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QUESTION NO:7

Which statement is true about TCN propagation?

A. The originator of the TCN immediately floods this information through the network.

B. The TCN propagation is a two step process.

C. A TCN is generated and sent to the root bridge.

D. The root bridge must flood this information throughout the network.

Answer: C

Explanation:

Explanation

New Topology Change Mechanisms

When an 802.1D bridge detects a topology change, it uses a reliable mechanism to first notify the

root bridge.

This is shown in this diagram:

Once the root bridge is aware of a change in the topology of the network, it sets the TC flag on the

BPDUs it sends out, which are then relayed to all the bridges in the network. When a bridge

receives a BPDU with the TC flag bit set, it reduces its bridging-table aging time to forward delay

seconds. This ensures a relatively quick flush of stale information. Refer to Understanding

Spanning-Tree Protocol Topology Changes for more information on this process. This topology

change mechanism is deeply remodeled in RSTP. Both the detection of a topology change and its

propagation through the network evolve.

Topology Change Detection

In RSTP, only non-edge ports that move to the forwarding state cause a topology change. This

means that a loss of connectivity is not considered as a topology change any more, contrary to

802.1D (that is, a port that moves to blocking no longer generates a TC). When a RSTP bridge

detects a topology change, these occur:

It starts the TC While timer with a value equal to twice the hello-time for all its non-edge

designated ports and its root port, if necessary.

It flushes the MAC addresses associated with all these ports.

Note: As long as the TC While timer runs on a port, the BPDUs sent out of that port have the TC

bit set.

BPDUs are also sent on the root port while the timer is active.

Topology Change Propagation

When a bridge receives a BPDU with the TC bit set from a neighbor, these occur:

It clears the MAC addresses learned on all its ports, except the one that receives the topology

change.

It starts the TC While timer and sends BPDUs with TC set on all its designated ports and root port

(RSTP no longer uses the specific TCN BPDU, unless a legacy bridge needs to be notified).

This way, the TCN floods very quickly across the whole network. The TC propagation is now a one

step process. In fact, the initiator of the topology change floods this information throughout the

network, as opposed to 802.1D where only the root did. This mechanism is much faster than the

802.1D equivalent. There is no need to wait for the root bridge to be notified and then maintain the

topology change state for the whole network for seconds.

In just a few seconds, or a small multiple of hello-times, most of the entries in the CAM tables of

the entire network (VLAN) flush. This approach results in potentially more temporary flooding, but

on the other hand it clears potential stale information that prevents rapid connectivity restitution.

Reference

http://www.cisco.com/en/US/tech/tk389/tk621/technologies_white_paper09186a0080094cfa.shtml

QUESTION NO:15

Which three options are considered in the spanning-tree decision process? (Choose three.)

A. lowest root bridge ID

B. lowest path cost to root bridge

C. lowest sender bridge ID

D. highest port ID

E. highest root bridge ID

F. highest path cost to root bridge

Answer: A,B,C

Explanation:

Configuration bridge protocol data units (BPDUs) are sent between switches for each port.

Switches use s four step process to save a copy of the best BPDU seen on every port. When a

port receives a better BPDU, it stops sending them. If the BPDUs stop arriving for 20 seconds

(default), it begins sending them again.

Step 1 Lowest Root Bridge ID (BID)

Step 2 Lowest Path Cost to Root Bridge

Step 3 Lowest Sender BID

Step 4 Lowest Port ID

Reference

Cisco General Networking Theory Quick Reference Sheets

QUESTION NO:26

Refer to the exhibit.

Why is AS 65333 in parentheses?

A. It is an external AS.

B. It is a confederation AS.

C. It is the AS of a route reflector.

D. It is our own AS.

E. A route map has been applied to this route.

F. The BGP next hop is unreachable.

Answer: B

Explanation:

QUESTION NO:34

In order to maintain security, with which hop count are IPv6 neighbor discovery packets sent?

A. 0

B. 1

C. 255

D. 256

Answer: C

Explanation:

QUESTION NO:29

What is the first thing that happens when IPv6 is enabled on an interface on a host?

A. A router solicitation is sent on that interface.

B. There is a duplicate address detection on the host interface.

C. The link local address is assigned on the host interface.

D. A neighbor redirect message is sent on the host interface.

Answer: B

Explanation:

Duplicate address detection (DAD) is used to verify that an IPv6 home address is unique on the

LAN before assigning the address to a physical interface (for example, QDIO). z/OS

Communications Server responds to other nodes doing DAD for IP addresses assigned to the

interface.

Reference

http://publib.boulder.ibm.com/infocenter/zos/v1r12/index.jsp?topic=/com.ibm.zos.r12.hale001

/ipv6d0021002145.htm

QUESTION NO:14

Which three options are features of VTP version 3? (Choose three.)

A. VTPv3 supports 8K VLANs.

B. VTPv3 supports private VLAN mapping.

C. VTPv3 allows for domain discovery.

D. VTPv3 uses a primary server concept to avoid configuration revision issues.

E. VTPv3 is not compatible with VTPv1 or VTPv2.

F. VTPv3 has a hidden password option.

Answer: B,D,F

Explanation:

Key Benefits of VTP Version 3

Much work has gone into improving the usability of VTP version 3 in three major areas:

The new version of VTP offers better administrative control over which device is allowed to update

other devices’ view of the VLAN topology. The chance of unintended and disruptive changes is

significantly reduced, and availability is increased. The reduced risk of unintended changes will

ease the change process and help speed deployment.

Functionality for the VLAN environment has been significantly expanded. Two enhancements are

most beneficial for today’s networks:

QUESTION NO:17

Which three combinations are valid LACP configurations that will set up a channel? (Choose

three.)

A. On/On

B. On/Auto

C. Passive/Active

D. Desirable/Auto

E. Active/Active

F. Desirable/Desirable

Answer: A,C,E

Explanation:

QUESTION NO:13

Which two statements are true about traffic shaping? (Choose two.)

A. Out-of-profile packets are queued.

B. It causes TCP retransmits.

C. Marking/remarking is not supported.

D. It does not respond to BECN and ForeSight Messages.

E. It uses a single/two-bucket mechanism for metering.

Answer: A,C

Explanation:

QUESTION NO:27

Refer to the exhibit.

What triggered the first SPF recalculation?

A. changes in a router LSA, subnet LSA, and external LSA

B. changes in a router LSA, summary network LSA, and external LSA

C. changes in a router LSA, summary network LSA, and summary ASBR LSA

D. changes in a router LSA, summary ASBR LSA, and external LSA

Answer: B

Explanation:

OSPFv2

Is built around links, and any IP prefix change in an area will trigger a full SPF. It advertises IP

information in Router and Network LSAs. The routers thus, advertise both the IP prefix information

(or the connected subnet information) and topology information in the same LSAs. This implies

that if an IP address attached to an interface changes, OSPF routers would have to originate a

Router LSA or a Network LSA, which btw also carries the topology information. This would trigger

a full SPF on all routers in that area, since the same LSAs are flooded to convey topological

change information. This can be an issue with an access router or the one sitting at the edge,

since many stub links can change regularly.

Only changes in interarea, external and NSSA routes result in partial SPF calculation (since type

3, 4, 5 and 7 LSAs only advertise IP prefix information) and thus IS-IS

QUESTION NO:22

Refer to the exhibit.

Which path is selected as best path?

A. path 1, because it is learned from IGP B.

path 1, because the metric is the lowest C.

path 2, because it is external

D. path 2, because it has the higher router ID

Answer: B

Explanation:

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