Test by Bikram | Computer Organization and Architecture | Test 1 | Question: 4

Question

A direct-mapped cache has  $2^{10}$ cache lines with  $2^4$ bytes of data per cache line. If the cache is used to store blocks for a byte addressable memory of size  $2^{30}$ bytes, then how many bytes of space will be required for storing the tags?

• A.  $2^{15}$ bytes
• B.  $2^{11}$ bytes
• C.  $2^6$ KB
• D.  $2^7$ KB

Answer 1

No. of main memory blocks per cache block $=\frac{2^{30}/2^4}{2^{10}} = 2^{16}$.

So, we need 16 bits per cache line to identify which of the $2^{16}$ blocks is currently present in it. And this should be present for each of the $2^{10}$ cache lines making a total tag memory size of $16 \times 2^{10} = 2^{14}$ bits = $2^{11}$ bytes.

Comments

sir here i think like this here  cache lines = 2¹⁰ = 10 bit

  block size = 2⁴ = 4 bit

total = Tag + line offset + WO

  30  = tag + 10 +4

tag = 16 bit = 2⁶ KB  option C .......plz check where i mistake?????

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Last line has the mistake..16bit != 2^6 KB

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1) Block size===> 2^4 bit

2) No of cache blocks==> 2^10 

3) No of cells in main memory==> 2^30

We know that ,

Total bit = Tag bit + Line offset + Word offset

ie.

30= Tag bit + 10 bit (line offset) + 4 bit (word offset)

SO, Tag bit = 30-10+4

ie Tag bit=16 bit.

Tag Memory = Tag size in bit * no of cache blocks

Tag memory = 16 bit * 2^10 ==> 2^4 * 2^10 bit= 2^14 bit.

NOTE (important):

The question is asking about “space” required to store tags in terms of "BYTES", ie they want you to find tag memory in terms of bytes.

so dont just  stop by finding tag memory in terms of bit, find it in terms of byte ,that is what you have to do.

we have,

Tag memory= 2^14 bit

ie .Tag memory (in BYTE)= (2^14) / 8        .....(BYTES)

ie. Tag memory in BYTES= 2^11 BYTES.

Thank you. 

 

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Question is asked about tag directory size.