Consider the following cache system: Memory 256 or 21 bytes, Cache size 128 or 2

Question

Consider the following cache system: Memory 256 or 21 bytes, Cache size 128 or 2^7 bytes, and Block size 16 or 2^4 bytes (a) Memory data [0x26] is positioned in the "2"-nd (actually 2 + 1 or 3^rd) block in memory. Likewise, memory data [0xA6] is positioned m the "A"-th (actually A + l or 11^th) block m memory. Mark the position of memory data addressed by [0x26] and [0xA6] m memory (figure on the left). (b) Mark the position of memory data addressed by [0x26] and [0xA6] m cache (figure on the right). (c) Since a cache block is 16 bytes, the "offset" within the block (or called block offset) can be 0 (0000) ~15 (1111). In other words, the block offset is represented by the last __ bit(s) of a memory address. Block offset of memory data [0x26] (00100110) and [0xA6] (10100110) is therefore ___ and ____, respectively, in binary format. (d) Since there are 8 cache blocks, the "block index' ranges from 0 (000) to 7 (111). In other words, the block index is represented by the next ___ bit(s) of a memory address. Block index of memory data [0x26] (00100110) and [0xA6] (10100110) is therefore ___ and ___, respectively, in binary format.

Explanation / Answer

Memory = 256 Bytes or 28 Bytes

Cache = 128 Bytes or 27 Bytes

Block Size = 16 Bytes or 24 Bytes

so, last 4 bits are offset and bits number 1-3 are represent their location in Cache Block.

a. 0x26 - 20 Block of memory

0xA6 - A0 Block of memory

b. 0x26 - 2 Block of Cache (Bits representation - 0010 0110)

0xA6 - 2 Block of Cache(bits representation - 1010 0110)

As both 1-3 bits are 010 -- Cache block number 010 i.e 2

c. last 4    bits,   0110 and   0110 respectively

d. next   3 bits, therefore   010 and   010 respectively.

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