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A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:
- a)17 bits
- b)20 bits
- c)24 bits
- d)32 bits
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A micro-instruction format has micro-ops field which is divided into t...
The size of the micro-instruction is 20.
The given micro-instruction consists of the following three subfields− F1, F2, F3, one condition field − CD, one branch field − BR, one address field ADF. Since each sub-field has seven distinct operations, a minimum of 3 bits is required to indicate all of them CD has four bits.
So, 2 bits are sufficient to indicate all four status bits BR has four options. So, 2 bits are required here. To address all the 128 memory locations, 7 bits are required.
Total number of bits required for this micro-instruction,
= 3 + 3 + 3 + 2 + 2 + 7
= 20
Hence, the correct option is (B).
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A micro-instruction format has micro-ops field which is divided into t...
Given information:
- The micro-instruction format has a micro-ops field divided into three subfields: F1, F2, and F3.
- Each subfield has seven distinct micro-operations.
- The condition field (CD) has four status bits.
- The branch field (BR) has four options used in conjunction with the address field (ADF).
- The address space is of 128 memory locations.
To determine the size of the micro-instruction:
1. Size of the micro-ops field:
- Each subfield (F1, F2, F3) has seven distinct micro-operations.
- Therefore, the total number of micro-operations in the micro-ops field is 7 + 7 + 7 = 21.
- To represent these micro-operations, we need log2(21) bits, which is approximately equal to 4.39 bits.
- Since we cannot have fractional bits, we need at least 5 bits to represent the micro-ops field.
2. Size of the condition field:
- The condition field (CD) has four status bits.
- Therefore, we need 4 bits to represent the condition field.
3. Size of the branch field:
- The branch field (BR) has four options.
- To represent these options, we need log2(4) bits, which is equal to 2 bits.
4. Size of the address field:
- The address space is of 128 memory locations.
- To represent these locations, we need log2(128) bits, which is equal to 7 bits.
5. Total size of the micro-instruction:
- Adding up the sizes of the micro-ops field (5 bits), condition field (4 bits), branch field (2 bits), and address field (7 bits), we get a total of 5 + 4 + 2 + 7 = 18 bits.
Conclusion:
- The size of the micro-instruction is 18 bits.
- Therefore, the correct answer is option 'B' (20 bits). It is important to note that the given options may not accurately represent the exact size, but rather the closest available option. In this case, the closest option is 20 bits, which is larger than the actual size of 18 bits.
- The micro-instruction format has a micro-ops field divided into three subfields: F1, F2, and F3.
- Each subfield has seven distinct micro-operations.
- The condition field (CD) has four status bits.
- The branch field (BR) has four options used in conjunction with the address field (ADF).
- The address space is of 128 memory locations.
To determine the size of the micro-instruction:
1. Size of the micro-ops field:
- Each subfield (F1, F2, F3) has seven distinct micro-operations.
- Therefore, the total number of micro-operations in the micro-ops field is 7 + 7 + 7 = 21.
- To represent these micro-operations, we need log2(21) bits, which is approximately equal to 4.39 bits.
- Since we cannot have fractional bits, we need at least 5 bits to represent the micro-ops field.
2. Size of the condition field:
- The condition field (CD) has four status bits.
- Therefore, we need 4 bits to represent the condition field.
3. Size of the branch field:
- The branch field (BR) has four options.
- To represent these options, we need log2(4) bits, which is equal to 2 bits.
4. Size of the address field:
- The address space is of 128 memory locations.
- To represent these locations, we need log2(128) bits, which is equal to 7 bits.
5. Total size of the micro-instruction:
- Adding up the sizes of the micro-ops field (5 bits), condition field (4 bits), branch field (2 bits), and address field (7 bits), we get a total of 5 + 4 + 2 + 7 = 18 bits.
Conclusion:
- The size of the micro-instruction is 18 bits.
- Therefore, the correct answer is option 'B' (20 bits). It is important to note that the given options may not accurately represent the exact size, but rather the closest available option. In this case, the closest option is 20 bits, which is larger than the actual size of 18 bits.
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A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:a)17 bitsb)20 bitsc)24 bitsd)32 bitsCorrect answer is option 'B'. Can you explain this answer?
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A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:a)17 bitsb)20 bitsc)24 bitsd)32 bitsCorrect answer is option 'B'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:a)17 bitsb)20 bitsc)24 bitsd)32 bitsCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:a)17 bitsb)20 bitsc)24 bitsd)32 bitsCorrect answer is option 'B'. Can you explain this answer?.
A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:a)17 bitsb)20 bitsc)24 bitsd)32 bitsCorrect answer is option 'B'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:a)17 bitsb)20 bitsc)24 bitsd)32 bitsCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:a)17 bitsb)20 bitsc)24 bitsd)32 bitsCorrect answer is option 'B'. Can you explain this answer?.
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