A national retail chain needs to design an IP addressing scheme to support a nationwide network. The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet. Working with only one Class B address, which of the following subnet masks will support an appropriate addressing scheme? (Choose two.)
A. 255.255.255.0 B. 255.255.255.128 C. 255.255.252.0 D. 255.255.255.224 E. 255.255.255.192 F. 255.255.248.0
Correct Answers: B,E
Explanation:
Subnetting is used to break the network into smaller more efficient subnets to prevent excessive rates of Ethernet packet collision in a large network. Such subnets can be arranged hierarchically, with the organization’s network address space (see also Autonomous System) partitioned into a tree-like structure. Routers are used to manage traffic and constitute borders between subnets.
A routing prefix is the sequence of leading bits of an IP address that precede the portion of the address used as host identifier. In IPv4 networks, the routing prefix is often expressed as a “subnet mask”, which is a bit mask covering the number of bits used in the prefix. An IPv4 subnet mask is frequently expressed in quad-dotted decimal representation, e.g., 255.255.255.0 is the subnet mask for the 192.168.1.0 network with a 24-bit routing prefix (192.168.1.0/24).
A network administrator is adding two new hosts to SwitchA. Which three values could be used for the configuration of these hosts? (Choose three.)
A. host A IP address: 192.168.1.79 B. host A IP address: 192.168.1.64 C. host A default gateway: 192.168.1.78 D. host B IP address: 192.168.1.128 E. host B default gateway: 192.168.1.129 F. host B IP address: 192.168.1.190
Correct Answers: A,C,F
Explanation:
It’s a “router-on-a-stick” configuration. Which means each host in the VLAN must corresponds with the VLAN configured on the sub-interfaces.
VLAN 10 is configured on fa0/0.10 and VLAN 20 on fa0/0.20. So each hosts in VLAN 10 must use fa0/0.10 IP address as their default gateway, each hosts must also be in the same subnet as fa0/0.10 IP – same with hosts in VLAN 20.
So find out the usable IP addresses on each sub-interfaces – for 192.168.1.78 /27: 192.168.1.65 – .94 and for 192.168.1.130 /26: 192.168.1.128 – .190.
Host A (using port 6 – VLAN 10) must use IP 192.168.1.79, default gateway 192.168.1.78.
Host B (using port 9 – VLAN 20) must use IP 192.168.1.190, default gateway 192.168.1.130.
Which address range efficiently summarizes the routing table of the addresses for router Main?
A. 172.16.0.0./21 B. 172.16.0.0./20 C. 172.16.0.0./16 D. 172.16.0.0/18
Correct Answer: B
Explanation:
The 172.16.0.0./20 network is the best option as it includes all networks from 172.16.0.0 – 172.16.16.0 and does it more efficiently than the /16 and /18 subnets. The /21 subnet will not include all the other subnets in this one single summarized address.
A. 2001:0db8:0000:130F:0000:0000:08GC:140B B. 2001:0db8:0:130H::87C:140B C. 2031::130F::9C0:876A:130B D. 2031:0:130F::9C0:876A:130B
Correct Answer: D
Explanation:
An IPv6 address is represented as eight groups of four hexadecimal digits, each group representing 16 bits (two octets). The groups are separated by colons (:). An example of an IPv6 address is 2001:0db8:85a3:0000:0000:8a2e:0370:7334. The leading 0’s in a group can be collapsed using ::, but this can only be done once in an IP address.
Which of these represents an IPv6 link-local address?
A. FE80::380e:611a:e14f:3d69 B. FE81::280f:512b:e14f:3d69 C. FEFE:0345:5f1b::e14d:3d69 D. FE08::280e:611:a:f14f:3d69
Correct Answer: A
Explanation:
In the Internet Protocol Version 6 (IPv6), the address block fe80::/10 has been reserved for link-local unicast addressing. The actual link local addresses are assigned with the prefix fe80::/64. They may be assigned by automatic (stateless) or stateful (e.g. manual) mechanisms.
The network administrator is asked to configure 113 point-to-point links. Which IP addressing scheme defines the address range and subnet mask that meet the requirement and waste the fewest subnet and host addresses?
A. 10.10.0.0/16 subnetted with mask 255.255.255.252 B. 10.10.0.0/18 subnetted with mask 255.255.255.252 C. 10.10.1.0/24 subnetted with mask 255.255.255.252 D. 10.10.0.0/23 subnetted with mask 255.255.255.252 E. 10.10.1.0/25 subnetted with mask 255.255.255.252
Correct Answer: D
Explanation:
We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128). The network used for point-to-point connection should be /30. So our initial network should be 30 – 7 = 23. So 10.10.0.0/23 is the correct answer. You can understand it more clearly when writing it in binary form: /23 = 1111 1111.1111 1110.0000 0000 /30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)