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3. (20 points) a. A host in an organization has an IP address of 158.78.118.108

ID: 3737161 • Letter: 3

Question

3. (20 points) a. A host in an organization has an IP address of 158.78.118.108 with a network mask of "/20" What is the network address of the subnet that this IP address belongs to? What is the range of the IP addresses that an individual host can have on this subnet? Justify your answers. A host in another organization has an IP address of 204.14.174.54 with a network mask of What is the network address of the supernet that this IP address belongs to? How many Class-C address blocks does this supernet include? Justify your answers. b. 20

Explanation / Answer

3 ( a )

Address:      158.78.118.108        10011110.01001110.0111 0110.01101100

Netmask is   /20

So the the netmask address would be as follows.

Netmask:   255.255.240.0 = 20    11111111.11111111.1111 0000.00000000     ( 20 MSB bits set to 1 )

The network address would be calculated as follows.

Address       158.78.118.108

Netmask     255.255.240.0

===========================

Network     158.78.112.0

Network:     158.78.112.0/20       10011110.01001110.0111 0000.00000000 (Class B range 128 - 191)

Broadcast:   158.78.127.255        10011110.01001110.0111 1111.11111111

The range of host will be as follows.

The minimum host will the next host address from the network address.

The maximum host will the previous host address of the Broadcast address.

HostMin:     158.78.112.1          10011110.01001110.0111 0000.00000001

HostMax:   158.78.127.254      10011110.01001110.0111 1111.11111110

Total Hosts/Net: 2 ^ 12 = 4096 - 2 = 4094

Total hosts = (2 to the power of no. of host bits) - 2

Total hosts is subtracted by 2 as the network address and broadcast address of the subnet cannot be used by any other hosts. So these two are not included in total hosts range.

3 (b)        

Address:   204.14.174.54         11001100.00001110.1010 1110.00110110

Netmask is    /20

So the the netmask address would be as follows.

Netmask:   255.255.240.0 = 20    11111111.11111111.1111 0000.00000000      ( 20 MSB bits set to 1 )

The network address would be calculated as follows.

Address       204.14.174.54

Netmask     255.255.240.0

=============================

Network     204.14.160.0

Network:   204.14.160.0/20       11001100.00001110.1010 0000.00000000 (Class C Range = 192 – 223 )

Broadcast: 204.14.175.255        11001100.00001110.1010 1111.11111111

The range of host will be as follows.

The minimum host will the next host address from the network address.

The maximum host will the previous host address of the Broadcast address.

HostMin:    204.14.160.1           11001100.00001110.1010 0000.00000001

HostMax:   204.14.175.254        11001100.00001110.1010 1111.11111110

Total Hosts/Net: (2 ^ 12) - 2 = 4096 - 2 = 4094

Total hosts = (2 to the power of no. of host bits) - 2

Total hosts is subtracted by 2 as the network address and broadcast address of the supernet cannot be used by any other hosts. So these two are not included in total hosts range.

                

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