GO Classes DPP | GATE CS | Digital Logic | Two's Complement Overflow

Answer 1

In $16$-bit $2$’s complement representation, the range is $-32768$ to $32767$.

Both numbers have sign bit $1$, so both numbers are negative.

Overflow in addition of two negative numbers happens only if the sum becomes less than $-32768$.

Now we need to check the most negative possible value of both numbers.

For a negative $2$’s complement number, the value becomes most negative when the remaining bits are as small as possible. So, to get the most negative case, put every $x=0$.

First number becomes:

$1010010110100000$

This is unsigned value $42400$.

So, its $16$-bit $2$’s complement value is $42400-65536=-23136$.

Second number becomes:

$1101101001100000$

This is unsigned value $55904$.

So, its $16$-bit $2$’s complement value is $55904-65536=-9632$.

Now add the most negative possible values:

$-23136+(-9632)=-32768$

The minimum possible sum is exactly $-32768$, which is still inside the valid $16$-bit $2$’s complement range.

If any $x$ becomes $1$, the unsigned value increases, so the negative number becomes less negative. Therefore, the sum cannot become smaller than $-32768$.

So, overflow is not possible.

Final Answer: $0$

Comments

so the main concept that I get from this in 2's complement if we want higher value in -ve then put as much as 0 is possible other then msb bit and for msb bit it should be 1

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Am i wrong?

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@Fardeen Overflow condition:-

If both operands are +ve and the sum is -ve

or both operands are -ve, and the sum is +ve

No Overflow:-

If both operands are +ve and the sum is +ve,

both operands are -ve, and the sum is -ve, or or operands have different signs

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see the rightmost bit are 1 , 1 , it was casue ripple carry till MSB so result is always 11 at last in this case , disccard 1 from there , so we got -ve + -ve = -ve of not possible

Answer 2

The 11th bit in both binary representations is 1, so adding results in a carry of 1, whether the x before it produces carry or not. The given binary representations carry forward until the first bit from the left, so at the first bit, it's always 1+1+1=11, so the sum is never positive, so no overflow. (in 2's and 1's complement when addition of numbers of same msb produces different msb then overflow)