Oracle 1Z0-468 Exam (page: 1)
Oracle Cloud Application Foundation Certified Implementation Specialist
Updated on: 11-Nov-2025

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What three components are not part of EM and require separate installations? (Choose two.)

  1. BTM Engine
  2. RUEI Engine
  3. JVM Diagnostics Engine
  4. Application Discovery and Performance Management Engine
  5. Ops Center

Answer(s): D,E



What is the difference between Entry Processors and Aggregators?

  1. Entry Processors require a key or a set of keys, whereas Aggregators require filters.
  2. Entry Processors can skip the network by executing against near caches directly, whereas Aggregators must execute on storage members.
  3. Entry Processors are better suited to Map-Reduce operations than Aggregators.
  4. Aggregators are optimized for read-only access, whereas Entry Processors can modify cacheentries.

Answer(s): D



Which statement about JRockit Discovery Protocol (JDP) is true?

  1. JDP enables network auto discovery of JVMs.
  2. JDP is a subset of HTTP.
  3. JDP can be used by application to exchange serialized data between JVMs.
  4. JDP was initially introduced in the HotSpot VM 1.5.

Answer(s): A



How would you create an SLA based on the "JVM Runtime Up" metric for a specific WLS server?

  1. Deploy JVM Diagnostics and set a threshold on the JVMD agent up time.
  2. Create a Service target that uses the WLS Domain as its system, and the JVM Runtime as itsperformance metric. Configure SLA by using that metric.
  3. Define a threshold on the metric on the All Metrics page, and define a rule to create an Event for that incident. Define an SLA on the count of Incidents.
  4. You cannot create an SLA on JVM-related metrics.

Answer(s): C



Your application will store data in two partitioned caches, each configured for one backup. You have calculated the total amount of data to be 10 GB for the first cache, and 3 GB for the second cache.
You plan to run 10 cache servers in total.
What is the minimum recommended JVM heap size for each cache server?

  1. 512 GB
  2. 1 GB
  3. 2 GB
  4. 4 GB

Answer(s): C

Explanation:

When considering heap size, it is important to find the right balance. The lower bound is determined by per-JVM overhead (and also, manageability of a potentially large number of JVMs). For example, if there is a fixed overhead of 100MB for infrastructure software (for example, JMX agents, connection pools, internal JVM structures), then the use of JVMs with 256MB heap sizes results in close to 40% overhead for non-cache data. The upper bound on JVM heap size is governed by memory management overhead, specifically the maximum duration of GC pauses and the percentage of CPU allocated to GC (and other memory management tasks).



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