Average Memory Access Time

Question

Suppose cache with hit ratio 'C_h' and access time 'C_t' is given and main memory with hit ratio 'M_h' and access time 'M_t' is given and a disk with access time 'D_t' is given. If with only this information average memory access time is asked, does the access time of main memory include time of cache and memory also.

For e.g. should the equation be -

Average memory access time = C_h * C_t + (1-C_h) * M_h * ( C_t + M_t ) + (1-C_h) * (1-M_h) * ( C_t + M_t + D_t)

If so, why the addition and is it implicit always ? 

Answer 1

Yes , CPU takes the blocks data from cache

if their is a cache miss then from main memory we get the block data and place it in cache and then given to CPU

if their is a miss in main memory then from disk ----> main memory -------->cache the block is placed and then the lock is taken by CPU

So Average memory access time = C_h * C_t + (1-C_h) * M_h * ( C_t + M_t ) + (1-C_h) * (1-M_h) * ( C_t + M_t + D_t)

Comments

It's for hierarchical access Right?