Subnet in Your Head
Transcript of Subnet in Your Head
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WELCOME
Subnet in your head in 90 minutes!Todd Lammle
TO THE PATH OF SUCCESS
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About Todd LammleAbout Todd Lammle
Sybex author More than 40 titles published on Cisco; Microsoft; and,
soon, wireless certification.
President, GlobalNet Training Inc.
Cisco, Microsoft, Security and Wireless certificationhands-on courses. www.globalnettraining.com.
CEO, RouterSim, LLC Cisco and Microsoft certification software products.
www.routersim.com.
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I Promise..You can subnet in your head..I Promise..You can subnet in your head..
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WWW.GLOBALNETTRAINING.COM
IP AddressingIP Addressing
and Subnettingand Subnetting
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IP AddressingIP Addressing
Hierarchical Addressing Framework.
Network.node addressing, 32 bits (fourbytes).
The Hierarchical advantage is increased
ability of addresses.
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Binary NumberingBinary Numbering
Defining basic IP addressing terms:
Bit = One digit (one or a zero).
Byte = Seven or eight bits (depends onparity).
Octet = Always eight bits.
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Binary to Decimal ConversionBinary to Decimal Conversion
Converting binary to decimal examples:
128 64 32 16 8 4 2 1: Bit values
0 0 0 0 0 0 0 0 = 0
1 1 1 1 1 1 1 1 = 255
0 0 0 0 1 1 1 1 = 150 1 0 1 0 1 0 1 = 85
1 0 0 0 0 0 1 1 = 1310 0 0 1 0 1 1 0 = 22
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Binary (Cont.)Binary (Cont.)
Bits Binary Decimal
0 00000000 0
1 10000000 1282 11000000 192
3 11100000 224
4 11110000 240
5 11111000 248
6 11111100 2527 11111110 254
8 11111111 255
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Math to Memorize!Math to Memorize!
Subnet mask First Subnet
256-192 64256-224 32
256-240 16
256-248 8
256-252 4
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Powers of 2Powers of 2
21=2
22=4
23=8
24=16
25=32
26=64
27=128
28=256
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IP AddressingIP Addressing
32 bits
Network Host
255
1 byte
255
1 byte
255
1 byte
255
1 byte
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IP AddressingIP AddressingIP AddressingIP Addressing
32 bits
255 255 255 255
Network Host
1
28
64
32
16 8 4 2 1
11111111 11111111 11111111 11111111
1 8 9 16 17 24 25 32
1
28
64
32
16 8 4 2 1
1
28
64
32
16 8 4 2 1
1
28
64
32
16 8 4 2 1
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IP Address ClassesIP Address Classes
8 bits 8 bits8 bits8 bits
Class A:
Class B:
Class C:
Class D: Multicast
Class E: Research
NetworkNetwork HostHost HostHost HostHost
NetworkNetwork NetworkNetwork HostHost HostHost
NetworkNetwork NetworkNetwork NetworkNetwork HostHost
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IP Address ClassesIP Address Classes
0NNNNNNN0NNNNNNN 00000000 = 0
01111111 = 127
Class A:Range (1-126)
10000000 =128
10111111 = 191
10NNNNNN10NNNNNN
Class B: Range (128-191)
Class C: 11000000 = 192
11011111 = 223
110NNNNN110NNNNN
Range (192-223)
Note: Class D range is 224-239
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IP Address RangesIP Address Ranges
10.0.0.0 Network Address
10.0.0.1 First valid host
10.255.255.254 Last valid host
10.255.255.255 Broadcast Address
172.16.0.0 192.168.10.0
172.16.0.1 192.168.10.1
172.16.255.254 192.168.10.254
172.16.255.255 192.168.10.255
Add i Wi h S b
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Addressing Without SubnetsAddressing Without Subnets
172.16.0.1 172.16.255.254172.16.0.2
Network 172.16.0.0
Add i Wi h S bAdd i With S b t
Add i Wi h S bAdd i With S b t
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Addressing With SubnetsAddressing With SubnetsAddressing With SubnetsAddressing With Subnets
Network 172.16.0.0
172.16.30.0
172.16.40.0
172.16.10.0 172.16.20.0
S b t M kS b t M k
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Subnet MasksSubnet Masks
Class A default 255.0.0.0
Class B default 255.255.0.0
Class C default 255.255.255.0
H D Y D t i th M k?H D Y D t i th M k?
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How Do You Determine the Mask?How Do You Determine the Mask?
Count the number of subnets in the
network and remember to think aboutgrowth.
Count the number of hosts in eachsubnet and remember to think about
growth.
Aft Y Ch S b t M k
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After You Choose a Subnet Mask
Just ask Six Questions!Just ask Six Questions!
