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2021 Dec ccie written dumps 400-101:
Q201. DRAG DROP
Drag each IS-IS command on the left to its effect on the right.
Answer:
Q202. Which three options are components of an EEM CLI policy? (Choose three.)
A. Safe-Tcl
B. applet name
C. Fast Tcl
D. event
E. action
F. Tcl bytecode
Answer: B,D,E
Explanation:
The Embedded Event Manager (EEM) monitors events that occur on your device and takes action to recover or troubleshoot these events, based on your configuration.
EEM consists of three major components:
Event statements — Events to monitor from another Cisco NX-OS component that might require some action, workaround, or notification.
Action statements — An action that EEM can take, such as sending an e-mail, or disabling an interface, to recover from an event.
Policies — An applet name paired with one or more actions to troubleshoot or recover from the event.
Reference:
http://www.cisco.com/c/en/us/td/docs/switches/datacenter/nexus5500/sw/system_manage ment/6x/b_5500_System_Mgmt_Config_6x/b_5500_System_Mgmt_Config_6x_chapter_01 0011.html
Q203. Which type of port would have root guard enabled on it?
A. A root port
B. An alternate port
C. A blocked port
D. A designated port
Answer: D
Explanation:
The root guard feature provides a way to enforce the root bridge placement in the network. The root guard ensures that the port on which root guard is enabled is the designated port. Normally, root bridge ports are all designated ports, unless two or more ports of the root bridge are connected together. If the bridge receives superior STP Bridge Protocol Data Units (BPDUs) on a root guard-enabled port, root guard moves this port to a root-inconsistent STP state. This root-inconsistent state is effectively equal to a listening state. No traffic is forwarded across this port. In this way, the root guard enforces the position of the root bridge.
Reference: http://www.cisco.com/c/en/us/support/docs/lan-switching/spanning-tree-protocol/10588-74.html
Q204. Refer to the exhibit.
Which statement about the R1 configuration is true?
A. It supports the service timestamps log uptime command to display time stamps.
B. The logging buffer command was used to increase the default of the buffer.
C. The logging of warning messages is disabled.
D. Log message sequence numbering is disabled.
Answer: D
Explanation:
To disable sequence numbers, use the no service sequence-numbers global configuration command.
This example shows part of a logging display with sequence numbers enableD. 000019: %SYS-5-CONFIG_I: Configured from console by vty2 (10.34.195.36)
In this example we see the absence of sequence numbers on the log messages.
Not A. In this example there are no time stamps or uptimes shown in the logs.
Not B. The default buffer size is 4096 bytes.
Not C. The logging level in this example is informational (level 6), which will display levels 0-6 in the logs. Warnings are level 4.
Q205. Which three TLVs does LLDP use to discover network devices? (Choose three.)
A. Management address
B. Port description
C. Network policy
D. System name
E. Location information
F. Power management
Answer: A,B,D
Explanation:
Basic Management TLV Set
This set includes the following five TLVs used in LLDP:
. Port description TLV: Provides a description of the port in an alpha-numeric format. The value equals the ifDescr object, if the LAN device supports RFC 2863.
. System name TLV: Provides the system's assigned name in an alpha-numeric format. The value equals the sysName object, if the LAN device supports RFC 3418.
. System description TLV: Provides a description of the network entity in an alpha-numeric format. This includes system's name and versions of hardware, operating system and networking software supported in the device. The value equals the sysDescr object, if the LAN device supports RFC 3418.
. System capabilities TLV: Indicates the primary function(s) of the device and whether or not these functions are enabled in the device. The capabilities are indicated by two octects. Bits 0 through 7 indicate Other, Repeater, Bridge, WLAN AP, Router, Telephone, DOCSIS cable device and Station respectively. Bits 8 through 15 are reserved.
. Management address TLV: Indicates the addresses of the local LLDP agent. Other remote managers can use this address to obtain information related to the local device.
Reference: http://www.eetimes.com/document.asp?doc_id=1272069
Far out ccie dumps 400-101:
Q206. In a nonbackbone OSPF area, all traffic that is destined to the Internet is routed by using a default route that is originated by the ABR. Which change in the configuration of the OSPF area type causes traffic from that area that is destined to the Internet to be dropped?
A. The OSPF area changes from NSSA to totally stubby area.
B. The OSPF area changes from NSSA to regular area.
C. The OSPF area changes from stub area to totally stubby area.
D. The OSPF area changes from stub area to NSSA.
Answer: D
Explanation:
The ABR for the NSSA generates the default route, but not by default. To force the ABR to generate the default route, use the area <area id> nssa default-information originate command. The ABR generates a Type 7 LSA with the link-state ID 0.0.0.0 and is advertised inside the NSSA. This default route will be propagated inside the NSSA as Type 7 LSA
Reference: http://www.cisco.com/c/en/us/support/docs/ip/open-shortest-path-first-ospf/13692-21.html#nssas
Q207. Which two types of traffic are blocked when the storm control threshold for multicast traffic is reached on a port? (Choose two.)
A. BPDU
B. OSPF
C. CDP
D. IS-IS
E. LLDP
Answer: B,D
Q208. What is Nagle's algorithm used for?
A. To increase the latency
B. To calculate the best path in distance vector routing protocols
C. To calculate the best path in link state routing protocols
D. To resolve issues caused by poorly implemented TCP flow control.
Answer: D
Explanation:
Silly window syndrome is a problem in computer networking caused by poorly implemented TCP flow control. A serious problem can arise in the sliding window operation when the sending application program creates data slowly, the receiving application program consumes data slowly, or both. If a server with this problem is unable to process all incoming data, it requests that its clients reduce the amount of data they send at a time (the window setting on a TCP packet). If the server continues to be unable to process all incoming data, the window becomes smaller and smaller, sometimes to the point that the data transmitted is smaller than the packet header, making data transmission extremely inefficient. The name of this problem is due to the window size shrinking to a "silly" value. When there is no synchronization between the sender and receiver regarding capacity of the flow of data or the size of the packet, the window syndrome problem is created. When the silly window syndrome is created by the sender, Nagle's algorithm is used. Nagle's solution requires that the sender sends the first segment even if it is a small one, then that it waits until an ACK is received or a maximum sized segment (MSS) is accumulated.
Reference: http://en.wikipedia.org/wiki/Silly_window_syndrome
Q209. The OSPF database of a router shows LSA types 1, 2, 3 and 7 only. Which type of area is this router connected to?
A. backbone area
B. totally stubby area
C. stub area
D. not-so-stubby area
Answer: D
Q210. Which two statements about the C-bit and PW type are true? (Choose two.)
A. The C-bit is 1 byte and the PW type is 15 bytes.
B. The PW type indicates the type of pseudowire.
C. The C-bit is 3 bits and the PW type is 10 bits.
D. The C-bit set to 1 indicates a control word is present.
E. The PW type indicates the encryption type.
Answer: B,D
Explanation:
The control word carries generic and Layer 2 payload-specific information. If the C-bit is set to 1, the advertising PE expects the control word to be present in every pseudowire packet on the pseudowire that is being signaled. If the C-bit is set to 0, no control word is expected to be present. Pseudowire Type—PW Type is a 15-bit field that represents the type of pseudowire.
Reference: http://www.ciscopress.com/articles/article.asp?p=386788&seqNum=2