Perform subnetting with the given address 180.180.0.0/16 as the address for the following topology:
Solution:
To perform subnetting with VLSM (concerning minimal wastage of host address) we will assign network address for various lans (in the given topology) sequentially, as follows:
a. Lan 1 : 8000 hosts
b. Lan 4 : 4000 hosts
c. Lan 5 : 4000 hosts
d. Lan 2 : 2000 hosts
e. Lan 3 : 1000 hosts
f. Lan 6: 500 hosts
g. Lan 7 : 254 hosts
h. Lan 8 : 50 hosts
a. Lan 1 : 8000 hosts:
We will use base address 180.180.0.0/16 to subnet for supporting 8000 hosts. We have to borrow 3 bits from the host portion for such subnetting. There will be 23 =8 subnet each of which will have the subnet mask: 255.255.224.0.
So, Block size = (256-224) = 32 (network size).
The subnets are:
Subnet 0: 180.180.0.0/19
Subnet 1: 180.180.32.0/19
Subnet 2: 180.180.64.0/19
Subnet 3: 180.180.96.0/19
Subnet 4: 180.180.128.0/19
Subnet 5: 180.180.160.0/19
Subnet 6: 180.180.192.0/19
Subnet 7: 180.180.224.0/19
We will assign 180.180.0.0/19 (subnet 0) as the network address for lan 1.
1st valid host address for lan 1: 180.180.0.1/19.
Last valid host address for lan 1: 180.180.31.254/19.
b. Lan 4: 4000 hosts:
We will use subnet 1: 180.180.32.0/19 as the base address to subnet for supporting 4000 hosts. We have to borrow 1 bit from the host portion for such subnetting. There will be 21=2 subnets, each of which will have the subnet mask: 255.255.240.0.
So, Block size = (256-240) =16 (network size).
Subnets are:
Subnet 10: 180.180.32.0/20
Subnet 11: 180.180.48.0/20
We will assign 180.180.32.0/20 (subnet 10) as the network address for lan 4.
1st valid host address for lan 4: 180.180.32.1/20.
Last valid host address for lan 4: 180.180.47.254/20.
c. Lan 5 : 4000 hosts:
Subnet 11: 180.180.48.0/20 can support up to 212-2 = 4096-2=4094 valid hosts.
So, we will assign 180.180.48.0/20 (subnet 11) as the network address for lan 5.
1st valid host address for lan 5: 180.180.48.1/20.
Last valid host address for lan 5: 180.180.63.254/20.
d. Lan 2: 2000 hosts:
Now we will use subnet 2: 180.180.64.0/19 as the base address to subnet for supporting 2000 hosts. So, we have to borrow 2 bits from the host portion for such subnetting. There will be 22 = 4 subnets, each of which will possess the subnet mask: 255.255.254.0
So, Block size = (256-248) =8 (network size)
The subnets are:
Subnet 20: 180.180.64.0/21
Subnet 21: 180.180.72.0/21
Subnet 22: 180.180.80.0/21
Subnet 23: 180.180.88.0/21
We will assign 180.180.64.0/21 (subnet 20) as the network address for lan 2.
1st valid host address for lan 2: 180.180.64.1/21.
Last valid host address for lan 2: 180.180.71.254/21.
e. Lan 3: 1000 hosts:
we will use subnet 21: 180.180.72.0/21 as the base address to subnet for supporting 1000 hosts. So, we have to borrow 1 bit from the host portion for such subnetting. There will be 21=2 subnets, each of which will possess the subnet mask: 255.255.252.0
So, Block size = (256-252) = 4 (network size).
Subnets are:
Subnet 210: 180.180.64.0/22
Subnet 211: 180.180.72.0/22
Now, we will assign 180.180.72.0/22 (subnet 210) as the network address for lan 3.
1st valid host address for lan 3: 180.180.72.1/22.
Last valid host address for lan 3: 180.180.75.254/22.
f. Lan 6: 500 hosts:
We will use subnet 2111: 180.180.76.0/22 as the base address to subnet for supporting 500 hosts. So, we have to borrow 1 bit from the host portion for such subnetting. There will be 21=2 subnets, each of which will possess the subnet mask: 255.255.254.0
So, Block size = (256-254) = 2 (network size).
