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?

The correct answer is option 2.

Concept:

Stored numbers in registers R1 and R2 are in 2's complement form. Register size is 4 bits. The range of numbers in 2's complement form is -8 to +7. If R1 + R2, the result is out of the above range, then it is overflow.

The given data,

Given two four-bit registers R1 and R2.

Option 1: R1 = 1011 and R2 = 1110

False,  

R1      =  1 0 1 1 = -(0101)= -5

+ R2   =  1 1 1 0 = -(0010)= -2

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               1 0 0 1  =           = -7        

Here No overflow occurred, because sign bit is same for (R1 + R2 ).

Option 2: R1 = 1100 and R2 = 1010

True,

R1      =  1 1 0 0 = -(0100)= -4

+ R2   =  1 0 1 0 = -(0110)= -6

  --------------------------------------------

              0 1 1 0 =            = -10       

Here Overflow occurred because the sign bit is different for (R1 + R2 ).

Option 3: R1 = 0011 and R2 = 0100

False,

R1      =   0 0 1 1 = +(0011)= +3

+ R2   =   0 1 0 0 = +(0100)= +4

  --------------------------------------------

                0 1 1 1                 =   +7       

Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).

Option 4: R1 = 1001 and R2 = 1111

False, 

R1      =   1 0 0 1 = -(0111)  = -7

+ R2   =   1 1 1 1 = -(0001) = -1

  --------------------------------------------

                1 0 0 0 =               = -8

Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).

Hence the correct answer is R1 = 1100 and R2 = 1010.