#Computer networks #Packet tracer #Task B only

I need Task B only. I m also attaching Task A results in the below table.

Task A Full answer and results are attached below:

#Computer networks

ANSWER

According to my network topology I need to do sub-netting of given Network IP 192.168.10.0/24 to accommodates 4 Networks and links.

Using VLSM for sub-netting of IP Addresses for my networks 192.168.10.0/24

1- Number of required host in Site-A for Subnet = 100 Hosts + 1 Router (For Connectivity) = 100 +1 = 101 Hosts

Required host = 2n-2= 27-2=128-2 =126 (This IP range can full fill requirement)

n=7 (Means 7 bits of sub-netting octet should be 0s for hosts)

Mask in Binary Notation = 11111111.11111111.11111111.10000000

Mask in Decimal Notation 255.255.255.128

Network ID = 192.168.10.0

First IP Address 192.168.10.1

Last IP Address 192.168.10.127

2- Number of required host in Site - B for Subnet = 60 Hosts + 1 Router (For Connectivity) = 60 +1 = 61 Hosts

Required host = 2n-2= 26-2=64-2 =62 (This IP range can full fill requirement)

n=6 (Means 6 bits of sub-netting octet should be 0s for hosts)

Mask in Binary Notation = 11111111.11111111.11111111.11000000

Mask in Decimal Notation 255.255.255.192

Network ID = 192.168.10.128

First IP Address 192.168.10.129

Last IP Address 192.168.10.191

3- Number of required host in Site-C for Subnet = 24 Hosts + 1 Router (For Connectivity) = 24 +1 = 25 Hosts

Required host = 2n-2= 25-2=32-2 =30 (This IP range can full fill requirement)

n=5 (Means 5 bits of sub-netting octet should be 0s for hosts)

Mask in Binary Notation = 11111111.11111111.11111111.11100000

Mask in Decimal Notation 255.255.255.224

Network ID = 192.168.10.192

First IP Address 192.168.10.192

Last IP Address 192.168.10.223

4- Suppose Number of host in Site-D for Subnet = 10 Hosts + 1 Router (For Connectivity) = 10 +1 = 11 Hosts

Required host = 2n-2= 24-2=16-2 =14 (This IP range can full fill requirement)

n=4 (Means 4 bits of sub-netting octet should be 0s for hosts)

Mask in Binary Notation = 11111111.11111111.11111111.11110000

Mask in Decimal Notation 255.255.255.240

Network ID = 192.168.10.224

First IP Address 192.168.10.225

Last IP Address 192.168.10.239

5- Number of required host for connectivity between Site-A and Site-B Subnet = 2 Hosts

Required host = 2n-2= 22-2=4-2 =2 (This IP range can full fill requirement)

n=2 (Means 2 bits of sub-netting octet should be 0s for hosts)

Mask in Binary Notation = 11111111.11111111.11111111.11111100

Mask in Decimal Notation 255.255.255.252

Network ID = 192.168.10.240

First IP Address 192.168.10.241

Last IP Address 192.168.10.243

6- Number of required host for connectivity between Site-B and Site-C Subnet = 2 Hosts

Required host = 2n-2= 22-2=4-2 =2 (This IP range can full fill requirement)

n=2 (Means 2 bits of sub-netting octet should be 0s for hosts)

Mask in Binary Notation = 11111111.11111111.11111111.11111100

Mask in Decimal Notation 255.255.255.252

Network ID = 192.168.10.244

First IP Address 192.168.10.245

Last IP Address 192.168.10.247

7- Number of required host for connectivity between Site-C and Site-D Subnet = 2 Hosts

Required host = 2n-2= 22-2=4-2 =2 (This IP range can full fill requirement)

n=2 (Means 2 bits of sub-netting octet should be 0s for hosts)

Mask in Binary Notation = 11111111.11111111.11111111.11111100

Mask in Decimal Notation 255.255.255.252

Network ID = 192.168.10.248

First IP Address 192.168.10.249

Last IP Address 192.168.10.251

8- Number of required host for connectivity between Site-D and Site-A Subnet = 2 Hosts

Required host = 2n-2= 22-2=4-2 =2 (This IP range can full fill requirement)

n=2 (Means 2 bits of sub-netting octet should be 0s for hosts)

Mask in Binary Notation = 11111111.11111111.11111111.11111100

Mask in Decimal Notation 255.255.255.252

Network ID = 192.168.10.252

First IP Address 192.168.10.253

Last IP Address 192.168.10.255

Table: Network Name, Network ID, Usable IP Range, Number of Host, Broadcast IP, Subnet Mask.