The subnets are:
Subnet 21110: 180.180.76.0/23
Subnet 21111: 180.180.78.0/23
Now, we will assign 180.180.76.1/23 (subnet 21110) as the network address for lan 6.
1st valid host address for lan 6: 180.180.76.1/23.
Last valid host address for lan 6: 180.180.77.254/23.
g. Lan 7: 254 hosts:
We will use subnet 21111: 180.180.78.0/23 as the base address to subnet for supporting 500 hosts. So, we have to borrow 1 bit from the host portion for such subnetting. There will be 21=2 subnets, each of which will possess the subnet mask: 255.255.255.0
So, Block size = (256-255) = 1 (network size).
The subnets are:
Subnet 211110: 180.180.78.0/24
Subnet 211111: 180.180.79.0/24
Now, we will assign 180.180.78.0/24 (subnet 211110) as the network address for lan 7.
1st valid host address for lan 7: 180.180.78.1/24.
Last valid host address for lan 7: 180.180.78.254/24.
h. Lan 8: 50 hosts:
We will use subnet 211111: 180.180.79.0/24 as the base address to subnet for supporting 50 hosts. So, we have to borrow 2 bits from the host portion for such subnetting. There will be 22 = 4 subnets, each of which will possess the subnet mask: 255.255.255.192.
So, Block size = (256-192) = 64 (network size).
The subnets are:
Subnet 2111110: 180.180.79.0/26
Subnet 2111111: 180.180.79.64/26
Subnet 2111112: 180.180.79.128/26
Subnet 2111113: 180.180.79.192/26
Now, we will assign 180.180.79.0/26 (subnet 2111110) as the network address for lan 8.
1st valid host address for lan 8: 180.180.79.1/26.
Last valid host address for lan 8: 180.180.79.62/26.
Network address for WAN:
We will use subnet 2111111: 180.180.78.0/24 to support 2 valid addresses. So, we have to borrow 4 bits from the host portion for such subnetting. There will be 24=16 subnets, each of which will possess the subnet mask: 255.255.255.252.
So, Block size = (256-252) = 4
a. 180.180.79.64/30
b. 180.180.79.68/30
c. 180.180.79.72/30
d. 180.180.79.76/30
e. 180.180.79.80/30
f. 180.180.79.84/30
g. 180.180.79.88/30
h. 180.180.79.92/30
i. 180.180.79.96/30
j. 180.180.79.100/30
k. 180.180.79.104/30
l. 180.180.79.108/30
m. 180.180.79.112/30
n. 180.180.79.116/30
o. 180.180.79.120/30
p. 180.180.79.124/30
We will assign:
a. 180.180.79.64/30 for network between Router A & Router B.
b. 180.180.79.68/30 network between Router B & Router C.
c. 180.180.79.72/30 network between Router A & Router C.
So, the resultant topology with network address will be:
Network address summary table:
| Network Name | Network address | 1st valid host | Last valid host | ||
| LAN 1 | 180.180.0.0/19 | 180.180.0.1/19 | 180.180.31.254/19 | ||
| LAN 2 | 180.180.64.0/21 | 180.180.64.1/21 | 180.180.71.254/21 | ||
| LAN 3 | 180.180.72.0/22 | 180.180.72.1/22 | 180.180.75.254/22 | ||
| LAN 4 | 180.180.32.0/20 | 180.180.32.1/20 | 180.180.47.254/20 | ||
| LAN 5 | 180.180.48.0/20 | 180.180.48.1/20 | 180.180.63.254/20 | ||
| LAN 6 | 180.180.76.0/23 | 180.180.76.1/23 | 180.180.77.254/23 | ||
| LAN 7 | 180.180.78.0/24 | 180.180.78.1/24 | 180.180.78.254/24 | ||
| LAN 8 | 180.180.79.0/26 | 180.180.79.1/26 | 180.180.79.62/26 | ||
| Network between Router A & Router B: | 180.180.79.64/30 | 180.180.79.65/30 | 180.180.79.66/30 | ||
| Network between Router B & Router C: | 180.180.79.68/30 | 180.180.79.69/30 | 180.180.79.70/30 | ||
| Network between Router A & Router C: | 180.180.79.72/30 | 180.180.79.73/30 | 180.180.79.74/30 | ||
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