GATE CSE 2016 Set 2 | Question: 07

Question

Consider an eight-bit ripple-carry adder for computing the sum of $A$ and $B$, where $A$ and $B$ are integers represented in $2$'s complement form. If the decimal value of $A$ is one, the decimal value of $B$ that leads to the longest latency for the sum to stabilize is ___________

Comments

A=00000001=1(in 2's complement form ),B=-1=11111111(in 2's complement form)

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Very Tricky.

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In case of smallest latency the answer would be 0.

Answer 1

Numbers are being represented  in 2s Complement form 

Now A is given as 1 (0000 0001)

2’s complement of A (+1)= 1111 1111

Now what no will be added to make sum longest to stabilize (Carry at each stage)

is 1111 1111

so Decimal value of B is -1

Answer 2

-1

In case of −1−1  we get bit sequence 1111111111111111 adding this we get a carry upto carry flag, so largest time to ripple!

Answer 3

Longest Latency in Ripple-Carry Adders:

In a ripple-carry adder, each full adder must wait for the carry output from the previous adder, creating a sequential chain of delays.
The longest latency occurs when carries propagate through every stage of the adder, from the least significant bit (LSB) to the most significant bit (MSB).

Maximizing Carry Propagation:

To maximize carry propagation, we need to create a situation where carry-ins are consistently generated at each stage.
This happens when the corresponding bits of A and B differ (one is 0, the other is 1), resulting in a carry-out from each full adder.
 

Given A = 1:

A in binary is 0000 0001.
To maximize carry propagation, we need B to have 1s in all bits except for the LSB, which should be 0.
This ensures carries are generated at each stage, leading to the longest latency.

B = -1:

B in binary is 1111 1111 (two's complement representation of -1).
Adding A and B results in 0000 0000 (with carries throughout), taking the longest possible time in a ripple-carry adder.
Therefore, the decimal value of B that leads to the longest latency is -1.