Difference between Equalance class partion and boundary
value analysis?
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EQUIVALENCE CLASS PARTITION : THIS IS A TECHNIQUE USED IN
THE BLACK BOX TESTING METHOD, USED TO REFINE THE DATA INTO
VALID AND INVALID CLASSES
FOR EXAMPLE : A TEXT BOX ACCEPTS ONLY ALPHABETS THEN CHECK
WITH VALID : ENTRING a-z,A-Z THEN WITH INVALID :!@&* AND
SPACES,12 ETC
BOUNDARY VALUE ANALYSIS: THIS IS A TECHNIQUE USED IN THE
BLACK BOX TESTING METHOD,USE TO CHECK THROUGH RANGE TO
MINIMIZE THE TIME , RATHER THAN TESTING WITH WHOLE DATA
FOR EXAMPLE :(1-1000)
HERE NO NEED OF TESTING RIGHT FROM START IE FROM 1 TO 100
INSTEAD, CHECK WITH THE VALUES 0,1,2,99,100,101
| Is This Answer Correct ? | 41 Yes | 10 No |
Answer / kanad
The main difference between EP and BVA is that EP
determines the number of test cases to be generated for a
given scenario where as BVA will determine the effectivenss
of those generated test cases.
Equivalence partitioning :
Equivalence Partitioning determines the number of test
cases for a given scenario.
Equivalence partitioning is a black box testing technique
with the following goal:
1.To reduce the number of test cases to a necessary
minimum.
2.To select the right test cases to cover all possible
scenarios.
EV is applied to the inputs of a tested component. The
equivalence partitions are usually derived from the
specification of the component's behaviour.
An input has certain ranges which are valid and other
ranges which are invalid. This may be best explained at the
following example of a function which has the pass
parameter "month" of a date. The valid range for the month
is 1 to 12, standing for January to December. This valid
range is called a partition. In this example there are two
further partitions of invalid ranges. The first invalid
partition would be <= 0 and the second invalid partition
would be >= 13.
... -2 -1 0 1 .............. 12 13 14 15 .....
------- -------------------- -----------------
invalid
partition 1 valid partition invalid partition 2
The testing theory related to equivalence partitioning says
that only one test case of each partition is needed to
evaluate the behaviour of the program for the related
partition. In other words it is sufficient to select one
test case out of each partition to check the behaviour of
the program. To use more or even all test cases of a
partition will not find new faults in the program. The
values within one partition are considered to
be "equivalent". Thus the number of test cases can be
reduced considerably.
Equivalence partitioning is no stand alone method to
determine test cases. It has to be supplemented by boundary
value analysis. Having determined the partitions of
possible inputs the method of boundary value analysis has
to be applied to select the most effective test cases out
of these partitions.
Boundary Value Analysis :
Boundary Value Analysis determines the effectiveness of
test cases for a given scenario.
To set up boundary value analysis test cases the tester
first has to determine which boundaries are at the
interface of a software component. This has to be done by
applying the equivalence partitioning technique. Boundary
value analysis and equivalence partitioning are inevitably
linked together.
For the example of the month a date would have the
following partitions:
... -2 -1 0 1 .............. 12 13 14 15 .....
--------------|-------------------|-------------------
--
invalid partition 1 valid partition invalid
partition 2
By applying boundary value analysis we can select a test
case at each side of the boundary between two partitions .
In the above example this would be 0 and 1 for the lower
boundary as well as 12 and 13 for the upper boundary. Each
of these pairs consists of a "clean" and a "dirty" test
case.
A "clean" test case should give a valid operation result of
the program. A "dirty" test case should lead to a correct
and specified input error treatment such as the limiting of
values, the usage of a substitute value, or in case of a
program with a user interface, it has to lead to warning
and request to enter correct data.
