Question 17

In an 8-bit ripple-carry adder, each 1-bit full adder $i$ ($0 \le i \le 7$) computes:

The latency is maximal when a carry is generated at the least significant bit ($i = 0$) and ripples through all subsequent bits up to bit 7 ($C_1 \to C_2 \to \dots \to C_8$).

Given $A = 1$, its 8-bit representation is:

So $A_0 = 1$ and $A_i = 0$ for all $1 \le i \le 7$.

To achieve maximum propagation delay:

1. At bit 0: With $C_0 = 0$, a carry must be generated: $C_1 = A_0 B_0 = 1 \cdot B_0 = 1 \implies B_0 = 1$.
2. At each bit $i$ ($1 \le i \le 7$): Since $A_i = 0$, the carry propagation condition is:
   $$C_{i+1} = C_i (A_i \oplus B_i) = C_i (0 \oplus B_i) = C_i B_i$$
   To ensure $C_{i+1} = 1$ whenever $C_i = 1$, we must have $B_i = 1$ for all $i = 1, 2, \dots, 7$.

Thus, the required 8-bit value of $B$ is:

In 8-bit 2's complement representation:

Therefore, the decimal value of $B$ is -1.