Extract All requirement cases using Cause&Effect Analysis
Introduction
This article describes in detail an organized method to extract all possible cases from the requirement, this can help the system analyst in defining all the cases and also can help developers while preparing the unit TCs and of course will help the testers to write unified test cases.
First of all, suppose you have the following requirement:
A mailing is to be sent out to customers.
The content of the mailing is about the current level of discounting and potential levels of discounting. The content is different for different types of customers.
Customer Types A, B, and C get a normal letter except Customer Type C, who gets a special letter. Any customer with 2 or more current lines or with a credit rating of ‘X’ gets a special paragraph added with an offer to subscribe to another level of discounting.
The problem is if you give this requirement to any 4 different testers, the 4 will produce different no of TCs, I will explain how to derive a unified number of TCs using cause and effect analysis and a decision table.
Step 1: Break the specification down to cause and effect:
First, the analyst must break down the requirement to causes and effects
– Cause: is a distinct input condition or an equivalence class of input conditions (the cause which triggers the effect) in our above example Customer Type (A, B, C, Other) is a cause
– Effect: is an output condition. Determine the logical relationship between the cause and the effects, in our above example “Get a normal letter” is an effect for the cause customer type(A, B,C, Other )
Usually, the effect is an output action (Ex. Error message, application action, executing specific application job, etc…)
After this step you should have causes and effects like that:
C1: Customer Type A, B, C, Other
C2: 2 or more lines
C3: Credit Rating
E1: Normal Letter
E2: Special Letter
E4: Add a special paragraph
E5: Get a Normal letter
Step 2: Put Causes and Effects in the decision Table.
Then we should fill the decision table by putting the cause and the effects together in that table as a following:
| Cause\Effect | ||||
| C1: | ||||
| C2: | ||||
| C3: | ||||
| E1: | ||||
| E2: | ||||
| E3: | ||||
| E4: | ||||
| E5: |
Step 3: Calculate all possible Combinations
This is a very important step to know no of columns in our decision table, simply use this formula to know all no of combinations that can occur:
• If all causes are simply Y/N values:
2number of causes
• If 1 cause with 3 values and 3 with 2:
31 * 23 = 24
In our example, C1 has 4 values (A or B or C or Other) AND C2 has 2 values (YES or NO) also C3 has 2 values (YES or NO) so our formula should be :
14 * 22 = 16 combinations
So we should draw for the decision table 16 columns which present 16 different combinations.
Now the table should be like that:
| Cause\Effect | ||||||||||||||||
| C1: | ||||||||||||||||
| C2: | ||||||||||||||||
| C3: | ||||||||||||||||
| E1: | ||||||||||||||||
| E2: | ||||||||||||||||
| E3: | ||||||||||||||||
| E4: | ||||||||||||||||
| E5: |
Step 4: Fill Columns
We can fill columns 2 methods:
1- Logical method: in this method, you start filling column by column, start with the causes and end up with the effect (put into consideration that every column is a case).
2- Repeating Factor Algorithm method:
a. Determine Repeating Factor (RF): divide remaining combinations by the number of possible values for that cause
b. Write RF times the first value, then RF times the next, etc. until the row is full
Example for using Repeating Factor Algorithm: suppose you have the following decision table:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| C1(Y\N) | ||||||||
| C2(Y\N) | ||||||||
| C3(Y\N) | ||||||||
| E1 | ||||||||
| E2 |
Fill first row: RF = number of remaining combination (8) / possible value for this
Cause (2) = (8/2 = 4), so we will fill 4 from Y and 4 of N, the DT will be :
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| C1(Y\N) | Y | Y | Y | Y | N | N | N | N |
| C2(Y\N) | ||||||||
| C3(Y\N) | ||||||||
| E1 | ||||||||
| E2 |
Fill Second row: RF = number of remaining combination (4) / possible value for this
Cause (2) = (4/2 = 2), so we will fill 2 from Y and 2 of N, the DT will be :
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| C1(Y\N) | Y | Y | Y | Y | N | N | N | N |
| C2(Y\N) | Y | Y | N | N | Y | Y | N | N |
| C3(Y\N) | ||||||||
| E1 | ||||||||
| E2 |
Fill Third row: RF = number of the remaining combination (2) / possible value for this
Cause (2) = (2/2 = 1), so we will fill 1 from Y and 1 of N, the DT will be :
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| C1(Y\N) | Y | Y | Y | Y | N | N | N | N |
| C2(Y\N) | Y | Y | N | N | Y | Y | N | N |
| C3(Y\N) | Y | N | Y | N | Y | N | Y | N |
| E1 | ||||||||
| E2 |
Filing DT in example1:
