ECE-C490 Security in Computing
Winter Term 2005
Quiz #1 Solutions
Consider floating-point addition as a computer would perform it. Add the following two
numbers by following the steps below.
A: 0.112 x 103
B: - 0.840 x 102
1) Convert these numbers to the IEEE standard for floating-point numbers, with 1
sign bit, an 8-bit signed exponent in excess-127 representation, and a 23-bit
mantissa fraction. (Show all of your work!)
We begin by determining the largest power of 2 that is less than or equal to each
number. A is equal to 112, so the largest power of 2 less than or equal to 112 is
64, or 26. Thus the exponent will be 6 (but remember, we are working in excess127 format, so this will be represented as 10000101). To determine the mantissa,
we divide 112 by 64 and obtain 1.75 – we are only concerned about the fractional
portion of the result (the 1 is assumed present in IEEE format) so we convert 0.75
to binary and obtain 0.1100…, which becomes our mantissa.
We obtain the representation of 84 similarly. Again, the largest power of 2 less
than or equal to 84 is 64, so the exponent for B will be equal to the exponent of A.
84/64 gives us 1.3125, and converting 0.3125 to binary gives us 0.010100… This
will be our mantissa.
A is positive, so its sign bit will be 0. B’s will be 1.
A:
0
10000101
1100…0
B:
1
10000101
010100…0
2) Perform the addition using the rules for floating point arithmetic. Be sure to
normalize the result if necessary, and write your final answer in the box below:
Following the steps for floating point addition:



The exponents of the two numbers are equal, so we do not need to shift
either mantissa.
We set the exponent of the result equal to the larger exponent (again, they
are equal, so the resulting exponent will be set to 10000101).
We perform addition on the mantissas. In this case, we are in fact
subtracting .0101 from .1100. This leaves us with .0111. But remember –
in IEEE format, we always assume that the mantissa is of the format
1.xxxx, where xxxx is the fractional form displayed. In other words, the
leading ‘1’ is assumed. Therefore, we must normalize the result to ensure
that there is in fact a leading 1. In order to do so, we take the result:
0.0111
and shift it to the left two bits to obtain a result with a leading 1:
1.1100…
But each time we shift, we must reduce the exponent of the result by 1 to
compensate. Thus we reduce the value of the exponent twice and obtain a
final result of:
0 10000011
1100…0
You can check that this result is correct by doing the following:
The value of the exponent as-is = 131. Remember, however, that it is in
excess-127 format, so we subtract 131-127 to obtain 4. Therefore, we
multiply the mantissa by 24.
The mantissa has an assumed leading 1, and a fractional value equal to .11
– in decimal, .75. Therefore we multiply 1.75 by 24 and obtain 28, which
is the correct result (remember, we were adding 112 and –84).