 
Junior Member
Equivalence Partitioning Question:
A machine is designed to accept scores from two indpendent markers. Each script should have 4 marks, each out of 25, and a total for the script. The two markers scores are compared, differences greater than 3 or a total difference of 10 overall is flagged for further scrutiny.
Define a set of partitions for 100% coverage of partitions?
Does anyone have any idea what the answer to this would be? 
Junior Member
Re: Equivalence Partitioning Yes, I have an idea. But it looks like homework so I'm not going to give you the answer.
Why don't you tell us what you think the answer is and we can give you pointers if you get it wrong? 
Junior Member
Re: Equivalence Partitioning Haha ... It is not actually homework it is just a question from a past exam paper, I am doing revision. I understand how to do this when there is a single scenario  for example a function that takes in months, the values are 112, so there are 3 partitions:
Valid  >=1 and <=12
Invalid <=0 and >=13
What is confusing me with this one is the fact that there are 2 markers and then the difference of 3?
Thanks, 
Member
Re: Equivalence Partitioning You can have a difference of 3 if the first has 0 marker and second has 3 marks
so in this case the valid conditions should be
>= 3 and <=16.
It also test for total difference on 10. 
Junior Member
Re: Equivalence Partitioning OK, well you understand EP. You just need to apply the same logic to the differences.
The two markers are giving each section a score which also creates a total score. So the marks for one test might be:
Marker 1:
§1 25
§2 21
§3 13
§4 20
Total 79
Marker 2:
§1 28
§2 22
§3 15
§4 25
Total 90
Here §4 has a difference of 5 and total has a difference of 11 which would be flagged for scrutiny.
So you just need to write partitions that would cover the different outcomes in differences. You would need two partitions for both the section and total scores. 
Junior Member
Re: Equivalence Partitioning [ QUOTE ]
You can have a difference of 3 if the first has 0 marker and second has 3 marks
so in this case the valid conditions should be
>= 3 and <=16.
It also test for total difference on 10.
[/ QUOTE ]
Well, no because the first mark might be 4 and the second might be 12. There's a difference of 8 which should be flagged for scrutiny but they will still fall between 3 and 16 so your EP would see that as valid.
I think the problem looks difficult because it's about comparing values we don't know in advance. So instead we should be doing EP on the difference instead. For example "if the difference is more than 3"...
Hope this helps.
By the way, which cert is this for? 
Junior Member
Re: Equivalence Partitioning Thank for very helpful!!, this is for an Information Technology Degree. 
Junior Member
Re: Equivalence Partitioning let me give a try:
valid:
case 1: at least 1 dif >3 and Ttl dif >10 flagged
case 2: at least 1 dif >3 and Ttl dif <=10 flagged
case 3: all dif <= 3 and Ttl dif >10 flagged
invalid:
case 4: all dif <=3 and Ttl dif <=10 not flagged 
Junior Member
Re: Equivalence Partitioning Looks spot on, elyhe. I was thinking of testing the total and section differences separately but that wouldn't be a true test of 100% of the outcomes. 
Member
Re: Equivalence Partitioning [ QUOTE ]
example a function that takes in months, the values are 112, so there are 3 partitions:
Valid  >=1 and <=12
Invalid <=0 and >=13
[/ QUOTE ]
Sorry, but your solution in this example is too simplistic esp. since there is no context!!!
You neglected to use the 2 of the 4 heuristics for identifying equivalent partitions. Unique values would be February and Groups would be 30 day months and 31 day months.
Broadly lumping ranges of values without understanding how values within a range could be handled differently is a recipe for disaster.
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