Virtual Gate Test Series: CO & Architecture - Cache Memory

Question

Consider a $256k$   $4$- way set associative cache with block size $64$ Bytes. Main memory is $2Gb.$ The number of bits used for tag,set and word will be respectively?

1. $10,15,6$
2. $9,16,6$
3. $8,17,6$
4. $7,18,6$

I think answer is $15,10,6 $ $\text{(tag,set,word)}$

Comments

yes its comming 15, 10, 6. May be they have written it in the wrong order in first option.

---

Here Gb means Gigabit or GigaByte?

Answer 1

See pointing in the beginning only solution is as per 2Gb main memory not 2GB .

2Gb=2^31 bits ie 2^28B

So 28 bit physical address.

Given block size = 64 Bytes ie 2^6 =6 bit for offset inside the block.  

256K cache  ie 2^8.2^10 =2^18 B

Now each block is 64B .  

So no of lines =2^(18-6)=2^12

4 way set associative so no of sets =2^(12-2) =2^10  so 10 bits for set .

 

Total 28 b physical address so for tag 28-10-6=12 b .

Therefore ans should be 12,10,6.

 

 

Answer 2

ans is 9,16,6

16 bcoz ,num of block=256k

and setoffset =s=n/k

                          =256k/4=64k

                            =16bit required