
Assume that 453 scientists have been invited to attend a conference. If every scientist is to be assigned
a unique bit pattern, what is the minimum number of bits required to do this?
453 < 512 (2^{9}). So 9 bits.

How many more scientists can be invited to the conference, without requiring additional bits for each
person's unique id?
512  453 = 59
Problem 2 (8 points)
Using 7 bits to represent each number, write the representations of 25, 25 and 0 in signed magnitude,
1's complement and 2's complement notations.
Number  SignMagnitude  1's Complement  2's Complement 
25  001 1001  001 1001  001 1001 
25  101 1001  110 0110  110 0111 
0  000 0000 OR 100 0000  000 0000 OR 111 1111  000 0000 
Problem 3 (4 points)
The following binary numbers are 5bit 2's complement binary numbers. Which of the
following operations generate overflow? Justify your answers by translating the operands
and results into decimal.

00111 + 00110
01101 = 13
7 + 6 = 13
No overflow

10111  11110
11001 = 7
(9)  (2) = 7
No overflow

11000  00011
10101 = 11
(8)  3 = 11
No overflow

10110 + 10011
01001 = 9
(10) + (13) = 23
Overflow
Problem 4 (2 points)
Compute the followings:

(10100 OR 01101) AND (NOT(10101))
= 11101 AND 01010 = 01000

(10101 OR 00100) AND (NOT(00101) OR 01010)
= 10101 AND 11010 = 10000
Problem 5 (4 points)
Convert the following decimal numbers into 6bit 2's complement binary numbers. Explain any
problem that you encounter for these conversions.

23
01 0111

33
33 < 32 (largest negative integer that can be represented 2's complement form using 6 bits)
Therefore, cannot be represented.

32
32 > 31 (largest positive integer that can be represented 2's complement form using 6 bits)
Therefore, cannot be represented.

32
10 0000
Problem 6 (2 points)
Perform the specified arithmentic operation for the following 2's complement binary numbers:

11001 + 1011
11001 + 11011 = 10100

00111  010
00111  00010 = 00111 + 11110 = 00101
Note:
 All numbers beginning with 1 are negative.
 Don't forget to signextend when required.
Problem 7 (4 points)
Write the decimal equivalents for the following IEEE floating point numbers.

1 01111110 01000000000000000000000
(1)^{1} x 2^{126127} x (1.01)_{2} = 0.625

0 10000001 10100000000000000000000
(1)^{0} x 2^{129127} x (1.101)_{2} = 6.5
Problem 8 (2 points)
Write the IEEE floating point representation of the decimal number 2.50.
(2.5)_{10} = (10.1)_{2} = (1.01)_{2} x 2^{1} = (1)^{0} x 2^{(128127)} x (1.01)_{2}
0 10000000 01000000000000000000000