w 5 Black Box Testing

7/31/2019 w 5 Black Box Testing

http://slidepdf.com/reader/full/w-5-black-box-testing 1/59

28-04-2012

ICI500



Equivalence Partitioning Boundary Value Analysis

Use Case Testing

Decision Table Testing State Transition Testing

Questios for the class



Unfeasible exhaustive testing means we needa limited subset of all possible inputs.

How to select the best subset? Why EP?

◦ The concept of defect clustering Test case design by EP has three :

1. Identify the equivalence classes (EC)2. Define the test cases

3. Define expected output Coverage

◦ Number of tested EC/total number of EC



Employee receive a Christmas bonus equal to50% of their monthly income if they havebeen working for the company for more thanthree years, employees who have been

employed for more than five years receive a75% bones, and those with more than eightyears of employment are awarded a 100%bonus.

How many equivalence classes you see?



Employment in Years vEc1: 0 <= x <= 3vEc2: 3 < x <= 5vEc3: 5 < x <= 8vEc4: x > 8

iEc1: x < 0iEc2: x > 70

24712

-380



1. Identify the equivalence classes (EC)1. (*)Identify the domain for each input parameter

2. (*)Subdivide until no more partitions can be made

2. Define the test cases

1. Select representatives (one per each class)2. Create the test cases

1. (**)Multiply the Number of valid EC per parameter

2. (**)Sum the Number of invalid EC per parameter

3. Define the expected outcomes for each Test Case



The specification for calculating the price is thefollowing:◦ The starting point is minus , where

is the basic price of the product and is the reduced amount done by salesman

◦ A for a special type of product and the pricefor extra sub-products ( ) shall be added

◦ If 3 or more sub-products are chosen there is adiscount of 10% on these sub-products only. If 5 ormore are chosen then a 15% is applied

◦

The granted by the salesman is applied to theonly, but the discount on extras is applied toonly



Identify the domain for each input parameter

Baseprice vEc11: [MIN_DOUBLE, … , MAX_DOUBLE]

iEc11: NaNspecialprice vEc21: [MIN_DOUBLE, … , MAX_DOUBLE]

iEc21: NaN

extraprice vEc31: [MIN_DOUBLE, … , MAX_DOUBLE]iEc31: NaN

extras vEc41: [MIN_INT, … , MAX_INT]iEc41: NaN

discount vEc51: [MIN_DOUBLE, … , MAX_DOUBLE]iEc51: NaN



Subdivide until no more partitions can bemade◦ Parameter 1 to 3 are prices. Prices are always

positive. However, there is no limit on this.

◦ The value controls the discount of extrapricewhere 10% if extras>=3 and 15% if extras>=5.

must be positive, because you can’t buynegative items.

◦

The shall be a percentage. Therefore avalue between 0-100 must be defined. Consultationwith the client should be done



Baseprice vEc11: [0, … , MAX_DOUBLE]iEc11: [MIN_DOUBLE, … , 0[iEc12: NaN

specialprice vEc21: [0, … , MAX_DOUBLE]iEc21: [MIN_DOUBLE, … , 0[iEc22: NaN

extraprice vEc31: [0 … , MAX_DOUBLE]

iEc31: [MIN_DOUBLE, … , 0[ iEc32: NaN

extras vEc41: [0, 1 , 2]vEc42: [3, 4]vEc43: [5, … , MAX_INT]

iEc41: [MIN_INT, … , 0[ iEc42: NaN

discount vEc51: [0, … , 100]iEc51: [MIN_DOUBLE, … , 0[ iEc52: ]100, … , MAX_DOUBLE] iEc53: NaN



Baseprice vEc11: [0, … , MAX_DOUBLE]iEc11: [MIN_DOUBLE, … , 0[iEc12: NaN

20000.00-1.00“abc”

specialprice

vEc21: [0, … , MAX_DOUBLE]iEc21: [MIN_DOUBLE, … , 0[iEc22: NaN

3450.00-1.00“abc”

extraprice vEc31: [0 … , MAX_DOUBLE]

iEc31: [MIN_DOUBLE, … , 0[ iEc32: NaN

6000.00

-1.00“abc”

extras vEc41: [0, 1 , 2]vEc42: [3, 4]vEc43: [5, … , MAX_INT]

iEc41: [MIN_INT, … , 0[ iEc42: NaN

1320

-1“abc”

discount vEc51: [0, … , 100]iEc51: [MIN_DOUBLE, … , 0[ iEc52: ]100, … , MAX_DOUBLE]

iEc53: NaN

10.00-1.00101.00

“abc”



