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NEW QUESTION: 1
A tagged frame arrives on a switch port that is not a tagged member of the VLAN indicated by the frame ID. What do fundamental VLAN operating rules require the switch to do with that frame?
A. Attempt best effort delivery by flooding a copy of the frame to all other VLANs.
B. Ignore the VLAN tag and forward the frame, based on destination MAC address.
C. Forwarda copy of the frame to the default gateway.
D. Drop the frame.
Answer: D
NEW QUESTION: 2
Which two statements about SAN zoning on Cisco Nexus switches are true? (Choose two.)
A. A zone can only be a member one zone set.
B. Zoning must be administered from the primary SAN switch in the fabric.
C. Only one zone set can be activated at any time.
D. Zone configuration changes are nondisruptive.
E. Devices can only belong to one zone.
F. Zoning is enforced by examining the destination ID field.
Answer: C,D
Explanation:
Explanation
A zone set can be activated or deactivated as a single entity across all switches in the fabric. Only one zone set can be activated at any time. If zoning is not activated, all devices are members of the default zone. If zoning is activated, any device that is not in an active zone (a zone that is part of an active zone set) is a member of the default zone. Zoning can be administered from any switch in the fabric. When you activate a zone (from any switch), all switches in the fabric receive the active zone set. Additionally, full zone sets are distributed to all switches in the fabric, if this feature is enabled in the source switch.
Reference: http://www.cisco.com/en/US/docs/storage/san_switches/mds
9000/sw/san-os/quick/guide/qcg_zones.html
NEW QUESTION: 3
A. 10.10.1.0/25 subnetted with mask 255.255.255.252
B. 10.10.0.0/18 subnetted with mask 255.255.255.252
C. 10.10.0.0/16 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.1.0/24 subnetted with mask 255.255.255.252
Answer: D
Explanation:
Explanation
We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7
bits (because 2