Question

An 8 KB direct-mapped write-back cache is organized as multiple blocks, each of size 32-bytes. The processor generates 32-bit addresses. The cache controller maintains the tag information for each cache block comprising of the following.

1 Valid bit

1 Modified bit

As many bits as the minimum needed to identify the memory block mapped in the cache.

What is the total size of memory needed at the cache controller to store meta-data (tags) for the cache?

1. 4864 bits
2. 6144 bits
3. 6656 bits
4. 5376 bits

Answer 1

Correct Answer - Option 4 : 5376 bits

The correct answer is option 4

Data:

Main Memory size =2³² Byte

Cache size = 8KB = 2¹³ Byte

Block size = 2⁵ Byte

Valid = 1 bit

Modified = 1 bit

Formula:cachesizeblockscachesizeblocksizetotallinesl

number of bits = ⌈log2 n⌉ 

Number of lines in cache = \(\frac{cache\;size}{block\;size}\)totallineslinesinaset

MM = tag + index + block offset .....(in bits)

Total bits required to store meta-data of 1 line =  Valid bit +  Modified bit + tag bit

Calculation:

Number of cache lines =\({{Cache}\ {size} \over block\ size }=\frac{2^{13}B}{2^5B}= 2^8B\)

So, index bit = 8

MM = tag + index + block offset

32 = tag +8 + 5

tag =19 bits

Tag	index	Block Offset
19bits	8 bits	5 bits

Total bits required to store meta-data of 1 line =  Valid bit +  Modified bit + tag bit

Total bits required to store meta-data of 1 line = 1  + 1 + 19 =21 bits

Cache memory has 256 lines

So, Total memory required =21 bits × 256 = 5376 bits