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NEW QUESTION: 1
A technician is instructed to configure a new workstation to boot the OS from the network. The technician must enable which of the following FIRST?
A. PXE
B. UPNP
C. WOL
D. WiFi
Answer: A

NEW QUESTION: 2
Which tool can you use to create a Server Intelligence Agent?
A. Central Management Console
B. Upgrade Management Tool
C. Lifecycle Management Console
D. Central Configuration Manager
Answer: D

NEW QUESTION: 3
You are the purchasing manager for a company. You enter into a consignment agreement with a vendor.
Raw material R0001 is managed under the consignment agreement with the vendor.
You need to replenish raw material R0001 for consumption and update ownership.
Which four actions should you perform in sequence? To answer, move the appropriate actions from the list of actions to the answer area and arrange them in the correct order.

Answer:
Explanation:

NEW QUESTION: 4
Which two statements are true regarding VMware vSphere Flash Read Cache (vFRC)? (Choose two.)
A. Cache fills and cache evictions happen in the granularity of the disk block size.
B. vFRC caches data from both read and write I/Os, but write I/Os are always serviced by the underlying storage.
C. Cache fills and cache evictions happen in the granularity of a cache block size.
D. vFRC caches data from both read and write I/Os, but write I/Os are always serviced by the underlying cache data.
Answer: B,C
Explanation:
Explanation/Reference:
Explanation:
Cache fills and cache evictions happen in the granularity of a cache block size. This value ranges from
4KB to 1MB to enable you to best configure your cache block size based on the I/O size of workloads.
Even though cache fills and cache evictions happen in the granularity of a cache block size, actual read I/O serviced by the cache can be smaller than the cache block size. For example, if the cache block size is
64KB, and a 4KB read I/O request is issued by the guest virtual machine, and if the data is not available in the cache, a 4KB read is issued to the VMDK. When populating the cache, the vFRC algorithm looks for a
64KB region to place the new 4KB data. If no free space is available, a 64KB region is evicted and the space is used to hold the new 4KB data. The remaining 60KB region in the 64KB cache block is marked as invalid. The cache block size parameter therefore has profound effects on performance Reference: https://www.vmware.com/files/pdf/techpaper/vfrc-perf-vsphere55.pdf