Question

The main memory of the system consists of 16 MB, the cache memory can hold 64 KB and data is transferred in blocks of 4 bytes each. What is the tag size in main memory address for direct mapping cache?

• A. 10
• B. 8
• C. 14
• D. 24

Hint:

Tag bits = (main memory address bits – (index bits + block offset bits))

Answer 1

The Correct Option is B

Total main memory = \(2^{24}\) bytes (16 MB), block size = 4 bytes ⇒ number of blocks = \(2^{24}/2^2 = 2^{22}\) blocks.
Cache size = 64 KB = \(2^{16}\) bytes ⇒ number of lines = \(2^{16}/2^2 = 2^{14}\)
Tag bits = Total block address bits – cache index bits = \(22 - 14 = 8\)