Download - 2.0.Subnetting

Course 640-801 : CCNA 2.0 - Subnetting [Author: [email protected]]

Planning & DesigningObjective 2 Design an IP addressing scheme to meet design requirements

Implementation & OperationObjective 8 Configure IP addresses, subnet masks, and gateway addresses on routers and hosts

TroubleshootingObjective 21 Troubleshoot IP addressing and host configuration

Binary: powers of 2 (from right to left) Hexadecimal: powers of 16 (from right to left)2^8 2^7 2^6 2^5 2^4 2^3 2^1 2^0 16^3 16^1 16^0128 64 32 16 8 4 2 1 512 16 1

A host or node is a computer or device on a TCP/IP network. Every TCP/IP host is uniquely identified by its IP address. An IP address consists of a network ID and a host ID. If two different hosts belong to the same network, they have the same network ID. The two hosts will have different host IDs and can communicate with each other locally without going through a router. If two hosts have different network IDs, they belong to different segments on the network and they must communicate with each other remotely through a router or default gateway.

An IP address consists of 32 binary bits, where each bit is either a 0 or 1. We write the 32 bits into four 8 bit numbers (octets) separated by periods.

Developing a TCP/IP addressing schemeHosts for a LAN subnet mask One for each router interface One for each NIC on each hostSubnet mask Number of subnets on the network Number of host IDs on each subnet

IP Address ClassesThe Class of the address determines which part is the network address and which part is the host address. There are 5 different address classes. Classes can be distinguished by the decimal notation of the very first octet. The following Address Class table illustrates how you can determine to which class an address belongs.

Private IP AddressesYou can use certain IP addresses privately within you own Intranet as long as they are not seen by the global community.

Class Valid Network Numbers

Octet Miscellaneous Default subnet mask

ANetwork (8 bits)Host (24 bits)

1.x.x.x to 127.x.x.x 0 - 127 Private Use reserved 10.x.x.x

Loopback reserved 127.x.x.x (127.0.0.1 loopback test)

255.0.0.0

BNetwork (16 bits)

Host (16 bits)

128.x.x.x to 191.x.x.x 128 - 191 Autoconfig reserved 169.254.x.x

Private Use reserved 172.16-31.x.x

255.255.0.0

CNetwork (24 bits)

Host (8 bits)

192.x.x.x to 223.x.x.x 192 - 223 Private Use reserved 192.168.x.x 255.255.255.0

D 224 to 239 Multicasts n/a

E 240 to 255 Reserved for future InterNIC use n/a

Class A IP addresses allow the most number of hostsClass C IP addresses allow the least number of hosts

Using this table we can see the IP address in 193.10.30.2 is a Class C address.

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We can also see which part of that IP address is the Network ID and which is the Host ID.

Network ID: (first 3 Octets) = 193.10.30Host ID: (remaining Octets) = 2

Addressing types

Cisco IOS software support 3 types of addressing: Unicast

o Directed to a valid host address on a segment Multicast

o Directed to a group of devices on a segmento are sent to a particular network and are allowed to transit by a routero all 1’s in the host portion of the address.o e.g. If you want to send a broadcast to the third subnet of the 172.16 network the address would

be 172.16.3.255o e.g. If you want to send a broadcast to all the subnets of 172.16 network the address would be

172.16.255.255 Broadcast

o Directed to every device on the same segmento Flooded broadcasts are considered local and are represented by 255.255.255.255 or

FFFF.FFFF.FFFF

Whenever you want to refer to your entire network with an IP address, the host section is set to all 0s (binary=00000000) = 0 For example 172.16.0.0 specifies the network for the above address.

