cache memory

Question

THE size of memory required at cache controller to store the metadata is 2kbyte The metadata include tag bits 1 modified bit and 1 valid bit The cache contain 1KBlocks of 32 bytes each organized as directed mapped The size of Main memory is __ Mbytes?

Comments

Is it $16MB$ ?

---

I am getting 512 MB .

---

@Kabir5454 Tag = 9bits. But MM = 9+10+5 bits. Are you getting something different?

---

I am getting tag =14 .

2 byte tag information for each block .so ,(16-1-1) =14 tag bits.

---

Metadata = 2^11B. 11 bits cache controller containing ( $9$ tag bits,1 modified bit and 1 valid bit)

---

https://gateoverflow.in/2145/gate-cse-2011-question-43 see this . we should not interpret it like 2^11 bit cache controller . 

For each block we store some fixed amount of data like page table entry .From the above link hope you will find some reference.

---

@Kabir5454 My bad. Thanks.

$2^{11}*8$ = $(Tag+1+1)*2^{10} \implies Tag = 14 bits$

---

MM should be $14+10+5 = 29bits \ address\implies 512 MB$

---

Yes correct .

---

I am getting 8 Mb 

is my approach correct if not plz correct me

---

@Vaishnavi Gadhe First of all Size of metadata = 2^14 in bits.

We need to compare these 2^14 bits with #cache lines * (tag + 2), from which we will get tag = 14bits.

See my answer below.

---

@Abhrajyoti00 okk i understood my mistake 

Answer 1

Answer :  $512 MB$

The size of memory required at cache controller to store the metadata = $2KB = 2^{11}B$

Tag entry size in metadata = $x$ tag bits +$1$ modified bit +$1$ valid bit

#Cache Lines given = $1K = 2^{10}$

Block Size = $32 = 2^5B$

Given, Total size of metadata = $2^{11} * 8bits  = 2^{14}b$

Total size of metadata = Tag entry size ×Number of cache blocks = $(x+1+1)* 2^{10}$

Hence, $2^{14} = (Tag+1+1)*2^{10} \implies Tag = 14 bits$

Thus Size of Main Memory address = Tag + Line Offset + Block Size = $14 + 10+5 = 29bits$

Therefore, the size of MM = $2^{29} = 512 MB$ (ANS)