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IP ADDRESS CLASSESTask 1: For a given IP address,Determine Network Information.Given :Host IP Address172.25.114.250
Network Mask255.255.0.0 (/16)
Find :
Network Address172.25.0.0
Network Broadcast Address172.25.255.255
Total Number of Host Bits65534
Number of Hosts16
Step 1 : Translate Host IP address and network mask into binary notation.FORMULA :2726252423222120
1286432168421
IP ADDRESS172 . 25 . 114 . 250Octet 1 Octet 2 Octet 3Octet 4Octet 12726252423222120
1286432168421
10101100
Let128+32+8+4= 172Octet 22726252423222120
1286432168421
00011001
Let16+8+1= 25
Octet 32726252423222120
1286432168421
01110010
Let64+32+16+2=144
Octet 42726252423222120
1286432168421
11111010
Let128+64+32+16+8+2=250
NETWORK MASK 255 . 255 . 0 . 0172 . 25 . 114 . 250Octet 1 Octet 2 Octet 3Octet 4Octet 1 2726252423222120
1286432168421
11111111
Let 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1= 255Octet 22726252423222120
1286432168421
11111111
Let128 + 64 + 32 + 16 + 4 + 2 + 1=255Octet 32726252423222120
1286432168421
00000000
LetThe value of network mask at octet 3 is 0 , so the binary number is still zero (0).
Octet 42726252423222120
1286432168421
00000000
LetThe value of network mask at octet 3 is 0 , so the binary number is still zero (0).
Step 2: Determine the network address. 17225114250IP Address10101100000110010111001011111010
Subnet Mask11111111111111110000000000000000
Network Address10101100000110010000000000000000
17225 0 0
EXP:IP Address10101100000110010111001011111010X MultiplySubnet Mask 11111111111111110000000000000000Network Address10101100000110010000000000000000 172 25 0 0 Answer
Step 3: Determine the broadcast address for the network addressThe network mask separates the network portion of the address from the host portion.The network address has all 0s in the host portion of the address and the broadcast address has all 1s in the host portion of the address. 172 25 0 0Network Add.10101100000110010000000000000000
Mask11111111111111110000000000000000
Broadcast.10101100000110011111111111111111
172 25 0 0
By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16`0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,5362n-2FORMULA = 65,536 2 = 65,536
Task 2: ChallengeFor all problem:Create a subnetting Worksheet to show and record all work for each problem.Problem 1Host Ip Address172.30.1.33
Network Mask255.255.0.0
Network Address172.16.0.0
Network Broadcast Address172.16.255.255
Total Number of Host Bits65534
Number of Hosts16
FORMULA :2726252423222120
1286432168421
IP ADDRESS172 . 30 . 1 . 33Octet 1 Octet 2 Octet 3Octet 4
Octet 12726252423222120
1286432168421
10101100
Let128+32+8+4=172
Octet 22726252423222120
1286432168421
00011110
Let16+8+4+2= 30Octet 32726252423222120
1286432168421
00000001
Let1= 1
Octet 42726252423222120
1286432168421
00100001
Let= 33
Step 2: Determine the network address. 17230 1 33IP Address10101100000111100000000100100001
Subnet Mask11111111111111110000000000000000
Network Address10101100000111100000000100000000
17230 0 0
EXP:IP Address10101100000111100000000100100001X MultiplySubnet Mask 11111111111111110000000000000000Network Address10101100000110010000000100000000 172 25 1 0 Answer
Step 3: Determine the broadcast address for the network addressThe network mask separates the network portion of the address from the host portion.The network address has all 0s in the host portion of the address and the broadcast address has all 1s in the host portion of the address. 172 30 0 0Network Add.10101100000111100000000000000000
Mask11111111111111110000000000000000
Broadcast.10101100000111101111111111111111
172 30 1 255
By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA = 65,536 2 = 65,536
Problem 2Host Ip Address172.30.1.33
Network Mask255.255.255.0
Network Address172.30.1.0
Network Broadcast Address172.30.1.255
Total Number of Host Bits254
Number of Hosts24
FORMULA :2726252423222120
1286432168421
IP ADDRESS172 . 30 . 1 . 33Octet 1 Octet 2 Octet 3Octet 4Octet 12726252423222120
1286432168421
10101100
Let128+32+8+4=172
Octet 22726252423222120
1286432168421
00011110
Let16+8+4+2= 30Octet 32726252423222120
1286432168421
00000001
Let1= 1Octet 42726252423222120
1286432168421
00100001
Let= 33
Step 2: Determine the network address. 17230 1 33IP Address10101100000111100000000100100001
Subnet Mask11111111111111110000000000000000
Network Address10101100000111100000000000000000
17230 0 0
EXP:IP Address10101100000111100000000100100001X MultiplySubnet Mask 11111111111111110000000000000000Network Address10101100000110010000000000000000 172 30 0 0 Answer
