Question

[show your work - no calculators of any sorts] Scenario 1 (Use the network 221.108.72.0 for the following questions.) 5) We need to subnet this network to support a minimum of 7 (seven) networks. How many bits do we need to use from the last octet to support this?| 6) How many hosts would each network be able to support? 7) What will the subnet mask be? 8) Give the network, IP range, and broadcast address for the first 3 networks: Network 1: Broadcast: Host IP Range: to Network 2: Broadcast: Host IP Range: to Network 3: Broadcast: Host IP Range: to

Answer 1

PART 5) The network given is 221.108. 72.0

The given IP address is of class C as in class C first octet ranges from 192 to 223

Here 1s t octet contains 221 which lies between 192 and 223 so the address belongs to class C

Class C contains 24 net id bits and 8 host I'd bits

We can write class C address as N. N.N.H

Where N = net bits each N represents 8 bits so

N+N+N=8+8+8=24 bits in Net id

H=8 bits in host I'd.

So now we have to subnet this network to support a minimum of seven networks.

Requirements of network =7

>= requirement

2

Where n is number of bits used from last octet

>7

2

Using n=3

>7

23

8>7

Condition is satisfied for n=3

3 bits are used for subnet of 7 network

3 bits are used from last octet to support a minimum of seven networks.

PART 6)

The number of hosts each network would able to support

Network address given = 221.108.72.0

The last octet has total 8 bits in which 3 bits are used to support seven networks as calculated in part 5 so

remaining bits for host = Total bits in last octet - subnet bits for seven networks

Remaining bit of host = 8-3=5

So 5 bits are used for host

Number of host for each network = $2^{5}$ =32

But the two host address of all 1's and all 0's are not allowed so subtracting 2 from 32

32-2=30

Each network would be able to support 30 hosts.

PART 7) subnet mask of network contains all 1's in net I'd and 0 in host I'd

Network address = 221.108.72.0

This is class C address which contain N. N. N. H

Net I'd =24 bits and host I'd = 8 bits

But for supporting 7 networks in subnet it uses 3 bits of host id  so total net id bits = 24+3=27 bits

Remaining host I'd bits = 5 bits

So putting all 1's in net I'd and 0 in host id

11111111.11111111.11111111.11100000

255.255.255.224

Subnet mask =255.255.255.224

PART 8)

(1) Network 1 address will have subnet bit =001 for 1 st network

221.108.72.00100000= 221.108.72.32

NETWORK 1 : 221.108.72.32

Broadcast address will contain all 1 in last 5 bits of host I'd

221.108.72.00111111=221.108.72.63

BROADCAST : 221.108.72.63

The host IP range will start from next address after subnet address (00001) represents 1st host of subnet and end at one less than broadcast address.(63-1=62)

Host IP range = 221.108.72.00100001 to 221.108.72.00111110

HOST IP RANGE = 227.108.72.33 to 221.108.72.62

(2)Network 2

Network 2 address contains subnet bit as 010( 2 in binary)

221.108.72.01000000=221.108.72.64

NETWORK 2 : 221.108.72.64

Broadcast address will contain all 1's in last 5 bits of host id

221.108.72.01011111= 221.108.72.95

BROADCAST : 221.108.72.95

Host IP range = 221.108.72.01000001 to 221.108.72.01011110

HOST IP RANGE=221.108.72.65 to 221.108.72.94

(3)Network 3 address will contain subnet bit =001(3 in binary)

221.108.72.01100000 =221.108.72.96

NETWORK 3 : 221.108.72.96

Broadcast address will contain all 1's in last 5 bits of host

221.108.72.01111111= 221.108.72.127

BROADCAST: 221.108.72.127

Host IP range = 221.108.72.01100001 to 221.108.72.01111110

HOST IP RANGE =221.108.72.97 to 221.108.72.126

NETWORK 1 : 221.108.72.32   BROADCAST : 221.108.72.63

HOST IP RANGE : 227.108.72.33 to 221.108.72.62

NETWORK 2 : 221.108.72.64 BROADCAST : 221.108.72.95

HOST IP RANGE : 221.108.72.65 to 221.108.72.94

NETWORK 3 : 221.108.72.96   BROADCAST: 221.108.72.127

HOST IP RANGE : 221.108.72.97 to 221.108.72.126

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