The boundary value analysis can have 6 test cases: n, n-1,
and n+1 for the upper limit; and n, n-1, and n+1 for the
lower limit.
| Is This Answer Correct ? | 33 Yes | 3 No |
Equalance partion is a method of black box testing.for eg
if the application is accept the values between 10 to 50.we
are giving valid and invalid inputs to the system and
check the status.
eg inputs valid:25,30
invalid:9,60
boundary value analyses also method of blackbox testing.in
this case we r taking boundary vaues for eg in abouve eg
upper boundary value is 50,and lower boundary value is 10.
inputs
upper boundary 49,51.
lower boundary 9,11
on boundary 10,50
| Is This Answer Correct ? | 32 Yes | 4 No |
Answer / raju
ecp is amethod by which we can deferentiate the test cases
into valid and invalid
where as in bva we have the 3 classes
1)min and maximum
2)min-1& max+1
3)min+1&max-1
| Is This Answer Correct ? | 25 Yes | 2 No |
Boundary value analysis and equivalent partitioning are
black box testing techniques.
Both are applied to validate the minimum and maximum limit.
BVA ex: if Username field accepts minimum 4 character and
maximum 12 characters, then we need to check with six
different data.
lets consider 'm' is minimum limit and 'n' is maximum limit,
then check with m-1, m, m+1 , n-1 , n , n+1.
note: m=4(minimum 4 character)
n=12(maximum 12 characters)
m-1= (4-1)=3 ( invalid data )
m=4 (valid data )
m+1= (4+1)=5 ( valid data )
n-1=(12-1)=11 (valid data)
n=12 (valid data)
n+1= 12+1=13 ( invalid data)
there fore , here we have 2 invalid data(-ve test case) and
4 valid data( +ve test case).
equivalent partitioning: this technique is used to reduce
the no. of test cases in to finite set. in this method input
data will be divided into 3 equivalent sets.
ex: if a input box accepts 1 to 1000 numbers then, it is not
necessary to write test cases for 1000 valid inputs and
other test case for invalid data.
instead validate the field with 3 different sets.
1. take one input data between 1 to 1000 as +ve test case.
if we testing for the other values between 1 to 1000 the
result is going to be same. so one test case is sufficient
for valid input data.
2. take one value less than 1 as invalid input (-ve test case)
3. take one input data above 1000 as invalid input ( -ve
test case).
There fore using this method we can reduce the no.of
testcases into finite set.
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / @mit
Both Equivalence Class Partion and Boundry Value Analysis
are the Test cases design technique under Black box Testing.
In Equivalence partion,break the data or value into equal
partion.
For Ex. suppose a field have a validation to accept the 1-
100 characters.
Then according to Equivalence partion,break the value
into equal classes such as :-
1-20
21-40
41-60
61-80
81-100
Preapre the Test cases by covering above stated classes.
In Boundary Value analysis,cover all boundry values (both
vslid & Invalid) in test case.
Like Max+1,Max-1,Min+1,Min-1
Then preapare the Test cases by covering all above stated
condition.
| Is This Answer Correct ? | 4 Yes | 4 No |
Answer / moderator
check the following link
http://www.allinterview.com/showanswers/118
| Is This Answer Correct ? | 3 Yes | 4 No |
Answer / k.sendhil kumar
1)Rather than selecting any element in an equi. class as
being, boundary - value analysis requires that one or more
elements be selected such that each edge of the equivalence
class
2)boundary val depends more on input test condition whereas
equi value test cases depend more on output classes
| Is This Answer Correct ? | 2 Yes | 3 No |
Answer / nagaraj.korke
ECP:here we will partition the input data into various valid
and invalid subclass
BVA:we will check the input datarange with max and min values.
| Is This Answer Correct ? | 3 Yes | 7 No |
Answer / sathish
Equivalence partitioning is the test case developed to
identify the maximum possible bugs, to ensure the
application is defect less.
Boundary value analysis is the test case to optimize the
Equivalence Partitioning. ie limit the number of inputs,
such that the application is behaving as expected.
| Is This Answer Correct ? | 5 Yes | 11 No |
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