Step 3: Determine the broadcast address for the network addressThe network mask separates the network portion of the address from the host portion.The network address has all 0s in the host portion of the address and the broadcast address has all 1s in the host portion of the address. 172 30 0 0Network Add.10101100000111100000000000000000
Mask11111111111111110000000000000000
Broadcast.10101100000111101111111111111111
172 25 0 0
By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA = 65,536 2 = 65,536
Problem 3Host Ip Address192.168.10.234
Network Mask255.255.255.0
Network Address192.168.10.0
Network Broadcast Address192.168.10.255
Total Number of Host Bits254
Number of Hosts28
FORMULA :2726252423222120
1286432168421
IP ADDRESS192 . 168 . 10 . 234Octet 1 Octet 2 Octet 3Octet 4Octet 12726252423222120
1286432168421
11000000
Let128+64= 192
Octet 22726252423222120
1286432168421
10101000
Let128+32+8= 168Octet 32726252423222120
1286432168421
00001010
Let8+2= 10Octet 42726252423222120
1286432168421
11111010
Let128+64+32+16+8+2= 234
Step 2: Determine the network address. 192168 10 234IP Address11000000101010000000101011111010
Subnet Mask11111111111111110000000000000000
Network Address11000000101010000000101011111010
192168 0 0
EXP:IP Address11000000101010000000101000100001X MultiplySubnet Mask 11111111111111110000000000000000Network Address11000000101010000000101011111010 192 168 0 0 Answer
Step 3: Determine the broadcast address for the network addressThe network mask separates the network portion of the address from the host portion.The network address has all 0s in the host portion of the address and the broadcast address has all 1s in the host portion of the address. 172 30 0 0Network Add.10101100000111100000000000000000
Mask11111111111111110000000000000000
Broadcast.10101100000111101111111111111111
172 25 0 0By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA =65,536 2 = 65,536
Problem 4Host Ip Address172.17.99.71
Network Mask255.255.0.0
Network Address172.17.0.0
Network Broadcast Address172.17.255.255
Total Number of Host Bits65534
Number of Hosts216
FORMULA :2726252423222120
1286432168421
IP ADDRESS172 . 17 . 99 . 71Octet 1 Octet 2 Octet 3Octet 4Octet 12726252423222120
1286432168421
10101100
Let128+32+8+4= 172
Octet 22726252423222120
1286432168421
00010001
Let16+1= 17
Octet 32726252423222120
1286432168421
01100011
Let64+32+2+1= 99Octet 42726252423222120
1286432168421
01000110
Let64+4+2= 71
Step 2: Determine the network address. 17217 99 71IP Address10101100000100010110001101000110
Subnet Mask11111111111111110000000000000000
Network Address11000000000100010000101011111010
17217 0 0
EXP:IP Address11000000101010000000101000100001X MultiplySubnet Mask 11111111111111110000000000000000Network Address11000000101010000000101011111010 172 17 0 0 Answer
Step 3: Determine the broadcast address for the network addressThe network mask separates the network portion of the address from the host portion.The network address has all 0s in the host portion of the address and the broadcast address has all 1s in the host portion of the address. 172 30 0 0Network Add.10101100000111100000000000000000
Mask11111111111111110000000000000000
Broadcast.10101100000111101111111111111111
172 25 0 0By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA = 65,536 2 = 65,536
Problem 5Host Ip Address192.168.3.219
Network Mask255.255.0.0
Network Address192.168.0.0
Network Broadcast Address192.168.255.255
Total Number of Host Bits65534
Number of Hosts216
FORMULA :2726252423222120
1286432168421
IP ADDRESS192 . 168 . 3 . 219Octet 1 Octet 2 Octet 3Octet 4Octet 12726252423222120
1286432168421
11000000
Let128+64= 192
Octet 22726252423222120
1286432168421
10101000
Let128+32+8= 168Octet 32726252423222120
1286432168421
00000011
Let2+1= 3Octet 42726252423222120
1286432168421
11011011
Let128+64+16+8+2+1= 219
Step 2: Determine the network address. 192168 3 219IP Address11000000101010000000001111011011
Subnet Mask11111111111111110000000000000000
Network Address11000000101010000000101011111010
192168 0 0
EXP:IP Address11000000101010000000101000100001X MultiplySubnet Mask 11111111111111110000000000000000Network Address11000000101010000000101011111010 192 25 0 0 Answer
Step 3: Determine the broadcast address for the network addressThe network mask separates the network portion of the address from the host portion.The network address has all 0s in the host portion of the address and the broadcast address has all 1s in the host portion of the address. 172 30 0 0Network Add.10101100000111100000000000000000
