Overflow condition can be written in any of the three ways –
- $C_{n} \oplus C_{n-1}=1$ which says that either 1st MSB or 2nd MSB should give carry but not both.
- $A_{n}B_{n}\bar{C_{n-1}} + \bar{A_{n}}\bar{B_{n}}C_{n-1}=1$ where $A_n$ and $B_n$ are the MSB (which are sign bits) and $C_{n-1}$ is the carry from the 2nd MSB.
- $A_{n}B_{n}\bar{R_{n}} + \bar{A_{n}}\bar{B_{n}}R_{n}=1$ where $R_{n}$ is the MSB (sign bit) of result of the addition.
From the third condition above, we can see that overflow will happen when:
1. $A_n,B_n$ is 0 (which means positive numbers) and $R_n$ is 1 (which means a negative number).
2. $A_n,B_n$ is 1 (which means both are negative numbers) and $R_n$ 0 (which means a positive number).
Conclusion: For overflow to happen, two positive numbers are added and the result is negative OR two negative numbers are added and the result is positive.
In the above question, one number is positive and the other is negative so there will be no overflow in the resulting number and the sum will be simply $0001$, Overflow = $0$, Carry = $1$.
PS: 2nd result can be derived from 1st result by replacing the value of $C_n$ with $A_nB_n + C_{n-1}(A_n \oplus B_n)$ and 3rd result can be derived from the 2nd result by thinking a bit. Let me know if you find it difficult to derive it.