HOMEWORK # 2 SOLUTIO

HOMEWORK # 2 SOLUTIO

Problem 1 (2 points)

a. There are 313 characters in the Tamil language. If every character is to be encoded into a unique bit pattern, what is the minimum number of bits required to do this?

8 bits can used to encode 28 = 256 characters and 9 bits can be used to encode 29 = 512 characters. So, we would need 9 bits.

b. How many more characters can be accommodated in the language without requiring additional bits for each character?

512 ? 313 = 199

Problem 2 (4 points)

Convert the following 2's complement binary numbers to decimal numbers. a. 1010 First bit is 1. So it is a ?ve number. 2's complement of 1010 = 0101 + 1 = 0110. So the answer is -6.

b. 0010 This is a +ve number since it starts with 0 Answer is 2.

c. 111111 This is a ?ve number since it starts with 1. Its 2's complement is 000000 + 1 = 000001. So the answer is -1

d. 011111 This is a +ve number since it starts with 0. The answer is 31.

Problem 3 (4 points)

a. What is the largest positive number one can represent in a 16-bit 2's complement code? Write your result in binary and decimal. 0111 1111 1111 1111 binary and 215 - 1 = 32767 decimal

b. What is the greatest magnitude negative number one can represent in a 16-bit 2's complement code? Write your result in binary and decimal.

1000 0000 0000 0000 binary and -215 = -32768 decimal

c. What is the largest positive number one can represent in a 16-bit signed magnitude code? Write your result in binary and decimal.

0111 1111 1111 1111 binary and 215 - 1 = 32767 decimal

d. What is the greatest magnitude negative number one can represent in a 16-bit signed magnitude code? Write your result in binary and decimal.

1111 1111 1111 1111 binary and ?(215 ? 1) = -32767 decimal

Problem 4 (2 points)

What are the 8-bit patterns used to represent each of the characters in the string "This Is Easy!"? (Only represent the characters between the quotation marks.)

T h i s Space I s Space E a s y !

Character

Hex (from ASCII table) 54 68 69 73 20 49 73 20 45 61 73 79 21

Binary equivalent 0101 0100 0110 1000 0110 1001 0111 0011 0010 0000 0100 1001 0111 0011 0010 0000 0100 0101 0110 0001 0111 0011 0111 1001 0010 0001

Problem 5 (4 points)

Convert the following decimal numbers to 8-bit 2's complement binary numbers. If there is problem while doing this, describe it.

a. 102

0110 0110

b. 64 0100 0000

c. 128 Does not fit in an 8-bit signed number

d. -128 1000 0000

Problem 6 (4 points)

The following binary numbers are 4-bit 2's complement binary numbers. Which of the following operations generate overflow? Justify your answers by translating the operands and results into decimal.

a. 0111 + 1101 No overflow. 0111 1101 -------10100

Answer is 0100 binary = 4 decimal [7 + (-3)] b. 1001 + 1110

Overflow. 1001 1110 -------10111 Answer is 0111 binary = 7 decimal. But actual answer is -9 [(-7) + (-2)]

c. 1111 + 1001 No overflow. 1111 1001 -------11000

Answer is 1000 binary = -8 decimal [(-1) + (-7)]

d. 0011 + 0101 Overflow. 0011 0101 -------1000

Answer is 1000 binary = -8 decimal. But actual answer is 8 [3 + 5]

Problem 7 (2 points)

A computer programmer wrote a program that adds two numbers. The programmer ran the program and observed that when 5 is added to 8, the result is the character m. Explain why this program is behaving erroneously. The error that is occurring here is that 5 and 8 are being interpreted as characters `5' and `8' respectively. As a result, the addition that is taking place is not 5 + 8; rather, it is `5' + `8'. If we look up values in the ASCII table, 5 is 0x35 and 8 is 0x38. 0x35 + 0x38 = 0x6d, which is the ASCII value for `m'.

Problem 8 (2 points)

Compute the following: a. OT(1011) OR (1011) NOT(1011) = 0100 Answer = (0100) OR (1011) = 1111 b. OT(1001 A D (0100 OR 0110)) 0100 OR 0110 = 0110 1001 AND 0110 = 0000 Answer = NOT(0000) = 1111

Problem 9 (4 points)

Write the decimal equivalents for these IEEE floating point numbers. a. 0 01111111 11000000000000000000000 Sign bit is 0 (+ve). Exponent = 127. Fraction = 1*2-1 + 1*2-2 = 0.75 Answer = (+) 1.fraction * 2exponent ? 127 = 1.75 * 20 = 1.75 b. 1 01111101 10000000000000000000000 Sign bit is 1 (-ve). Exponent = 125. Fraction = 1*2-1 = 0.5 Answer = (-) 1.fraction * 2exponent ? 127 = - 1.5 * 2-2 = - 0.375

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