1 20000.00 3450.00 6000.00 1 10.00

2 20000.00 3450.00 6000.00 3 10.00

3 20000.00 3450.00 6000.00 20 10.00

4 -1.00 3450.00 6000.00 1 10.00

5 “abc” 3450.00 6000.00 1 10.00

6 20000.00 -1.00 6000.00 1 10.00

7 20000.00 “abc” 6000.00 1 10.00

8 20000.00 3450.00 -1.00 1 10.00

9 20000.00 3450.00 “abc” 1 10.00

10 20000.00 3450.00 6000.00 -1 10.00

11 20000.00 3450.00 6000.00 “abc” 10.00

12 20000.00 3450.00 6000.00 1 -1.00

13 20000.00 3450.00 6000.00 1 101.00

14 20000.00 3450.00 6000.00 1 “abc”



1 20000.00 3450.00 6000.00 1 10.00 27450.00

2 20000.00 3450.00 6000.00 3 10.00 26850.00

3 20000.00 3450.00 6000.00 20 10.00 26550.004 -1.00 3450.00 6000.00 1 10.00 NOT_VALID

5 “abc” 3450.00 6000.00 1 10.00 NOT_VALID

6 20000.00 -1.00 6000.00 1 10.00 NOT_VALID

7 20000.00 “abc” 6000.00 1 10.00 NOT_VALID

8 20000.00 3450.00 -1.00 1 10.00 NOT_VALID

9 20000.00 3450.00 “abc” 1 10.00 NOT_VALID

10 20000.00 3450.00 6000.00 -1 10.00 NOT_VALID11 20000.00 3450.00 6000.00 “abc” 10.00 NOT_VALID

12 20000.00 3450.00 6000.00 1 -1.00 NOT_VALID

13 20000.00 3450.00 6000.00 1 101.00 NOT_VALID

14 20000.00 3450.00 6000.00 1 “abc” NOT_VALID



What level of equivalence class coverage didwe achieved?



A compiler for subset of the DIMENSIONstatement of the FORTRAN language

General rules◦ DIMENSION statements can span over multiple lines

◦ Items in italics indicate variables

◦ Brackets indicate optional items

◦ Ellipsis indicates succession

◦ Form of the dimension statement is:

DIMENSION ad [,ad ]…

◦ ad is an array descriptor of the form

n(d[,d]…)



n = symbolic name of the array d = dimension declarator

n can be 1-6 letter or digits, the first charmust be a letter

d is defined as follows:◦ [lb:]ub

◦ lb and ub are the lower and upper bounds

◦

Must be in the range -65534 to 65535◦ If lb not specified is 1

◦ ub >= lb



1. Identify restrictions and conditions2. For each restriction/condition create:

1. One valid and two invalid classes if the domain is: E.g. “the item count can be from 1 to 999”), identify one

valid equivalence class (1 < item count < 999) and two invalidequivalence classes (item count < 1 and item count > 999).

: E.g. “one through six owners can be listed for theautomobile”, identify one valid equivalence class and two invalidequivalence classes (no owners and more than six owners).

2. One valid class for each valid input and only one invalid if thedomain is:

: E.g. type of vehicle must be BUS, TRUCK,TAXICAB, PASSENGER, or MOTORCYCLE. Identify a valid equivalenceclass for each and one invalid equivalence class (“TRAILER”).

3. Only one valid and only one invalid class if the domain is:: E.g. first character of the identifier must be a

letter,” identify one valid equivalence class (it is a letter) and oneinvalid equivalence class (it is not a letter).



Differences with EP◦ Focus only in the boundaries

What it does?◦ Checks the “border” of the equivalence classes ◦ How many test cases for each boundary?

