GATE CS 2026 Set 2 Q43 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

To find the number of blocks required to hold the free disk block numbers, we calculate as follows:

Step 1: Calculate the total number of blocks on the disk

• Disk size = $16 \text{ GB} = 16 \times 2^{30} \text{ bytes}$
• Block size = $2 \text{ KB} = 2 \times 2^{10} \text{ bytes} = 2048 \text{ bytes}$
• Total number of blocks ($N$) = $\frac{16 \times 2^{30}}{2 \times 2^{10}} = 8 \times 2^{20} = 8,388,608$ blocks

Step 2: Calculate the capacity of a single index block

• Each block in the list uses $4$ bytes for a pointer to the next block.
• Remaining space in each block for storing block numbers = $2048 \text{ bytes} - 4 \text{ bytes} = 2044 \text{ bytes}$.
• Size of each block number = $32 \text{ bits} = 4 \text{ bytes}$.
• Number of block numbers stored per block = $\lfloor \frac{2044}{4} \rfloor = 511$.

Step 3: Calculate the number of blocks required

To store the addresses of all $8,388,608$ blocks: