CCNA 2 – An assignment on VLSM

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:

<!–[if !supportLists]–>a. <!–[endif]–>Lan 1 : 8000 hosts

<!–[if !supportLists]–>b. <!–[endif]–>Lan 4 : 4000 hosts

<!–[if !supportLists]–>c. <!–[endif]–>Lan 5 : 4000 hosts

<!–[if !supportLists]–>d. <!–[endif]–>Lan 2 : 2000 hosts

<!–[if !supportLists]–>e. <!–[endif]–>Lan 3 : 1000 hosts

<!–[if !supportLists]–>f. <!–[endif]–>Lan 6: 500 hosts

<!–[if !supportLists]–>g. <!–[endif]–>Lan 7 : 254 hosts

<!–[if !supportLists]–>h. <!–[endif]–>Lan 8 : 50 hosts

<!–[if !supportLists]–>a. <!–[endif]–>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.

1^st valid host address for lan 1: 180.180.0.1/19.

Last valid host address for lan 1: 180.180.31.254/19.

<!–[if !supportLists]–>b. <!–[endif]–>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.

1^st valid host address for lan 4: 180.180.32.1/20.

Last valid host address for lan 4: 180.180.47.254/20.

<!–[if !supportLists]–>c. <!–[endif]–>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.

1^st valid host address for lan 5: 180.180.48.1/20.

Last valid host address for lan 5: 180.180.63.254/20.

<!–[if !supportLists]–>d. <!–[endif]–>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.

1^st valid host address for lan 2: 180.180.64.1/21.

Last valid host address for lan 2: 180.180.71.254/21.

<!–[if !supportLists]–>e. <!–[endif]–>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.

1^st valid host address for lan 3: 180.180.72.1/22.

Last valid host address for lan 3: 180.180.75.254/22.

<!–[if !supportLists]–>f. <!–[endif]–>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.

1^st valid host address for lan 6: 180.180.76.1/23.

Last valid host address for lan 6: 180.180.77.254/23.

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<!–[if !supportLists]–>g. <!–[endif]–>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.

1^st valid host address for lan 7: 180.180.78.1/24.

Last valid host address for lan 7: 180.180.78.254/24.

<!–[if !supportLists]–>h. <!–[endif]–>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.

1^st 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

<!–[if !supportLists]–>a. <!–[endif]–>180.180.79.64/30

<!–[if !supportLists]–>b. <!–[endif]–>180.180.79.68/30

<!–[if !supportLists]–>c. <!–[endif]–>180.180.79.72/30

<!–[if !supportLists]–>d. <!–[endif]–>180.180.79.76/30

<!–[if !supportLists]–>e. <!–[endif]–>180.180.79.80/30

<!–[if !supportLists]–>f. <!–[endif]–>180.180.79.84/30

<!–[if !supportLists]–>g. <!–[endif]–>180.180.79.88/30

<!–[if !supportLists]–>h. <!–[endif]–>180.180.79.92/30

<!–[if !supportLists]–>i. <!–[endif]–>180.180.79.96/30

<!–[if !supportLists]–>j. <!–[endif]–>180.180.79.100/30

<!–[if !supportLists]–>k. <!–[endif]–>180.180.79.104/30

<!–[if !supportLists]–>l. <!–[endif]–>180.180.79.108/30

<!–[if !supportLists]–>m. <!–[endif]–>180.180.79.112/30

<!–[if !supportLists]–>n. <!–[endif]–>180.180.79.116/30

<!–[if !supportLists]–>o. <!–[endif]–>180.180.79.120/30

<!–[if !supportLists]–>p. <!–[endif]–>180.180.79.124/30

We will assign:

<!–[if !supportLists]–>a. <!–[endif]–>180.180.79.64/30 for network between Router A & Router B.

<!–[if !supportLists]–>b. <!–[endif]–>180.180.79.68/30 network between Router B & Router C.

<!–[if !supportLists]–>c. <!–[endif]–>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

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