GATE CS 1996 Q2.25 — ALU, Data-path and Control Unit

To find the minimum number of bits required in the control word, we can analyze the problem using two different design perspectives: the practical field-based encoding (which matches the given options) and the theoretical minimum encoding.

Method 1: Practical Field-Based Encoding (Standard Interpretation)

Since at most two control signals can be active during any microinstruction, we can divide the control word into two identical control fields. Each field is responsible for activating at most one control signal.

• Each field must be able to select any of the $25$ control signals, or represent an "inactive" (no signal) state.
• Therefore, each field needs to represent $25 + 1 = 26$ distinct states.
• The number of bits required for one field is: $$\lceil \log_2(26) \rceil = 5 \text{ bits}$$
• Since there are two such fields, the total number of bits required is: $$5 + 5 = 10 \text{ bits}$$

This matches Option C.

Method 2: Theoretical Minimum Encoding (Fully Compressed)

If we encode all valid combinations of active signals directly into a single binary value, we calculate the total number of valid states:

• 0 signals active: $\binom{25}{0} = 1$
• 1 signal active: $\binom{25}{1} = 25$
• 2 signals active: $\binom{25}{2} = \frac{25 \times 24}{2} = 300$

Total valid states = $1 + 25 + 300 = 326$.

The minimum number of bits required to represent $326$ states is: $$\lceil \log_2(326) \rceil = 9 \text{ bits}$$

Since $9$ is not present in the options, the question assumes the practical field-based design (Method 1), which requires 10 bits.