Why BVA? What is the disadvantage of this technique? Test case design by BVA has two :

1. Identify the boundary values (BVs)2. Define test cases

(Freely) combine valid boundary values in test cases Single test case for each invalid boundary value

Coverage◦ Number of tested BVs/ total number of BVs



The customer has given the followingadditional information:◦ The base price is between 10,000 and 150,000

◦ The special price for the extra items is between 800

and 3,500◦ The maximum possible extras is 25

◦ The extra prices are between 50 and 750

◦ The maximum salesman discount is 25%

What are the valid BVs and invalid BVs?



Baseprice 10000.00, 10000.01, 149999.99, 150000.00

Specialprice 800.00, 800.01, 3499.99, 3500.00

Extraprice 50.00, 50.01, 18749.99, 18750.00

Extras 0, 1, 2, 3, 4, 5, 6, 24, 25Discount 0.00, 0.01, 24.99, 25.00

Baseprice 9999.99, 150000.01

Specialprice 799.99, 3500.01

Extraprice 49.99, 18750.01

Extras -1, 26

Discount -0.01, 25.01



What is the minimum set of test cases we canbuild to get 100% BVA coverage?



1 10,000.00 800.00 50.00 0 0.00 10,850

2 10,000.01 800.01 50.01 1 0.01 10,849.03

… … … … … … …

10 9,999.99 1000.00 700.00 2 10.00 NOT_VALID

11 150,000.01 1000.00 700.00 2 10.00 NOT_VALID

… … … … … … …

19 150,00.00 1000.00 700.00 2 25.01 NOT_VALID



MTEST is a program that grades multiple-choice examinations. Theinput is a data file named OCR, with multiple records that are. Per the file specification, the first record is a title used

as a title on each output report. The next set of records describes thecorrect answers on the exam. These records contain a

. In the first record of this set, the number of questions is listedin columns . Columns 10–59 contain the correctanswers for questions 1–50 (any character is valid as an answer).

Subsequent records contain,.

The third set of records describes the answers of each student; each of these records contains a . For each student, the firstrecord contains the student’s name or number in (anycharacters); columns

. If the test has more than 50 questions, subsequent records for the

student contain answers , in columns 10–59.The .



4 Output records1. A report, sorted by student identifier,

showing each student’s grade (percentageof answers correct) and rank.

2. A similar report, but sorted by grade.

3. A report indicating the mean, median, andstandard deviation of the grades.

4. A report, ordered by question number,showing the percentage of studentsanswering each question correctly.



Empty input file 999 question exam

0 Students

200 Students



1. Empty input file.2. Missing title record.3. 1-character title.4. 80-character title.5. 1-question exam.

6. 50-question exam.7. 51-question exam.8. 999-question exam.9. 0-question exam.10. Number-of-questions field has nonnumeric value.11. 0-correct-answer.12. Too many correct-answer records.13. Too few correct-answer records.



14. 0 students.15. 1 student.

16. 200 students.

17. 201 students.

18.

A student has one answer record, but there are twocorrect- answer records.

19. The above student is the first student in the file.

20. The above student is the last student in the file.

21. A student has two answer records, but there is just

one correct-answer record.22. The above student is the first student in the file.

23. The above student is the last student in the file.



0 Students 1 Student

200 Students



24. All students receive the same grade.25. All students receive a different grade.26. Some (not all) students receive the same grade.

Why?27. A student receives a grade of null28. A student receives a grade of 1029. A student has the lowest possible identifier.

Why?30. A student has the highest possible identifier.

31. The number of students is such that the reportis just large enough to fit on one page. Why?32. The number of students is such that all

students but one fit on one page.



33. The standard deviation is at its maximum

All the other are already contained in Test Case24



34. All students answer question 1 correctly.35. All students answer question 1 incorrectly.36. All students answer the last question

correctly.

37. All students answer the last questionincorrectly.

38. The number of questions is such that thereport is just large enough to fit on one

page.39. The number of questions is such that all

questions but one fit on one page.



BVA works for INPUTS and OUTPUTS◦ E.g. An information-retrieval system displays the

most relevant paragraphs based on an inputrequest, but paragraphs

◦

Write test cases such that the program displays, and write a test

case that might cause the program to.

