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Exam Code: C2150-057Exam Name: AppScan Source Edition
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Exam Code: C2180-183Exam Name: IBM WebSphere Mesage Broker V7.0, Solution Development
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Exam Code: C2020-703Exam Name: IBM Cognos TM1 10.1 Administrator
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NO.1 Which Message Broker artifact type is not analyzed by Impact Analysis?
A..esql
B..mset
C..msgmap
D..msgflow
Answer: B
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NO.2 Consider the following ESQL code snippet from a Compute node.
What is the value of OutputRoot.XMLNSC?
A.NULL
B.<Winner/>
C.<Winner Name="Robert"/>
D.<Winner><Name>Rocky</Name></Winner>
Answer: D
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NO.3 When a message is received by an Input node in a message flow, the message assembly is
created.Which tree or trees are populated when an error free message is received by a FileInput node?
A.The Message tree only.
B.The Message tree and Environment tree.
C.The Message tree and LocalEnvironment tree.
D.The Message tree, Environment tree, LocalEnvironment tree, and ExceptionList tree.
Answer: C
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NO.4 In a message flow, a developer needs to enrich an employee's record with information from a
database.Consider this input message.
Which of the ESQL statements in the response options will produce the following output message?
A.SET OutputRoot.XMLNSC.Employee.Name[] = (SELECT ITEM E.EmpName from Database.Employee
as E where E.EmpNumber = InputBody.Employee.Number);
B.SET OutputRoot.XMLNSC.Employee.Name[] = (SELECT E.EmpName AS Name from
Database.Employee as E where E.EmpNumber = InputBody.Employee.Number);
C.SET OutputRoot.XMLNSC.Employee = THE(SELECT E.EmpName from Database.Employee as E
where E.EmpNumber = InputBody.Employee.Number);
D.SET OutputRoot.XMLNSC.Employee = THE(SELECT ITEM E.EmpName AS Name from
Database.Employee as E where E.EmpNumber = InputBody.Employee.Number);
Answer: A
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NO.5 A developer has a message flow that consists of an MQInput node, Compute node and MQOutput
node.The message is parsed by the XMLNSC domain.The Compute node will extract the name and age
of each player, then send them onto the register players application to have the team registered.The input
message that the Compute node will be processing is shown below.
What MUST be done to access the information from the first player?
A.Set OutputRoot.XMLNSC.RegPlayer.Player[NextCount] = InputBody.Body.Player[1]
B.Set OutputRoot.XMLNSC.RegPlayer.Player[NextCount] = InputBody.DetailedMsg.Body.Player[1]
C.Set OutputRoot.XMLNSC.RegPlayer.Player[NextCount] = InputRoot.DetailedMsg.Body.Player[0]C.Set
OutputRoot.XMLNSC.RegPlayer.Player[NextCount] = InputRoot.DetailedMsg.Body.Player[0]
D.Set OutputRoot.XMLNSC.RegPlayer.Player[NextCount] = InputRoot.DetailedMsg.Body.Player[1]D.Set
OutputRoot.XMLNSC.RegPlayer.Player[NextCount] = InputRoot.DetailedMsg.Body.Player[1]
Answer: B
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NO.6 A developer is tasked with designing a message flow, utilizing an XSLTransform node.Which location is
searched first by the broker, to determine the name of the style sheet to use?
A.The LocalEnvironment.
B.The BAR file overrides.
C.The XSLTransform node properties.
D.Inside the message XML data.
Answer: D
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NO.7 Which one of the following advantages does the Enterprise Service Bus deliver over traditional
point-to-point solutions?
A.Scalability
B.Adaptability
C.Redundancy
D.Performance
Answer: B
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NO.8 View the Exhibit:
The above flow processes a large incoming file line by line once every four hours.Performance is
unsatisfactory even though the server has spare memory and unused processors.Configuring additional
instances does not improve this.The following pair of flows was suggested by an architect.The first flow
will slice the incoming file into messages without any further processing, which is to be done by the
second flow.
What is the expected outcome if the second flow is configured with a large number of additional
instances?
A.Throughput will improve somewhat.Message sequence will not be affected.
B.Throughput will be unaffecte
D.Message sequence will most likely be affected.
C.Throughput will improve significantly.Message sequence will most likely be affected.
D.Throughput will deteriorate slightly due to the extra queuing.Message sequence will not be affected.
Answer: C
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