Equivalence Class Testing Use the mathematical concept of

Equivalence Class Testing • Use the mathematical concept of partitioning into equivalence classes to generate test cases for Functional (Black-box) testing • The key goals for equivalence class testing are similar to partitioning: 1. 2. completeness of test coverage lessen duplication of test coverage

Equivalence Class Test Cases • Consider a numerical input variable, i, whose values may range from -200 through +200. Then a possible partitioning of testing input variable by 4 people may be: – -200 to -100 – – – -101 to 0 1 to 100 101 to 200 • Define “same sign” as the equivalence relation, R, defined over the input variable’s value set, i = {-200 - -, 0, - -, +200}. Then one partitioning will be: – – – -200 to -1 (negative sign) 0 (no sign) 1 to 200 (positive sign)

Weak Normal Equivalence testing 1. Assumes the ‘single fault’ or “independence of input variables. ” – e. g. If there are 2 input variables, these input variables are independent of each other. 2. Partition the test cases of each input variable separately into different equivalent classes. 3. Choose the test case from each of the equivalence classes for each input variable independently of the other input variable

Example of : Weak Normal Equivalence testing Assume the equivalence partitioning of input X is: 1 to 10; 11 to 20, 21 to 30 and the equivalence partitioning of input Y is: 1 to 5; 6 to 10; 11; 15; and 16 to 20 X We have covered everyone of the 3 equivalence classes for input X. 30 20 10 1 Y 1 5 10 15 We have covered each of the 4 equivalence classes for input Y. 20 For ( x, y ) we have: (24, 2) (15, 8 ) ( 4, 13) (23, 17) General rule for # of test cases? What do you think? # of partitions of the largest set?

Strong Normal Equivalence testing • This is the same as the weak normal equivalence testing except for “multiple fault assumption” or “dependence among the inputs” • All the combinations of equivalence classes of the variables must be included.

Example of : Strong Normal Equivalence testing Assume the equivalence partitioning of input X is: 1 to 10; 11 to 20, 21 to 30 and the equivalence partitioning of input Y is: 1 to 5; 6 to 10; 11; 15; and 16 to 20 X 30 We have covered everyone of the 3 x 4 Cartesian product of equivalence classes 20 10 1 Y 1 5 10 15 20 General rule for # of test cases? What do you think?

Weak Robust Equivalence testing • Up to now we have only considered partitioning the valid input space. • “Weak robust” is similar to “weak normal” equivalence test except that the invalid input variables are now considered. A note about considering invalid input is that there may not be any definition “specified” for the various, different invalid inputs - - - making definition of the output for these invalid inputs a bit difficult at times. (but fertile ground for testing)

Example of : Weak Robust Equivalence testing Assume the equivalence partitioning of input X is 1 to 10; 11 to 20, 21 to 30 and the equivalence partitioning of input Y is 1 to 5; 6 to 10; 11; 15; and 16 to 20 X 30 We have covered everyone of the 5 equivalence classes for input X. 20 10 1 Y 1 5 10 15 20 We have covered each of the 6 equivalence classes for input Y.

Strong Robust Equivalence testing • Does not assume “single fault” - - - assumes dependency of input variables • “Strong robust” is similar to “strong normal” equivalence test except that the invalid input variables are now considered.

Example of : Strong Robust Equivalence testing Assume the equivalence partitioning of input X is: 1 to 10; 11 to 20, 21 to 30 and the equivalence partitioning of input Y is: 1 to 5; 6 to 10; 11; 15; and 16 to 20 X 30 We have covered everyone of the 5 x 6 Cartesian product of equivalence classes (including invalid inputs) 20 10 1 Y 1 5 10 15 20

Equivalence class Definition • Note that the examples so far focused on defining input variables without considering the output variables. • For the earlier “triangle problem, ” we are interested in 4 questions: – – Is it a triangle? Is it an isosceles? Is it a scalene? Is it an equilateral? • We may define the input test data by defining the equivalence class through “viewing” the 4 output groups: – – input sides <a, b, c> do not form a triangle input sides <a, b , c> form an isosceles triangle input sides <a, b, c> form a scalene triangle input sides <a, b, c> form an equilateral triangle

