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LAN 1 192168 10.16/28 2001.0B81.10-/64 GO/0 10.10.100/30 2001.088A1-/64 198.0.0.0/24 2001:DB8FF-/64 ISP1 GO/O PCA S1 G01 Corporate Network R1 G0/1 S0/0/1 $3 G0/2 101010 4/30 2001 DB8A.2 /64 S2 LAN 2 G0/0 192168 11.32/27 2001 DB8 1.11 /64 Custome 198.0.0 2001 088 ISP2 PC-B 1. Write the command to configure an IPV4 directly connected default static route on router R1 to forward the traffic through router ISP1. RI(config)# 2. Write the command to configure an IPV4 next-hop static route on router ISP1 to forward the traffic to LAN 1 network through router R1. ISPI(config)# 3. Write the command to configure an IPv6 next-hop flioating static default route on router R1 to forward the traffic through router ISP2 Configure the administrative distance of 5. R1(config)# A- B

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.

Find & define the IP addressing problem. 192.168.1.33 Default gateway 192.168.162 1900 FO/27 FO/26 192.168.1.66 Default gateway: 192.168.1.95 FOO 192.168.1.62/27 SO/O FO/2 50/0 2950 192.168.1.65 Default gateway 192.168.1.95 192.168.1.97/27 192.168.1.100/27 F0/3 FO/1 FOO 192.168.1.95/27

A supernet has the first address ( 192.168.16.0/22 ). The Supernet mask is Select one: a. 255.255.253.0 b. 255.255.252.0 C. 255.255.192.0 d. 255.255.0.0 The address 0.0.0.0 in the network 192.168.1.0 used:

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 Lab

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 answer

Write the Host Addresses Network Address Subnet Mask Broadcast Address 10.1.1.0 /24 10.2.0.0 /16 10.0.0.0 /16

IP = 198.12.56.40 %3D Host A MAC = 6611 %3D Router-1 Port = Ethernet 0=E0- Е 1 %3D Wire address 198.12.56.64 MAC = 3456 %3D Wire address IP = 198.12.56.33 IP = 198.12.56.65 198.12.56.32 E 2 MAC = 2244 %3D IP = 198.12.56.97 Wire address 198.12.56.96 Е1 MAC = 7755 P = 198.12.56.98 %3D Router-2 EO MAC = 8787 bridge port0 MAC = 8899 IP = 198.12.56.129 %3D %3D bridge bridge port1 MAC = 6789 %3D Wire address 198.12.56.128 MAC = 1234 %3D IP = 198.12.56.140 %3D Host B

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 Address

13 In the IPV4 addressing format, the number of networks allowed under Class C addresses is 2^7 2^24 2^14 2^21

Routing table.  Thankyou sir !

Identify the default subnet mask and CIDR value for 167.225.138.254 Identify the respective class, network address and host address of 167.225.138.254

Compare and contrast Layer 2 addresses with Layer 3 addresses. N# Layer 2 addresses Layer 3 addresses 1. 2. 3. 4. 5.

Computer Science

[15] IP ADDRESS DECODING: IP: 139.182.148.50 NM: 255.255.254.0 128 64 32 1 1 1 NA: NETWORK ADDRESS 16 8 1 1 OR BA: BROADCAST ADDRESS AND 255.255.254.0 = 11111111.11111111.11111110.00000000 10001010.10110110.10010100.00000000 139. 182. 148. 0 NA: 4 1 BA: 2 1 1 1 139.182.148.50=10001010.10110110.10010100.00110010 139.182.148.50 = 10001010.10110110.10010100.00110010 255.255.254.000000000.00000000.00000001.11111111 PRACTICE: PLEASE CALCULATE [NA: Network Address] AND [BA: Broadcast Address] IP: 172.16.0.100 ● NM: 255.255.252.0 10001010.10110110.10010101.11111111 139. 182. 149 . 255 (First Address of the Network) (Last Address of the Network)

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       130

H.W(2) :- Complete the Following Table:- PC-PT PC-PT PC3 PCI 192. 168. 1.0/30 192. 168. 2.0/26 2950-24 Switch2 2950-24 Switch! 1841 1841 Router1 Router2 192. 168.3.0/25 PC-PT PC-PT PC4 PC2 Device Interface IP Address Subnet Mask Gateway Router1 Router2 PC1 PC2 РСЗ PC4

Write the Host Addresses in binary 175.0.0.0 (net) 10101111.00000000.00000000.00000000 255.0.0.0 (SNM) 11111111.00000000.00000000.00000000

Explain how to solve this table