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IBM Assessment: InfoSphere DataStage v8.5 Sample Questions:

1. In the exhibit, the Aggregator stage aggregates over CustomerID. The Join stage joins by CustomerID. When the score is created, DataStage will insert hash partitioners and tsort operators to ensure the correct results.

Which two choices will eliminate sort operators from being inserted at run-time? (Select two)

A) Add a Sort stage before the Copy stage and specify Auto partitioned for the links going into the Aggregator and Join stages.
B) Add a Sort stage before the Copy stage and specify Same partitioners for the links going into the Aggregator and Join stages.
C) Add Sort stages in front of the Join stage and specify the Same partitioner for the links going into the Join. Set the $APT_NO_SORT_INSERTION environment variable to "True".
D) Add a Sort stage before the Aggregator stage and specify the Same partitioner for the link going into the Aggregator Set the $APT_SORTJNSERTION_CHECK_ONLY environment variable to "True".

2. You have a dataset hash-partitioned and sorted on CustID. You need to create a Sequential File sorted on CustID from this dataset. Which technique is the most appropriate?

A) Use SortMerge collector for the Sequential File input link.
B) Insert Sort stage before the final Sequential File stage and use Ordered collector for the Sequential File stage input link.
C) Use Copy stage.
D) Insert Sort stage before the final Sequential File stage and use Round Robin collector for the Sequential File stage input link.

3. Which job design technique can be used to give unique names to sequential output files that are used in multi-instance jobs?

A) Generate unique file names by using a macro.
B) Use DSJobInvocationID to generate a unique filename.
C) Use a Transformer stage variable to generate the name.
D) Use parameters to identify file names.

4. Which condition is required to use the LastRowInGroup() function to determine a key break on a field?

A) The input data has to be sorted by the key field and DataStage must be able to detect that the data is sorted, for example, by the presence of a Sort stage.
B) The LastRowInGroup() function requires the data to be grouped by the key field, but the data does not have to be sorted.
C) The LastRowInGroup() function does not require the data to be either sorted or grouped by the key field. If necessary the LastRowInGroup() function will perform a sort.
D) The input data has to be sorted by the key field, but it is not required that DataStage be able to detect this. For example, the input data might be sorted before it is input to the job. (However, the job will abort at runtime if the data isn't sorted.)

5. Which Oracle data type conversion is correct?

A) Oracle data type RAW converts to RAW in Oracle Enterprise stage.
B) Oracle data type NUMBER(15, 0) converts to INT32 in Oracle Enterprise stage.
C) Oracle data type NUMBER converts to DECIMAL(38, 0) in Oracle Enterprise stage.
D) Oracle data type NUMBER(6, 0) converts to INT32 in Oracle Enterprise.

Solutions: