Question

Explain the instruction format in detail as per the no of address in the instruction with examples?

Answer 1

Instruction size = 32-bit

Number of instruction n = 98

a. number of bits for opcode

= $ceil\left(\log{n}\right)$

= $ceil\left(\log{98}\right)$

7 bits for opcode will be required

b.

number of bit in operand = (number of bit for instruction) - (bit required in opcode)

= $32 - 7 = 25$

= 25 bits

c.

since there is only one Address and operand bit in instruction are 25

$\therefore maximum\ memory\ address\ that\ can\ be \ can\ be\ accessed \ is\ 2^{25}$

d.

yes it is because

In 7 bit opcode $2^7$ = 128 instruction possible, but only 98 instructions are used in this set.

Total number of instruction may additionally use = $\left(128\ -\ 98\right)$ $= 30$