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azamat
3 years ago
12

How many total bits are required for a direct mapped cache with 16 KB of data and 4 word blocks Assuming a 32 bit address

Computers and Technology
1 answer:
Wittaler [7]3 years ago
8 0

Answer:

147 Kibibits or 18.4 KiB

Explanation:

Given that KiB = kibibyte = 2^10 bytes

Hence 16 KiB is 16384 (2^14) bytes = 4096 (2^12) words.

Therefore, with a block size of 4 words (2^2), we have 1024 (2^10) blocks

Given 32 bits address

We have 4 x 32 = 128 bits of data

The total bits is calculated as (number of blocks) * (data (32 * 4) + tag and validation bits)

Where index part = 10 bits, Offset = 2 bits

2^10 * (4 x 32 + (32–10–2 - 2) + 1)

=> 2^10 * 147 = 147 kibibits

=> 18.4 KiB

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3 years ago
Convert the following numbers. (Please show all steps; no marks will be awarded if no steps are shown) [1.5 x 4 = 6 marks]
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Explanation:

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ii. = Octal is taken in pair of 3 bits , Writing binary of each number in 3 bits we get  = 100 101 001 in BASE 2

iii. = 1 x 52 + 2 x 51 + 3 x 50 + 3 x 5-1   = 38.6 in decimal

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.6 x 8 =  4.8

.8 x 8 = 6.4

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3 years ago
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By the information given, Windows Vista was the last to be released.

Best of Luck!

8 0
3 years ago
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