Networking Help Assume the Internet Network Information center assigns you the r

Question

Networking Help

Assume the Internet Network Information center assigns you the range of IP address: 172.200.0.0/16 (show your work) a. How many bits do you need to barrow from n the host portion for 1024 subnets? b. How many host addresses can be generated for each subnet? c. What is the subnet mask? d. What subnet and broadcast address is the IP address 172.200.1.190 a member of? e. A router receives packet on an interface with a destination address of 172.200.1.127, what a will the router do with this packet?

Explanation / Answer

So, here you are supposed to make 1024 subnets.

210 = 1024.

So, if we use 10 bits from the host part, we can get different 1024 addresses.

0000000000

0000000001

0000000010

.

.

.

1111111101

1111111110

1111111111

But, from those, 1st and last addresses are not available for use. Because always the first address is used to determine the network id and the last address is broadcast address.

So, only 1022 addresses are used to define subnets. But this does not serve our need as we need 1024 subnets.

So, we will use 1 more bit so that we have 211 = 2048 possible addressees and from these, maximum 2046 are available for subnets.

So, we can have 11 bits and we will use only 1024 from the maximum 2046 available subnets.

So, 11 bits will be needed to borrow from the host portion to define 1024 subnet.

Now, 16 bits are already use for network part and more 11 bits are used in subnetting.

So, total 27 bits are used for networking and only 5 bits are remaining for host addresses.

25 = 32 and from those 32, we can not use the first and the last address, the first address is to identify the network and the last address is used for broadcasting.

So, 30 host addresses can be generated for each subnet.

From 00000 to 11101

The subnet mask is of 27-bits as I wrote above. So, let’s write it in binary first.

11111111.11111111.11111111.11100000

So, writing it in octet format, we get 255.255.255.224

So, the subnet mask is 255.255.255.224.

Given IP address is 172.200.1.190. To find the required details, let’s write it in binary first.

10101100.11001000.00000001.10111110 / 27

The first 16 bits are network id, then next 11 bits are determining the subnet.

The next 11 digits are : 00000001101. So in decimal it becomes 7. So, this IP address belongs to 7th subnet as the first subnet address was 00000000001 i.e. 1 in decimal.

So, the network address of this subnet is : 10101100.11001000.00000001.10100000.

So the network address of subnet 7 to which given address belongs is 172.200.1.160/27.

The broadcast address means write 1 in every bit of host. So, replace the last 5 with 1, we get 10101100.11001000.00000001.10111111.

So in decimal format the broadcast address becomes 172.200.1.191.

Now the router receives a packet with destination address 172.200.1.127 so let’s first write it in binary.

10101100.11001000.00000001.01111111

From this, firstly it will identify the network id which is of 27-bits from above which is 10101100.11001000.00000001.01100000 (172.200.1.96/27 in decimal format)

Here, subnet representing 11 bits are 00000001011 which in decimal means 11. So this destination is in the subnet 11.

And the last 5 bits representing host part are written 1 which means you need to broadcast on this subnet network.

So, the router send this packet to every host (total there are 30 hosts in that subnet) to subnet 11 which is having IP address 172.200.1.96/27.

The answers are written above in Bold Letters along with the description.

Do comment if there is any query. Thank you. :)

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