Given 10.0.0.0/8 as the Network Prefix, and the IP requirements of the subnet are as follows:
Largest Subnetwork - FTMK: 500 Hosts
Quantity of host: 2H - 2 ≈ 29 - 2 = 510(Usable Address)
Therefore, we need 9 hosts bits.
Subnet Mask = /23 (32-23 = 9)
So, the situation is 10.0.0000 000X . XXXX XXXX
Subnet 0: 10.0.0000 000X . XXXX XXXX
Subnet 1: 10.0.0000 001X . XXXX XXXX
So, the IP for subnet 1:
For 500 Hosts:
10.0.0000 000X . XXXX XXXX
Then:
Network Address: 10.0.0000 0000.0000 0000
IP Range: 10.0.0000 0000.0000 0001- 10.0.0000 0001.1111 1110
Broadcast Address: 10.0.0000 0001.1111 1111
Decimal Form:
Network Address: 10.0.0.0
IP Range: 10.0.0.1 - 10.0.1.254
Broadcast Address: 10.0.1.255
Second Largest Subnetwork FKE: 300 Hosts
Quantity of host: 2H - 2 ≈ 29 - 2 = 510(Usable Address)
Therefor, we need 9 hosts bits.
Subnet Mask = /23 (32-23 = 9)
Subnet 0: 10.0.0000 001X . XXXX XXXX
Subnet 1: 10.0.0000 010X . XXXX XXXX
So, the IP for subnet 2:
For 300 Hosts:
10.0.0000 001X . XXXX XXXX
Then:
Network Address: 10.0.0000 0010.0000 0000
IP Range: 10.0.0000 0010.0000 0001- 10.0.0000 0011.1111 1110
Broadcast Address: 10.0.0000 0011.1111 1111
Network Address: 10.0.2.0
IP Range: 10.0.2.1 - 10.0.3.254
Broadcast Address: 10.0.3.255
Third Largest Subnetwork FKEKK: 250 Hosts
Quantity of host: 2H - 2 ≈ 28 - 2 = 254(Usable Address)
Therefor, we need 9 hosts bits.
Subnet Mask = /24 (32-24 = 8)
Subnet 0: 10.0.0000 0100 . XXXX XXXX
Subnet 1: 10.0.0000 0101 . XXXX XXXX
So, the IP for subnet 3:
For 250 Hosts:
10.0.0000 0100 . XXXX XXXX
Then:
Network Address: 10.0.0000 0100.0000 0000
IP Range: 10.0.0000 0100.0000 0001- 10.0.0000 0100.1111 1110
Broadcast Address: 10.0.0000 0100.1111 1111
Network Address: 10.0.4.0
IP Range: 10.0.4.1 - 10.0.4.254
Broadcast Address: 10.0.4.255
Fourth Largest Subnetwork FKM: 124 Hosts
Quantity of host: 2H - 2 ≈ 27 - 2 = 126(Usable Address)
Therefore, we need 7 hosts bits.
Subnet Mask = /25 (32-25 = 7)
Subnet 0: 10.0.0000 0101 . 0XXX XXXX
Subnet 1: 10.0.0000 0101 . 1XXX XXXX
So, the IP for subnet 4:
For 124 Hosts:
10.0.0000 0101 . 0XXX XXXX
Then:
Network Address: 10.0.0000 0101.0000 0000
IP Range: 10.0.0000 0101.0000 0001- 10.0.0000 0101.0111 1110
Broadcast Address: 10.0.0000 0101.0111 1111
Network Address: 10.0.5.0
IP Range: 10.0.5.1 - 10.0.5.126
Broadcast Address: 10.0.5.127
Fifth Largest Subnetwork FKP: 54 Hosts
Quantity of host: 2H - 2 ≈ 26 - 2 = 62(Usable Address)
Therefor, we need 6 hosts bits.
Subnet Mask = /26 (32-23 = 6)
Subnet 0: 10.0.0000 0101 . 10XX XXXX
Subnet 1: 10.0.0000 0101 . 11XX XXXX
So, the IP for subnet 5:
For 54 Hosts:
10.0.0000 0101 . 10XX XXXX
Then:
Network Address: 10.0.0000 0101.1000 0000
IP Range: 10.0.0000 0101.1000 0001- 10.0.0000 0101.1011 1110
Broadcast Address: 10.0.0000 0101.1011 1111
Network Address: 10.0.5.128
IP Range: 10.0.5.129 - 10.0.5.190
Broadcast Address: 10.0.5.191
Sixth Largest Subnetwork Post Graduate Center: 22 Hosts
Quantity of host: 2H - 2 ≈ 25 - 2 = 30(Usable Address)
Therefor, we need 5 hosts bits.
