Associative Mapping - Solved Examples

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  • เผยแพร่เมื่อ 4 ต.ค. 2024

ความคิดเห็น • 15

  • @rajeshprajapati1851
    @rajeshprajapati1851 3 ปีที่แล้ว +16

    Direct Mapping :
    1. Many to One Relation and Direct Formula (Block Number % Number of Cache Lines) -> One Comparison.
    2. P.A. Bit Split - Tag Bits, Line Number and Block / Line Offset.
    3. Less Search Time because of One Comparison.
    Drawback - Conflict Miss
    Associative Mapping :
    1. Many to Many Relation and No Direct Formula -> Many Comparisons.
    2. P.A. Bit Split - Tag Bits / Block Number and Block / Line Offset.
    3. More Search Time because of Many Comparisons.
    Drawback - No clue where a particular block will be placed inside the Cache, there is hardware price for judging all the tag bits.

    • @kaushik5572
      @kaushik5572 2 ปีที่แล้ว +1

      Hi Rajesh....I think by Search Time you mean to say Hit Latency....I think search time will be similar for both of the cases, only OR gate is getting added for Associative Mapping which may increase the Hit Latency for the associative mapping little extra addition to the nbit comparator as no of Tag bits are more here. Thank you for your comparison.

    • @rajeshprajapati1851
      @rajeshprajapati1851 2 ปีที่แล้ว

      @@kaushik5572 Yes ! Search time is Hit Latency here.

  • @aphelmusonda5253
    @aphelmusonda5253 2 หลายเดือนก่อน

    Thank you so much sir this is very helpful

  • @sachinsudaraka6397
    @sachinsudaraka6397 3 ปีที่แล้ว

    Thank you sir

  • @aniketsamudre4895
    @aniketsamudre4895 2 ปีที่แล้ว +4

    In second question
    We can calculate cache size using the formula
    No of tag bits = log to the base 2 of (MM size/cache size)
    And then we can calculate cache line and tag directory size
    Is this correct sir ??

    • @harshagrawal2722
      @harshagrawal2722 10 หลายเดือนก่อน

      same question

    • @manavshah1844
      @manavshah1844 10 หลายเดือนก่อน +1

      I think your formula works only for Direct Mapping

  • @FatimaSarwar105
    @FatimaSarwar105 2 ปีที่แล้ว

    You're the best.

  • @abdulhamoothm4752
    @abdulhamoothm4752 ปีที่แล้ว +1

    we have tag feild size =20 then
    no.of tag bits =log2(MM size/chache size), then log2 comes this side
    2^no.of tag bits=MM size/cache size
    2^20=2^34/cache size
    cache size=2^34/2^20
    cache size=2^(34-20)
    cache size=2^14=16384
    tag directory size=16384 * 20= 327680
    in bytes=327680/8=40960 Bytes
    then tag directory size = 40960 Bytes
    is this correct sir.

    • @iam_bantu
      @iam_bantu 11 หลายเดือนก่อน

      Wrong... You can't multiply with the cache size in tag directory.. you need to multiple with the cache lines.. So it's wrong.. Tag directory becomes 20 bits only

  • @chernettaye8557
    @chernettaye8557 2 ปีที่แล้ว +1

    Thanks so much
    እናመሰናለን

  • @gatecomputerscience1484
    @gatecomputerscience1484 ปีที่แล้ว

    😊

  • @aakashbandhakavi8559
    @aakashbandhakavi8559 3 ปีที่แล้ว +1

    please upload the videos a little faster and it'll be really helpful if covered control unit design topic as well.

    • @HappyHumbleHopefulHelpKey
      @HappyHumbleHopefulHelpKey 10 หลายเดือนก่อน

      Please forgive me, I spoke bad about you in my mind. Please forgive ❤ कृपया मुझे क्षमा करें, मैंने मन में आपके बारे में बुरा कहा। कृपया क्षमा करें ❤