IPv4 ADDRESS CLASS OF IP ADDRESS DEFAULT SUBNET MASK NETWORK / HOST ADDRESS ex: 108.117.55.36 A 255.0.0.0 108.0.0.0 / 0.117.55.36 215.125.87.181 25.235.118.12 152.7.241.195 195.204.150.10 190.12.211.125
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IPv4 ADDRESS |
CLASS OF IP ADDRESS |
DEFAULT SUBNET MASK |
NETWORK / HOST ADDRESS |
ex: 108.117.55.36 |
A |
255.0.0.0 |
108.0.0.0 / 0.117.55.36 |
215.125.87.181 |
|
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|
25.235.118.12 |
|
|
|
152.7.241.195 |
|
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|
195.204.150.10 |
|
|
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190.12.211.125 |
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- Required Resources: 1 Router (Cisco 1841 with Cisco IOS). 1 Switch (Cisco 2950-24 with Cisco IOS). 3 PCs (with terminal emulation program). • Console cable to configure the Cisco IOS devices via the console ports. Ethernet cables as shown below in the topology. Topology: PC-PT PCO 2960-24TT 1841 PC-PT Switcho Router0 PC2 PC-PT PC1Default Subnet Masks Write the correct default subnet mask for each of the following addresses: 177.100.18.4 255 255.0.0 255.0.0.0 119.18.45.0 191.249.234.191 223.23.223.109 10.10.250.1 126.123.23.1 223,69 230.250 192.12.35.105 77.251.200.51 189.210.50 1 88.45.65.35 128.212.250.254 193.100.77.83 125.125.250.1 1.1.10.50 220.90.130 45 134 125.34 9 95.250.91.99Variable Length Subnet Mask Let's say that you are given a network address block as 192.168.16.0/24, and are asked to divide it into five different networks. The largest network will have 80 hosts. What would be the subnet mask for this network? Oa. 255.255.255.0 or / 24 Ob. 255.255.255.252 or /30 OG 255.255.255.128 or/25 Od. 255.255.255.224 or /27
- Variable Length Subnet Mask Let's say that you are given a network address block as 192.168.16.0/24, and are asked to divide it into five different networks. The largest network will have 80 hosts. What would be the subnet mask for this network? Oa. 255.255.255.0 or / 24 Ob. 255.255.255.252 or /30 O. 255.255.255.128 or/25 Od. 255.255.255.224 or /27PC-1 Switch-1 Fa0 Fa0/1 Fa0/2 IP Address: 10.0.0.1 Subnetmask: 255.255.255.0 Default gateway: 10.0.0.254 Gig0/0 Router-1 From the above exhibit, the address 10.0.0.254 should be configured onDescription Subnet A Subnet B First IP address 172.16.1.0 172.16.1.63 Last IP address 172.16.1.63 172.16.1.127 Maximum number of hosts 64 64 Record the IP address information for each device: Device IP address Subnet Mask Gateway PC-A R1-G0/0/0 R1-G0/0/1 S1 PC-B
- a- Given the following screenshot, some errors are preventing PC-C from accessing the Redundant Internet Connection. Find the errors and correct them. Router0(config)# ip route 0.0.0.0 0.0.0.0 209.0.0.1 Router1(config)# router rip Router0(config)# router rip Router1(config-router)# passive-interface s0/0/1 Router0(config-router)# network 172.16.1.4 Router1(config-router)# network 172.16.1.0 Router0(config-router)# auto-summary Router1(config-router)# no auto-summary Your answer b- After correcting the errors, PC-C is able to reach the Primary Internet. Which command is used to view the routing table on Router 1?* Your answerIdentify the address type for the following IPV6 addresses. i. :1/128 ii. ff02:1 ii. 2001:660:7307:6666:3797:f3f4:7500:24b6/64 iv. fda9:44c3:2f5e::10/64 fe80:5054:ff:fe20:1506/64 V.Write the Host Addresses Network Address Subnet Mask Broadcast Address 192.168.1.0 255.255.255.0 192.168.0.0 255.255.0.0 192.168.0.0 255.255.255.0
- uuter620Networks%20-%20HandoutLpdf Network Addresses Using the IP address and subnet mask shown write out the network address 188.10.18.2 188. 10.0.0 255.255.0.0 10. 10.48.0 10.10.48.80 255.255.255.0 192.149.24.191 255.255.255.0 150.203.23.19 255.255.0.0 10.10.10.10 255.0.0.0 186.13.23.110 255.255.255.0 223.69.230.250 255.255 0.0 200.120.135.15 255.255.255.0 27.125.200 151 255.0.0.0 199.20.150 35 255 255.2550 191.55 165.135 255.255.255.0 28 212.250 254 25525500150. IPv6 address have a size of a. 32 bits b. 64 bits c. 128 bits d. 265 bitsHost IP Address Class Subnet Mask # of Subnet Number of Number of Host Bits Subnets Addresses 200.50.7.219 255.255.255.224 Subnet address of this subnet or Range of host addresses for this Broadcast address of this subnet "wire" subnet