Consider: Weak Normal Equivalence Test Cases for Triangle Problem “valid” inputs: 1<= a <= 200 1<= b <= 200 1<= c <= 200 and for triangle: a<b+c b<a+c c<b+a inputs output a b c Not triangle 35 10 4 Equilateral 35 35 35 Isosceles 24 24 7 18 24 Valid Inputs to get Equilateral Not Triangle Isosceles Scalene 35

Strong Normal Equivalence Test Cases for Triangle Problem • Since there is no further sub-intervals inside the valid inputs for the 3 sides a, b, and c, Strong Normal Equivalence is the same as the Weak Normal Equivalence

Weak Robust Equivalence Test Cases for Triangle Problem <200, 200> Now, on top of the earlier 4 normal test cases, include the “invalid” inputs Valid Inputs <1, 1, 1> Equilateral Not Triangle Isosceles Scalene Include 6 invalid test case in addition to Weak Normal above: below: <201, 45, 50 > <45, 204, 78 > <50, 78, 208 > < -5, 76, 89 > < 56, -20, 89 > < 56, 89, 0 >

Strong Robust Equivalence Test Cases for Triangle Problem • Similar to Weak robust, but all combinations of “invalid” inputs must be included to the Strong Normal. • Look at the “cube” figure and consider the corners (two diagonal ones) a) Consider one of the corners <200, 200> : there should be (2 3 – 1) = 7 cases of “invalids” < 201, < 201, 50 , 201 > 50 > < 50 , 201, 50 > < 50 , 201 > < 50, 50 , 201 > b) There will be 7 more “invalids” when we consider the other corner , <1, 1, 1 >: < 0, 0, 0 > < 0, 0, 5 > < 0, 10, 0 > < 0, 8, 10> <7, 0, 9 > <8, 0, 0 > <8, 9, 0 >

Next Day Program example • Here we have 3 input variables and may choose to have the sets defined as (without partitioning of days, month, or year): – Day : 1 through 31 days – Month: 1 through 12 – Year: 0001 through 3000 • In this case, for weak normal equivalence testing only needs 1 test case from each input: – (year; month; day) : (2001, 9, 23) • Clearly, this “non-partitioning” of the input gives very limited test case!

Next Day Program example (cont. ) • A more useful situation is to partition the 3 inputs. As an example: – – Day: 1 through 28 Day : 29 Day: 30 Day: 31 4 partitions of days – Month: those that have 31 days or {1, 3, 5, 7, 8, 10, 12} – Month: those that have 30 days or {4, 6, 9, 11} – Month: that has less than 30 days or {2} – Year: leap years between 0001 and 3000 – Year: non-leap years between 0001 and 3000 3 partitions of months 2 partitions of years

Next Day Problem example (cont. ) • Weak Normal equivalence Test Cases: • Number of test cases is driven by the # of partitions of the largest set, which in this case is the Days – has 4 partitions: (year, month, day): – – (leap year, 10, 8) (leap year, 4, 30) (non-leap year, 2, 31) (non-leap year, 7, 29) Without considering any relationship among the inputs and just mechanically following the rule How good is this set of test cases ? ? How valuable is the “generic” Weak Normal test Cases? --- Not Much!

Next Day Problem example (cont. ) • What about the Strong Normal case where we consider all the permutations of the previously partitioned 3 inputs? – We should have (2 years x 3 months x 4 days) = 24 test cases (leap year, 10, 5) (leap year, 10, 30) (leap year, 10, 31) (leap year, 10, 29) (non-leap year, 10, 5) (non-leap year, 10, 30) (non-leap year, 10, 31) (non-leap year, 10, 29) (leap year, 6, 5) (leap year, 6, 30) (leap year, 6, 31) (leap year, 6, 29) (non-leap year, 6, 5) (non-leap year, 6, 30) (non-leap year, 6, 31) (non-leap year, 6, 29) (leap year, 2, 5) (leap year, 2, 30) (leap year, 2, 31) (leap year, 2, 29) (non-leap year, 2, 5) (non-leap year, 2, 30) (non-leap year, 2, 31) (non-leap year, 2, 29) A little better than previous?