Subnet Mask = /27 (32-27 = 5)
Subnet 0: 10.0.0000 0101 . 110X XXXX
Subnet 1: 10.0.0000 0101 . 111X XXXX
So, the IP for subnet 6:
For 22 Hosts:
10.0.0000 0101 . 110X XXXX
Then:
Network Address: 10.0.0000 0101.1100 0000
IP Range: 10.0.0000 0101.1100 0001- 10.0.0000 0101.1101 1110
Broadcast Address: 10.0.0000 0101.1101 1111
Network Address: 10.0.5.192
IP Range: 10.0.5.193 - 10.0.5.222
Broadcast Address: 10.0.5.223
Seventh Largest Subnetwork Canselory Office: 10 Hosts
Quantity of host: 2H - 2 ≈ 24- 2 = 14(Usable Address)
Therefore, we need 5 hosts bits.
Subnet Mask = /28 (32-28 = 4)
Subnet 0: 10.0.0000 0101 . 1110 XXXX
Subnet 1: 10.0.0000 0101 . 1111 XXXX
So, the IP for subnet 7:
For 10 Hosts:
10.0.0000 0101 . 1110 XXXX
Then:
Network Address: 10.0.0000 0101.1110 0000
IP Range: 10.0.0000 0101.1110 0001- 10.0.0000 0101.1110 1110
Broadcast Address: 10.0.0000 0101.1110 1111
Network Address: 10.0.5.224
IP Range: 10.0.5.225 - 10.0.3.238
Broadcast Address: 10.0.5.239
Eight Largest Subnetwork Registrar Office: 5 Hosts
Quantity of host: 2H - 2 ≈ 23- 2 = 6(Usable Address)
Therefor, we need 4 hosts bits.
Subnet Mask = /29 (32-29 = 3)
Subnet 0: 10.0.0000 0101 . 1111 0XXX
Subnet 1: 10.0.0000 0101 . 1111 1XXX
So, the IP for subnet 8:
For 5 Hosts:
10.0.0000 0101 . 1111 0XXX
Then:
Network Address: 10.0.0000 0101.1111 0000
IP Range: 10.0.0000 0101.1111 0001- 10.0.0000 0101.1111 0110
Broadcast Address: 10.0.0000 0101.1111 1011
Network Address: 10.0.5.240
IP Range: 10.0.5.241 - 10.0.5.246
Broadcast Address: 10.0.5.247
Ninth Largest Subnetwork Library: 4 Hosts
Quantity of host: 2H - 2 ≈ 23 - 2 = 6(Usable Address)
Therefor, we need 3 hosts bits.
Subnet Mask = /29 (32-29 = 3)
Subnet 0: 10.0.0000 0101 . 1111 1XXX
Subnet 1: 10.0.0000 0110 . 0000 0XXX
So, the IP for subnet 9:
For 4 Hosts:
10.0.0000 0101 . 1111 1XXX
Then:
Network Address: 10.0.0000 0101.1111 1000
IP Range: 10.0.0000 0101.1111 1001- 10.0.0000 0101.1111 1110
Broadcast Address: 10.0.0000 0101.1111 1111
Network Address: 10.0.5.248
IP Range: 10.0.5.249 - 10.0.5.254
Broadcast Address: 10.0.5.255
Tenth Largest Subnetwork Sports Center: 2 Hosts
Quantity of host: 2H - 2 ≈ 22 - 2 = 2(Usable Address)
Therefore, we need 2 hosts bits.
Subnet Mask = /30 (32-30 = 2)
Subnet 0: 10.0.0000 0110 . 0000 00XX
Subnet 1: 10.0.0000 0110 . 0000 01XX
So, the IP for subnet 10:
For 2 Hosts:
10.0.0000 0110 . 0000 00XX
Then:
Network Address: 10.0.0000 0110.0000 0000
IP Range: 10.0.0000 00110.0000 0001- 10.0.0000 0110.0000 0010
Broadcast Address: 10.0.0000 0110.0000 0011
Network Address: 10.0.6.0
IP Range: 10.0.6.1 - 10.0.6.2
Broadcast Address: 10.0.6.3
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