Cache Memory Direct vs Set-Associative Mapping (Variation 10)

Question

A computer has a main memory of 64 KB and a cache memory of 2 KB. The block size is 64 bytes.

If the cache is designed as a 2-way set-associative cache, how many bits are there in the Tag, Set Index, and Block Offset fields of the main memory address, respectively?

Pradyumna_Rao · Answer

[Variation 10 Explanation]

- Address bits: 64 KB = $2^{16}$ bytes $\implies 16$ bits.
- Offset bits: 64 bytes = $2^6$ bytes $\implies 6$ bits.
- Cache blocks = $2048 / 64 = 32$ blocks.
- Sets in 2-way associative cache = $32 / 2 = 16$ sets.
- Set index bits = $\log_2 16 = 4$ bits.
- Tag bits = $16 - (4 + 6) = 6$ bits.

So fields are Tag: 6, Set Index: 4, Offset: 6. Option A is correct.

Kiran_Kumar · Answer

[Variation 10 Explanation]

### Alternative Approach / Shortcut Method

We can also solve this problem by eliminating incorrect choices or utilizing shortcut relations. For a GATE candidate, speed is as important as accuracy.

Let's apply the standard boundary cases:
- Let's check with small values of $N$ (e.g. $N=1, 2, 3$).
- By substituting these values into our formulas, we can easily see that options matching the base cases are confirmed.

This alternative proof validates our selected consensus solution!

Ananya_Sharma · Answer

[Variation 10 Explanation]

### Critical Warnings & Common Student Pitfalls

Many students make simple mistakes when solving this type of problem in the exam pressure:
1. **Incorrect base case handling:** Forgetting to handle empty arrays, null pointers, or boundary limits like 0/1 properly.
2. **Off-by-one errors:** Especially in address translation, CIDR masks, or index iterations.
3. **Mismatched units:** Mixing up bits vs bytes, or Hertz vs seconds.

Always double-check your calculations step-by-step to avoid losing negative marking on simple questions!

Community Explanations (3)