Remember valid BVs can be freelycombined. However, invalid ones can’t



Tests can be derived from use cases Use cases might be described at an abstract level

or at the system level Each use case has: preconditions, post-

conditions, mainstream scenario and alternative

scenarios Useful for

◦ designing acceptance tests with customer/userparticipation

◦ system testing. Why?◦

uncover integration defects. Why? Coverage

◦ At least one test case per use case



GetMoney

PINQuery

Eat CardCondition:{3rd wrong PIN input}

Client

PaymentUnit

Bank



One weakness of boundary-value analysisand equivalence partitioning is that they donot explore combinations of inputcircumstances

Decision tables are a good way to capturesystem requirements that contain logicalconditions

Coverage is to have

in the table Strength is that it exercises conditions that

will not be covered otherwise



If Student A and Student B are friends and if Student (A) don’t go to lectures, then Student(B) will need to take additional notes

If Student A and Student B are friends and if

Student (B) don’ go to lectures, then Student(A) will need to take additional notes

If neither Student (A) nor Student (B) go tolectures and is the final session, then both

will fail the final exam If both Students go to lectures and is the final

session, then both will pass the exam



Conditions R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R 10 R 11 R 12 R 13 R 14 R 15 R 16

A goes T T T T T T T T F F F F F F F F

B goes T T T T F F F F T T T T F F F F

A and B are friends T T F F T T F F T T F F T T F F

Is Final Session T F T F T F T F T F T F T F T F

Actions

A takes additional notes F F F F F T F F F F F F F F F F

B takes additional notes F F F F T F F F T T F F F F F F

Both pass the exam T F T F F F F F F F F F F F F F



A system may exhibit a different responsedepending on current conditions or

A state table shows the relationship between thestates and inputs, and can highlight possibletransitions that are valid

Test cases can be designed to cover:◦ a common sequence of states◦ every state◦

every transition◦ specific sequence of transitions

◦ invalid transitions



State: Is a in which a system isfor one or more . States

“remember” inputs the system has received inthe past and define how the system should

respond to subsequent events when theyoccur.

Transition: Represents a from onestate to another caused by an event



Event: within/outside the systemthat the system to .Events can be independent or causally related

Action: initiated because of a state

change. Actions betweenstates



Represent a Reservation process by using aState Transition diagram

Entry point will be a

Exit point will be a

Start by and



Source: Copeland 2004



Using the above diagram, can you build astate transition table? Use the followingtemplate:

null giveInfo startPayTimer Made




Why are STT useful?



1. Create test cases that atleast once

2. Create test cases that atleast once

3. Create test cases that (becareful with the )

4. Create test cases that

at least once5. Create from the state

transition table



Visit States Trigger Events




Execute all paths



Exercise transitions




Invalid transitions using the table (see above)



Which of the followingstatements about thegiven state table is TRUE?◦ a)

◦ b)

◦ c)

◦ d)

Source: ISTQB 2011



Which of thefollowingstatements areTRUE?◦ A. The test case

table exercisesthe shortestnumber of

transitions.◦ B. The test case

gives only thevalid statetransitions.

◦ C. The test casegives only theinvalid statetransitions.

◦ D. The test caseexercises thelongest numberof transitions.

Source: ISTQB 2011



An employee’s bonus is to be calculated. Itcannot become negative, but it can be calculatedto zero. The bonus is based on the duration of the employment. An employee can be employed

for less than or equal to 2 years, more than 2years but less than 5 years, 5 to 10 years, orlonger than 10 years. Depending on this periodof employment, an employee will get either nobonus or a bonus of 10%, 25% or 35%.

How many equivalence partitions are needed totest the calculation of the bonus?

Source: ISTQB 2011



Which of the following statements about the benefits of derivingtest cases from use cases are most likely to be true?A. Deriving test cases from use cases is helpful for system andacceptance testing.B. Deriving test cases from use cases is helpful only for automatedtesting.

C. Deriving test cases from use cases is helpful for componenttesting.D. Deriving test cases from use cases is helpful for testing the

interactionbetween different components of the system. a) A and D are true; B and C are false.

b) A is true; B, C and D are false. c) A and B are true; C and D are false. d) C is true; A, B and D are false.

Source: ISTQB 2011

w 5 Black Box Testing

Documents

Transcript of w 5 Black Box Testing