Q: Would like to know how to solve Networking Basics exercise, details will defi
ID: 3857185 • Letter: Q
Question
Q: Would like to know how to solve Networking Basics exercise, details will definitely help in understanding, thanks for taking time and helping me.
Q2: Given an IPv4 address 10.20.30.0/23, design and apply an IP addressing scheme for the topology shown in Figure 1. You must subnet the subnets to provide enough address space to use. Each LAN or WAN requires enough IP addresses to support the number of hosts specified in Table 1.
Building
Number of Addresses
Building A
160
Building B
100
Building C
50
Building D
10
Building E
10
WAN-1
2
WAN-2
2
Table 1 Number of Hosts in Each Building
Complete the table below
Subnet #
Subnet Address
Address Range
Broadcast Address
Network/Prefix
# of Wasted Host Addresses
0
10.20.30.0
10.20.30.1-
10.20.30.254
10.20.30.255
10.20.30.0/24
94
1
2
3
4
5
6
7
Building
Number of Addresses
Building A
160
Building B
100
Building C
50
Building D
10
Building E
10
WAN-1
2
WAN-2
2
Explanation / Answer
The network given is : -
10.20.30.0/23
This network have 510 total number of hosts (by formula 232-n - 2)
For subnetting, we need to divide these hosts into 7 subnets . Note that, we can divide hosts only in power of 2, i.e. 2,4,8,16,32,64,128,256 hosts..
Below is the table generates using above information
Subnetting
Subnet #
Subnet Address
Address Range
Broadcast Address
Network/Prefix
# of Wasted Host Addresses
0
10.20.30.0
10.20.30.1-
10.20.30.254
10.20.30.255
10.20.30.0/24
94
1
10.20.31.0
10.20.31.1 - 10.20.31.126
10.20.31.127
10.20.31.0/25
26
2
10.20.31.128
10.20.31.129 - 10.20.31.190
10.20.31.191
10.20.31.128/26
12
3
10.20.31.192
10.20.31.193 - 10.20.31.206
10.20.31.207
10.20.31.192/28
4
4
10.20.31.208
10.20.31.209 - 10.20.31.222
10.20.31.223
10.20.31.208/28
4
5
10.20.31.224
10.20.31.225 - 10.20.31.226
10.20.31.227
10.20.31.224/30
0
6
10.20.31.228
10.20.31.229 - 10.20.31.230
10.20.31.231
10.20.31.228/30
0
7
10.20.31.232
10.20.31.233 - 10.20.31.238
10.20.31.239
10.20.31.232/29
16
Note that, number of subnets that can be generated can also only be in power of 2. So here 7 were needed so 8 subnets were generate.
Observe that in last row , the left over subnet was generated. After creating all the subnets i.e. first 7 rows. 24 addresses were left. But with the network address only 6 addresses were possible. So 24-8 = 16 addresses were wasted.
Let me know if you have any concern in this.
Subnetting
Subnet #
Subnet Address
Address Range
Broadcast Address
Network/Prefix
# of Wasted Host Addresses
0
10.20.30.0
10.20.30.1-
10.20.30.254
10.20.30.255
10.20.30.0/24
94
1
10.20.31.0
10.20.31.1 - 10.20.31.126
10.20.31.127
10.20.31.0/25
26
2
10.20.31.128
10.20.31.129 - 10.20.31.190
10.20.31.191
10.20.31.128/26
12
3
10.20.31.192
10.20.31.193 - 10.20.31.206
10.20.31.207
10.20.31.192/28
4
4
10.20.31.208
10.20.31.209 - 10.20.31.222
10.20.31.223
10.20.31.208/28
4
5
10.20.31.224
10.20.31.225 - 10.20.31.226
10.20.31.227
10.20.31.224/30
0
6
10.20.31.228
10.20.31.229 - 10.20.31.230
10.20.31.231
10.20.31.228/30
0
7
10.20.31.232
10.20.31.233 - 10.20.31.238
10.20.31.239
10.20.31.232/29
16
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.