GATE CS 2004 Q67 — ALU, Data-path and Control Unit

We are given that the total width of each microinstruction is $26$ bits, divided into three fields:

• Micro-operation field: $13$ bits
• Next address field ($X$): $x$ bits
• MUX select field ($Y$): $y$ bits

This gives us the relation:
$$\text{Total Width} = 13 + x + y = 26 \implies x + y = 13$$

The multiplexer (MUX) has $8$ status bits as inputs. To select one of these $8$ inputs, the MUX select field ($Y$) must have:
$$y = \log_2(8) = 3\text{ bits}$$

Substituting $y = 3$ into the relation:
$$x = 13 - 3 = 10\text{ bits}$$

Since the next address field ($X$) is $10$ bits wide, the control memory can address up to $2^{10} = 1024$ words. Thus, the size of the control memory is $1024$ words.

Therefore, the values of $X$, $Y$, and the control memory size are $10, 3, 1024$ respectively.