GATE CS 2006 Q62 — Memory Management and Virtual Memory

To find the minimum size of the TLB tag, we can break down the virtual address generated by the CPU:

1. Virtual Address Breakdown:

• Virtual Address Size: $32\text{ bits}$
• Page Size: $4\text{ KB} = 2^{12}\text{ bytes}$
• Page Offset: $\log_2(2^{12}) = 12\text{ bits}$

The Virtual Page Number (VPN) is the remaining part of the virtual address:

2. TLB Structure:

The TLB is $4$-way set associative and holds a total of $128$ entries.

• Number of Sets in TLB: $$\text{Number of Sets} = \frac{\text{Total Entries}}{\text{Associativity}} = \frac{128}{4} = 32\text{ sets}$$
• TLB Set Index Bits: $$\log_2(32) = 5\text{ bits}$$

3. TLB Tag Size Calculation:

The Virtual Page Number (VPN) is split into the TLB Tag and the TLB Index:

Thus, the minimum size of the TLB tag is 15 bits.