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Lelechka [254]
2 years ago
5

Contact Us Save

Computers and Technology
1 answer:
svetoff [14.1K]2 years ago
3 0
All except drinking caffeinated beverages:)))
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Give me facts on john john son
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Answer: Story -

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Explanation:

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3 years ago
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 1024-byte 4-way set as
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Answer:

The answer is The Cache Sets (S) = 32, Tag bits (t)=24, Set index bits(s) = 5 and Block offset bits (b) = 3

Explanation:

Solution

Given Data:

Physical address = 32 bit (memory address)

Cache size = 1024 bytes

Block size = 8 bytes

Now

It is a 4 way set associative mapping, so the set size becomes 4 blocks.

Thus

Number of blocks = cache size/block size

=1024/8

=128

The number of blocks = 128

=2^7

The number of sets = No of blocks/set size

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= 32

Hence the number of sets = 32

←Block ←number→

Tag → Set number→Block offset

←32 bit→

Now, =

The block offset = Log₂ (block size)

=Log₂⁸ = Log₂^2^3 =3

Then

Set number pc nothing but set index number

Set number = Log₂ (sets) = log₂³² =5

The remaining bits are tag bits.

Thus

Tag bits = Memory -Address Bits- (Block offset bits + set number bits)

= 32 - (3+5)

=32-8

=24

So,

Tag bits = 24

Therefore

The Cache Sets = 32

Tag bits =24

Set index bits = 5

Block offset bits = 3

Note: ←32 bits→

Tag 24 → Set index  5→Block offset 3

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3 years ago
Commercial applications are never free<br><br> -True<br><br> -False
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False

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The cord of a bow string drill was used for
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