[2017-Mar-NEW] CCIE R&S Written New Cisco 400-101 Vce Valid Exam Certification Training Collection

Exam Code: 400-101 
Exam Name: CCIE Routing and Switching Written v5.0
Exam Number：400-101 CCIE
Associated Certifications：CCIE Routing and Switching
Duration：120 Minutes (90 – 110 questions)
Available Languages：English
Register：Pearson VUE
Exam Policies：Read current policies and requirements
Exam Tutorial：Review type of exam questions
Updated: Mar 03, 2017
Q&As: 1276
Exam Information：http://www.pass4itsure.com/400-101.html
Download Complete List of Topics in PDF format:

• 1.0 Network Principles 10%
• 2.0 Layer 2 Technologies 13%
• 3.0 Layer 3 Technologies 37%
• 4.0 VPN Technologies 13%
• 5.0 Infrastructure Security 5%
• 6.0 Infrastructure Services 12%

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2017 Cisco 400-101 VCE (#1-16) from Pass4itsure:

400-101 vce Question No : 7 – (Topic 1)  How many hash buckets does Cisco Express Forwarding use for load balancing?
A. 8
B. 16
C. 24
D. 32
Answer: B
Explanation: In order to understand how the load balance takes place, you must first see how the tables
relate. The Cisco Express Forwarding table points to 16 hash buckets (load share table),
which point to the adjacency table for parallel paths. Each packet to be switched is broken
up into the source and destination address pair and checked against the loadshare table.

400-101 vce  Question No : 8 – (Topic 1)  Which statement is true regarding the UDP checksum?
A. It is used for congestion control.
B. It cannot be all zeros.
C. It is used by some Internet worms to hide their propagation.
D. It is computed based on the IP pseudo-header.
Answer: D
Explanation: The method used to compute the checksum is defined in RFC 768: “Checksum is the 16-bit one’s complement of the one’s complement sum of a pseudo header of information from the IP header, the UDP header, and the data, padded with zero
octets at the end (if necessary) to make a multiple of two octets.” In other words, all 16-bit words are summed using one’s complement arithmetic. Add the 16-bit values up. Each time a carry-out (17th bit) is produced, swing that bit around and
add it back into the least significant bit. The sum is then one’s complemented to yield the value of the UDP checksum field.
If the checksum calculation results in the value zero (all 16 bits 0) it should be sent as the one’s complement (all 1s).

400-101 vce  Question No : 9 – (Topic 1)
Which two Cisco Express Forwarding tables are located in the data plane? (Choose two.)
A. the forwarding information base
B. the label forwarding information base
C. the IP routing table
D. the label information table
E. the adjacency table
Answer: A,B
Explanation:
The control plane runs protocols such as OSPF, BGP, STP, LDP. These protocols are needed so that routers and switches know how to forward packets and frames. The data plane is where the actual forwarding takes place. The data plane is populated based on the protocols running in the control plane. The Forwarding Information Base (FIB) is used for IP traffic and the Label FIB is used for MPLS.

400-101 vce  No : 10 – (Topic 1)
Which two mechanisms can be used to eliminate Cisco Express Forwarding polarization?
(Choose two.)
A. alternating cost links
B. the unique-ID/universal-ID algorithm
C. Cisco Express Forwarding antipolarization
D. different hashing inputs at each layer of the network
Answer: B,D
Explanation:
This document describes how Cisco Express Forwarding (CEF) polarization can cause suboptimal use of redundant paths to a destination network. CEF polarization is the effect when a hash algorithm chooses a particular path and the redundant paths remain completely unused.How to Avoid CEF Polarization Alternate between default (SIP and DIP) and full (SIP + DIP + Layer4 ports) hashing inputs configuration at each layer of the network. Alternate between an even and odd number of ECMP links at each layer of the network.The CEF load-balancing does not depend on how the protocol routes are Question

