Question

For a direct-mapped cache, 4 bits are used for the tag field and 12 bits are used to index into a cache block. The size of each cache block is one byte. Assume that there is no other information stored for each cache block. Which ONE of the following is the CORRECT option for the sizes of the main memory and the cache memory in this system (byte addressable), respectively?

• A. {64 KB and 4 KB}
• B. {128 KB and 16 KB}
• C. {64 KB and 8 KB}
• D. {128 KB and 6 KB}

Hint:

In direct-mapped cache, the number of index bits determines the number of cache blocks, while the tag bits determine how memory addresses are mapped uniquely.

Answer 1

The Correct Option is A

- The total address size in bits is the sum of the tag and index bits: \(4 + 12 = 16\). This means the main memory size is \(2^{16} = 64\) KB. - The number of cache blocks is \(2^{12} = 4096\), and since each block stores 1 byte, the cache memory size is \(4096 = 4\) KB. Thus, the correct option is (A).