GATE CS 2015 Set 2 Q25 — Memory Management and Virtual Memory

To find the minimum length of the TLB tag, we can analyze how the virtual address is split for TLB lookup.

1. Analyze the Virtual Address Structure

• Virtual Address Size: $40\text{ bits}$
• Page Size: $8\text{ KB} = 8 \times 1024\text{ bytes} = 2^{13}\text{ bytes}$

The page offset requires $\log_2(2^{13}) = 13\text{ bits}$.

Therefore, the size of the Virtual Page Number (VPN) is: $$\text{VPN size} = 40\text{ bits} - 13\text{ bits} = 27\text{ bits}$$

2. Analyze the TLB Structure

• Total TLB entries: $128$
• Organization: $4\text{-way}$ set-associative with $32$ sets ($32 \times 4 = 128$ entries)

To select one of the $32$ sets, we need a set index. The number of index bits is: $$\text{TLB Index bits} = \log_2(32) = 5\text{ bits}$$

3. Calculate the TLB Tag Length

During a TLB lookup, the Virtual Page Number (VPN) is split into the TLB Set Index and the TLB Tag:

$$\text{VPN} = \text{TLB Tag} + \text{TLB Index}$$

$$\text{TLB Tag size} = 27\text{ bits} - 5\text{ bits} = 22\text{ bits}$$

Thus, the minimum length of the TLB tag is $22$ bits.