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100-105 PDF DEMO:

QUESTION NO: 1
Under which circumstance should a network administrator implement one-way NAT?
A. when traffic that originates outside the network must be routed to internal hosts
B. when the network has few public IP addresses and many private IP addresses require outside access
C. when the network must route UDP traffic
D. when traffic that originates inside the network must be routed to internal hosts
Answer: A
Explanation:
NAT operation is typically transparent to both the internal and external hosts. Typically the internal host is aware of the true IP address and TCP or UDP port of the external host. Typically the NAT device may function as the default gateway for the internal host. However the external host is only aware of the public IP address for the NAT device and the particular port being used to communicate on behalf of a specific internal host.
NAT and TCP/UDP
"Pure NAT", operating on IP alone, may or may not correctly parse protocols that are totally concerned with IP information, such as ICMP, depending on whether the payload is interpreted by a host on the "inside" or "outside" of translation. As soon as the protocol stack is traversed, even with such basic protocols as TCP and UDP, the protocols will break unless NAT takes action beyond the network layer. IP packets have a checksum in each packet header, which provides error detection only for the header. IP datagrams may become fragmented and it is necessary for a NAT to reassemble these fragments to allow correct recalculation of higher-level checksums and correct tracking of which packets belong to which connection. The major transport layer protocols, TCP and
UDP, have a checksum that covers all the data they carry, as well as the TCP/UDP header, plus a
"pseudo-header" that contains the source and destination IP addresses of the packet carrying the
TCP/UDP header. For an originating NAT to pass TCP or UDP successfully, it must recompute the
TCP/UDP header checksum based on the translated IP addresses, not the original ones, and put that checksum into the TCP/UDP header of the first packet of the fragmented set of packets. The receiving
NAT must recompute the IP checksum on every packet it passes to the destination host, and also recognize and recompute the TCP/UDP header using the retranslated addresses and pseudo-header.
This is not a completely solved problem. One solution is for the receiving NAT to reassemble the entire segment and then recompute a checksum calculated across all packets.
The originating host may perform Maximum transmission unit (MTU) path discovery to determine the packet size that can be transmitted without fragmentation, and then set the don't fragment (DF) bit in the appropriate packet header field. Of course, this is only a one- way solution, because the responding host can send packets of any size, which may be fragmented before reaching the NAT.

QUESTION NO: 2
Drag and Drop Question
Drag and drop the descriptions of routing protocols from the left onto the correct routing types on the right.
Answer:

QUESTION NO: 3
Which IPv6 address type is a public address?
A. Multicast
B. Link local
C. Unique-local
D. Global unicast
Answer: D

QUESTION NO: 4
Drag and Drop Question
Drag and drop the PDUs from the left onto the correct TCP/IP layers on the right.
Answer:
Explanation:
http://knutsonco.tripod.com/id9.html

QUESTION NO: 5
Drag and Drop Question
Drag and drop the IPv6 IP addresses from the left onto the correct IPv6 address types on the right.
Answer:

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Updated: May 28, 2022

100-105 Exam Vce Format - Cisco Interconnecting Cisco Networking Devices Part 1 (ICND1 V3.0) Reliable Test Sample

PDF Questions & Answers

Exam Code: 100-105
Exam Name: Cisco Interconnecting Cisco Networking Devices Part 1 (ICND1 v3.0)
Updated: June 15, 2025
Total Q&As:832
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Exam Code: 100-105
Exam Name: Cisco Interconnecting Cisco Networking Devices Part 1 (ICND1 v3.0)
Updated: June 15, 2025
Total Q&As:832
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Exam Code: 100-105
Exam Name: Cisco Interconnecting Cisco Networking Devices Part 1 (ICND1 v3.0)
Updated: June 15, 2025
Total Q&As:832
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