Homework SourseI hope you to help me with the question cde and filling the
I hope you to help me with the question c,d,e and filling the table!
Network Topology B
The network topology from Part 1 has expanded to accommodate the addition of router R3 and its accompanying network, as illustrated in the following topology. Use the 192.168.10.0/25 network address to provide addresses to the network devices, and then design a new addressing scheme to support the additional network requirement.
Step 1: Determine the number of subnets in Network Topology B.
a. How many subnets are there? _____4______
b. How many bits should you borrow to create the required number of subnets? __2_______
c. How many usable host addresses per subnet are in this addressing scheme? ___________
d. What is the new subnet mask in dotted decimal format? _________________________
e. How many subnets are available for future use? _______________
Step 2: Record the subnet information.
Fill in the following table with the subnet information:
Subnet Number
Subnet Network Address
Broadcast Address
First Usable Host Address
Last Usable Host Address
0
1
2
3
4
5
6
7
| Subnet Number | Subnet Network Address | Broadcast Address | First Usable Host Address | Last Usable Host Address |
| 0 | ||||
| 1 | ||||
| 2 | ||||
| 3 | ||||
| 4 | ||||
| 5 | ||||
| 6 | ||||
| 7 |
Solution
Dear Student,
here i have provided you the answer...
==============================================================
The given IP adress is 192.168.10.0/25.
Considering the network topology B we have follwing information...
a. How many subnets are there? _____4______
b. How many bits should you borrow to create the required number of subnets? __2_______
If we borrow 2 bits from the Host bits then subnet mask of Class C network become = 255.255.255.192
in dotted binary = 11111111.11111111.11111111.11000000
Network bits = 26
host bits = 6
number of subnets = 22 = 4
host per subnets = 26 = 64
becuase 2 address are reserved one for network and 1 broadcast
hence final number of host per nwtwork = 64 - 2 = 62
c. How many usable host addresses per subnet are in this addressing scheme? = 62
d. What is the new subnet mask in dotted decimal format? = 255.255.255.192
e. How many subnets are available for future use? = 0
Subnet Number
Subnet Network Address
Broadcast Address
First Usable Host Address
Last Usable Host Address
0
192.168.10.0
192.168.10.63
192.168.168.10.1
192.168.10.62
1
192.168.10.64
192.168.10.127
192.168.10.65
192.168.10.126
2
192.168.10.128
192.168.10.191
192.168.10.129
192.168.10.190
3
192.168.10.192
192.168.10.255
192.168.10.193
192.168.10.254
--------------------------------------------------------------------------------------------------------------------------------------
Kindly Check and Verify Thanks...!!!
| Subnet Number | Subnet Network Address | Broadcast Address | First Usable Host Address | Last Usable Host Address |
| 0 | 192.168.10.0 | 192.168.10.63 | 192.168.168.10.1 | 192.168.10.62 |
| 1 | 192.168.10.64 | 192.168.10.127 | 192.168.10.65 | 192.168.10.126 |
| 2 | 192.168.10.128 | 192.168.10.191 | 192.168.10.129 | 192.168.10.190 |
| 3 | 192.168.10.192 | 192.168.10.255 | 192.168.10.193 | 192.168.10.254 |
