Cloudera Certification Practice Test

Cloudera Certification Practice Test Exam with answers

1. You have decided to develop a custom composite key class that you will use for keys emitted during the map phase to be consumed by a reducer. Which interface must this key implement?
Correct Answer: WritableComparable

Explanation

All keys and values must implement the Writable interface so that they can be wriiten to disk. In addition, keys emitted during the map phase must implement WritableComparable so that the keys can be sorted during the shuffle and sort phase. WritableComparable inherits from Writable.

Further Reading

• For more information, see the Yahoo! Developer Network Apach Hadoop Tutorial, Custom Key Types.
• Hadoop: the Definitive Guide, chapter four, in the Serialization: The Writable Interface section.

2. You’ve built a MapReduce job that denormalizes a very large table, resulting in an extremely large amount of output data. Which two cluster resources will your job stress? (Choose two).
Correct Answer: network I/O , disk I/O

Explanation

When denormalizing a table, the amount of data written by the map phase will far exceed the amount of data read by the map phase. All of the data written during the map phase is first written to local disk and then transferred over the network to the reducers during the beginning of the reduce phase. Writing a very large amount of data in the map phase will therefore create a large amount of local disk I/O on the machines running map tasks and network I/O. Because map output is stored in a fixed size buffer that is written periodically to disk, this operation will not tax the memory of the machines running the map tasks. Denormalizing a table is not a compute-intesive operation, so this operation will not tax the processors of the machines running the map tasks.

Further Reading

• Hadoop: the Definitive Guide, chapter six, Shuffle and Sort: The Map Side section includes more information on the process for writing map output to disk.
• Hadoop: the Definitive Guide, chapter six, Shuffle and Sort: The Reduce Side section explains further how data is transferred to the reducers
• Denormalizing a table is a form of join operation. You can read more about performing joins in MapReduce in Join Algorithms using Map/Reduce

3. When is the earliest that the reduce() method of any reduce task in a given job called?
Correct Answer: Not until all map tasks have completed

Explanation

No reduce task&rquo;s reduce() method is called until all map tasks have completed. Every reduce task&rquo;s reduce() method expects to receive its data in sorted order. If the reduce() method is called before all of the map tasks have completed, it would be possible that the reduce() method would receive the data out of order.

Further Reading

• Hadoop: The Definitive Guide, chapter six includes a detailed explanation of the shuffle and sort phase of a MapReduce job.

4. You have 50 files in the directory /user/foo/example. Each file is 300MB. You submit a MapReduce job with /user/foo/example as the input path.

How much data does a single Mapper processes as this job executes?

Correct Answer: A single input split

Explanation

An input split is a unit of work (a chunk of data) that is processed by a single map task in a MapReduce program (represented by the Java interface InputSplit). The InputFormat you specify for MapReduce program determines how input files are split into records and read. Each map task processes a single split; each split is further comprised of records (key-value pairs) which the map task processes.

A MapReduce program run over a data set is usually called a MapReduce “job.” By splitting up input files, MapReduce can process a single file in parallel; if the file is very large, this represents a significant performance improvement. Also, because a single file is worked on in splits, it allows MapReduce to schedule those processes on different nodes in the cluster, nodes that have that piece of data already locally stored on that node, which also results in significant performance improvements. An InputSplit can span HDFS block boundaries.

Further Reading

• Hadoop: The Definitive Guide, chapter two includes an excellent general discussion of input splits

Hadoop Administrator

5. In the execution of a MapReduce job, where does the Mapper place the intermediate data of each Map task?

Correct Answer: The Mapper stores the intermediate data on the underlying filesystem of the local disk of the machine which ran the Map task

Explanation

Intermediate data from a Mapper is stored on the local filesystem of the machine on which the Mapper ran. Each Reducer then copies its portion of that intermediate data to its own local disk. When the job terminates, all intermediate data is removed.

• See Hadoop: The Definitive Guide, 3rd Edition, Chapter 2, under the section "Data Flow"

6. A client wishes to read a file from HDFS. To do that, it needs the block IDs (their names) which make up the file. From where does it obtain those block IDs?
Correct Answer: The NameNode reads the block IDs from memory and returns them to the client.

