KEY TERMS, REVIEW QUESTIONS,AND PROBLEMS

Answer Kuis >>>

9.10
Assume numbers are represented in 8-bit twos complement representation. Show the calculation of the following: 

a. 6+13 = 00000110 + 00001101 = 00011011 

b. -6+13 = 11111010 + 00001101 = (1)00000111 

c. 6-13 = 00000110 + 11110011 = 11111001 

d. -6-13 = 11111010 + 11110011 = (1)11101101 

9.11 
Find the following differences using twos complement arithmetic 

a. 111000 

  -110011 

     111000 

    +001101 

 =(1)000101 

b. 11001100 

   - 101110 

101110 = 010010 

 11001100 

+  010010 

=11011110 

c. 111100001111 

  -110011110011 

110011110011 = 001100001101 

    111100001111 

 +  001100001101 

 =(1)001000011100 

d. 11000011 

  -11101000 

11101000 = 00011000 

  11000011 

+ 00011000 

= 11011011 

9.23 
Express the following numbers in IEEE 32-bit floating point format: 

a. -5 

5 = 101 = 1.01 x 22 

exponent = 127 + 2 = 129 = 10000001 

1 10000001 01000000000000000000000 

b. -6 

6 = 110 = 1.10 x 22 

exponent = 127 + 2 = 129 = 10000001 

1 10000001 10000000000000000000000 

c. -1.5 

1.5 = 1.1 = 1.1 x 20 

exponent = 127 = 01111111 

1 01111111 10000000000000000000000 

d. 384 

384 = 384-256 = 128 -128 = 0; = 110000000 = 1.1 x 28 

exponent = 8 + 127 = 135 = 10000111 

0 10000111 10000000000000000000000 

e. 1/16 

1/16 = .0001 = 1.0 x 2-4 

exponent = 127 – 4 = 123 = 01111011 

0 01111011 00000000000000000000000 

f. -1/32 

1/32 = .00001 = 1.0 x 2-5 

exponent = 127 – 5 = 122 = 01111010 

1 01111010 00000000000000000000000 

9.24
The following numbers use the IEEE 32-bit floating-point format. What is the equivalent decimal value? 

a. 1 10000011 11000000000000000000000 

sign = - 

exponent = 131 – 127 = 4 

1.11 x 24 = 11100 = 28 

-28 

b. 0 01111110 10100000000000000000000 

sign = + 

exponent = 126 – 127 = -1 

1.101 x 2-1 = .1101 = ½ + ¼ + 1/16 = .8125 

.8125 

c. 0 10000000 00000000000000000000000 

sign = + 

exponent = 128 – 127 = 1 

1.0 x 21 = 10 = 2 

2

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