Need help with a problem!

127.0.0.1

My sister got this problem for homework. I am a CCENT and I was not able to figure it out! Can anyone solve it and explain to me how it is solved?

Rewrite this forwarding table using the a.b.c.d/x notation instead of the binary string notation.

Destination Address Range | Link Interface

Destination Address Range: 11001000 00010000 00000000 through 11001000 00010111 00010111 11111111 Link Interface: 0

Destination Address Range: 11001000 00010111 00011000 00000000 through 11001000 00010111 00011000 11111111 Link Interface: 1

Destination Address Range: 11001000 00010111 00011001 00000000 through 11001000 00010111 00011001 00000000 through 11001000 00010111 0001111 11111111 Link Interface: 2

Destination Address Range: otherwise Link Interace: 3

dtlokee

I can solve it...

Convert the binary values to decimal then use that to derive the subnet mask, start there

yuriz43

It will take some time to convert the binary but it is a relatively easy problem to solve.

To make life easier, create a table with the following columns.

128 - 64 -32 -16 - 8 - 4 - 2 - 1



Use this to help you determine the value of each binary string. Once you have that figured out it should be easy.

127.0.0.1

I already know how to convert from binary to decimal format. Can someone at least solve one of these and show me how it is done?

dtlokee

Post the conversions then I will show you the next step, I won't do your (or your sister's) homework for you.

127.0.0.1

1. 200.16.0 - 200.20.22.255 Link Interface: 0

Ok, what do I do after this? (This is for the first one).

tiersten

They didn't cover this in CCENT at all?

wbosher

There is an octet missing in the first question!! This is covered in CCENT, without actually answering any of the above questions I'll give you an example.

11000000 10101000 00000000 00000000 - 11000000 10101000 00000011 11111111 in decimal this is 192.168.0.0 - 192.168.3.255.

Take a look at the third octet where the increment is 4, so the mask would be 255.255.252.0 or /22.

So the network address would be 192.168.0.0/22

Hope this makes sense.