Routing, Switching & Virtual LANs

1. Specify the number of bits allowed for the host part of each CIDR IP address below and the number of hosts the network can address.
     
   1. 200.10.50.4/26
   2. 170.13.45.77/18
   3. 116.54.31.6/10
                                                      

  
       Q1 Answer

1. 26 bits are for the network address and so 6 bits are left for the hosts. Thus, 26 - 2 = 62 useable host addressess.
    
2. 18 bits are for the network address and so 14 bits are left for the hosts. Thus, 214- 2 = 16382 useable host addressess.
      
3. 10 bits are for the network address and so 22 bits are left for the hosts. Thus, 222 - 2 = 4194302 useable host addressess.

1. Suppose an ISP maintains the following four Class B address blocks:
     
   • 150.12.0.0/16
   • 150.13.0.0/16
   • 150.14.0.0/16
   • 150.15.0.0/16
       
   Aggregate these four address blocks into a single route.

  
      Q1 Answer

The four network addresses in binary are...

10010110.00001100.00000000.00000000
10010110.00001101.00000000.00000000
10010110.00001110.00000000.00000000
10010110.00001111.00000000.00000000

The common digits starting from the left are...

10010110.000011

Since there are 14 common digits then the single route is...

150.12.0.0/14

1. Suppose we have four IP subnets on the four LAN interfaces of our router:

• 210.6.0.0/24
• 210.6.1.0/24
• 210.6.2.0/24
• 210.6.3.0/24

Summarize these networks into a single route that can be advertised across the WAN.

  
       Q1 Answer

The four subnets in binary are...

11010010.00000110.00000000.00000000
11010010.00000110.00000001.00000000
11010010.00000110.00000010.00000000
11010010.00000110.00000011.00000000

The common digits starting from the left are...

11010010.00000110.000000

Since there are 22 common digits then the single route is...

210.6.0.0/22

1. Calculate the supernet address for the networks...
         
   • 172.19.0.0/16
   • 172.20.0.0/16
   • 172.21.0.0/16
      
2. Summarize the following network addresses..
         
   • 172.16.169.0/24
   • 172.16.170.0/24
   • 172.16.171.0/24
       
3. Summarize the following network addresses..
         
   • 200.198.48.0/24
   • 200.198.52.0/24
   • 200.198.56.0/24

  
       Q1 Answer

The three networks in binary are...

10101100.00010011.00000000.00000000
10101100.00010100.00000000.00000000
10101100.00010101.00000000.00000000

The common digits starting from the left are...

10101100.00010

Since there are 13 common digits then the summary route is...

172.19.0.0/13

       Q2 Answer

The three networks in binary are...

10101100.00010000.10101001.00000000
10101100.00010000.10101010.00000000
10101100.00010000.10101011.00000000

The common digits starting from the left are...

10101100.00010000.101010

Since there are 22 common digits then the summary route is...

172.6.169.0/22

       Q3 Answer

The three networks in binary are...

11001000.11000110.00110000.00000000
11001000.11000110.00110100.00000000
11001000.11000110.00111000.00000000

The common digits starting from the left are...

11001000.11000110.0011

Since there are 20 common digits then the summary route is...

200.198.48.0/20