Question

A computer uses memory unit of 512 words of 16 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code, a register code to specify one of 8registers and an address part. How many bits are there in each part?

Answer 1

well your question is not clear enough but i will try to get the meaning.
..
wait
indirect bit = 1
adress part = 2^9(512)=9 bits
register = 2^3 (8) = 3 bits
operation code =3 bits

Answer 2

Answer:

Required number of bits are 4 bits.

Explanation:

Given,

Memory unit = 512 K words

1 word = 32 bits = 4 B

$1 \: word = 32 \: bits = 4 \: B$

Binary instruction code stored in 1 word of memory

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 $= (1, 4, 8, 19)$

Required number of bits are 4 bits.