Question

In a paging system, the page table is stored in memory. If memory access time is 100 ns, what is the effective access time if one page table lookup is needed per access?

• A. 100 ns
• B. 150 ns
• C. 200 ns
• D. 50 ns

Hint:

Always double the memory access time if the system uses memory-resident page tables without a TLB (Translation Lookaside Buffer).

Answer 1

The Correct Option is C

In a computer system that uses paging, the logical address is first translated to a physical address using the page table.
If the page table is stored in main memory, every access to data needs:
- One access to get the page table entry (i.e., frame number), and
- Another access to retrieve the actual data from the calculated physical address.
Hence, each memory access will require two memory operations.
Given that memory access time = 100 ns
\[ \text{Effective Access Time (EAT)} = 100\,\text{ns (for page table)} + 100\,\text{ns (for actual data)} = 200\,\text{ns} \] Thus, the correct effective access time is 200 ns.