Question

A digital computer has 2 memory unit 64k*8.5
a)how many data input and data output lines does it have
b)how many address lines does it have
c)what is the memory capacity in bytes​

Answer 1

Answer:

Main memory has 64K = 64 x 1024 = 2^6 x 2^10 = 2^16 words

Cache memory has 1K = 1024 = 2^10 words

Cached address consists of Index and Tag part. Index and Tag together make main memory address. Index part addresses cache memory and Tag part represents the rest of the main memory address.

In this case to address main memory we need 16 bits (2^16) and to address cache memory we need 10 bits (2^10).

So Index is 10 bits wide and Tag is 6 bits wide (16 - 10 = 6).

When using blocks the the Index is divided into the "Block" part and the "Word" part. Block part addresses blocks of cache memory and Word part addresses individual words in a block.

In this case the block is 4 words long for what we need 2 bits (2^2) and that leaves 8 bits (10 - 2 = 8) for addressing blocks.

Summary:

Tag = 6 bits

Index = 10 bits

Block = 8 bits

Word = 2 bits

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