CS/ECE 252 Introduction to Computer Engineering

1. 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 (2⁹). So 9 bits.
2. 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	Sign-Magnitude	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 5-bit 2's complement binary numbers. Which of the following operations generate overflow? Justify your answers by translating the operands and results into decimal.

1. 00111 + 00110
   01101 = 13
   7 + 6 = 13
   No overflow
2. 10111 - 11110
   11001 = -7
   (-9) - (-2) = -7
   No overflow
3. 11000 - 00011
   10101 = -11
   (-8) - 3 = -11
   No overflow
4. 10110 + 10011
   01001 = 9
   (-10) + (-13) = -23
   Overflow

Problem 4 (2 points)

Compute the followings:

1. (10100 OR 01101) AND (NOT(10101))
   = 11101 AND 01010 = 01000
2. (10101 OR 00100) AND (NOT(00101) OR 01010)
   = 10101 AND 11010 = 10000

Problem 5 (4 points)

Convert the following decimal numbers into 6-bit 2's complement binary numbers. Explain any problem that you encounter for these conversions.

1. 23
   01 0111
2. -33
   -33 < -32 (largest negative integer that can be represented 2's complement form using 6 bits)
   Therefore, cannot be represented.
3. 32
   32 > 31 (largest positive integer that can be represented 2's complement form using 6 bits)
   Therefore, cannot be represented.
4. -32
   10 0000

Problem 6 (2 points)

Perform the specified arithmentic operation for the following 2's complement binary numbers:

1. 11001 + 1011
   11001 + 11011 = 10100
2. 00111 - 010
   00111 - 00010 = 00111 + 11110 = 00101

1. All numbers beginning with 1 are negative.
2. Don't forget to sign-extend when required.

Problem 7 (4 points)

Write the decimal equivalents for the following IEEE floating point numbers.

1. 1 01111110 01000000000000000000000
   (-1)¹ x 2^126-127 x (1.01)₂ = -0.625
2. 0 10000001 10100000000000000000000
   (-1)⁰ x 2^129-127 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 2¹ = (-1)⁰ x 2^(128-127) x (1.01)₂
0 10000000 01000000000000000000000