Exam 1z0-908 topic 1 question 102 discussion

Actual exam question from Oracle's 1z0-908

Question #: 102
Topic #: 1

You have semi-synchronous replication configured and working with one slave. rpl_semi_sync_master_timeout has never been reached.
You find that the disk system on the master has failed and as a result, the data on the master is completely unrecoverable.
Which two statements are true? (Choose two.)

A. No committed transactions are lost.
B. The slave automatically identifies that the master is unreachable and performs any required actions so that applications can start using the slave as the new master.
C. As soon as the incident happens, application can read data from the slave and rely on it to return a full and current set of data.
D. Reads from the slave can return outdated data until the value of the rpi_semi_sync_master_timeout variable is reached.
E. Reads from the slave can return outdated data for some time, until it applies all transactions from its relay log.
F. A small amount of committed transactions may be lost in case they were committed just before the disk failure.

Show Suggested Answer

Suggested Answer: DF 🗳️

by Dhanushka at Aug. 2, 2023, 8:47 a.m.

Comments

Submit Cancel

EdyCv

2 months, 2 weeks ago

Selected Answer: AE

The source waits until at least one replica has received and logged the events (the required number of replicas is configurable), and then commits the transaction. The source does not wait for all replicas to acknowledge receipt, and it requires only an acknowledgement from the replicas, not that the events have been fully executed and committed on the replica side.

upvoted 1 times

...

FelipeK

11 months, 3 weeks ago

Selected Answer: AE

A - With semi-synchronous replication, once the source commits a transaction, it waits until at least one replica has received and logged the events before confirming the commit. Therefore, if the source crashes, all committed transactions have been transmitted to at least one replica. E - The slave applies transactions from its relay log, and until it catches up with all transactions, there might be a delay in reading the most up-to-date data from the slave. https://dev.mysql.com/doc/refman/8.0/en/replication-semisync.html

upvoted 1 times