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CSc 252 Homework 1 - Binary Arithmetic and Conversions - Prof. Homer, Study notes of Computer Architecture and Organization

University of Arizona (UA)Computer Architecture and Organization
Prof. Homer

Solutions to problems related to binary arithmetic and conversions from decimal to hexadecimal, octal, and two's complement for a given set of binary numbers. It covers the computation of hexadecimal and octal equivalents, decimal equivalents for signed numbers, binary addition and subtraction, and turning a given bit pattern into its decimal equivalent.

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2013/2014

Uploaded on 05/07/2014

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CSc 252 — Homework 1!Version 1.0!January 19, 2014
Solutions to Problems Turned In!
1. — Assume that 16-bit, two’s complement, binary numbers are being used in this problem.
Some binary arithmetic and conversions. For each of the pairs of numbers below, compute:!
3 points for each part. 15 points for iv, 15 points for v. 30 points for question 1.!
a.) hexadecimal (base 16) equivalents for both a and b, assume unsigned numbers!
b.) octal (base 8) equivalents for both a and b, assume unsigned numbers!
c.) decimal (base 10) equivalents for both a and b, assume signed numbers!
d.) a + b and indicate if overflow occurs, assume signed numbers!
e.) a - b by negating b and adding, indicate if overflow occurs, assume signed numbers!
iv.) turnin this one !
!a = 0110 1001 0001 1101!
!b = 0011 1010 1100 1000! !
a.) hexadecimal:!
!a = 0110 1001 0001 1101 = 69 1Dhex !
!b = 0011 1010 1100 1000 = 3A C8hex!
b.) octal!
!a = 0 110 100 100 011 101 = 64,435octal !
!b = 0 011 101 011 001 000 = 35,310octal!
c.) decimal:!
a is a positive number!
0110 1001 0001 1101 = !
214 + 213 + 211 + 28 + 24 + 23 + 22 + 20 = !
16,384 + 8,192 + 2,048 + 256 + 16 + 8 + 4 + 1 =!
1 33!
16,384!
8,192!
2,048!
256!
16!
8!
4!
1!
26,909!
26,909ten !
b is a positive number:!
0011 1010 1100 1000 =!
213 + 212 + 211 + 29 + 27 + 26 + 23 = !
8,192 + 4,096 + 2,048 + 512 + 128 + 64 + 8 = !
!
11 33!
8,192!
4,096!
2,048!
! " of " 1 8
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Download CSc 252 Homework 1 - Binary Arithmetic and Conversions - Prof. Homer and more Study notes Computer Architecture and Organization in PDF only on Docsity!

Solutions to Problems Turned In

1. — Assume that 16-bit, two’s complement, binary numbers are being used in this problem.

Some binary arithmetic and conversions. For each of the pairs of numbers below, compute:

3 points for each part. 15 points for iv, 15 points for v. 30 points for question 1.

a.) hexadecimal (base 16) equivalents for both a and b, assume unsigned numbers

b.) octal (base 8) equivalents for both a and b, assume unsigned numbers

c.) decimal (base 10) equivalents for both a and b, assume signed numbers

d.) a + b and indicate if overflow occurs, assume signed numbers

e.) a - b by negating b and adding, indicate if overflow occurs, assume signed numbers

iv.) turnin this one

a = 0110 1001 0001 1101 b = 0011 1010 1100 1000!

a.) hexadecimal:

! a = 0110 1001 0001 1101 = 69 1Dhex! ! b = 0011 1010 1100 1000 = 3A C8hex!

b.) octal

! a = 0 110 100 100 011 101 = 64,435octal! ! b = 0 011 101 011 001 000 = 35,310octal!

c.) decimal:

a is a positive number

26,909ten!

b is a positive number:

15,048ten!

d.) a + b

1010 0011 1110 0101 overflow!

When adding a positive to a positive, the answer should be positive. Since this answer

came out negative, there was overflow.

e.) a - b

! a = 0110 1001 0001 1101! ! b = 0011 1010 1100 1000!

Find the two's complement of b, then add.

Two’s complement of b:

1100 0101 0011 0111 Invert step!

  • 1 add 1! 1100 0101 0011 1000

Now, add a and the two’s complement of b:

0010 1110 0101 0101 no overflow

Positive - Positive can never produce overflow.

v.) turnin this one

! a = 1011 0011 1110 0110! b = 1100 1010 0101 1000!

! a.) hexadecimal

a = 1011 0011 1110 0110 = B3 E6hex! b = 1100 1010 0101 1000 = CA (^58) hex

b.) octal

a = 1 011 001 111 100 110 = 131,746octal b = 1 100 101 001 011 000 = 145,130octal

c.) decimal

a is a negative number; find two’s complement first:

0100 1100 0001 1001 Invert step!

  • 1 add 1!

Now, add a and the two’s complement of b:!

1110 1001 1000 1110 no overflow.

When subtracting negative - negative, overflow is not possible.

5. turnin this one (15 points, 5 points each part) — This exercise is similar to the

previous one, but this time, use the bit pattern:

What does it represent, assuming that it is

a.) a two’s complement integer?

b.) an unsigned integer?

c.) a single precision floating-point number?

d.) four ascii characters?

a.) The sign bit is 1 , so this is a negative number. Start by finding the two’s complement:

0011 0110 1000 1100 1001 0100 1001 1110 invert step!

  • 1! 0011 0110 1000 1100 1001 0100 1001 1111!

Answer:! -915,182,751 (^) ten!

b.) Assume an unsigned number:

Answer:! 3,379,784,545 (^) ten!

c.)! S Expo nent Sig nifi cand! 1 1001 0001 111 0011 1110 1011 0110 0001! 1 1001 0010 111 0011 0110 1011 0110 0001 new!

S = 1, a negative number!

Exponent = 1001 0010 = 27 + 24 + 20 = 128 + 16 + 2 = 146!

11,342ten = 10 1100 0100 1110two

Convert the fractional part to a binary number. Do this by subtracting successive negative

powers of two. When the number cannot be subtracted, put a 0 in. When the number can

be subtracted, put in a 1.

- 0.5 2 -1^!

2 -2^ is 0.25 and is too large. Put in 0

2 -3^ is 0.125 and is too large. Put in 0.

Subtract 2-4:

- 0.0625 2 -4^!

Subtract 2-5:

- 0.03125 2 -5^!

Subtract 2-6:

- 0.015625 2 -6^!

0.0 done!!

The fraction part is 0.

The binary equivalent is 10 1100 0100 1110.100111 =

sign is 1, since the number is negative! exponent = 13 + 127 = 140 = 1000 1100! significand = 011 0001 0011 1010 0111! ! S Expo nent Sig nifi cand! 1 1000 1100 011 0001 0011 1010 0111 0000!

b.)

There are two differences between single- and double-precision:

The exponent uses 11 digits and a bias of 1,023.

There are more digits in the significand.

To convert the exponent:

subtract the bias: 140 - 127 = 13

Now, use a bias of 1,023:

The result is:

S Exp onen t Significand! 1 100 0000 1100 0110 0010 0111 0100 1110 0000 0000! 0000 0000 0000 0000 0000 0000!