1. How many subnets?
2. How many hosts per subnet?
3. What are the subnets?
4. Whats the broadcast address?
5. Whats the first valid host address?
6. Whats the last valid host address?
N G t Si A !N G t Si A !
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Now, Get Six Answers!Now, Get Six Answers!
1. 2subnet bits -2= amount of subnets.
2. 2host bits-2 = amount of hosts per subnet.
3. 256-subnet mask = base number.
4. Broadcast address = next subnet 1.5. First valid host = subnet + 1.
6. Last valid host = broadcast 1.(Valid hosts must notbe all 0s or all 1s).
E S b ttiEasy Subnetting
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Easy SubnettingEasy Subnetting
192.168.10.0
255.255.255.192
22-2=2
26-2=62
256-192= 64 128 Network
65 129 First Host
126 190 Last Host127 191 Broadcast
Now Implement ItNow Implement It
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Now Implement It.Now Implement It.
.131.130 .132.66 .67
.65 .129
.68
192.168.10.64 192.168.10.128
Easy SubnettingEasy Subnetting
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Easy SubnettingEasy Subnetting
192.168.10.0
255.255.255.224
23-2=625-2=30
256-224= 32 64 96 128 160 19233 65 97 129 161 193
62 94 126 158 190 222
63 95 127 159 191 223
Now Implement ItNow Implement It
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Now Implement It.Now Implement It.
192.168.10.32
192.168.10.64
.97
.33
.65
.66
.34 .35 .36
.130 .131 .132
192.168.10.128.129
.98 .99 .100
192.168.10.96
Easy SubnettingEasy Subnetting
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Easy SubnettingEasy Subnetting
192.168.10.0
255.255.255.240
24-2=1424-2=14
256-240= 16 32 48 64 224 Network17 33 49 65 225 First Host
30 46 62 78 238 Last Host31 47 63 79 239 Broadcast
Easy SubnettingEasy Subnetting
Easy SubnettingEasy Subnetting
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Easy SubnettingEasy SubnettingEasy SubnettingEasy Subnetting
192.168.10.0
255.255.255.248
25-2=30
23-2=6
256-248= 8 16 24 32 240 Network
9 17 25 33 241 First Host
14 22 30 38242 Last Host15 23 31 39 247 Broadcast
Easy SubnettingEasy Subnetting
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Easy SubnettingEasy Subnetting
192.168.10.0
255.255.255.252
26-2=6222-2=2
256-252= 4 8 12 16 248 Network5 9 13 17 249 First Host
6 10 14 18 250 Last Host
7 11 15 19 251 Broadcast
Now Implement ItNow Implement It
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Now Implement It.Now Implement It.
192.168.10.8
.13
.5
.14
.10
.9
.18
192.168.10.16
.17
192.168.10.12
.6
192.168.10.4
Very Easy SubnettingVery Easy Subnetting
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Very Easy SubnettingVery Easy Subnetting
192.168.10.68 255.255.255.192
When viewing an IP address and subnetmask, just answer three easy questions:
Whats the valid subnet?
Whats the broadcast address?
Whats the valid host range?
AnswerAnswer
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AnswerAnswer
256-192= 64 128 Subnet65 129 First Host
126 190 Last Host
127 191 Broadcast
Very Easy SubnettingVery Easy Subnetting
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Very Easy SubnettingVery Easy Subnetting
192.168.10. 65 255.255.255.224
AnswerAnswer
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AnswerAnswer
192.168.10. 65 255.255.255.224
256-224=32, 64, 96Subnet = 64
Broadcast = 95
Very Easy SubnettingVery Easy Subnetting
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Very Easy SubnettingVery Easy Subnetting
192.168.10.38 255.255.255.240
Answer:Answer:
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Answer:Answer:
192.168.10.38 255.255.255.240
256-240=16, 32, 48
Subnet = 32Broadcast = 47
Very Easy SubnettingVery Easy Subnetting
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Very Easy SubnettingVery Easy Subnetting
192.168.10. 26 255.255.255.248
Answer:Answer:
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Answer:
192.168.10. 26 255.255.255.248
256-248=8, 16, 24, 32
Subnet = 24Broadcast = 31
Very Easy SubnettingVery Easy Subnetting
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e y asy Sub ett gy y g
192.168.10.13 255.255.255.252
Answer:Answer:
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192.168.10.13 255.255.255.252
256-252=4, 8, 12, 16
Subnet = 12Broadcast = 15
Very Easy SubnettingVery Easy Subnetting
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y y gy y g
192.168.10.99 255.255.255.248
Very Easy SubnettingVery Easy Subnetting
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y y gy y g
192.168.10.99 255.255.255.192
Very Easy SubnettingVery Easy Subnetting
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y y gy y g
192.168.10.69 255.255.255.224
Very Easy SubnettingVery Easy Subnetting
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y y g
192.168.10.25 255.255.255.252
Subnetting QuestionSubnetting Question
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Which mask would you assign if you
had a class C address with a maximumof 35 hosts per network?
