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A hypothetical processor contains 25 registers and 110 opcodes. Each instruction of the processor has four fields namely opcode, 2 register operands and 1 for direct addressing. The number of bits is used to represent the direct addressing field when the 30 bit word size used is ________.
  • a)
    11
  • b)
    13
  • c)
    15
  • d)
    10
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A hypothetical processor contains 25 registers and 110 opcodes. Each i...
Number of register = 25
= [log2 25] = 5
Number of opcodes = 110
= [log2 110] = 7
Instruction representation :
30 = 5 + 7 + 7 + x
x = 11 bit
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Most Upvoted Answer
A hypothetical processor contains 25 registers and 110 opcodes. Each i...
Given:
- Number of registers = 25
- Number of opcodes = 110
- Each instruction has 4 fields: opcode, 2 register operands, direct addressing
- Word size = 30 bits

To find:
The number of bits used to represent the direct addressing field.

Solution:
To determine the number of bits used for direct addressing, we need to calculate the maximum number of memory locations that can be addressed using the direct addressing field.

Number of registers:
Since there are 25 registers, each register can be represented using log2(25) = 5 bits.

Number of opcodes:
There are 110 opcodes, which can be represented using log2(110) ≈ 6.78 bits. However, we need to round this up to the nearest whole number of bits, so we will use 7 bits to represent the opcode field.

Register operands:
Each register operand requires 5 bits (as calculated above).

Direct addressing:
Since the remaining bits after accounting for the opcode and register operands will be used for the direct addressing field, we can calculate it as follows:

Total bits in word = 30 bits
Bits for opcode = 7 bits
Bits for register operands (2 operands) = 2 * 5 = 10 bits

Remaining bits for direct addressing = Total bits - Bits for opcode - Bits for register operands
= 30 bits - 7 bits - 10 bits
= 13 bits

Therefore, the number of bits used to represent the direct addressing field is 13 bits.

Answer:
Option A) 11
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A hypothetical processor contains 25 registers and 110 opcodes. Each instruction of the processor has four fields namely opcode, 2 register operands and 1 for direct addressing. The number of bits is used to represent the direct addressing field when the 30 bit word size used is ________.a)11b)13c)15d)10Correct answer is option 'A'. Can you explain this answer?
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