✅ Given Information:

The CPU generates 32-bit virtual addresses with a 4 KB page size. The TLB holds 128 page table entries and is 4-way set associative.

✅ Step 1: Virtual Address Breakdown:

We split the 32-bit virtual address into:

• Page Offset: 12 bits (because 4 KB = 2¹²)
• Page Number: 20 bits (32 - 12)

✅ Step 2: TLB Breakdown:

The TLB has 128 entries, and with 4-way set associativity, there are:

128 / 4 = 32 sets, so we need 5 bits for the index.

✅ Step 3: Tag Calculation:

The total number of bits for the virtual address is 32:

• Page Offset: 12 bits
• Index: 5 bits (because we have 32 sets)
• Tag: Remaining bits = 32 - 12 - 5 = 15 bits

✅ Final Answer:

The minimum size of the TLB tag is: 15 bits

TLB Tag Size Calculation

Given:

• Virtual Address = 32 bits
• Page Size = 4 KB = $2^{12}$ → Offset = 12 bits
• VPN = 32 − 12 = 20 bits
• TLB entries = 128, 4-way set associative → 32 sets

Set index bits: $\log_2(32) = 5 \text{ bits}$

TLB Tag = VPN − Set Index = 20 − 5 = 15 bits

✅ Final Answer: 15 bits