GATE CS 1997 Q5.3 — ALU, Data-path and Control Unit

To find the minimum number of bits in the micro-instruction, we analyze the two parts of the specification separately:

1. Part (a): None or one of the three micro-operations of one kind
   Since we can select at most one micro-operation out of three, these operations are mutually exclusive. Including the "none" (no operation) case, there are a total of $1 + 3 = 4$ possible states:
   • State 0: No operation
   • State 1: Micro-operation 1
   • State 2: Micro-operation 2
   • State 3: Micro-operation 3
   To encode $4$ mutually exclusive states, the number of bits required is: $$\lceil \log_2(4) \rceil = 2 \text{ bits}$$


2. Part (b): None or up to six micro-operations of another kind
   The phrase "none or up to six" implies that any combination of these $6$ micro-operations can be executed simultaneously (compatible/horizontal microprogramming). Since they can be active independently, we must allocate $1$ control bit for each of the $6$ micro-operations. $$\text{Bits required} = 6 \text{ bits}$$

Total minimum number of bits required:

Hence, the correct option is C.