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Q21. - (Topic 3)
An administrator is working with the 192.168.4.0 network, which has been subnetted with a /26 mask. Which two addresses can be assigned to hosts within the same subnet? (Choose two.)
Increment: 64 (/26 = 11111111.11111111.11111111.11000000)
The IP 192.168.4.0 belongs to class C. The default subnet mask of class C is /24 and it has
been subnetted with a /26 mask so we have 2(26-24).= 22.= 4 sub-networks:
1st subnet: 192.168.4.0 (to 192.168.4.63)
2nd subnet: 192.168.4.64 (to 192.168.4.127)
3rd subnet: 192.168.4.128 (to 192.168.4.191)
4th subnet: 192.168.4.192 (to 192.168.4.225)
In all the answers above, only answer C and D are in the same subnet.
Therefore only IPs in this range can be assigned to hosts.
Q22. - (Topic 5)
Refer to the exhibit.
The network administrator is testing connectivity from the branch router to the newly installed application server. What is the most likely reason for the first ping having a success rate of only 60 percent?
A. The network is likely to be congested, with the result that packets are being intermittently dropped.
B. The branch router had to resolve the application server MAC address.
C. There is a short delay while NAT translates the server IP address.
D. A routing table lookup delayed forwarding on the first two ping packets.
E. The branch router LAN interface should be upgraded to FastEthernet.
Initially the MAC address had to be resolved, but once it is resolved and is in the ARP table of the router, pings go through immediately.
Q23. - (Topic 3)
Which option is a valid IPv6 address?
IPv6 Address Notation IPv6 addresses are denoted by eight groups of hexadecimal quartets separated by colons in between them. Following is an example of a valid IPv6 address: 2001:cdba:0000:0000:0000:0000:3257:9652 Any four-digit group of zeroes within an IPv6 address may be reduced to a single zero or altogether omitted. Therefore, the following IPv6 addresses are similar and equally valid: 2001:cdba:0000:0000:0000:0000:3257:9652 2001:cdba:0:0:0:0:3257:9652 2001:cdba::3257:9652
Q24. - (Topic 1)
Refer to the exhibit.
A network device needs to be installed in the place of the icon labeled Network Device to accommodate a leased line attachment to the Internet. Which network device and interface configuration meets the minimum requirements for this installation?
A. a router with two Ethernet interfaces
B. a switch with two Ethernet interfaces
C. a router with one Ethernet and one serial interface
D. a switch with one Ethernet and one serial interface
E. a router with one Ethernet and one modem interface
Only a router can terminate a leased line attachment access circuit, and only a router can connect two different IP networks. Here, we will need a router with two interfaces, one serial connection for the line attachment and one Ethernet interface to connect to the switch on the LAN.
Q25. - (Topic 5)
A receiving host has failed to receive all of the segments that it should acknowledge. What can the host do to improve the reliability of this communication session?
A. decrease the window size
B. use a different source port for the session
C. decrease the sequence number
D. obtain a new IP address from the DHCP server
E. start a new session using UDP
Answer: A Explanation:
The Window bit in the header determines the number of segments that can be sent at a time. This is done to avoid overwhelming the destination. At the start of the session the window in small but it increases over time. The destination host can also decrease the window to slow down the flow. Hence the window is called the sliding window. When the source has sent the number of segments allowed by the window, it cannot send any further segments till an acknowledgement is received from the destination. On networks with high error rates or issues, decreasing the window size can result in more reliable transmission, as the receiver will need to acknowledge fewer segments. With a large window size, the sender will need to resend all the frames if a single one is not received by the receiver.
Q26. - (Topic 3)
What is the subnet address for the IP address 172.19.20.23/28?
From the /28 we can get the following:
Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 172.19.20.16 (because 16 < 23)
Broadcast address: 172.16.20.31 (because 31 = 16 + 16 – 1)
Q27. - (Topic 3)
Refer to the exhibit.
The two routers have had their startup configurations cleared and have been restarted. At a minimum, what must the administrator do to enable CDP to exchange information between R1 and R2?
A. Configure the router with the cdp enable command.
B. Enter no shutdown commands on the R1 and R2 fa0/1 interfaces.
C. Configure IP addressing and no shutdown commands on both the R1 and R2 fa0/1 interfaces.
D. Configure IP addressing and no shutdown commands on either of the R1 or R2 fa0/1 interfaces.
If the no shut down commands are not entered, then CDP can exchange information between the two routers. By default, all Cisco device interfaces and ports are shut down and need to be manually enabled.
Q28. - (Topic 7)
Which component of the routing table ranks routing protocols according to their preferences?
A. administrative distance
B. next hop
D. routing protocol code
Administrative distance - This is the measure of trustworthiness of the source of the
route. If a router learns about a destination from more than one routing protocol,
administrative distance is compared and the preference is given to the routes with lower
administrative distance. In other words, it is the believability of the source of the route.
Q29. - (Topic 7)
When a router makes a routing decision for a packet that is received from one network and destined to another, which portion of the packet does if replace?
A. Layer 2 frame header and trailer
B. Layer 3 IP address
C. Layer 5 session
D. Layer 4 protocol
Router Switching Function (220.127.116.11)A primary function of a router is to forward packets toward their destination. This is accomplished by using a switching function, which is the process used by a router to accept a packet on one interface and forward it out of another interface. A key responsibility of the switching function is to encapsulate packets in the appropriate data link frame type for the outgoing data link. NOTE In this context, the term “switching” literally means moving packets from source to destination and should not be confused with the function of a Layer 2 switch. After the router has determined the exit interface using the path determination function, the router must encapsulate the packet into the data link frame of the outgoing interface. What does a router do with a packet received from one network and destined for another network? The router performs the following three major steps:
. Step 1. De-encapsulates the Layer 3 packet by removing the Layer 2 frame header and trailer. . Step 2. Examines the destination IP address of the IP packet to find the best path in the routing table. . Step 3. If the router finds a path to the destination, it encapsulates the Layer 3 packet into a new Layer 2 frame and forwards the frame out the exit interface.
Q30. - (Topic 7)
Under which circumstance should a network administrator implement one-way NAT?
A. when the network must route UDP traffic
B. when traffic that originates outside the network must be routed to internal hosts
C. when traffic that originates inside the network must be routed to internal hosts
D. when the network has few public IP addresses and many private IP addresses require outside access
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.