Let R1 and R2 be two 4-bit registers that store numbers in 2's complement form. For the operation R1+R2, which one of the following values of R1 and R2 gives an arithmetic overflow?

In 4-bit 2's complement representation, the representable range of signed integers is:

An arithmetic overflow occurs when adding two numbers of the same sign produces a result outside this range (i.e., adding two negative numbers yields a positive result, or adding two positive numbers yields a negative result).

Let us evaluate each option:

• Option (A):

  • $\text{R1} = 1011_2 = -5$

  • $\text{R2} = 1110_2 = -2$

  • $\text{Sum} = -5 + (-2) = -7$

  • Since $-7 \in [-8, +7]$, no overflow occurs.

• Option (B):

  • $\text{R1} = 1100_2 = -4$

  • $\text{R2} = 1010_2 = -6$

  • $\text{Sum} = -4 + (-6) = -10$

  • Since $-10 < -8$, the result is outside the representable range.

  • In 4-bit binary addition: $1100_2 + 1010_2 = 10110_2$. Taking the lower 4 bits gives $0110_2 = +6$, which has the wrong sign. The carry into MSB is $C_{\text{in}} = 0$ and carry out is $C_{\text{out}} = 1$, giving overflow $V = C_{\text{in}} \oplus C_{\text{out}} = 1$. Thus, overflow occurs.

• Option (C):

  • $\text{R1} = 0011_2 = +3$

  • $\text{R2} = 0100_2 = +4$

  • $\text{Sum} = 3 + 4 = +7 \in [-8, +7]$ (No overflow).

• Option (D):

  • $\text{R1} = 1001_2 = -7$

  • $\text{R2} = 1111_2 = -1$

  • $\text{Sum} = -7 + (-1) = -8 \in [-8, +7]$ (No overflow).

Hence, (B) is the correct option.