Question

Note: Show All the steps of your work.

1. Convert the following decimal real number to a binary number with six places to the right of the binary point.

57.553

2. Represent the following decimal integer numbers in binary using 8-bit signed magnitude, one’s complement, and two’s complement representations:

(a) 65 (b) -24

3. What is the value of the 8-bit binary number 10011110 in decimal assuming the following representation: (a) unsigned, (b) sign-magnitude, (c) one’s complement, (d) two’s complement.

4. Using arithmetic shifting, perform the following operations assuming the numbers are expressed using 8-bit signed two’s complement representation:

(a) Multiply the value 00111101 by 8

(b) Divide the value 11001010 by 4

5. Assume we are using a simple model for floating-point representation similar to the one used in Chapter 2 of the textbook which has a 14-bit format, 5 bits for the exponent with a bias of 15 (not 16 as in the textbook), a normalized mantissa of 8 bits, and a single sign bit.

(a) Show how the computer would represent the numbers 100.0 and 0.25 using this representation.

(b) Show how the computer would add the two floating-point numbers in part (a) by changing one of the numbers so they are both expressed using the same power of 2.

(c) Show how the computer would represent the sum in part (b) using the given floating point

representation. What decimal value for the sum is the computer actually storing? Explain.

Answer 1

Answer 1

57.553

=> 57 (integer part)

• 57/2 = 28   57%2 = 1 (LSB)
• 28/2 = 14 28%2 = 0
• 14/2 = 7 14%2 = 0
• 7/2 = 3 7%2 = 1
• 3/2 = 1 3%2 = 1
• 1/2 = 0 1%2 = 1 (MSB)

=> .553 (fraction part)

• 0.553 x 2 = 1.106 => 1 (MSB)
• 0.106 x 2 = 0.212 => 0
• 0.212 x 2 = 0.424 => 0
• 0.424 x 2 = 0.848 => 0
• 0.848 x 2 = 1.696 => 1
• 0.696 x 2 = 1.392 => 1 (LSB)

111001.100011

Answer 2

1's complement and 2's complement is calculated for negative numbers. for positive numbers 1's and 2's complement is same.

(a) 65

• 65/2 = 32   65%2 = 1 (LSB)
• 32/2 = 16   32%2 = 0
• 16/2 = 8   16%2 = 0
• 8/2 = 4 8%2 = 0
• 4/2 = 2 4%2 = 0
• 2/2 = 1 2%2 = 0
• 1/2 = 0 1%2 = 1 (MSB)

65₁₀ = 1000001

1. 8-bit signed magnitude = 01000001  
2. one’s complement = 01000001  
3. two’s complement representations = 01000001  

(b) -24

• 24/2 = 12 24%2 = 0 (LSB)
• 12/2 = 6 12%2 = 0
• 6/2 = 3    6%2 = 0
• 3/2 = 1 3%2 = 1
• 1/2 = 0 1%2 = 1 (MSB)

24₁₀ = 11000

1. 8-bit signed magnitude = sign-bit and number = 10011000 (-24)
2. one’s complement = 1's of 00011000= 11100111   (flipping all the bits)
3. two’s complement representations = 1's complement + 1 = 11100111 + 1 = 11101000

Answer 3

10011110

• Unsigned => 2⁷ + 2⁴+ 2³+ 2²+ 2¹ = 128 + 16 + 8 + 4 + 2 = 158₁₀
• Sign-magnitude =>
  • Sign bit = 1 (negative)
  • Decimal part = 0011110 = 2⁴+ 2³+ 2²+ 2¹ = 30₁₀
  • - 30
• One’s complement (negative number)
  • decimal = 1's of 10011110 = 01100001 = 2⁶+ 2⁵+ 2⁰ = 97
  • - 97
• Two’s complement (negative number)
  • decimal = 2's of 10011110 = 01100010 = 2⁶+ 2⁵+ 2¹ = 98
  • - 98

Answer 4

4. Using arithmetic shifting, perform the following operations assuming the numbers are expressed using 8-bit signed two’s complement representation:

(a) Multiply the value 00111101 by 8

• 8 = 2³
• 00111101 << 3 = 11101000

(b) Divide the value 11001010 by 4

• 4 = 2²
• 11001010 >> 2 = 11110010