Answer:Answer:
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255.255.255.192=62 hosts per subnet
255.255.255.224=30 hosts per subnet 255.255.255.240=14 hosts per subnet
255.255.255.248=6 hosts per subnet 255.255.255.252=2 hosts per subnet
QuestionQuestion
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How many subnets and hosts?
192.168.10.0/26
Answer:Answer:
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/24 = 255.255.255.0/25 = 255.255.255.128
/26 = 255.255.255.192/27 = 255.255.255.224
/28 = 255.255.255.240
/29 = 255.255.255.248/30 = 255.255.255.252
192.168.10.0 255.255.255.192
Two subnets with 62 hosts each
Class B SubnettingClass B Subnetting
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172.16.0.0
255.255.192.0
22-2=2
214-2=16,382
256-192= 64.0 128.0 Network
64.1 128.1 First Host
127.254 191.254 Last Host127.255 191.255 Broadcast
Class B SubnettingClass B Subnetting
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172.16.0.0
255.255.240.0
24-2=14212-2=4094
256-240= 16.0 32.0 224.0 Network
16.1 32.1 224.1 First Host
31.254 47.254 239.254 Last Host
31.255 47.255 239.255 Broadcast
Class B SubnettingClass B Subnetting
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172.16.0.0
255.255.248.025-2=30
211-2=2046
256-248= 8.0 16.0 240.0 Network
8.1 16.1 240.1 First Host
15.254 23.254 247.254 Last Host15.255 23.255 247.255 Broadcast
Class B SubnettingClass B Subnetting
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172.16.0.0
255.255.254.027-2=126
29-2=510
256-254= 2.0 16.0 252.0 Network
2.1 16.1 252.1 First Host
3.254 17.254 253.254 Last Host3.255 17.255 253.255 Broadcast
Class B SubnettingClass B Subnetting
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172.16.0.0
255.255.255.028-2=254
28-2=254
256-255= 1.0 2.0 254.0 Network
1.1 2.1 254.1 First Host
1.254 2.254 254.254 Last Host1.255 2.255 254.255 Broadcast
Very Easy SubnettingVery Easy Subnetting
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172.16.10.90
255.255.255.192
AnswerAnswer
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172.16.10.90
255.255.255.192
256-192= 10.64 10.128Subnet = 10.64
Broadcast = 10.127
Very Easy SubnettingVery Easy Subnetting
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172.16.10.66
255.255.255.224
AnswerAnswer
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172.16.10.66
255.255.255.224
256-224= 32, 64, 96Subnet = 10.64
Broadcast = 10.95
Very Easy SubnettingVery Easy Subnetting
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172.16.10.33
255.255.255.240
AnswerAnswer
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172.16.10.33
255.255.255.240
256-240= 16, 32, 48Subnet = 10.32
Broadcast = 10.47
Very Easy SubnettingVery Easy Subnetting
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172.16.10.33
255.255.255.248
AnswerAnswer
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172.16.10.33
255.255.255.248
256-248= 8, 16, 24, 32, 40Subnet = 10.32
Broadcast = 10.39
Very Easy SubnettingVery Easy Subnetting
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172.16.10.13
255.255.255.252
AnswerAnswer
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172.16.10.13
255.255.255.252
256-252= 4, 8, 12, 16Subnet = 10.12
Broadcast = 10.15
Subnet QuestionSubnet Question
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If you had a class B address of
172.16.0.0, which mask would you useto provide a maximum of 100 hosts with
more than 300 subnets?
Answer:Answer:
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255.255.255.192=1022 subnets, 62 hosts 255.255.255.128 =510 subnets, 126 hosts
255.255.255.0 = 254 subnet, 254 hostsStart at 255.255.255.0, which provides 254
subnets, each with 254 hosts.
Then move right if you need more subnets, orleft if you need more hosts. In this example,
we move right with the subnet bits becausewe need more subnets.
QuestionQuestion
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The network 172.16.0.0 needs to be
divided into subnet where you haveover 400 hosts in each subnet. What is
your mask?
Answer:Answer:
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172.16.0.0 with over 400 hosts per subnet.Start at 255.255.255.0, which provides 254
subnets, each with 254 hosts.Then move right if you need more subnets, or
left if you need more hosts. In this example,
we move left with the subnet bits becausewe need more hosts.
255.255.254.0126 subnets, each with 510 hosts
Questions?Questions?
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Please see:
www.lammle.com www.lammleprep.com
www.routersim.com www.globalnettraining.com
http://www.lammle.com/http://www.routersim.com/http://www.routersim.com/http://www.lammle.com/http://www.lammle.com/