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NEW QUESTION: 1

A. Option C
B. Option A
C. Option D
D. Option B
Answer: D
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.

NEW QUESTION: 2
Refer to the exhibit.

The network is converged.After link-state advertisements are received from Router_A, what information will Router_E contain in its routing table for the subnets 208.149.23.64 and
208.149.23.96?
A. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:16, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
B. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, FastEthemet0/0
2 08.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
C. 208.149.23.64[110/3] via 190.172.23.10, 00:00:07, Serial1/0
2 08.149.23.96[110/3] via 190.173.23.10, 00:00:16, Serial1/0
D. 208.149.23.64[110/1] via 190.172.23.10, 00:00:07, Serial1/0
2 08.149.23.96[110/3] via 190.173.23.10, 00:00:16, FastEthemet0/0
Answer: B
Explanation:
Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A through FastEthernet interface. The interface cost is calculated with the formula 108 /
Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 / 100,000,000 = 1. Therefore the cost is 12 (learned from Router_A) + 1 = 13 for both subnets ->
The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps =
64; T3 cost = 108 / 45 Mbps = 2) so surely OSPF will choose the path through T3 link ->
Router_E will choose the path from Router_A through FastEthernet0/0, not Serial1/0.
In fact, we can quickly eliminate answers B, C and D because they contain at least one subnet learned from Serial1/0 -> they are surely incorrect.

NEW QUESTION: 3
ウイルス対策ソフトウェアを選択する際に最も重要な基準は次のうちどれですか？
A. 新しいウイルスのアラート通知と影響評価
B. 製品市場シェアと年間コスト
C. メンテナンスの容易さと更新の頻度
D. 侵入検知システム（IDS）ソフトウェアおよびファイアウォールとのインターフェース機能
Answer: C
Explanation:
説明
ソフトウェアが効果的であるためには、それを維持し、最新の状態に保つことが容易でなければなりません。市場シェアと年間コスト、侵入検知システム（IDS）へのリンク、および自動通知は、本質的にすべて二次的なものです。

NEW QUESTION: 4
What information is required when creating a new field in the data dictionary?
Note: There are 3 correct answers to this question
A. Repeating Field
B. Name
C. Type
D. Description
E. Label
Answer: A,B,C