Mask11111111111111110000000000000000
Broadcast.10101100000111101111111111111111
172 25 0 0By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA = 65,536 2 = 65,536
Problem 6Host Ip Address192.168.3.219
Network Mask255.255.0.0
Network Address192.168.0.0
Network Broadcast Address192.168.255.255
Total Number of Host Bits65534
Number of Hosts216
FORMULA :2726252423222120
1286432168421
IP ADDRESS192 . 168 . 3 . 219Octet 1 Octet 2 Octet 3Octet 4Octet 12726252423222120
1286432168421
11000000
Let128+64= 192
Octet 22726252423222120
1286432168421
10101000
Let128+32+8= 168Octet 32726252423222120
1286432168421
00000011
Let2+1= 3Octet 42726252423222120
1286432168421
11011011
Let128+64+16+8+2+1= 219
Step 2: Determine the network address. 192168 3 219IP Address11000000101010000000001111011011
Subnet Mask11111111111111110000000000000000
Network Address11000000101010000000101011111010
192168 0 0
EXP:IP Address11000000101010000000101000100001X MultiplySubnet Mask 11111111111111110000000000000000Network Address11000000101010000000101011111010 172 25 0 0 Answer
Step 3: Determine the broadcast address for the network addressThe network mask separates the network portion of the address from the host portion.The network address has all 0s in the host portion of the address and the broadcast address has all 1s in the host portion of the address. 172 30 0 0Network Add.10101100000111100000000000000000
Mask11111111111111110000000000000000
Broadcast.10101100000111101111111111111111
172 25 0 0By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA = 65,536 2 = 65,536
TASK/EXERCISETASK 1: For a Given IP Adress and Subnet Mask , Determine Subnet Information.Given:Host IP Address172.25.114.250
Network Mask255.255.0.0 (/16)
Subnet Mask255.255.255.192 (/26)
Find:Number of Subnet Bits10
Number of Subnets210=1,024 subnets
Number of Host Bits per Subnet6 bit
Number of Usable Hosts per Subnet26=64 - 2= 62 hosts per subnet
Subnet Address for this IP Address172.25.114.192
IP Address of First Host on this Subnet 172.25.114.193
IP Address of Last Host on this Subnet172.25.114.254
Broadcast Address for this Subnet172.25.114.255
Step 1 : Translate Host IP address and network mask into binary notation.FORMULA :2726252423222120
1286432168421
172 25 114 250IP Address10101100000110010111001011111010X MultiplySubnet Mask 11111111111111111111111111000000Subnet Address10101100000110010111001011000000Subnet address for 172 25 114 192 AnswerThis IP Address =IP Address of First Host 192 + 62 Hosts IP Address of Last Host
Broadcast Address172.25.144.255172.25.114.193 = 254 172.25.114.254
Determining the Number of IP Subnets and HostsStep 1-Use the first octet of the IP address to determine the class of address (A,B or C).172.25.114.250255.255.255.192Step 2Use the class of the address to determine which octets are available for hosts.Network. Network. Host. Host. 172 . 25 . 114 . 250 255 . 255 . 255 . 192Step 3Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed to step four.172.25.114.250255.255.255.192 = 11111111 11000000Step 4Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 11000000 = 10 ones.210 = 1,024 2 = 1,022 usable subnets created.Step 5Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 11000000 = 6 zeros. 26 = 64 2 = 62 usable host addresses created.-Number of Host Bit per Subnet - 6 bit Number of Usable Hosts per Subnet 26 = 64 2 = 62 hosts per subnet Problem 1Host IP Address172.30.1.33
Subnet Mask255.255.255.0
Number of Subnet Bits8
Number of Subnets28 = 256 subnets
Number of Host Bits per Subnet8 bit
Number of Usable Hosts per Subnet28 = 256 2 = 254 hosts per subnet
Subnet Address for this IP Address172.30.1.0
IP Address of First Host on this Subnet172.30.1.1
IP Address of Last Host on this Subnet172.30.1.254
Broadcast Address for this Subnet172.30.1.255
FORMULA :2726252423222120
1286432168421
172 30 1 33IP Address10101100000111100000000100100001X MultiplySubnet Mask 11111111111111111111111100000000Subnet Address10101100000111100000000100000000Subnet address for 172 30 1 0 AnswerThis IP Address =IP Address of First Host 0 + 254 Hosts IP Address of Last Host
Broadcast Address172.25.144.255172.25.114.193 = 254 172.25.114.254
Determining the Number of Subnets and Hosts
Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed to step four.172.25.114.250255.255.255.192 = 11111111 11000000
Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 11000000 = 10 ones.210 = 1,024 2 = 1,022 usable subnets created.
Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 00000000 = 8 zeros. 28 = 256 2 = 254 usable host addresses created.-Number of Host Bit per Subnet 8 bit Number of Usable Hosts per Subnet 28 = 256 2 = 254 hosts per subnet Number of Subnet bit 8 Number of Subnets 28= 256 subnets (all 0s used,all 1s not used)
Problem 2Host IP Address172.30.1.33
Subnet Mask255.255.255.252
Number of Subnet Bits14
Number of Subnets214 = 16384 subnets
Number of Host Bits per Subnet4 bit
Number of Usable Hosts per Subnet22 = 4 2 = 2hosts per subnet
Subnet Address for this IP Address172.30.1.32
IP Address of First Host on this Subnet172.30.1.33
IP Address of Last Host on this Subnet172.30.1.34
Broadcast Address for this Subnet172.30.1.35
FORMULA :2726252423222120
1286432168421
172 30 1 33IP Address10101100000111100000000100100001X MultiplySubnet Mask 11111111111111111111111100000000Subnet Address10101100000111100000000100100000Subnet address for 172 30 1 32 AnswerThis IP Address =IP Address of First Host 32 + 2 Hosts IP Address of Last Host
Broadcast Address172.30.144.255172.30.114.193 = 34 172.30.114.32
Determining the Number of Subnets and Hosts
Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed to step four.172.30.1.33255.255.255.252 = 11111111 11111100 (host octets only)
Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 11000000 = 14 ones. 214 = 16384 usable subnets created.Number of Subnet bit14Number of Subnets1214 = 16384 subnets(all 0s used, all 1s not used)
Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 11111100 = 2 zeros. 22 = 4 2 = 2 usable host addresses created.-Number of Host Bit per Subnet 8 bit Number of Usable Hosts per Subnet 22 = 4 2 = 2 hosts per subnet
Problem 3Host IP Address192.192.10.234
Subnet Mask255.255.255.0
Number of Subnet Bits8
Number of Subnets28 = 256 subnets
Number of Host Bits per Subnet8 bit
Number of Usable Hosts per Subnet28 = 256 2 = 254 hosts per subnet
Subnet Address for this IP Address192.192.10.0
IP Address of First Host on this Subnet192.192.10.1
IP Address of Last Host on this Subnet192.192.10.254
Broadcast Address for this Subnet192.192.10.255
FORMULA :2726252423222120
1286432168421
192 192 10 234IP Address11000000110000000000101011101010X MultiplySubnet Mask 11111111111111111111111100000000Subnet Address11000000110000000000101000000000Subnet address for 192 192 10 0 AnswerThis IP Address =IP Address of First Host 0 + 254 Hosts IP Address of Last Host
Broadcast Address192.192.10.255192.192.10.1 = 254 192.192.10.254
Determining the Number of Subnets and Hosts
Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed the step .192.192.10.230255.255.255.0 = 11111111 00000000 (host octets only)
Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 00000000 = 8 ones. 28 = 256 usable subnets created.Number of Subnet bit8Number of Subnets28 = 256 subnets(all 0s used, all 1s not used)
Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 00000000= 8 zeros. 28 = 256 2 = 254 usable host addresses created.-Number of Host Bit per Subnet 8 bit Number of Usable Hosts per Subnet 28 = 256 2 = 254 hosts per subnet
Problem 4Host IP Address172.17.99.71
Subnet Mask255.255.0.0
Number of Subnet Bits0
Number of Subnets20 = 1 subnets
Number of Host Bits per Subnet16 bit
Number of Usable Hosts per Subnet216 = 65536 2 = 65536 hosts per subnet
Subnet Address for this IP Address172.17.0.0
IP Address of First Host on this Subnet172.17.0.1
IP Address of Last Host on this Subnet172.17.255.254
Broadcast Address for this Subnet172.17.255.255
FORMULA :2726252423222120
1286432168421
