A processor has 16 integer registers (R0, R1, .. , R15) and 64 floating point registers (F0, F1,… , F63)

Q. A processor has 16 integer registers (R0, R1, .. , R15) and 64 floating point registers (F0, F1,… , F63). It uses a 2-byte instruction format. There are four categories of instructions: Type-1, Type-2, Type-3, and Type-4. Type-1 category consists of four instructions, each with 3 integer register operands (3Rs). Type-2 category consists of eight instructions, each with 2 floating point register operands (2Fs). Type-3 category consists of fourteen instructions, each with one integer register operand and one floating point register operand (1R+1F). Type-4 category consists of N instructions, each with a floating point register operand (1F).

The maximum value of N is ________

(A) 32
(B) 64
(C) 256
(D) 512

Ans: 32

Sol:

Given, size of instruction format is 2 byte (= 16 bits), therefore number of instruction encoding = 2¹⁶
Also, total number of bits in integer operand = log₂(16 integer registers) = 4
Total number of bits in floating point operand = log₂(64 floating point registers) = 6

So, number of encoding consumed:

By type 1 instructions = 4×2^3×4 = 2¹⁴
By type 2 instructions = 8×2^2×6 = 2¹⁵
By type 3 instructions = 14×2⁽⁴⁺⁶⁾ = 14336

Now, number of encoding left for type 4 instructions = 2¹⁶ − (2¹⁴ + 2¹⁵ + 14336) = 2048
Therefore, total number of different instructions of type 4 instructions = 2048 /64 = 32

Please note that there is difference between number of different instructions and number of different encoding, a single instruction can have different encodings when the address part differs.

So, answer is 32.