A Computer uses a memory unit with 256K word of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in operation code, the register code part and the address part?

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UGC NET Computer Science (Paper 2) Dec 2018 Paper

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Answer 1

Data:

1 word = 32 bits = 4 bytes

Memory size = 256K word = 2¹⁸ word

Indirect bits = 1

Number of registers = n = 64

Formula:

Number of bits needed to present a register = ⌈log₂ n⌉

Calculation:

Number of bits needed to present a register = ⌈log₂ 64⌉ = 6

Binary Instruction (32 bit):

Indirect bit (1-bit)

Operation Code (x-bits)

Register Code (6-bits)

Address Part (18-Bits)

1 + x + 6 + 18 = 32

∴ x = 7

Therefore 7,6 and 18 bits are there in operation code, the register code part and the address part respectively.