Testbook test series

Question

Can anyone please solve this?

Comments

block size 32B, therefore 5 bits for offset.
Total blocks = 256KB / 32B = 2^13, 13 bits for indexing.

Logical address space <tag bits, index bits, offset bits>
LAS size = 32
<tag bits, 13 , 5>, tag bits = 14.

14+1+1+1 = 17bits required per entry as meta data.
Total memory for meta data: 17 * 2^13 bits or 17*2^10 B or 17KB.

---

@palashbehra5 Ohh silly mistake I forgot to convert bits into bytes :(

---

B R U H!

Answer 1

Cache size=256KB= 2^18

Block size=32 KB=2^5

Number of blocks=(2^18)/(2^5)=2^13 blocks

14 bit(Tag)

13bit(index)

5 bit(Offset)

Total entry=14+1+1+=17bit

Total size of tag memory=17bit*2^13=136Kb=136/8=17KB