A block of addresses is granted to a small organization. We know that one of the

Question

A block of addresses is granted to a small organization. We know that one of the addresses and it is related to your student ID. For example, if your student ID is 1013298 then the known address is 192.101.32.98/26 What are the first and the last addresses of this block? Determine the number of addresses granted to this small organization. The network administrator of this small organization wants to create three subnets that have 5, 14, and 21 hosts respectively. Create table showing the followings for each subnet: Subnet mask Network Address First and last hosts addresses Broadcast address

Explanation / Answer

Dear Student,

here is the complete answer of the above asked question...

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The given IP address = 192.101.32.98/26

The given IP address belong to Class C address.

The default subnet mask for class C is 255.255.255.0

Binary notation of 255.255.255.0 is 11111111.11111111.11111111.00000000.

but we have /26 notation it means we have 26 bits which are known as network bits.

so

Subnet Mask = 255.255.255.192

Binary notation of 255.255.255.192 is 11111111.11111111.11111111.11000000.

It means that 26 bit for network and rest bit for host(bit=1 i.e Network, if bit=0 i.e Host).

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Answer....1

What are the first and the last addreesses of this block.?

Here we have borrowed 2 bits then the mask will be 11111111.11111111.11111111.11000000

Subnet = 2N=22=4 subnet
Block size (256 - subnet mask) = 256 - 192 = 64.
Valid subnets ( Count blocks from 0) = 0,64,128,192

So here we have 4 subnets and each subnet has block size = 64.

The address 192.101.32.98 comes in the block range of 64 to 128 that is the second subnet.

hence

First address of second subnet = 192.101.32.65/26

Last address of second subnet = 192.101.32.126/26

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Answer No -2:

Number of addresses granted to this small organization?

Total hosts (2H) in a subnet =26= 64

Valid hosts per subnet ( Total host - 2 ) = 64 - 2 = 62

So we have Total number of address granted = 62 * 4 = 248 address    Ans..

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Answer-3--

In the subnet 1 administrator want to connect 5 hosts per subnet. So Host are less. So create a network in which we can connet 5 PCs.We can borrow 5 bits from the Host bit to create subnetwork.

IF i borrow 5 bit then the mask will be 11111111.11111111.11111111.11111000

Subnet = 2N=25=32 subnets
Block size (256 - subnet mask) = 256 - 248 = 8.
Valid subnets ( Count blocks from 0) = 0,8,16.......256.
Total hosts (2H) =23= 8
Valid hosts per subnet ( Total host - 2 ) = 8 - 2 = 6

So in this subnets we can connect maximum 6 hosts per subnet. So our requirement is to connect 5 Hosts right?

Subnet Mask = 255.255.255.248.0

The network address for Subnet 1 which can connect 5 host is:   192.101.32.0/29

The first usable address for subnet 1 is: 192.101.32.1/29

The last usable address for Subnet 1 is : 192.101.32.6/29

The broadcast address for Subnet 1 is : 192.101.32.7/29

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Subnet:2

In the subnet 2 administrator want to connect 14 hosts per subnet. So Host are less. So create a network in which we can connet 14 PCs.We can borrow 4 bits from the Host bit to create subnetwork.

If i borrow 4 bit then the mask will be 11111111.11111111.11111111.11110000

Subnet = 2N=24=16 subnets
Block size (256 - subnet mask) = 256 - 240 = 16.
Valid subnets ( Count blocks from 0) = 0,16,32.......256.
Total hosts (2H) =24= 16
Valid hosts per subnet ( Total host - 2 ) = 16 - 2 = 14

So in this subnets we can connect maximum 14 hosts per subnet. So our requirement is to connect 14 Hosts right?

Subnet Mask = 255.255.255.240.0

The network address for Subnet 2 which can connect 5 host is:   192.101.32.0/28

The first usable address for subnet 2 is: 192.101.32.1/28

The last usable address for Subnet 2 is : 192.101.32.14/28

The broadcast address for Subnet 2 is : 192.101.32.15/28

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For Subnet: 3

In the subnet 3 administrator want to connect 21 hosts per subnet. So Host are less. So create a network in which we can connet 21 PCs.We can borrow 3 bits from the Host bit to create subnetwork.

IF i borrow 3 bit then the mask will be 11111111.11111111.11111111.11100000

Subnet = 2N=23=8 subnets
Block size (256 - subnet mask) = 256 - 224 = 32.
Valid subnets ( Count blocks from 0) = 0,32,64.......256.
Total hosts (2H) =25= 32
Valid hosts per subnet ( Total host - 2 ) = 32 - 2 = 30

So in this subnets we can connect maximum 32 hosts per subnet. So our requirement is to connect 21 Hosts right?

Subnet Mask = 255.255.255.224.0

The network address for Subnet 3 which can connect 21 host is:   192.101.32.0/27

The first usable address for subnet 3 is: 192.101.32.1/27

The last usable address for Subnet 3 is : 192.101.32.30/27

The broadcast address for Subnet 3 is : 192.101.32.31/27

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Here is the complete Table....which list the Subnet mask, Network address, First and last hosts address and Broad cast address....

Subnet Mask

Network

Address

First Address

Last Address

Broadcast Address

Kindly Check and Verify Thanks...!!!

Subnetting Table Subnet-1 Subnet-2 Subnet-3

Subnet Mask

255.255.255.248.0 255.255.255.240.0 255.255.255.224.0

Network

Address

192.101.32.0/29 192.101.32.0/28 192.101.32.0/27

First Address

192.101.32.1/29 192.101.32.1/28 192.101.32.1/27

Last Address

192.101.32.6/29 192.101.32.14/28 192.101.32.30/27

Broadcast Address

192.101.32.7/29 192.101.32.15/28 192.101.32.31/27

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