Explanation

When a client wishes to read a file from HDFS, it contacts the NameNode and requests the names and locations of the blocks which make up the file. For rapid access, the NameNode has the block IDs stored in RAM.

Further Reading

See Hadoop Operations, under the section "The Read Path."

7. Your cluster has slave nodes in three different racks, and you have written a rack topology script identifying each machine as being in rack1, rack2, or rack3. A client machine outside of the cluster writes a small (one-block) file to HDFS. The first replica of the block is written to a node on rack2. How is block placement determined for the other two replicas?
Correct Answer: Either both will be written to nodes on rack1, or both will be written to nodes on rack3.

Explanation

For the default threefold replication, Hadoop’s rack placement policy is to write the first copy of a block on a node in one rack, then the other two copies on two nodes in a different rack. Since the first copy is written to rack2, the other two will either be written to two nodes on rack1, or two nodes on rack3.

Further Reading

• For more information on rack topology and block placement policy, see Hadoop Operations: A Guide for Developers and Administrators page 130.
• See the Apache Docs on Data Replication

8. A slave node in your cluster has 24GB of RAM, and 12 physical processor cores on hyperthreading-enabled processors. You set the value of mapred.child.java.opts to -Xmx1G, and the value of mapred.tasktracker.map.tasks.maximum to 12. What is the most appropriate value to set for mapred.tasktracker.reduce.tasks.maximum?
Correct Answer: 6

Explanation

For optimal performance, it is important to avoid the use of virtual memory (swapping) on slave nodes. From the configuration shown, the slave node could run 12 Map tasks simultaneously, and each one will use 1GB of RAM, resulting in a total of 12GB used. The TaskTracker daemon itself will use 1GB of RAM, as will the DataNode daemon. This is a total of 14GB. The operating system will also use some RAM -- a reasonable estimate would be 1-2GB. Thus, we can expect to have approximately 8-9GB of RAM available for Reducers. So the most appropriate of the choices presented is that we should configure the node to be able to run 6 Reducers simultaneously.

Further Reading

Hadoop: The Definitive Guide, 3rd Edition, Chapter 9, under the section “Environment Settings”
HBase

9. Your client application needs to scan a region for the row key value 104. Given a store file that contains the following list of Row Key values:

100, 101, 102, 103, 104, 105, 106, 107 What would a bloom filter return?
Correct Answer: Confirmation that 104 may be contained in the set

Explanation

A Bloom filter is a kind of membership test using probability -- it tells you whether an element is a member of a set. It is quick and memory-efficient. The trade-off is that it is probabilistic where false positives are possible though false negatives are not; thus if your Bloom Filter returns true, it confirms that a key may be contained in a table. If Bloom Filter returns false, it confirms that a key is definitely not contained in a table.

Enabling Bloom Filters may save your disk seek and improve read latency.

Further Reading

• Wikipedia on Bloom Filters
• HBase Documentation on Bloom Filters section 12.6.4. Bloom Filters includes:

Bloom Filters can be enabled per ColumnFamily. Use HColumnDescriptor.setBloomFilterType(NONE | ROW | ROWCOL) to enable blooms per ColumnFamily. Default = NONE for no bloom filters. If ROW, the hash of the row will be added to the bloom on each insert. If ROWCOL, the hash of the row + column family + column family qualifier will be added to the bloom on each key insert.

10. You have two tables in an existing RDBMS. One contains information about the products you sell (name, size, color, etc.) The other contains images of the products in JPEG format. These tables are frequently joined in queries to your database. You would like to move this data into HBase. What is the most efficient schema design for this scenario?
Correct Answer: Create a single table with two column families

Explanation

Access patterns are an important factor in HBase schema design. Even though the two tables in this scenario have very different data sizes and formats, it is better to store them in one table if you are accessing them together most of the time.

Column families allow for separation of data. You can store different types of data and format into different column families. Attributes such as compression, Bloom filters, and replication are set on per column family basis. In this example, it is better to store product information and product images into two different column families and one table.

