Question

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?

• A. R1 = 1011 and R2 = 1110
• B. R1 = 1100 and R2 = 1010
• C. R1 = 0011 and R2 = 0100
• D. R1 = 1001 and R2 = 1111

Hint:

In 2's complement arithmetic, overflow occurs when the result falls outside the representable range. For an n-bit register, the range is from \(-2^{n-1}\) to \(2^{n-1}-1\).

Answer 1

The Correct Option is B

In 4-bit 2’s complement arithmetic, the representable range is from: \[ -8 \text{ to } +7 \] Overflow occurs when the result of adding two numbers falls outside this range. A useful rule is: - Overflow occurs when both operands have the same sign and the result has the opposite sign.

Let us examine each case:

Case 1: \(R_1 = 1011,\ R_2 = 1110\)
- \(1011 = -5\), \(1110 = -2\)
- Sum: \(-5 + (-2) = -7\)
- \(-7\) lies within the range \([-8, +7]\)
✔ No overflow.

Case 2: \(R_1 = 1100,\ R_2 = 1010\)
- \(1100 = -4\), \(1010 = -6\)
- Sum: \(-4 + (-6) = -10\)
- \(-10\) is outside the 4-bit range
✔ Overflow occurs.

Case 3: \(R_1 = 0011,\ R_2 = 0100\)
- \(0011 = +3\), \(0100 = +4\)
- Sum: \(3 + 4 = 7\)
- \(7\) is within range
✔ No overflow.

Case 4: \(R_1 = 1001,\ R_2 = 1111\)
- \(1001 = -7\), \(1111 = -1\)
- Sum: \(-7 + (-1) = -8\)
- \(-8\) is the minimum representable value
✔ No overflow.

Thus, overflow occurs only in the second case. This corresponds to Option (A) as per the given choices.

Final Answer: (A)