Subnetting

SUBNETTING CREATES MORE NETWORKS WITH FEWER HOSTS AVAILABLE (on each network)

subnet bits are borrowed from the Node/Host portion.

you may use all but 2 bits when subnetting (can’t borrow all bits from the host portion of an address)

Network address 207.103.56.105 with a SM of 255.255.255.240 is a class C address with 4 bits borrowedThis creates a^4 = 16-2 means you can have 14 valid subnets with the same number of hosts per subnets

Directed/only subnet broadcast address – N.N.S.255 (this is for class B)Flooded/total network broadcast address – N.N.255.255 (this is for class B)

Subnetting is the process used to divide the total available IP addressed (hosts) for your Network into smaller subnetworks (subnets). For example, the Network ID we used in the discussion above (193.10.30.0). This network would consist of 256 possible IP addresses (193.10.30.0 - 193.10.30.255).

We know this because in a Class C address, only the last octet is available for host IDs (0000000 11111111) or (0 255). Since 0 is used to identify the whole network and 255 is reserved for broadcasts, that leaves us with 254 possible hosts (193.10.30.1 193.10.30.254).

Suppose we wanted to divide those 254 addresses up into 6 smaller subnets. This can be done by using what is referred to as a Subnet Mask. By looking at the above table we can see Class C addresses all have a default subnet mask of 255.255.255.0. Since the last octet of the subnet mask is 0, it means that the host IDs have not been subdivided into smaller subnets. However, if we choose to divide our network into a few smaller segments (subnets), then we would change the default subnet mask by replacing the last octet with one of the valid subnet masks.

255.255.255.240 = 11111111.11111111.11111111.11110000 (/ 20)

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Class Network address range address format valid host ids------------------------------------------------------------------------------------A 0|000.0000 = 0 N.H.H.H N. 0. 0. 1

0|111.1111 = 127 N.H.H.H N.255.255.254range is 1-127

B 10|00.0000 = 128 N.N.H.H N. N. 0. 110|11.1111 = 191 N.N.H.H N. N.255.254

range is 128-191

C 110|0.0000 = 192 N.N.N.H N. N. N. 1110|1.1111 = 223 N.N.N.H N. N. N.254

range is 192-223

SUBNETTING CLASS C – 8 bits to use for subnetting

Binary decimal # of bits shorthand # subnets # hosts----------------------------------------------------------------------------------1000.0000 255.255.255.128 /25 <----- invalid1100.0000 255.255.255.192 2 /26 2 62 1110.0000 255.255.255.224 3 /27 6 301111.0000 255.255.255.240 4 /28 14 141111.1000 255.255.255.248 5 /29 30 61111.1100 255.255.255.252 6 /30 62 21111.1110 255.255.255.254 /31

class C is an exception whereby we remove 2 hosts from each of the ranges because we are already subnetting the 4th octet and consequently don’t have room for the network and broadcast addresses – weird heh?

exampleSubnet 192.168.10.0SM 255.255.255.128 /25

Cannot use this subnet mask as it involves only using one bit for subnetting

exampleSubnet 192.168.10.0SM 255.255.255.192 (/26)

A. how many valid subnets?192 is 2 masked-bits and 6 host-bits2 ^ MaskedBits - 2 = 2^2-2 = 2

B. how many valid hosts/subnet?2 ^ host-bits - 2 = 2^6-2 = 62

C. first valid subnet and incremental value256 – 192 = 64, 128... until 192 (which is invalid as it is the SM)

D. broadcast address for each subnet ?i.e. the number before the next subnet

1st is 64, therefore 128-1=1272nd is 128, therefore 192-1=191

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E. valid hosts ?

Network First host Last host Broadcast addressinvalid 0

1st 64 65 126 1272nd 128 129 190 191

invalid 192

exampleSubnet 192.168.10.0SM 255.255.255.224 (/27)

A. how many valid subnets?224 is 3 MB and 5 HB2 ^ MB - 2 = 2^3-2 = 6

B. how many valid hosts/subnet?2 ^ HB - 2 = 2^5-2 = 30

C. first valid subnet and incremental value256 – 224 = 32, 64, 96, 128, 160, 192... until 224 (which is invalid as it is the SM)


1st is 32, therefore 32-1= 312nd is 64, therefore 64-1= 633rd is 96, therefore 96-1= 954th is 128, therefore 128-1=1275th is 160, therefore 160-1=1596th is 192, therefore 192-1=1917th is 224, therefore 224-1=223

