Q6 ch-5 me workbook

Question

Suppose that a direct mapping cache has 2⁹  lines, with 2⁴ bytes per cache line. If cache items of a byte addressable memory space of 2²⁹ bytes. How many bits of space will be required for storing tags( do not include bits for validity or other tags; only consider the cost of tags themselves) [in bits].

A. 2⁸        b. 2¹³      C. 2¹¹    

 D. none of these

 

 

Answer 1

A tag is used to identify which of the memory block is currently present in a cache block. So, here you have to fin the no. of distinct memory block (a block is the size of cache line- 16 bytes in this question) which can be mapped to a given cache line. Say, this is x- now we need $\lg x$ tag bits as we store tag in binary. Now, this tag is needed for each cache line- so multiply by the total no. of cache lines to get the total tag storage needed in bits.

Don't look below until you get an answer.  

Calculations will be as follows:

No. of memory blocks = $\frac{2^{29}}{2^{4}} = 2^{25}$.

No. of memory blocks which can be mapped to a given cache line = Total no. of blocks/No. of cache lines (As direct mapping is used)

$ = \frac{2^{25}}{2^9} = 2^{16}$.

So, we need 16 tag bits for each cache line.

So, total tag storage required $= 2^9 \times 16 = 2^{13} $bits.

Comments

I  had solved upto no. Of tag bits which ws 16 bit... And didnt know wht to do next...thank you sir

Answer 2

Tag=16 bits

line number=9bits

4bits for B-offset

Tag directory =number of cache lines x tag bits=16 x 2^9=2^13 bits