Further Reading

HBase Documentation on Column Family Section 5.6.especially the part:

Physically, all column family members are stored together on the filesystem. Because tunings and storage specifications are done at the column family level, it is advised that all column family members have the same general access pattern and size characteristics.

11. You need to create a Blogs table in HBase. The table will consist of a single Column Family calledContent and two column qualifiers, Author and Comment. What HBase shell command should you use to create this table?
Correct Answer: create 'Blogs', 'Content'

Explanation

When you create a HBase table, you need to specify table name and column family name.
In the Hbase Shell, you can create a table with the command:
create 'myTable', 'myColumnFamily'

For this example:

• Table name: 'Blogs'
• ColumnFamily: 'Content'

We can create the table and verify it with describe table command.
hbase> create 'Blogs', 'Content' hbase> describe 'Blogs' {Name => 'Blogs', FAMILIES => [{NAME => 'Content', ....

Further Reading

see the HBase Shell commands for create
​
Create table; pass table name, a dictionary of specifications per column family, and optionally a dictionary of table configuration. Dictionaries are described below in the GENERAL NOTES section.
Examples:
hbase> create 't1', {NAME => 'f1', VERSIONS => 5} hbase> create 't1', {NAME => 'f1'}, {NAME => 'f2'}, {NAME => 'f3'} hbase> # The above in shorthand would be the following: hbase> create 't1', 'f1', 'f2', 'f3' hbase> create 't1', {NAME => 'f1', VERSIONS => 1, TTL => 2592000, BLOCKCACHE => true}

12. From within an HBase application you are performing a check and put operation with the following code:table.checkAndPut(Bytes.toBytes("rowkey"),Bytes.toBytes("colfam"),Bytes.toBytes("qualifier"), Bytes.toBytes("barvalue"), newrow));

Which describes this check and put operation?

Correct Answer: Check if rowkey/colfam/qualifier exists and has the cell value "barvalue". If so, put the values in newrow and return "true".

Explanation

The method checkAndPut returns "true" if a row with specific column family, and column qualifier value matches the expected value; if "true" is returned, it executes the put with new value. If the specific value is not present in a row, it returns "false" and the put is not executed.

Further Reading

HBase Documentation on checkAndPut:

Data Science Essentials (DS-200)

13. What is the best way to evaluate the quality of the model found by an unsupervised algorithm likek-means clustering, given metrics for the cost of the clustering (how well it fits the data) and its stability (how similar the clusters are across multiple runs over the same data)?
Correct Answer: The lowest cost clustering subject to a stability constraint

Explanation

There is a tradeoff between cost and stability in unsupervised learning. The more tightly you fit the data, the less stable the model will be, and vice versa. The idea is to find a good balance with more weight given to the cost. Typically a good approach is to set a stability threshold and select the model that achieves the lowest cost above the stability threshold.

• Yale Algorithms in Data Mining - Lecture 10: k-means clustering
• Evaluation of Stability of k-means Cluster Ensembles with Respect to Random Initialization

14. A sandwich shop studies the number of men, and women, that enter the shop during the lunch hour from noon to 1pm each day. They find that the number of men that enter can be modeled as a random variable with distribution Poisson(M), and likewise the number of women that enter asPoisson(W). What is likely to be the best model of the total number of customers that enter during the lunch hour?
Correct Answer: Poisson(M+W)

Explanation

The total number of customers is the sum of the number of men and women. The sum of two Poisson random variables also follows a Poisson distribution with rate equal to the sum of their rates. The Normal and Binomial distribution can approximate the Poisson distribution in certain cases, but the expressions above do not approximate Poisson(M+W).

15. Which condition supports the choice of a support vector machine over logistic regression for a classification problem?
Correct Answer: The test set will be dense, and contain examples close to decision boundary learned from the training set

Explanation

The SVM algorithm is a maximum margin classifier, and tries to pick a decision boundary that creates the widest margin between classes, rather than just any boundary that separates the classes. This helps generalization to test data, since it is less likely to misclassify points near the decision boundary, as the boundary maintains a large margin from training examples.

SVMs are not particularly better at multi-label clasification. Linear separability is not required for either classification technique, and does not relate directly to an advantage of SVMs. SVMs are not particularly more suited to low dimensional data.