Please do Task B with proper explanation. It will be a great help and I will surely give you a thumbs up.

Site A Site C Site B Site D Fig. 1: Physical topology of the corporate network Number of hosts at each site are shown below. Number of hosts at site A = 100 Number of hosts at site B = 60 Number of hosts at site C = 24 Number of hosts at site D = 10 You are given an IP address block of 192.168.10.0/24. Task B: Configuration and implementation using Cisco IOS (20 Marks) This task requires configuration of the IP Addressing design produced in Task A using Packet Tracer (or Cisco Routers). Provide your configuration steps and final configuration at each router to achieve full IP connectivity. You must assume that initially all routers are running a version of Cisco IOS and there is no running configuration. Any viable method to achieve full IP connectivity is acceptable. Task B Deliverables: 1) A brief discussion about your approach to configure the network. 2) Final configuration of all routers in the network s Network No. Name Network ID Usable IP Range Number Broadcast IP of Host Subnet Mask 1 SITE 1 192.168.10.0 126 192.168.10.1 192.168.10.127 192.168.10.127 255.255.255.128 2 SITE 2 62 192.168.10.128 192.168.10.129 - 192.168.10.191 192.168.10.191 255.255.255.192 3 SITE 3 30 192.168.10.192 192.168.10.193 - 192.168.10.223 19217210.223255.255.255.224 4 SITE 4 192.168.10.224 14 192.168.10.225 192.168.10.239 192172.10.239 255.255.255.240 2 192.172.10.243255.255.255.252 LINK SITE 192.168.10.240 192.168.10.241 - -ATO 192.168.10.243 SITE-B 2 192172.10.247 255.255.255.252 LINK SITE 192.168.10.244 192.168.10.245 - 192.168.10.247 SITE-C 2 19217210 251 255.255.255.252 LINK SITE 192.168.10 248 192.168.10.249 - 192.168.10.251 SITE-D 8 192.168.10.252 2 LINK SITE -D TO SITE-A 192.168.10.253 - 192.168.10.255 192.172.10.255 255.255.255.252 Task A: IP Addressing Design of a Corporate Network (20 Marks) This task is related to design of IP networks. You are given a task to design an IP Addressing scheme for a corporate network consisting of 4 sites. The physical topology of the corporate network is shown in Fig. 1. The connections between different sites are made using point- to-point serial links. Site A Site C Site B Site D Fig. 1. Physical topology of the corporate network Number of hosts at each site are shown below. Number of hosts at site A = 100 Number of hosts at site B = 60 Number of hosts at site C = 24 Number of hosts at site D = 10 You are given an IP address block of 192.168.10.0/24. Task A Deliverable: You are required to design the IPv4 Addressing based on the host requirements well as connectivity between sites (as shown in Fig. 1). The addressing scheme should be designed to conserve addresses. Discuss the approach you will use to tackle this task. For each subnetwork, provide all details including, the network address, broadcast, and valid hosts. Show working of your steps. Task B: Configuration and implementation using Cisco IOS (20 Marks) This task requires configuration of the IP Addressing design produced in Task A using Packet Tracer (or Cisco Routers). Provide your configuration steps and final configuration at each router to achieve full IP connectivity. You must assume that initially all routers are running a version of Cisco IOS and there is no running configuration. Any viable method to achieve full IP connectivity is acceptable. Task B Deliverables: 1) A brief discussion about your approach to configure the network. 2) Final configuration of all routers in the network