Download - MELJUN CORTES ERRORS Rm104tr-12
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7/29/2019 MELJUN CORTES ERRORS Rm104tr-12
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Lesson 12 - 1
Year 1
CS113/0401/v1
LESSON 12
Errors
Common Mistakes
Lose track of the source
documents Enter data wrongly
Invalid datatype
( e.g. enter alphabets instead of
numerical data ) Out of the valid data range
( e.g. enter 23322 while the
actual data is 23223 )
Transportation of characters( e.g. 000100101 is transmitted
along a cable and on the way
due to voltage surge, the data
becomes 101001101
Mistakes are due to humancarelessness
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Lesson 12 - 2
Year 1
CS113/0401/v1
Example :
If B is represented as 100 0010,
we can use even parity
conversion to produce a parity bit
IF odd number of 1s THEN
parity bit 1
ELSE
parity bit 0
ENDIF
PARITY BIT CHECK
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Lesson 12 - 3
Year 1
CS113/0401/v1
CHECK DIGIT PRINCIPLES
Select a weight for each digit inthe number
Multiply each digit by its weight
Sum the results
Select and divide by a modulus
The modulus less the remainder
is the check digit
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Lesson 12 - 4
Year 1
CS113/0401/v1
Example for 831 405 -8 x 7 = 56
3 x 6 = 18
1 x 5 = 5
4 x 4 = 16
0 x 3 = 0
5 x 2 = 10
Total = 105
Divide by modulus (11)= 9 remainder 6
Check digit is 11-6=5 831 405 - 5
CHECK DIGIT
CALCULATION
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Lesson 12 - 5
Year 1
CS113/0401/v1
NCC 1/94
Question 8 c)
i) Using mod 11 and weightings of 2,
3 and 4 for the units, tens and
hundreds columns, append acheck digit to 974.
(2m)
ii) Show how a transposition erroroccurring in the number 974 wold
be detected.
(2m)
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Lesson 12 - 6
Year 1
CS113/0401/v1
I) 9 x 4 = 36 65 11=5 r 10 [1]
7 x 3 = 21 11 - 10 = 1
4 x 2 = 8 9741 required answer
65 [ 1 ]
ii) Let error give to 9471. Follow through on
candidates valid transportation error
e.g. 4971 etc. [1] for recognizing what atransportation error is.
9 x 4=36
4 x 3=12 [1] for appropriate method.
7 x 2=14
1 x 1= 1
63 Not exactly divisible by 11
Error in data
Answer
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Lesson 12 - 7
Year 1
CS113/0401/v1
ERROR
Inherent Error
It is error that already exist by
itself in the measurement scale.
E.g. Ruler measurement
If the length fall between 5.1cm
and 5.2cm, the smallest divisionis 0.1cm, then the reported
reading may be 5.15cm and the
inherent error is 0.05cm.
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Lesson 12 - 8
Year 1
CS113/0401/v1
Induced Error It is error which is brought in from
outside due to external factors
Example: In a fixed point 8-bit
register in computer, the implied
point to the right of 5 bits
If a data 1010.01110 is to be
stored, rounding or truncation is
necessary.
Rounding : The above data isstored as
Thus the error is brought in dueto the limitation of computer
0 1 0 1 0 1 0 0
ERROR
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Lesson 12 - 9
Year 1
CS113/0401/v1
Absolute Error is: Value used - True value
The difference between the
number represented and its true
value (value as stored less true
value)
Relative Error is:
The proportion of the absoluteerror to the true value I.e.
absolute error
true value
ABSOLUTE AND RELATIVE
ERROR (1)
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Lesson 12 - 10
Year 1
CS113/0401/v1
ABSOLUTE AND RELATIVE
ERROR (2)
From a previous example: Absolute error is
= Value used - True value
0.296875 - 0.3
= -0.003125
Relative error is
= Absolute error / True value
-0.003125 / 0.3
= -0.0104166 ( or -1.04166% )
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Lesson 12 - 11
Year 1
CS113/0401/v1
NCC 2/94
Question 1(g)
Two values are recorded as 8.7 and -4.3,
both correct to 1D. What is the MAXIMUMABSOLUTE ERROR when they are added
together?
(2m)
Answer
8.65 x 8.75
-4.35 y - 4.25
4.3 x 4.5
Recorded value is 4.4
Hence maximum absolute error is 0.1.
One mark for maximum individual error
0.05.
One mark for correct answer.
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Lesson 12 - 12
Year 1
CS113/0401/v1
ERROR PROPAGATION
Inherent errors in data values,may produce further errors as
they are operated on
arithmetically
These errors may be termed as
Induced errors
This spread of errors is called theError Propagation
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Lesson 12 - 13
Year 1
CS113/0401/v1
When data is first recorded in a
User Department When this data is input to the
computer or transcribed into
machine readable form
SINGLE TRANSPOSITION
ERRORS
237426 1546
234726 1645
WHEN DO TRANSACTION
ERRORS OCCUR?
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Lesson 12 - 14
Year 1
CS113/0401/v1
TRUNCATION ERRORS
Calculate: (a + b) / (c - d)
Where:a = 3.62841 b = 5.38634
c = 8.32174 d = 8.31079
If actual values used:
9.01475 / 0.01095 = 823.26484
If values corrected to 3 decimals places:
9.015 / 0.011 = 819.54545
Notice that the resulting error after the
division is much larger than the original
rounding off
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Lesson 12 - 15
Year 1
CS113/0401/v1
CONVERSION ERRORS
Decimal input 0.3
Binary equivalent 0.010011001
Equivalent in 6 bits 0.010011
Which is equivalent 0.296875
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Lesson 12 - 16
Year 1
CS113/0401/v1
DEALING WITH ROUNDING
ERRORS
Never truncate or round up all thetime (round off instead)
Always used the maximum space
available for sorting intermediateresults
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Lesson 12 - 17
Year 1
CS113/0401/v1
ADDING IN DESCENDING
ORDER OF MAGNITUDE
0.154308 + 0.019276 = 0.173584= 1.174
0.174000 + 0.003574 = 0.177574
= 0.178
0.178000 + 0.002807 = 0.180807
= 0.181
Exact answer is 0.179965
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Lesson 12 - 18
Year 1
CS113/0401/v1
ADDING IN ASCENDING
ORDER OF MAGNITUDE
0.002807 + 0.003574 = 0.006381= 0.00638
0.006380 + 0.019276 = 0.025656
= 0.0257
0.025700 + 0.154308 = 0.180008
= 0.180
Exact answer is 0.179965
Previous answer was 0.181