Topic 2 Tutorial Solutions PDF

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ITC161/544 Computer Systems/Computer Organisation and Architecture Topic 2 Tutorial Solutions (Data Representation)

1. Convert the following decimal numbers into binaries: i) 54, ii) 255, iii) 1026 Answer: i) 110110 ii) 11111111 iii) 10000000010 2. Convert the following binary numbers into decimals: i) 11001011, ii) 10001001, iii) 11001100 Answer: i) 203 ii) 137 iii) 204 3. Convert the following decimal fractions to binaries: i) 237.25, ii) 32.32, iii) 100.11 Answer: i) 11101101.01 ii) 100000.010100011... iii) 1100100.000111000010...

4. Convert the following binary fractions into decimals i) 1101.11, ii) 101.001, iii) 0.0110 Answer: i) 13.75 ii) 5.125 iii) 0.375 5. Represent the number -92 and -256 in 8-bit 2's complement Answer: i) 10100100 ii) Not possible 6.1 Represent the following decimal numbers in binary using 8-bit signed magnitude, one's complement and two's complement: a. 77

b. −42

c. 119

d. −107

ITC161/544 Topic 2 Tutorial Solutions

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SCM, CSU201530

Ans. a.

Signed magnitude: 01001101 One's complement: 01001101 Two's complement: 01001101

b.

Signed magnitude: 10101010 One's complement: 11010101 Two's complement: 11010110

c.

Signed magnitude: 01110111 One's complement: 01110111 Two's complement: 01110111

d.

Signed magnitude: 11101011 One's complement: 10010100 Two's complement: 10010101

6.2 Generalize the range of values (in decimal) that can be represented in any given x number of bits using: a. Signed magnitude b. One's complement c. Two's complement Ans. a. −(2x−1−1) to +(2x−1 − 1) b. −(2x−1 − 1) to +(2x−1 − 1) c. −(2x−1) to +(2x−1 − 1)

7. What is the highest and lowest values can be stored using 6 bits 2’s complement method? Answer: Highest= 31 Lowest=-32 8. Convert the following numbers from unsigned binary notation to decimal notation, and from 6-bit 2's complement notation to decimal notation: i) 110011, ii) 001101, iii) 101101 Answer: Unsignedi) 51 ii) 13 iii) 45 2’s complementi) -13 ii) 13 iii) -19 9. Convert the following numbers from hexadecimal to binary and decimal notations: i) 6AB2, ii) FF9A, iii) FACE Answer: i) 110101010110010, 27314 ii) 1111111110011010, 65434 iii) 1111101011001110, 64206

ITC161/544 Topic 2 Tutorial Solutions

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SCM, CSU201530

10. Using a "word" of 4 bits, list all of the possible signed binary numbers and their decimal equivalents that are representable in: a. Signed magnitude b. Two's complement Answer: a. 0111 to 1111, or +7 to −7 b. 0111 to 1000, or +7 to −8

11. Show how each of the following floating point values would be stored using IEEE-754 single precision (be sure to indicate the sign bit, the exponent, and the significand fields): a. 12.5 b. −1.5 c. 0.75 d. 26.625 Answer: a. 12.5 = 1.1001x23 3+127 = 130 = 10000010

0

10000010

1001000...0

b.

−1.5 = −1.1x20 0 + 127 = 127 = 01111111

1

01111111

1000000...0

c.

0.75 = 1.1x2−1 −1 + 127 = 126 = 01111110

0

01111110

1000000...0

d.

26.625 = 1.1010101x24 4 + 127 = 131 = 10000011

0

10000011

1010101...0

12. Given the following two binary numbers: 11111100 and 01110000. a. Which of these two numbers is the larger unsigned binary number? b. Which of these two is the larger when it is being interpreted on a computer using signed-two’s complement representation? c. Which of these two is the smaller when it is being interpreted on a computer using signed-magnitude representation? Answer: a. 11111100 b. 01110000 c. 11111100 13. What is the most common representation used in most computers to store signed integer values and why? Answer: 2’s complement representation. The algorithm for adding and subtracting numbers is easier, it has the best representation for zero, it has a larger range of negative numbers, and it easily extends to larger numbers.

ITC161/544 Topic 2 Tutorial Solutions

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SCM, CSU201530

14. The following is a representation of a decimal floating value using IEEE-754 single precision. Find out the value in decimal. 0 10000011 10101000000...0 Answer: S=0 => this is a positive number 10000011 = 131 e = 131 – 127 = 4 mantissa = .10101 = 0.5+0.125+.03125 = 0.65625 Thus, 1.65625 X 24 = 26.5

ITC161/544 Topic 2 Tutorial Solutions...