Question

tion 3 15 Points) Consider the following Internet site. (2001.16.0/24 ( 200.1.17.0/24 (200.1.20.0/24 200 1.21.0/24 Route Advertisement 200 1.22.0/24 Internet R3 R1 Route Advertisement Advertisemen Advertisemel 200 1.23.0/24 Route R4 200.1.18.0/24 2001.19.0/24 The site has 8 class C IP addresses, from 200.1.16.0/24 to 200.1.23.0/24. To reduce the size of the routing tables, address summarization is used. Determine the routes that should be advertised by routers R1,R2 R3 and R4 in order to minimize the number of advertised routes. Over-summarization is not allowed.

Answer 1

1. R2 has two networks

200.1.16.0/24

200.1.17.0/24

Lets' convert them in binary

200.1.16.0 = 11001000.00000001.00010000.00000000

200.1.17.0 = 11001000.00000001.00010001.00000000

Common Bits = 11001000.00000001.00010000.00000000

Common Bits In Decimal = 200.1.16.0

NetMask For Common Bits(All Common Bits as 1 and other as 0) = 11111111.11111111.11111110.00000000(23 1's)

So Advertised Route by R2 = 200.1.16.0/23

2. R3 has four networks

200.1.20.0/24

200.1.21.0/24

200.1.22.0/24

200.1.23.0/24

Lets' convert them in binary

200.1.20.0 = 11001000.00000001.00010100.00000000

200.1.21.0 = 11001000.00000001.00010101.00000000

200.1.22.0 = 11001000.00000001.00010110.00000000

200.1.23.0 = 11001000.00000001.00010111.00000000

Common Bits = 11001000.00000001.00010100.00000000

Common Bits In Decimal = 200.1.20.0

NetMask For Common Bits(All Common Bits as 1 and other as 0) = 11111111.11111111.11111100.00000000(22 1's)

So Advertised Route by R3 = 200.1.20.0/22

3. R4 has two networks

  200.1.18.0/24

  200.1.19.0/24

Lets' convert them in binary

200.1.18.0 = 11001000.00000001.00010010.00000000

200.1.19.0 = 11001000.00000001.00010011.00000000

Common Bits = 11001000.00000001.00010010.00000000

Common Bits In Decimal = 200.1.18.0

NetMask For Common Bits(All Common Bits as 1 and other as 0) = 11111111.11111111.11111110.00000000(23 1's)

So Advertised Route by R3 = 200.1.18.0/23

4. R1 has three networks(As calculated in the above 3 examples)

200.1.16.0/23

200.1.20.0/22

200.1.18.0/23

Lets' convert them in binary

200.1.16.0 = 11001000.00000001.00010000.00000000

200.1.20.0 = 11001000.00000001.00010100.00000000

200.1.18.0 = 11001000.00000001.00010010.00000000

Common Bits = 11001000.00000001.00010000.00000000

Common Bits In Decimal = 200.1.16.0

NetMask For Common Bits(All Common Bits as 1 and other as 0) = 11111111.11111111.11111000.00000000(21 1's)

So Advertised Route by R1 = 200.1.16.0/21

Trick:

If you want to avoid binary, this might work

Advertised Route IP: Smallest IP of Network

So How to Find Net Mask?

/24 Means 24 1's

and 1 bit can be used to represent two networks

Ex.

-> If Two Network then decrease one 1's, because of 2¹ = 2 => /23(Ex. 1 & 3)

-> If more than two but less than or equal to four networks then decrease two 1's, because of 2² = 4 => /22 (Ex. 2)

-> If more than four but less than or equal to eight networks then decrease three 1's, because of 2³ = 8 => /21(Ex. 4)

-> Similarly, decrease n 1's where 2ⁿ are next bigger network than the number of networks => /(24-n)