Question

as required below using the base network: te subnets 1. Crea 192.168.20.0/24 (12 Points) Subnet Name Needed Size 80 30 14 2 2 ANS: VETİ NUMBER | NETWORK |SUBNET | ASSIGNABLE | BROADCAST E OF IPS 80 50 ADDRESS MASK IP RANGE REQUIRED 30 14 2

Answer 1

Number of IP required	network address	Subnet mask	assignable IP range	Broadcast
80	192.168.20.0	25	192.168.20.1-192.168.20.126	192.168.20.127
50	192.168.20.128	26	192.168.20.129-192.168.20.190	192.168.20.191
30	192.168.20.192	27	192.168.20.193-192.168.20.222	192.168.20.223
14	192.168.20.224	28	192.168.20.225-192.168.20.238	192.168.20.239
2	192.168.20.240	30	192.168.241-192.168.242	192.168.243
2	192.168.20.244	30	192.168.20.245-192.168.20.246	192.168.20.247

For subnet A ,number of IP required = 80

Number of host bits are = 2ⁿ - 2 = 80

n=7 (approx)

Therefore,prefix for subnet A = 32-7= 25

first IP address of the subnet = 192.168.20.0 (network address)

For last IP address, OR operation will be performed between first IP address and !Netmask

11000000.10101000.00010100.00000000 .(first IP address)

OR

00000000.00000000.00000000.01111111 (!Netmask)

11000000.10101000.00010100.01111111 (result)

Result (in decimal)= 192.168.20.127

Broadcast IP address = 192.168.20.126

As first IP address of the subnet is network ID and last IP address is broadcast address and both are not usable IP address therefore,

usable IP address range = 192.168.20.1 to 192.168.20.126

Now, for subnet B ,number of IP address required are 50

Number of host bits = 2ⁿ - 2 = 50

n= 6

prefix for subnet B = 32-6= 26

Carrying forward the previous subnet ,

first IP address of subnet B = 192.168.20.128 (network address)

For last IP address , OR operation will be performed

11000000.10101000.00010100.10 000000 (first IP address i.e 192.168.20.128)

OR

00000000.00000000.00000000.00111111 (!Netmask)

11000000.10101000.00010100.10111111 (result)

Result (in decimal)= 192.168.20.191

Broadcast IP address= 192.168.20.191

therefore, usable IP address range = 192.168.20.129 to 192.168.20.190

Now, subnet C requires 30

Number of host bits = 5

prefix for subnet C =32-5 =27

first IP address = 192.168.20.192 (network address)

for last IP ,OR operation will be performed

11000000.10101000.00010100.11000000 (first IP address i.e 192.168.20.192)

OR

00000000.00000000.00000000.00011111 (!Netmask)

​11000000.10101000.00010100.11011111 (result)

Result (in decimal) = 192.168.20.223

last IP address of the subnet = 192.168.20.223 (broadcast IP)

usable IP range = 192.168.20.192 to 192.168.20.223

Now, for subnet D ,it requires 14 IP address

Number of host bits required = 4

required prefix for subnet D = 32 -4 = 28

First IP address = 192.168.20.224

last IP address =

11000000.10101000.00010100.11011111 (IP address)

OR

00000000.00000000.00000000.00001111 (!Netmask)

00000000.00000000.00000000.00001111 (result)

Result (in decimal) = 192.168.20.239

last IP address of the subnet = 192.168.20.239 (broadcast IP)

usable IP address = 192.168.20.225 to 192.168.20.238

For both subnet E and F ,number of host bits required are

2ⁿ - 2 = 2

n=2

^{Required prefix for both subnet E and F are = 32 -2 =30}

^{first IP address of subnet E = 192.168.20.240 (network ID)}

^{last IP address =}

11000000.10101000.00010100.11110000 (first IP address)

OR

00000000.00000000.00000000.00000011 (!Netmask)

11000000.10101000.00010100.11110011 (result)

Result (in decimal) = 192.168.20.243

last IP address = 192.168.20.243 (broadcast IP)

usable IP address = 192.168.20.241 to 192.168.20.242

first IP for subnet F= 192.168.20.244

last IP address = 192.168.20.247 (broadcast IP)