Homework Solution: Consider a cache memory consisting of 128 blocks of 32 words each, for a total of 4K words. Assume that Main…

    Consider a cache memory consisting of 128 blocks of 32 words each, for a total of 4K words. Assume that Main memory is addressable by a 32 bit address. The MM has 4 Giga words, which we will view as 1 Mega Blocks of 4K words each. For simplicity we will assume that consecutive addresses refer to consecutive words.

    What is the Main Memory address format if Direct mapping is used

    please format as Tag|Block|Word

    Expert Answer

     
    Block size(Block off) = 32 words = 128 bytes = 2^7 bytes

    Consider a cache perpetuation consisting of 128 arrests of 32 suffrage each, restraint a aggregate of 4K suffrage. Usurp that Main perpetuation is orationable by a 32 morsel oration. The MM has 4 Giga suffrage, which we allure conception as 1 Mega Arrests of 4K suffrage each. Restraint artlessness we allure usurp that coherent orationes assign to coherent suffrage.

    What is the Main Perpetuation oration restraintmat if Plain mapping is used

    please restraintmat as Tag|Block|Word

    Expert Reply

     

    Arrest bulk(Arrest unstudied) = 32 suffrage = 128 bytes = 2^7 bytes

    cache bulk = 4K suffrage = 2^12 suffrage = 2^14 bytes

    Main perpetuation sixze = 4Giga suffrage = 2^32 suffrage = 2^34 bytes

    Given that we are using Plain mapping..

    no.of lines = cace bulk/ arrest bulk

    = 2^14 bytes/ 2^7 bytes

    = 2^7 bytes

    Therefore no.of lines = 7 morsels

    In plain mapping, no.of lines = abjuration.

    Therefore abjuration = 7 morsels

    And as-well given that Main perpetuation is orationable by a 32 morsel addres,

    Hnece tag = 32-index-arrest unstudied

    = 32-7-7

    = 18

    Therefore Main Perpetuation oration restraintmat is as

    Tag Index Blockoff
    18 7 7