Next Day Problem example (cont. ) • What if we consider the invalids, Weak Robust? – – – 31 < days < 1 12 < months < 1 3000 < year < 0001 Then we need to include the following to the Weak Normal : ( valid year, 5, 45) ( valid year, 5, -5) ( valid year, 22, 30) ( valid year, 0, 15) ( 3500, 7, 20 ) ( 0000, 7, 15) ( 3500, 22, 45) ( 0000, 0 , -5) Would you combine or leave them separately? Why?

Next Day Problem example (cont. ) • For Strong Robust, we need to consider and include all the permutations of the invalids to the Strong Normal: • 3 inputs, so there are 23 – 1=7 invalids each for high and low, making a total of 14 more. – – – – ( 3025, 45, 80) ( 3025, 45, 25) ( 3025, 7, 80) ( 3025, 7, 25) ( 2000, 45, 80) ( 2000, 45, 25) ( 2000, 7, 80) ( 0000, -2, 15) ( 0000, 4 , 0) ( 0000, 4, 10) ( 2000, -2, 10) ( 2000, 4, 0)

The Phone Company Example • A phone company system has a function that computes the bill (or charge) of a call depending on: • Call duration in minutes (Real number) • Destination country (integer code between 1 and 999) (e. g. USA: 1, Jordan: 962, etc. ) • The phone company of the receiver (integer code between 1 and 200) (e. g. Zain: 1, Orange: 2, etc. ) • The call time – If the call starts any time between midnight and 10: 00 then call charge is 0. 03 JD for each minute or fraction of minute. – If the call starts any time between 10. 01 and 17: 00 then call charge is 0. 04 JD for each minute or fraction of minute. – If the call starts any time between 17. 01 and 23. 59 then call charge is 0. 05 JD for each minute or fraction of minute.

Cont. – If destination country code < 500 then add 1. 00 JD to the call charge else add 0. 50 JD to the call charge. – If the receiving company code = 1 or 2 or 3 then add only 0. 30 JD to the call charge else add 2. 00 JD. – Call duration of seconds are rounded up to the next larger minute. – No call lasts more than 3 hours.

Cont. Apply the following Equivalence Class based testing techniques for this billing function: • Weak normal equivalence class testing • Strong normal equivalence class testing • Weak robust equivalence class testing • Strong robust equivalence class testing

Valid relations • R 1 = { Call Duration : 0. 1 =< cd <= 180. 0 } • R 2 = { Destination Country : 1 =< dest <= 499} • R 3 = { Destination Country : 500 =< dest <= 999 } • R 4 = { Phone Company : 1 =< comp <= 3 } • R 5 = { Phone Company : 4 =< comp <= 200 } • R 5 = { Call Time : 0. 0 =< ct <= 10. 00 } • R 6 = { Call Time : 10. 01 =< ct <= 17. 00 } • R 7 = { Call Time : 17. 01 =< ct <= 23. 59 }

Invalid relations • • R 8 = { Call Duration : cd < 0. 1 } R 9 = { Call Duration : cd > 180. 0 } R 10 = { Destination Country : dest < 1 } R 11 = { Destination Country : dest > 999 } R 12 = { Phone Company : comp < 1 } R 13 = { Phone Company : comp > 200 } R 14 = { Call Time : ct < 0. 0 } R 15= { Call Time : ct > 23. 59 }

1) Weak Normal EC Testing Case ID cd dest comp ct Expected Output WN 1 90. 0 250 2 5. 00 90. 0 * 0. 03 + 1. 00 + 0. 30 WN 2 90. 0 750 102 13. 30 90. 0 * 0. 04 + 0. 50 + 2. 00 WN 3 90. 0 750 102 20. 30 90. 0 * 0. 05 + 0. 50 + 2. 00

2) Strong Normal EC Testing

3) Weak Robust EC Testing

4) Strong Robust EC Testing SR 1. . SR 8 is the same as WR 1. . WR 6