Find & define the IP addressing problem. 192.168.1.33 Default gateway 192.168.1 62 1900 FO/27 FO/26 192.168.1.66 Default gateway 192.168.1.95 FOO 192.168.1.62/27 50/0 FO/2 50/0 2950 192.168.1.65 Default gateway 192.168.1.96 192.168.1.97/27 192.168.1.100/27 F0/3 FO/1 FOO 192.168.1.96/27
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- Refer the exhibit and the required number of hosts given. Subnet the address 192.168.50.0/24 for the required number of hosts using VLSM and complete the subnet table using the following instructions: NW: 192.168.50.0/24 NET-X: 45 USABLE HOSTS NET-Z: 2 USABLE HOSTS NET-Y: 25 USABLE HOSTS ------ GO/O R2 G0/0 GO/1 GO/1 R1 PC2 PC1 Assign the first subnet to network with the highest host requirement. Assign the second subnet to the network with the second highest host requirement. Assign the third subnet to the network with the third highest host requirement. Subnet Table: Subnet No. of Subnet First Last Usable Broadcast Subnet Name Usable Address Usable Host Address Mask hosts Host Address required AddressSUBNET ADDRESSING Consider the router and the two attached subnets below (A and B). The number of hosts is also shown below. The subnets share the 23 high-order bits of the address space: 144.145.110.0/23 : A 175 hosts QUESTION 2 OF 10 Answer How many hosts can there be in this address space? 144.145.110.0/23 B Assign subnet addresses to each of the subnets (A and B) so that the amount of address space assigned is minimal, and at the same time leaving the largest possible contiguous address space available for assignment if a new subnet were to be added. Then answer the questions below. 84 hosts19. ARP is used for: O packet path determination O finding of a mapping IPV6 → MAC address O finding of a mapping IPV4 → MAC address O finding of a mapping IPV4 → TCP address
- Find & define the IP addressing problem. 192.168.1.33 Default gateway 192.168.162 1900 FO/27 192.168.1.66 Default gateway: 192.168.1.95 F0/26 FOO 192.168.1.62/27 SO/0 FO/2 $0/0 2950 192.168.1.65 Default gateway 192.168.1.95 192.168.1.97/27 192.168.1.100/27 FO/3 FO/1 FOO 192.168.1.96/27 LabSUBNET ADDRESSING Consider the router and the two attached subnets below (A and B). The number of hosts is also shown below. The subnets share the 23 high-order bits of the address space: 144.145.110.0/23 175 hosts QUESTION 3 OF 10 A Answer What is the subnet address of subnet A? (CIDR notation) 144,145.110.0/23 Assign subnet addresses to each of the subnets (A and B) so that the amount of address space assigned is minimal, and at the same time leaving the largest possible contiguous address space available for assignment if a new subnet were to be added. Then answer the questions below. B 84 hostsComplete the given tables below: # of Subnets # of hosts per subnet Network Address of the first subnet Network Address of the last subnet Subnet Mask 1. 101.138.121.114 /22 --> /25 2. 221.17.27.41 /24 ---> /26 3. 65.120.65.162 /18 ---> 20 #hosts Subnet Mask in Binary Subnet Mask in Decimal Prefix 35 1289 130
- Block Address: 192.168.16.0/24 Port to Point Link WANI Usable Hof Hosts: 2 Users LAN: Usable #of Hosts: 25 Network Address: LANI Usable #of Hosts: 100 Users Network Address: 7 Point to Point Link WAN2 Usable 4of Hosts: 2 Users LAN2: Usable #of Hosts: 50 Network Address: The network given in the above figure consists of three local area networks and two wide area networks are connected with two serial links. With an ID range 192.168.15.0/24, design an IP plan for this network. Find the network address and broadcast address, and subnet mask for each network.Subnetting Complete the given tables below: # of Subnets # of hosts per subnet Network Address of the first subnet Network Address of the last subnet Subnet Mask 1. 101.138.121.114 /22 --> /25 2. 221.17.27.41 /24 ---> /26 3. 65.120.65.162 /18 ---> 20 #hosts Subnet Mask in Binary Subnet Mask in Decimal Prefix 35 1289 13013 In the IPV4 addressing format, the number of networks allowed under Class C addresses is 2^7 2^24 2^14 2^21
- 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.NAT translation table LAN side addr WAN side addr 135.122.200.215 10.0.1.15 10,0.1.13 address 135.122.200.215 S: 10,0,1.27 2 D: 1 D: S: S: D: S: 10,0.1.15 D: 135.122,200,215 10.०.1.22