Post on 23-Jun-2015
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
1.17Truncation, Rounding, Overflow, and
Conversion Error
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Goal
To be able to explain and demonstrate the concepts of truncation, rounding, overflow, and conversion error.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
The computer is imperfect
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
The computer is imperfect
No matter how large a computer is, it still has a limited amount of storage.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
The computer is imperfect
No matter how large a computer is, it still has a limited amount of storage.
Consider the result of dividing 2 by 3.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
The computer is imperfect
No matter how large a computer is, it still has a limited amount of storage.
Consider the result of dividing 2 by 3.
0.666666 is a repeating number.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
The computer is imperfect
No matter how large a computer is, it still has a limited amount of storage.
Consider the result of dividing 2 by 3.
0.666666 is a repeating number.
Regardless of how many bits we use to store this number, it will get “cut off” at some point. No computer
can accurately store this number.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Truncation
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Truncation
To truncate a number means to simply ignore the extra digits that the computer cannot store.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Truncation
To truncate a number means to simply ignore the extra digits that the computer cannot store.
Truncate the following to 3 significant digits
0.2349
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Truncation
To truncate a number means to simply ignore the extra digits that the computer cannot store.
Truncate the following to 3 significant digits
0.234
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Truncation
To truncate a number means to simply ignore the extra digits that the computer cannot store.
Truncate the following to 5 significant digits
0.666666666666
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Truncation
To truncate a number means to simply ignore the extra digits that the computer cannot store.
Truncate the following to 5 significant digits
0.66666
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Truncation
To truncate a number means to simply ignore the extra digits that the computer cannot store.
Truncate the following to 8 significant binary digits
0.1010101111000101
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Truncation
To truncate a number means to simply ignore the extra digits that the computer cannot store.
Truncate the following to 8 significant binary digits
0.10101011
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Real Numbers0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
Represent 0.02510 in IEEE standard1. 0.02510 = 0.00000112
2. normalized as 1.1001 x 2-6
3. set the sign bit4. store -6 in the exponent section as (-6 + 127 = 121) 011110012
5. store the normalized binary form
1.1001 x 2-60.0000011001100111.100110011001
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
Calculate 12610 + 1310
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
Calculate 12610 + 1310
12610 + 1310
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
Calculate 12610 + 1310
011111102
+000011012
12610 + 1310
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
Calculate 12610 + 1310
011111102
+000011012
100010112
12610 + 1310
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
Calculate 12610 + 1310
011111102
+000011012
100010112
12610 + 1310
sign
bit
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
Calculate 12610 + 1310
011111102
+000011012
100010112
12610 + 1310
sign
bit
this is -11710 in 2’s complement, but
the answer should be 13910.
What happened?
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
Calculate 12610 + 1310
011111102
+000011012
100010112
12610 + 1310
sign
bit
7 bits can only store up to 127.
Remember that the first bit is used as a sign
bit.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number).
Calculate 12610 + 1310
011111102
+000011012
100010112
12610 + 1310
sign
bit
7 bits can only store up to 127.
We have an overflow problem.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
2.53
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
2.53
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
2.53
Since 3 is less than 5, this will be like
truncation
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
2.5
Since 3 is less than 5, this will be like
truncation
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
2.5
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 3 significant digits
17.948
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 3 significant digits
17.948
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 3 significant digits
17.948
Since 4 is less than 5, this will be like
truncation
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 3 significant digits
17.9
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
-0.002463
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
-0.002463
Since we are concerned only
about the significant digits, we will only
consider these digits
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
-0.002463
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
-0.002463
Since 6 is greater than or equal to 5, we “round up” the 4 to its
left to 5
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 2 significant digits
-0.0025
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 5 significant digits
0.173
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 5 significant digits
0.173
Note that this is a repeating number.
We should expand to at least 6 significant digits before rounding to 5 significant
digits
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 5 significant digits
0.173737
Note that this is a repeating number.
We should expand to at least 6 significant digits before rounding to 5 significant
digits
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 5 significant digits
0.173737
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 5 significant digits
0.173737
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 5 significant digits
0.173737
Since 7 is greater than or equal to 5, we “round up” the 3 to its
left to 4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
2/3
Rounding
An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation
of the number.
Round the following to 5 significant digits
0.17374
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.1. 0.110 = 0.000112
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.1. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.1. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
1.10011 x 2-4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.1. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
3. set the sign bit
1.10011 x 2-4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.0
1. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
3. set the sign bit
1.10011 x 2-4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.0
1. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
3. set the sign bit4. store -4 in the exponent section
1.10011 x 2-4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.0
1. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
3. set the sign bit4. store -4 in the exponent section
1.10011 x 2-4
-4 + 127 = 123123 = 011110112
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.0 0 1 1 1 1 0 1 1
1. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
3. set the sign bit4. store -4 in the exponent section
1.10011 x 2-4
-4 + 127 = 123123 = 011110112
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.0 0 1 1 1 1 0 1 1
1. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
3. set the sign bit4. store -4 in the exponent section5. store the normalized binary form
1.10011 x 2-4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1.10011 x 2-41. 0.110 = 0.000112
2. normalize 0.000112 = 1.10011 x 2-4
3. set the sign bit4. store -4 in the exponent section5. store the normalized binary form
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1. since the sign bit is 0, we know this is a positive number
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1. since the sign bit is 0, we know this is a positive number2. the exponent section is 01111011 (= 123)
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1. since the sign bit is 0, we know this is a positive number2. the exponent section is 01111011 (= 123)3. the decimal exponent is 123 - 127 = -4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1. since the sign bit is 0, we know this is a positive number2. the exponent section is 01111011 (= 123)3. the decimal exponent is 123 - 127 = -44. from the number section, we have 1.10011001100110011001100
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1. since the sign bit is 0, we know this is a positive number2. the exponent section is 01111011 (= 123)3. the decimal exponent is 123 - 127 = -44. from the number section, we have 1.100110011001100110011005. therefore the number is 1. 10011001100110011001100 x 2-4
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1. since the sign bit is 0, we know this is a positive number2. the exponent section is 01111011 (= 123)3. the decimal exponent is 123 - 127 = -44. from the number section, we have 1.100110011001100110011005. therefore the number is 1. 10011001100110011001100 x 2-4
6. 1. 10011001100110011001100 = 1.59999990463 (approximately)
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1. since the sign bit is 0, we know this is a positive number2. the exponent section is 01111011 (= 123)3. the decimal exponent is 123 - 127 = -44. from the number section, we have 1.100110011001100110011005. therefore the number is 1. 10011001100110011001100 x 2-4
6. 1. 10011001100110011001100 = 1.59999990463 (approximately)7. 1.59999990463 x 2-4 = 0.099999994
MATH1003
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
0.1
Conversion Error
Now, let’s convert this back to decimal0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1. since the sign bit is 0, we know this is a positive number2. the exponent section is 01111011 (= 123)3. the decimal exponent is 123 - 127 = -44. from the number section, we have 1.100110011001100110011005. therefore the number is 1. 10011001100110011001100 x 2-4
6. 1. 10011001100110011001100 = 1.59999990463 (approximately)7. 1.59999990463 x 2-4 = 0.099999994
This is a conversion error:
0.099999994 ≠ 0.1