Hi, please explain step by step how to do it, I really don\'t understand. I have
ID: 3846056 • Letter: H
Question
Hi, please explain step by step how to do it, I really don't understand. I have included the solution to the question so that you know what the answer is. Please, explain it in details how to solve this :)
B:24 2p. onsider the following forwarding table. Destination network Interface 0.0.0.0/0 m0 172.58.128.0/17 ml 172.58.128.0/19 m2 172.58.160.0/19 m3 Specify the outgoing interface for each of the following destination addresses: 172.58.124.36, 172.58.169.18, 172.58.218.80, 172. 58.1 55. 112. Svar: 172.58.124.36 m0 172.58.169.18 m3 172.58.218.80 ml 172.58.155.112 m2Explanation / Answer
To explain this, we are going to represent each of the destination addresses in its binary form in 8 bits format and understand what each of it means:
m0: 0.0.0.0/0
Binary representation- 00000000.00000000.00000000.00000000/0, the '/0' means the number of bits that is assigned to the sub network of this destination. It means for a destination address to get forwarded to this interface the first 0 bits must match. Any ip of the form xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx can be forwarded to this interface.
m1: 172.58.128.0/17
Binary representation- 10101100.00111010.10000000.00000000. The '/17' means the number of bits that is assigned to the sub network of this destination. It means for a destination address to get forwarded to this interface first 17 bits must match. So any IP of the form 10101100.00111010.1xxxxxxx.xxxxxxxx, where x can be either 0 or 1 can be forwarded to this interface.
m2: 172.58.128.0/19
Binary representation- 10101100.00111010.10000000.00000000. The '/19' means the number of bits that is assigned to the sub network of this destination. It means for a destination address to get forwarded to this interface first 19 bits must match. So any IP of the form 10101100.00111010.100xxxxx.xxxxxxxx, where x can be either 0 or 1 can be forwarded to this interface.
m3: 172.58.160.0/19
Binary representation- 10101100.00111010.10100000.00000000. The '/19' means the number of bits that is assigned to the sub network of this destination. It means for a destination address to get forwarded to this interface first 19 bits must match. So any IP of the form 10101100.00111010.101xxxxx.xxxxxxxx, where x can be either 0 or 1 can be forwarded to this interface.
We now understand how the outgoing interface has been computed:
1) 172.58.124.36
Binary representation- 10101100.00111010.01111100.00100100. If we look at first 19 bits, it doesnt match m3 or m2 (the last 3 binary digits differ). If we look at the first 17, bits they do not match m1(the last bit differs). Hence It has to be forwarded to m0.
d: 10101100.00111010.01111100.00100100
m3: 10101100.00111010.101xxxxx.xxxxxxxx
m2: 10101100.00111010.100xxxxx.xxxxxxxx
m1: 10101100.00111010.1xxxxxxx.xxxxxxxx
m0: xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx
2) 172.58.169.18
Binary representation- 10101100.00111010.10101001.00010010. If we look at first 19 bits, it doesnt matches m3, which is '10101100.00111010.101', and hence the out going interface is m3.
d: 10101100.00111010.10101001.00010010
m3: 10101100.00111010.101xxxxx.xxxxxxxx
m2: 10101100.00111010.100xxxxx.xxxxxxxx
m1: 10101100.00111010.1xxxxxxx.xxxxxxxx
m0: xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx
3) 172.58.218.80
Binary representation- 10101100.00111010.11011010.01010000. If we look at first 19 bits, it doesnt match m3 or m2 (the last 3 binary digits differ). If we look at the first 17, bits they match m1(10101100.00111010.1). Hence It has to be forwarded to m1.
d: 10101100.00111010.11011010.01010000
m3: 10101100.00111010.101xxxxx.xxxxxxxx
m2: 10101100.00111010.100xxxxx.xxxxxxxx
m1: 10101100.00111010.1xxxxxxx.xxxxxxxx
m0: xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx
4) 172.58.155.112
Binary representation- 10101100.00111010.10011011.1110000. If we look at first 19 bits, it matches m2 . Hence It is forwarded to m2.
d: 10101100.00111010.10011011.1110000
m3: 10101100.00111010.101xxxxx.xxxxxxxx
m2: 10101100.00111010.100xxxxx.xxxxxxxx
m1: 10101100.00111010.1xxxxxxx.xxxxxxxx
m0: xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.