(b) A computer has 64 K Word RAM with each memory word of 16 bits. It has cache memory having 32 blocks having a size of 32 bits (2 memory words). Show how the main memory address (1AFC)h will be mapped to the cache address, if (i) Direct cache mapping is used (ii) Associative cache mapping is used (iii)Two way set associative cache mapping is used.

 To map the main memory address to the cache address under different cache mapping schemes, let's consider each scenario:

### Given Information:

- Main Memory Size: 64 K words

- Word Size: 16 bits

- Cache Size: 32 blocks

- Block Size: 32 bits (2 memory words)

### (i) Direct Cache Mapping:

1. Calculate Block Size

   - Block Size = Cache Line Size = 32 bits

2. Calculate Block Offset:

   - Block Offset = log2(Block Size) = log2(32) = 5 bits

3. Calculate Index Bits:

   - Index Bits = log2(Number of Blocks) = log2(32) = 5 bits

4. Calculate Tag Bits:

   - Tag Bits = Total Address Bits - (Index Bits + Block Offset) = 16 - (5 + 5) = 6 bits

5. Convert the Main Memory Address (1AFC)h to Binary:

   - Tag: \(000110\)   - \(0001 \, 1010 \, 1111 \, 1100\) (in 16 bits)

6. Extract Tag, Index, and Offset:

   - Index: \(10111\)

   - Offset: \(11100\)

7. Cache Address:

   - Concatenate Index and Offset: \(1011111100\) (binary)

(ii) Associative Cache Mapping (Fully Associative):

Since it's fully associative, there's no need to calculate indexes. The entire main memory address is used as the tag.

1. Convert the Main Memory Address (1AFC)h to Binary:

   - \(0001 \, 1010 \, 1111 \, 1100\) (in 16 bits)

2. Cache Address:

   - Entire Address is the Tag: \(0001101011111100\) (binary)

 (iii) Two-Way Set Associative Cache Mapping:

1. Calculate Sets:

   - Number of Sets = Number of Blocks / Associativity = 32 / 2 = 16 sets

2. Calculate Index Bits:

   - Index Bits = log2(Number of Sets) = log2(16) = 4 bits

3. Calculate Block Offset:

   - Block Offset = log2(Block Size) = log2(32) = 5 bits

4. Calculate Tag Bits:

   - Tag Bits = Total Address Bits - (Index Bits + Block Offset) = 16 - (4 + 5) = 7 bits

5. Convert the Main Memory Address (1AFC)h to Binary:

   - \(0001 \, 1010 \, 1111 \, 1100\) (in 16 bits)

6. Extract Tag, Index, and Offset:

   - Tag: \(0001101\)

   - Index: \(0111\)

   - Offset: \(11100\)

7. Cache Address:

   - Concatenate Index and Offset: \(011111100\) (binary)

In summary:

- (i) Direct Cache Mapping: \(1011111100\) (binary)

- (ii) Associative Cache Mapping: \(0001101011111100\) (binary)

- (iii) Two-Way Set Associative Cache Mapping: \(011111100\) (binary)