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Information Technology

ITEC 1000

Peter Khaiter

Winter

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AP/ITEC1000 Sections A and B
"Introduction to Information Technologies"
Fall 2011
Assignment # 2
Answers and Solutions
1)
a) Convert 121 to its binary represen10tion: 1212= 01111001 . Positive numbers are always
represented by themselves.
The answer: 01111001
b) Convert -121 to its binary represen10tion: -1212= -01111001 . Using the second method
for negative numbers, invert and then add 1.
The answer: 10000111
c) Convert 54 to its binary repres10tation: 542= 00110110 . Positive numbers are always
represented by themselves.
The answer: 00110110
2) Yes. The four least significant bits of the EBCDIC code match the four least significant digits
of ASCII.
3) Between -79999 and +79999. This is because 20 bits provide five four-bit locations. The
highest (left most) 4 bits, one bit is used for the sign, leaving 3 bits for digits. These 3 bits can
accommodate numbers between 0 and 7.
4)
a) 15765 is a positive number and is represented by itself
99999 - 8773 = 91226 b) 15765
+91226
106991 add the carry into the result
+ 1
06992
The leading digit in the complement is 0. Therefore, it represents a positive decimal
number: 6992.
c) There was no overflow condition because the inputs in the sum have opposite signs and
overflow cannot occur.
There was a carry condition because in the result, the number of digit exceeded the
specified number (i.e., five) and the carry was added to the result.
d) 15765 is a positive number and is represented by itself
100000 - 8773 = 91227
e) 15765
+91227
106992 Ignore the carry and the result is: 6992
f) Diminished radix complement arithmetic requires an additional step of adding the carry
bit to the result.
Radix complement arithmetic does not require this step, because the carry is simply
ignored. The latter is easier.
5) Replace each decimal digit by its 4-bit BCD representation and add the corresponding sign
pattern in the low-order bits.
a) -924 = 1001001001001101 b) 80356 = 100000000011010101101111 c)
+981 = 1001100000011100
6)
a) The leading digit is between 0 and 4 and, therefore, represents a positive number. Positive
numbers are represented by themselves.
The answer: 3589
b) The leading digit is between 5 and 9 and, therefore, represents a negative number. To get
its magnitude, subtract the complement from the base: 999
-541
458 The answer: -458
7)
First, convert each item in the sum to its floating point sign-and-magnitude and excess-50
representation:
96.775 = 0.96775 x 10 = 05296775
0
0.96775 = 0.96775 x 10 = 05096775
Align the exponents by shifting the lower mantissa in the second item two places right and
add the two mantissas:
05296775
0520096775
05297742(75)
Cut the mantissa to five digits and present the result in floating point format:
0.97742 * 10 = 97.742 (can be 97.743 with rounding)
The absolute precise result would be 97.74275. This difference is caused by the rounding due to
the limited space for mantissa.
8) a)
First, convert to the exponential notation: 111011.0101 = 111011.0101 x 2 . 0
Adjust exponent to get the n0mber in the normali5ed form by shifting the radix point five
bits to the left: 111011.0101 x 2 = 1.110110101 x 2 .
Represent exponent in the excess-127 notation: 127 + 5 = 132 = 10000100 2
Drop initial "1." in the mantissa, add sign bit as the left most bit and pad with zeros to get
single precision IEEE 754 format:
01000010011011010100000000000000
Convert the answer to hexadecimal format: 426D4000
b) 0
First, convert to the exponential notation: -1001.101101 = -1001.101101 x 2 .
Adjust exponent to get the number in the normalized form by shifting the radix point three
0 3
bits to the left: -1001.101101 x 2 = -1.001101101 x 2
Represent exponent in the excess-127 notation: 127 + 3 = 130 = 10000010 2
Drop initial "-1." in the mantissa, add sign bit as the left most bit and pad with zeros to get
single precision IEEE 754 format:
11000001000110110100000000000000
Convert the answer to hexadecimal format: C11B4000
c)
0
First, convert to the exponential notation: -0.00101011 = -0.00101011 x 2 .
Adjust exponent to get the number in the normalized form by shifting the radix point
three bits to the right: -0.00101011 x 2 = -1.01011 x 2 -3
Represent exponent in the excess-127 notation: 127 + (-3) = 124 = 01111100 2
Drop initial "-1." in the mantissa, add sign bit as the left most bit and pad with zeros to
get single precision IEEE 754 format:
10111110001011000000000000000000
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