172 17 99 71IP Address10101100000100010110001101000111X MultiplySubnet Mask 11111111111111110000000000000000Subnet Address10101100000100010000000100000000Subnet address for 172 17 0 0 AnswerThis IP Address =IP Address of First Host 0 + 254 Hosts IP Address of Last Host
Broadcast Address192.192.10.255172.17.0.1 = 254 172.17.255.254
Determining the Number of Subnets and Hosts
Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed the step .172.17.99.71255.255.0.0 = 00000000 00000000 (host octets only)
Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.00000000 00000000 = 0 ones. 20 = 1 usable subnets created.Number of Subnet bit0Number of Subnets20 = 1 subnets(all 0s used, all 1s not used)
Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 00000000 00000000= 16 zeros. 216 = 65536 2 = 65534 usable host addresses created.-Number of Host Bit per Subnet 16 bit Number of Usable Hosts per Subnet 216 = 65536 2 = 65534 hosts per subnet
Problem 5Host IP Address192.168.3.219
Subnet Mask255.255.255.0
Number of Subnet Bits8
Number of Subnets28 = 256 subnets
Number of Host Bits per Subnet8 bit
Number of Usable Hosts per Subnet28 = 256 2 = 254 hosts per subnet
Subnet Address for this IP Address192.168.3.0
IP Address of First Host on this Subnet192.168.3.1
IP Address of Last Host on this Subnet192.168.3.254
Broadcast Address for this Subnet192.168.3.255
FORMULA :2726252423222120
1286432168421
192 168 3 219IP Address11000000101010000000001111011011X MultiplySubnet Mask 11111111111111111111111100000000Subnet Address11000000101010000000001100000000Subnet address for 192 168 3 0 AnswerThis IP Address =IP Address of First Host 0 + 254 Hosts IP Address of Last Host
Broadcast Address192.192.10.255192.168.3.1 = 254 192.168.3.254
Determining the Number of Subnets and Hosts
Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed the step .192.168.3.219255.255.255.0 = 11111111 00000000 (host octets only)
Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 00000000 = 8 ones. 28 = 256 usable subnets created.Number of Subnet bit8Number of Subnets28 = 256 subnets(all 0s used, all 1s not used)
Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 00000000= 8 zeros. 28 = 256 2 = 254 usable host addresses created.-Number of Host Bit per Subnet 8 bit Number of Usable Hosts per Subnet 28 = 256 2 = 254 hosts per subnet
Problem 6Host IP Address192.168.3.219
Subnet Mask255.255.255.252
Number of Subnet Bits14
Number of Subnets214 = 16384 subnets
Number of Host Bits per Subnet2 bit
Number of Usable Hosts per Subnet22 = 4 2 = 2 hosts per subnet
Subnet Address for this IP Address192.168.3.216
IP Address of First Host on this Subnet192.168.3.217
IP Address of Last Host on this Subnet192.168.3.218
Broadcast Address for this Subnet192.168.3.219
FORMULA :2726252423222120
1286432168421
192 168 3 219IP Address11000000101010000000001111011011X MultiplySubnet Mask 11111111111111111111111111111100Subnet Address11000000101010000000001111011000Subnet address for 192 168 3 216 AnswerThis IP Address =IP Address of First Host 216 + 2 Hosts IP Address of Last Host
Broadcast Address192.192.10.255192.168.3.217 = 218 192.168.3.218
Determining the Number of Subnets and Hosts
Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed the step .192.168.3.219255.255.255.252 = 11111111 11111100 (host octets only)
Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 11111100 = 14 ones. 214 = 16384 usable subnets created.Number of Subnet bit14Number of Subnets214 = 16384 subnets(all 0s used, all 1s not used)
Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 11111100= 2 zeros. 22 = 4 2 = 2 usable host addresses created.-Number of Host Bit per Subnet 2 bit Number of Usable Hosts per Subnet 22 = 4 2 = 2 hosts per subnet
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