A computer uses a memory unit of 512 K words of 32 bits each. A binary instruction code is stored in one word of the memory. The instruction has four parts: an addressing mode field to specify one of the two-addressing mode (direct and indirect), an operation code, a register code part to specify one of the 256 registers and an address part. How many bits are there in addressing mode part, opcode part, register code part and the address part?

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UGC NET Computer Science (Paper 2) Dec 2019 Paper

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Answer 1

Data:

Memory unit = 512 K words

1 word = 32 bits = 4 B

Binary instruction code stored in 1 word of memory

Instruction divided as follows,

Addressing mode

Operation code

Register code

Address part

Calculation:

Addressing mode = 1 bit for direct or indirect

Register code = ln (256) = 8 bits

Since, registers are already addressed, only memory needs to be addressed

Address part = ln (512 K) = 19 bits

Operation mode = 32 – 1 – 8 – 19 = 4 bits

Therefore, bits for addressing mode part, opcode part, register code part and the address part = (1, 4, 8, 19)