✑ ✑ ✑ ✑
inserted in the routing table. Therefore, the OSPF routes exhibit the same behavior
as EIGRP. In a hierarchical network where there are several routers that perform
load-sharing in a row, they all use same algorithm to load-share.
The hash algorithm load-balances this way by default:
1: 1
2: 7-8
3: 1-1-1
4: 1-1-1-2
5: 1-1-1-1-1
6: 1-2-2-2-2-2
7: 1-1-1-1-1-1-1
8: 1-1-1-2-2-2-2-2
The number before the colon represents the number of equal-cost paths. The number after
the colon represents the proportion of traffic which is forwarded per path.
This means that:
For two equal cost paths, load-sharing is 46.666%-53.333%, not 50%-50%.
For three equal cost paths, load-sharing is 33.33%-33.33%-33.33% (as expected).
For four equal cost paths, load-sharing is 20%-20%-20%-40% and not 25%-25%- 25%-25%. This illustrates that, when there is even number of ECMP links, the traffic is not load balanced. Cisco IOS introduced a concept called unique-ID/universal-ID which helps avoid CEF polarization. This algorithm, called the universal algorithm (the default in current Cisco IOS versions), adds a 32-bit router-specific value to the hash function (called the universal ID – this is a randomly generated value at the time of
the switch boot up that can can be manually controlled). This seeds the hash function on each router with a unique ID, which ensures that the same source/destination pair hash into a different value on different routers along the path. This process provides a better network-wide load-sharing and circumvents the polarization issue. This unique -ID concept does not work for an even number of equal-cost paths due to a hardware limitation, but it works perfectly for an odd number of equal-cost paths. In order to overcome this problem, Cisco IOS adds one link to the hardware adjacency table when there is an even number of equal
cost paths in order to make the system believe that there is an odd number of equal-cost links.

400-101 vce  Question No : 11 DRAG DROP – (Topic 1)  Drag and drop the argument of the mls ip cef load-sharing command on the left to the
function it performs on the right.

400-101 vce  Question No : 12 – (Topic 1)  Refer to the exhibit.

Which two are causes of output queue drops on FastEthernet0/0? (Choose two.)
A. an oversubscribed input service policy on FastEthernet0/0
B. a duplex mismatch on FastEthernet0/0
C. a bad cable connected to FastEthernet0/0
D. an oversubscribed output service policy on FastEthernet0/0
E. The router trying to send more than 100 Mb/s out of FastEthernet0/0
Answer: D,E
Explanation: Output drops are caused by a congested interface. For example, the traffic rate on the outgoing interface cannot accept all packets that should be sent out, or a service policy is applied that is oversubscribed. The ultimate solution to resolve the problem is to increase the line speed. However, there are ways to prevent, decrease, or control output drops when you do not want to increase the line speed. You can prevent output drops only if output drops are a consequence of short bursts of data. If output drops are caused by a constant high-rate flow, you cannot prevent the drops. However, you can control them.

400-101 vce  Question No : 13 – (Topic 1)  Which two mechanisms provide Cisco IOS XE Software with control plane and data plane
separation? (Choose two.)
A. Forwarding and Feature Manager
B. Forwarding Engine Driver
C. Forwarding Performance Management
D. Forwarding Information Base
Answer: A,B
Explanation:
Control Plane and Data Plane Separation
IOS XE introduces an opportunity to enable teams to now build drivers for new Data Plane
ASICs outside the IOS instance and have them program to a set of standard APIs which in
turn enforces Control Plane and Data Plane processing separation.
IOS XE accomplishes Control Plane / Data Plane separation through the introduction of the
Forwarding and Feature Manager (FFM) and its standard interface to the Forwarding
Engine Driver (FED). FFM provides a set of APIs to Control Plane processes. In turn, the
FFM programs the Data Plane via the FED and maintains forwarding state for the system.
The FED is the instantiation of the hardware driver for the Data Plane and is provided by
the platform.

400-101 vce  Question No : 14 – (Topic 1)  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.

400-101 vce  Question No : 15 – (Topic 1)  Which technology can create a filter for an embedded packet capture?
A. Control plane policing
B. Access lists
C. NBAR
D. Traffic shaping
Answer: B
Explanation:
A filter can be applied to limit the capture to desired traffic. Define an Access Control List (ACL) within config mode and apply the filter to the buffer: ip access-list extended BUF-FILTER
permit ip host 192.168.1.1 host 172.16.1.1
permit ip host 172.16.1.1 host 192.168.1.1

400-101 vce  Question No : 16 – (Topic 1) Which statement about MSS is true?
A. It is negotiated between sender and receiver.
B. It is sent in all TCP packets.
C. It is 20 bytes lower than MTU by default.
D. It is sent in SYN packets.
E. It is 28 bytes lower than MTU by default.
Answer: D
Explanation:
The maximum segment size (MSS) is a parameter of the Options field of the TCP header that specifies the largest amount of data, specified in octets, that a computer or communications device can receive in a single TCP segment. It does not count the TCP header or the IP header. The IP datagram containing a TCP segment may be self contained within a single packet, or it may be reconstructed from several fragmented pieces; either way, the MSS limit applies to the total amount of data contained in the final, reconstructed TCP segment. The default TCP Maximum Segment Size is 536. Where a host wishes to set the maximum
segment size to a value other than the default, the maximum segment size is specified as a TCP option, initially in the TCP SYN packet during the TCP handshake. The value cannot be changed after the connection is established.

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Official Sites:http://www.cisco.com/c/en/us/training-events/training-certifications/exams/current-list/ccie-routing-switching.html