PART III: Subnetting Complete the given tables below: 1. 101.138.121.114 /22 -->/25 2. 221.17.27.41 /24 ---> /26 3. 65.120.65.162/18 ---> 20 # of Subnets # of hosts per subnet Network Address of the first subnet Network Address of the last subnet Subnet Mask #hosts Subnet Mask in Binary Subnet Mask in Decimal Prefix 35 1289 130
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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 |
Step by step
Solved in 2 steps
- 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 130Design and implement an addressing scheme with a given network address: 172.32.0.0/18. Subnet the allocated prefix into 5 networks with 8 addressable host on the first 4 subnetworks, and 2 addressable host on the 5th subnetwork.SUBNET 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 hosts
- 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 hostsRefer to the following exhibit. R1(config)# aaa new-model R1 (config)# R1(config)# radius server Server-R R1(config-radius-server)# address ipv4 192.168.1.101 auth-port 1812 acct-port 1813 R1(config-radius-server) R1(config-radius-server) R1(config)# # key RADIUS-Pa55w0rd #exit Explain the reason(s) of specifying the ports being used for RADIUS and the key in the configuration. How could you include the RADIUS server as a AAA server?
- • IPV6 addresses use 128 bits to represent an address which includes bits to be used for subnetting. The second half of the address (least significant 64 bits) is always used for hosts only. Therefore, there is no compromise if we subnet the network. Use the Figure below to Create five subnets using route prefix 2001:db:acad::/48 (1 Marks) Routing Prefix Subnet ID Interface ID 48 Bits 16 Bits 64 BitsSubnetting in a network is important because it allows a large network to be divided into a smaller network that is easier to manage and maintain. Determine the subnet of class C network address for 101.101.101.0 using the subnet mask as follows. a) Determine subnet mask of 255.255.255.128. b) Determine hosts per subnet. c) Determine the valid subnet. d) Determine the broadcast address for each subnet. e) Determine the valid hosts. f) With the aid of a diagram, recommend any TWO (2) ways to assign Internet Protocol (IP) addresses to the host and devices.Design a classfull network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 25% growth in all areas. IP Address 172.16.0.0 FO/0 S/0/1 FO/O Sales Administrative Router A FO/1 SO/0/0 185 Hosts 30 Hosts Marketing 50 Hosts
- 4 LANS are connected to a router (R1) with each LAN compares of several hosts connected to a switch. Each LAN cotains the following number of hosts (end devices and the gateway interface), • LANI à 30 Hosts • LAN2 à 15 Hosts • LAN3 à 20 Hosts • LAN4 à 3 Hosts You have received the address space 200.16.1.0/24. Use the address space to create a subnetting scheme of four subnets with equal size in a way to accommodate the number of hosts required. Consider the following points in your answer, / Explain the procedure of your subnetting scheme / Highlight the suitable prefix and the subnet mask / Highlight the number of wasted hosts in each subnet after determining the most suitable prefix / Use the table below to provide the subnet details for each LAN Subnet Address First Host address Last Host address Broadcast address 1 LANI 2 LAN2 3 LAN3 4 LAN4Design a classfull network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 25% growth in all areas. IP Address 172.16.0.0 FO/O S//0/1 FO/O Sales Administrative FO/1 185 Hosts 30 Hosts SO/0/0 Marketing 50 Hosts,2. The following illustration shows a transmission from Host A to Host B using Network Address Translation (NAT). Host A is located inside a Local Area Network (LAN) with an IP address of 192.168.1.2. The LAN is connected to the Wide Area Network (WAN), i.e. the Internet, through a router R1 that conducts NAT. R1 has two interfaces with an internal LAN IP address (192.168.1.1) and and external WAN IP address (127.45.10.35). Host B is located outside the LAN and is a well-known server, e.g., Google's server. Host B has an IP address of 130.160.23.45. Port numbers for the application components running at both hosts are provided in the figure. For a single message M1 being sent from Host A to Host B, please fill in the blanks in the illustration below with the source IP address and Port number at Host A and at the Router R1. Any information not provided may be assumed, however, please state your assumption clearly. Host A Local Area Network (LAN) IP: 192.168.1.2 Port: 5252 Source IP…