A student has configured a software router and a workstation on a private LAN. H

Question

A student has configured a software router and a workstation on a private LAN. He has set up the network with IP address 10.0.0.0 with subnetting that borrows bits 15, 16 and 17 (ONLY bits 15, 16 and 17, counting from the left) to create eight subnets (six useable). The student has configured a workstation on the second useable subnet as the lowest available host IP address. The router has the highest usable address on this subnet for its interface connection to this network. The student has set up the workstation using the 138th useable IP address on this subnet. What are these three IP addresses in binary? What are these addresses in dotted decimal?

Explanation / Answer

Network Id is: 10.0.0.0

So this is the address of Class A.

Its subnet mask is 255.0.0.0 and in binary it will be..
11111111.00000000.00000000.00000000

Now, student is using bits 15, 16 and 17 for subnetting. Representing those bits in above subnet mask using X..

11111111.000000XX.X0000000.00000000

As there are 3 bits, 23 = 8 combinations are possible.

11111111.00000000.00000000.00000000
11111111.00000000.10000000.00000000
11111111.00000001.00000000.00000000
11111111.00000001.10000000.00000000

11111111.00000010.00000000.00000000
11111111.00000010.10000000.00000000
11111111.00000011.00000000.00000000
11111111.00000011.10000000.00000000

From these 8, as it is written, only 6 are usable subnets. As the first one is used as a network id and the last one will be used in broadcasting.

So the 6 usable subnet addresses are:

11111111.00000000.10000000.00000000
11111111.00000001.00000000.00000000
11111111.00000001.10000000.00000000
11111111.00000010.00000000.00000000
11111111.00000010.10000000.00000000
11111111.00000011.00000000.00000000

Now, student is using second subnet from these. Which is
11111111.00000001.00000000.00000000

Now, this subnet mask is having till 17 bits as network part and so the remaining 15 bits will be used for the host part.

The host part here will be : (denoted by X)
11111111.00000001.0XXXXXXX.XXXXXXX

So, the addresses of these combinations are: (total 215 addresses)

11111111.00000001.00000000.00000000
.
.
.
11111111.00000001.01111111.11111111

From these, the first address is used as the network id and the last one will be used for the broadcast address.

So, the remaining addresses will be used for assigning to the workstations and routers, PCs etc..

Now, 10 is the first octant of the IP address so its binary will be 00001010. This will be the first octant of every IP addresses.

From these, the lowest available host
00001010.00000001.00000000.00000001

So, student has configured the workstation on this address.

The router has the highest usable address which is just 1 above the broadcast address. So the address for the router becomes
00001010.00000001.01111111.11111110

As broadcast address for the same subnet is
00001010.00000001.01111111.11111111

So, just the above address of it is used for router.

The 138th usable IP address is: (binary of 138 at the end)
00001010.00000001.00000000.10001010

This is assigned to the workstation.

So, the 3 addresses used here are:

workstation 1 (lowest possible address) : 00001010.00000001.00000000.00000001
router (highest possible address) : 00001010.00000001.01111111.11111110
workstation 2 (138th address) : 00001010.00000001.00000000.10001010

The addresses in the dotted decimal form are:

workstation 1 : 10.1.0.1/17
Router : 10.1.127.254/17
workstation 2 : 10.1.0.138/17

Here 17 represents the (total) network part as the subnet mask for all these addresses is : 11111111.00000001.00000000.00000000. or 255.1.0.0/17.

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

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