GATE CS 2010 Q49 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

To find the total transfer time, we determine the data path (bus) width from the $L1$ cache block size, which is $4$ words.

Step 1: Transfer from Main Memory to $L2$ Cache

• The block size of $L2$ is $16$ words.
• With a $4$-word bus width, the transfer requires $\frac{16}{4} = 4$ transactions.
• Each transaction involves reading from Main Memory and writing to $L2$:
  $\text{Time per transaction} = t_{\text{MM}} + t_{L2} = 200\text{ ns} + 20\text{ ns} = 220\text{ ns}$
• Total time for MM to $L2$ transfer:
  $T_{\text{MM} \to L2} = 4 \times 220\text{ ns} = 880\text{ ns}$

Step 2: Transfer from $L2$ Cache to $L1$ Cache

• The block size of $L1$ is $4$ words, requiring exactly $1$ transaction.
• This transaction involves reading from $L2$ and writing to $L1$:
  $T_{L2 \to L1} = t_{L2} + t_{L1} = 20\text{ ns} + 2\text{ ns} = 22\text{ ns}$

Step 3: Total Time Taken

$$\text{Total Time} = T_{\text{MM} \to L2} + T_{L2 \to L1} = 880\text{ ns} + 22\text{ ns} = 902\text{ ns}$$