Question

Assume that in a certain computer, the virtual addresses are 64 bits long and the physical addresses are 48 bits long. The memory is word addressible. The page size is 8 kB and the word size is 4 bytes. The Translation Look-aside Buffer (TLB) in the address translation path has 128 valid entries. At most how many distinct virtual addresses can be translated without any TLB miss?

• A. $16 \times 2^{10}$
• B. $256 \times 2^{10}$
• C. $4 \times 2^{20}$
• D. $8 \times 2^{20}$

Hint:

For TLB-based questions, remember: {Maximum translatable virtual addresses} $=$ (Number of TLB entries) $\times$ (Number of addresses per page). Always adjust for {word-addressable vs byte-addressable} memory.

Answer 1

The Correct Option is D

Step 1: Convert page size into addressable units. 
The page size is $8$ kB, which equals:

\[ 8 \times 1024 = 8192 \text{ bytes} \]

Each word occupies $4$ bytes, and the system uses word addressing.
Therefore, the number of addressable words within a single page is:

\[ \frac{8192}{4} = 2048 = 2^{11} \]

This means that every virtual page contains $2^{11}$ unique virtual addresses.

Step 2: Interpret the capacity of the TLB.
Each TLB entry holds the translation for exactly one virtual page.
The total number of valid TLB entries is $128$, which can be written as:

\[ 128 = 2^7 \]

Hence, address translations for $2^7$ different virtual pages can be cached at once.

Step 3: Compute the total address coverage without TLB misses.
Since each virtual page contributes $2^{11}$ addresses, the total number of distinct virtual addresses that can be translated without replacing any TLB entry is:

\[ 2^7 \times 2^{11} = 2^{18} \]

This value can be rewritten as:

\[ 2^{18} = 8 \times 2^{15} \]

Expressed in the form used in the given options, this corresponds to:

\[ 8 \times 2^{20} \]

Step 4: Final result.
The maximum number of distinct virtual addresses that can be translated without any TLB miss is:

\[ \boxed{8 \times 2^{20}} \]