In example 1 we have 3 causes and 5 effects with the total number of combinations = 16 combinations
Fill first row: RF = number of remaining combination (16) / possible value for this
Cause (4) = (16/4 = 4) , so we will fill 4 from every type (A,B,C,O) , the DT will be :
| Cause\Effects | ||||||||||||||||
| C1(A,B,C,O) | A | A | A | A | B | B | B | B | C | C | C | C | O | O | O | O |
| C2(Yes\No) | ||||||||||||||||
| C3(yes\No) | ||||||||||||||||
| E1 | ||||||||||||||||
| E2 | ||||||||||||||||
| E3 | ||||||||||||||||
| E4 | ||||||||||||||||
| E5 |
Fill Second row: RF = number of remaining combination (2*2) / possible value for this
Cause (2) = (4/2 = 2), so we will fill 2 from every Y and 2 from N, the DT will be :
| Cause\Effects | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
| C1(A,B,C,O) | A | A | A | A | B | B | B | B | C | C | C | C | O | O | O | O |
| C2(Yes\No) | Y | Y | N | N | Y | Y | N | N | Y | Y | N | N | Y | Y | N | N |
| C3(yes\No) | ||||||||||||||||
| E1 | ||||||||||||||||
| E2 | ||||||||||||||||
| E3 | ||||||||||||||||
| E4 | ||||||||||||||||
| E5 |
Fill Third row: RF = number of the remaining combination (2) / possible value for this
Cause (2) = (2/2 = 1), so we will fill 1 from every Y and 1 from N, the DT will be :
| Cause\Effects | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
| C1(A,B,C,O) | A | A | A | A | B | B | B | B | C | C | C | C | O | O | O | O |
| C2(Yes\No) | Y | Y | N | N | Y | Y | N | N | Y | Y | N | N | Y | Y | N | N |
| C3(yes\No) | Y | N | Y | N | Y | N | Y | N | Y | N | Y | N | Y | N | Y | N |
| E1 | ||||||||||||||||
| E2 | ||||||||||||||||
| E3 | ||||||||||||||||
| E4 | ||||||||||||||||
| E5 |
Step 5: Reduce Combinations
1- In our simple example find a don’t care “-” sign and place it,
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| C1(Y\N) | Y | Y | Y | Y | N | N | N | N |
| C2(Y\N) | Y | Y | N | N | Y | Y | N | N |
| C3(Y\N) | Y | N | Y | N | Y | N | Y | N |
| E1 | ||||||||
| E2 |
In the above DT suppose in column 3 and column 4, when the C2 become “N” the program end, so we will put in C3 the sign “-” which mean done care, the DT will become as follow :
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| C1(Y\N) | Y | Y | Y | Y | N | N | N | N |
| C2(Y\N) | Y | Y | N | N | Y | Y | N | N |
| C3(Y\N) | Y | N | – | – | Y | N | – | – |
| E1 | ||||||||
| E2 |
2- After placing the “-” don’t care sign, now remove the repeated columns, column 3 and 4 now become identical so will remove one of them, also column 7 and 8 now identical so I will remove one of them too, so the DT will be :
| 1 | 2 | 3 | 5 | 6 | 8 | |
| C1(Y\N) | Y | Y | Y | N | N | N |
| C2(Y\N) | Y | Y | N | Y | Y | N |
| C3(Y\N) | Y | N | – | Y | N | – |
| E1 | ||||||
| E2 |
After removing the columns you must do a checksum for the table to assure that the DT still
cover all the original combination = 8
To do checksum:
• For each column calculate the combinations it represents
• A ‘-’ represents as many combinations as the cause has
• Multiply for each ‘-’ down the column
• Add up the total and compare with step 2
So the checksum will be as follow:
| 1 | 2 | 3 | 5 | 6 | 8 | ||
| C1(Y\N) | Y | Y | Y | N | N | N | |
| C2(Y\N) | Y | Y | N | Y | Y | N | |
| C3(Y\N) | Y | N | – | Y | N | – | |
| E1 | |||||||
| E2 | |||||||
| Checksum | 1 | 1 | 2 | 1 | 1 | 2 | 8 |
Step 6: Add effects to table
• Read column by column and determine the effects
• One effect can occur in multiple test combinations
| Cause\Effects | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | |
| C1(A,B,C,O) | A | A | A | A | B | B | B | B | C | C | C | C | O | O | O | O | |
| C2(Yes\No) | Y | Y | N | N | Y | Y | N | N | Y | Y | N | N | Y | Y | N | N | |
| C3(yes\No) | Y | N | Y | N | Y | N | Y | N | Y | N | Y | N | Y | N | Y | N | |
| E1: Normal letter | X | X | X | X | X | X | X | X | ? | ? | ? | ? | |||||
| E2: Special Letter | X | X | X | X | ? | ? | ? | ? | |||||||||
| E3: Add paragraph | ? | X | X | ? | X | X | ? | X | X | ? | ? | ? | ? | ||||
| E4: No Letter | |||||||||||||||||
| Checksum | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 16 |
As we see in the above table after adding effects to the table you can:
1- Determine if you have a missing requirement or not as in the case we place “?”, the requirement is missing if the customer type is other than the 3 types A, B,C
2- You can determine if there is a redundant requirement, example in columns 6 and 7 by looking into the table you will find the 2 columns have the same effects, so you can ask the client or the analyst about this situation and if it doesn’t matter to have a credit line or not in case of customer type B
3- Finally you will be sure that you cover all possible scenarios by doing this analysis.
Finally, In QLeaders we help you to analyze your requirement to ensure the maximum coverage of the requirement in the testing
QLeaders, Your Quality Gate!
Ahmed Ibrahim.
