Answer
Each 1 -CE circuit contains two NOT gates, two AND gates, and one OR
gate. This will require $(2 \times 1)+(2 \times 3)+(1 \times 3)=2+6+3=11$
transistors. The 32 -bit compare-for-equality circuit of Figure 4.28 has
the following components:
32 1-CE circuits $=32 \times 11=352$ transistors
31 AND gates $=31 \times 3=93$ transistors
Total $=352+93=445$ transistors
Work Step by Step
Each 1 -CE circuit contains two NOT gates, two AND gates, and one OR
gate. This will require $(2 \times 1)+(2 \times 3)+(1 \times 3)=2+6+3=11$
transistors. The 32 -bit compare-for-equality circuit of Figure 4.28 has
the following components:
32 1-CE circuits $=32 \times 11=352$ transistors
31 AND gates $=31 \times 3=93$ transistors
Total $=352+93=445$ transistors
