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NEW QUESTION: 1

A. Option B
B. Option D
C. Option C
D. Option A
Answer: B
Explanation:
Structure and Organization of FC Data In an FC network, data transport is analogous to a conversation between two people, whereby a frame represents a word, a sequence represents a sentence, and an exchange represents a conversation.

Exchange: An exchange operation enables two node ports to identify and manage a set of information units. Each upper layer protocol has its protocol-specific information that must be sent to another port to perform certain operations. This protocol-specific information is called an information unit. The structure of these information units is defined in the FC-4 layer. This unit maps to a sequence. An exchange is composed of one or more sequences. Sequence: A sequence refers to a contiguous set of frames that are sent from one port to another. A sequence corresponds to an information unit, as defined by the ULP.
Frame: A frame is the fundamental unit of data transfer at Layer 2. An FC frame consists of five parts: start of frame (SOF), frame header, data field, cyclic redundancy check (CRC), and end of frame (EOF). The SOF and EOF act as delimiters. The frame header is 24 bytes long and contains addressing information for the frame. The data field in an FC frame contains the data payload, up to 2,112 bytes of actual data-in most cases, the SCSI data. The CRC checksum facilitates error detection for the content of the frame. This checksum verifies data integrity by checking whether the content of the frames was received correctly. The CRC checksum is calculated by the sender before encoding at the FC-1 layer. Similarly, it is calculated by the receiver after decoding at the FC-1 layer. EMC E10-001 Student Resource Guide. Module 5: Fibre Channel Storage Area Network (FC SAN)

NEW QUESTION: 2
An administrator enables DRS and sets the automation level to Fully Automated. DRS is only making initial placement recommendations.
What is causing this?
A. HA is disabled on the cluster
B. DRS is set to aggressive
C. vMotion network is misconfigured
D. The VMs have virtual flash reservations
Answer: C
Explanation:
Explanation
https://geek-university.com/vmware-esxi/vsphere-distributed-resource-scheduler-drs-requirements/

NEW QUESTION: 3
You are using the performance Schema to investigate replication on a slave:
Mysql> SELECT THREAD_ID threads.NAME, SUM (COUNT_STAR) AS Totalcount, SUM (SUM_TIMER_WAIT) AS Totaltime
FROM performance_schema.events_waits_summary_by_thread_by_event_name
INNER JOIN performance_schema_threads USING (THREAD_ID)
WHERE threads .NAME LIKE 'thread/sql/slave\-%'
GROUP BY THREAD_ID, threads.NAME;

Assume that all instruments and consumers are enabled and all threads are instrumented.
Which two facts can be concluded from the given output?
A. At most two schemas are being updated concurrently.
B. The slave cannot process the relay log fast enough to use all threads.
C. The slave is configured with slave_parallel_workers = 4.
D. The server needs more cores to use all slave threads.
Answer: A,B
Explanation:
* To see which instruments have been executed the most times or have taken the most wait time, sort the events_waits_summary_global_by_event_name table on the COUNT_STAR or SUM_TIMER_WAIT column, which correspond to a COUNT(*) or SUM(TIMER_WAIT) value, respectively, calculated over all events
* slave_parallel_workers
Sets the number of slave worker threads for executing replication events (transactions) in parallel.
Setting this variable to 0 (the default) disables parallel execution. The maximum is 1024.
Reference: 22.1 Performance Schema Quick Start ; 17.1.4.3 Replication Slave Options and Variables

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