UGC NET CSE | December 2019 | Part 2 | Question: 10

Question

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?

• A. $1, 3, 9, 19$
• B. $1, 4, 9, 18$
• C. $1, 4, 8, 19$
• D. $1, 3, 8, 20$

Answer 1

There are 2 addressing modes.

So we need $\lceil log_22 \rceil=1$ bit for addressing mode part.

There are 256 registers.

So we need $\lceil log_2{256} \rceil=8$ bits for identifying a register. i.e register code part needs 8 bits.

There are 512K words.i.e $512*2^{10}=2^{19}$ words

So we need $\lceil log_2{2^{19}} \rceil=19$ bits for identifying a word. i.e,address part needs 19 bits.

So remaining bits of 32 bits are used to identify a instruction. i.e opcode.

So opcode has $32-(1+8+19)=32-28=4$ bits

Hence the computer can support upto 16 instructions.

So option c is the correct answer.