University of Illinois at Chicago
CS 266: Computer Architecture I Fall 2012
Homework 5 Solutions
Question 1
The following binary numbers have a sign in the leftmost position and, if negative, are in 2s complement form. Perform
the indicated arithmetic operations and verify answers. Indicate whether overflow occurs for each computation.
a. [10] 100010+111110
Carries:
1
1
1
1
1
1
0
0
0
1
0
+
1
1
1
1
1
0
1
0
0
0
0
0
The carry out into sign bit position is the same as carry out of the sign bit position. Therefore there is no
overflow. Carry out of the sign bit position is discarded.
Verifying the answer:
The 2s complement of 100010 is 011110, which in decimal is 2
+2
+2
+2
=16+8+4+2 =30.
Therefore the decimal equivalent of 100010 is −30.
The 2s complement of 111110 is 000010, which in decimal is 2. Therefore the decimal equivalent of 111110
is −2.
The 2s complement of 100000 is 0100000, which in decimal is 2
=32. Therefore the decimal equivalent of
100000 is −32.
Therefore we have −30+−2=−32.
b. 10001111+100010
Carries:
1
1
1
0
0
1
1
1
1
+
1
0
0
0
1
0
1
1
0
0
0
1
No overflow occurred.
Verifying the answer:
001111
=2
+2
+2
+2
=8+4+2+1 =15
The binary number 100010 is negative, since the leftmost bit is 1. Its 2s complement is 011110.
011110
=2
+2
+2
+2
=16+8+4+2 =30
The binary number 110001 is negative. Its 2s complement is 001111.
001111
=2
+2
+2
+2
=8+4+2+1 =15
Therefore we have 15+−30= −15
c. 10101101–111011
The 2s complement of 111011 is 000101. Therefore we may do the addition 101101+000101, which is
equivalent to the original subtraction.
Carries:
1
1
1
1
0
1
1
0
1
+
0
0
0
1
0
1
1
1
0
0
1
0
There is no overflow since carry into the sign bit position and carry out of the sign bit position are equal.
