

    so Total cache size = 25+218+20

    = 223

    =8388608 bytes
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