Consider a processor with 64 registers and an instruction set of size twelve. Each instruction has five distinct fields, namely, opcode, two source register identifiers, one destination register identifier, and a twelve-bit immediate value. Each instruction must be stored in memory in a byte-aligned fashion. If a program has 100 instructions, the amount of memory (in bytes) consumed by the program text is _______.

To find the size of each instruction, we calculate the number of bits required for each field:

1. Opcode: For 12 instructions, bits needed = $\lceil \log_2 12 \rceil = 4\text{ bits}$.

2. Source Register 1: For 64 registers, bits needed = $\lceil \log_2 64 \rceil = 6\text{ bits}$.

3. Source Register 2: Bits needed = $\lceil \log_2 64 \rceil = 6\text{ bits}$.

4. Destination Register: Bits needed = $\lceil \log_2 64 \rceil = 6\text{ bits}$.

5. Immediate Value: Given as $12\text{ bits}$.

Total bits per instruction:

Since instructions must be stored in a byte-aligned fashion (multiples of 8 bits / 1 byte):

For a program of 100 instructions: