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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 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 _________.
consumed by the program text is _________.
Correct answer is '=> 500 bytes'. Can you explain this answer?
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Verified Answer
Consider a processor with 64 registers and an instruction set of size ...
Answer => 500 bytes
Number of register = 64
Number of bits to address register = [log264] = 6 bits
Number of Instruction = 12
Opcode size = [log212] = 4
Opcode(4) reg1(6) reg2(6) reg2(6) Immediate(12)
Total bits per instruction = 34
Total bytes per instruction= 4.25
Due to byte alignment we cannot store 4.25 bytes, without wasting 0.75 bytes ,
So Total bytes per instruction = 5
Total instruction = 100
Total size = Number of instruction x Size of instruction 100 x 5= 500 Byes
Number of register = 64
Number of bits to address register = [log264] = 6 bits
Number of Instruction = 12
Opcode size = [log212] = 4
Opcode(4) reg1(6) reg2(6) reg2(6) Immediate(12)
Total bits per instruction = 34
Total bytes per instruction= 4.25
Due to byte alignment we cannot store 4.25 bytes, without wasting 0.75 bytes ,
So Total bytes per instruction = 5
Total instruction = 100
Total size = Number of instruction x Size of instruction 100 x 5= 500 Byes
Most Upvoted Answer
Consider a processor with 64 registers and an instruction set of size ...
Calculation of Memory Consumed by the Program Text
To calculate the memory consumed by the program text, we need to consider the size of each instruction and the number of instructions in the program.
Size of Each Instruction
Each instruction has five distinct fields:
1. Opcode: This field determines the operation to be performed by the instruction. Since the instruction set size is twelve, the opcode field requires log2(12) = 4 bits.
2. Two Source Register Identifiers: These fields identify the source registers for the instruction. Since there are 64 registers, each identifier requires log2(64) = 6 bits.
3. Destination Register Identifier: This field identifies the destination register for the instruction. Again, it requires 6 bits.
4. Twelve-bit Immediate Value: This field stores a 12-bit immediate value. Therefore, it requires 12 bits.
The total size of each instruction can be calculated as follows:
Opcode (4 bits) + Source Register Identifiers (2 x 6 bits) + Destination Register Identifier (6 bits) + Immediate Value (12 bits) = 34 bits = 4.25 bytes
Number of Instructions in the Program
The program has 100 instructions.
Calculation
To calculate the total memory consumed by the program text, we multiply the size of each instruction by the number of instructions:
Total Memory Consumed = Size of Each Instruction x Number of Instructions
Total Memory Consumed = 4.25 bytes/instruction x 100 instructions = 425 bytes
However, the question specifies that each instruction must be stored in memory in a byte-aligned fashion. This means that each instruction must be aligned on a byte boundary. Since the size of each instruction is not a multiple of 8 (1 byte), we need to round it up to the nearest multiple of 8.
The rounded-up size of each instruction is 5 bytes (4.25 bytes rounded up to the nearest multiple of 8).
Therefore, the total memory consumed by the program text is:
Total Memory Consumed = Rounded-up Size of Each Instruction x Number of Instructions
Total Memory Consumed = 5 bytes/instruction x 100 instructions = 500 bytes
To calculate the memory consumed by the program text, we need to consider the size of each instruction and the number of instructions in the program.
Size of Each Instruction
Each instruction has five distinct fields:
1. Opcode: This field determines the operation to be performed by the instruction. Since the instruction set size is twelve, the opcode field requires log2(12) = 4 bits.
2. Two Source Register Identifiers: These fields identify the source registers for the instruction. Since there are 64 registers, each identifier requires log2(64) = 6 bits.
3. Destination Register Identifier: This field identifies the destination register for the instruction. Again, it requires 6 bits.
4. Twelve-bit Immediate Value: This field stores a 12-bit immediate value. Therefore, it requires 12 bits.
The total size of each instruction can be calculated as follows:
Opcode (4 bits) + Source Register Identifiers (2 x 6 bits) + Destination Register Identifier (6 bits) + Immediate Value (12 bits) = 34 bits = 4.25 bytes
Number of Instructions in the Program
The program has 100 instructions.
Calculation
To calculate the total memory consumed by the program text, we multiply the size of each instruction by the number of instructions:
Total Memory Consumed = Size of Each Instruction x Number of Instructions
Total Memory Consumed = 4.25 bytes/instruction x 100 instructions = 425 bytes
However, the question specifies that each instruction must be stored in memory in a byte-aligned fashion. This means that each instruction must be aligned on a byte boundary. Since the size of each instruction is not a multiple of 8 (1 byte), we need to round it up to the nearest multiple of 8.
The rounded-up size of each instruction is 5 bytes (4.25 bytes rounded up to the nearest multiple of 8).
Therefore, the total memory consumed by the program text is:
Total Memory Consumed = Rounded-up Size of Each Instruction x Number of Instructions
Total Memory Consumed = 5 bytes/instruction x 100 instructions = 500 bytes
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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 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 _________.Correct answer is '=> 500 bytes'. Can you explain this answer?
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