E. valid hosts ?Network First host Last host Broadcast address

invalid 01st 32 33 62 632nd 64 65 94 953rd 96 97 126 1274th 128 129 158 1595th 160 161 190 1916th 192 193 222 223

invalid 224

exampleSubnet 192.168.10.0 / 28SM 255.255.255.240

A. how many valid subnets?240 is 4 MB and 4 HB2 ^ MB - 2 = 2^4-2 = 14

B. how many valid hosts/subnet?2 ^ HB-2 = 2^4-2 = 14

C. first valid subnet and incremental value256 – 240 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240... until 240 (which is invalid as it is the SM)

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31, 47, 63, 79, 95, 111, 127, 143, 175, 191, 207, 223, 239


invalid 01st 16 17 30 312nd 32 33 46 473rd 48 49 62 634th 64 65 78 795th 80 81 94 956th 96 97 110 1117th 112 113 126 1278th 128 129 142 1439th 144 145 158 15910th 160 161 174 17511th 176 177 190 19112th 192 193 206 20713th 208 209 222 22314th 224 225 238 239

invalid 240

exampleSM 255.255.255.248 /29Subnet 192.168.10.0

A. how many valid subnets?248 is 5 masked-bits and 3 host-bits2^MB-2 = 2^5-2 = 30

B. how many valid hosts/subnet?2^HB-2 = 2^3-2 = 6

C. first valid subnet and incremental value256-248 = 8, 16, 24, 32, 40, 48, ... 240... until 248 (which is invalid as it is the SM)


15, 23, 31, 39, ... 247


invalid 01st 8 9 14 152nd 16 17 22 233rd 24 25 30 314th 32 33 38 395th 40 41 46 47... ... ... ... ...30th 240 241 246 247

invalid 248

exampleSubnet 192.168.10.0 /30

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SM 255.255.255.252

A. how many valid subnets?252 is 6 MB and 2 HB2^MB-2 = 2^6-2 = 62

B. how many valid hosts/subnet?2^host-bits-2 = 2^2-2 = 2

C. first valid subnet and incremental value256-252 = 4, 8, 16, 32, ... 248... until 252 (which is invalid as it is the SM)

4 is the first subnet and the base number


7, 11, 15, ... 251

E. valid hosts ?

Network First host Last host Broadcast addressinvalid 0

1st 4 5 6 72nd 8 9 10 113rd 12 13 14 154th 16 17 18 195th 20 21 22 23... ... ... ... ...

invalid 252

SUBNETTING CLASS B – 16 bits to use for subnetting

Binary decimal shorthand--------------------------------------------------------------1000.0000 0000.0000 255.255.128.0 /17 <----- invalid1100.0000 0000.0000 255.255.192.0 /181110.0000 0000.0000 255.255.224.0 /191111.0000 0000.0000 255.255.240.0 /201111.1000 0000.0000 255.255.248.0 /211111.1100 0000.0000 255.255.252.0 /221111.1110 0000.0000 255.255.254.0 /231111.1111 0000.0000 255.255.255.0 /241111.1111 1000.0000 255.255.255.128 /251111.1111 1100.0000 255.255.255.192 /261111.1111 1110.0000 255.255.255.224 /271111.1111 1111.0000 255.255.255.240 /281111.1111 1111.1000 255.255.255.248 /291111.1111 1111.1100 255.255.255.252 /30

Given a class B address 172.16 from NIC.A typical subnetting scheme for a class B network would be to use an 8 bit subnet mask. Since the 3rd octet is the first free octet for class B, you will start there. So, if an 8 bit subnet mask would be 255.255.255.0, this means you have 254 subnets available and 254 addresses for nodes per subnet (there are only 254 subnets

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available and not 256 because you cannot use subnet’s 0 or subnet 1, as these are broadcast addresses.

exampleSubnet 172. 16. 0.0 /18SM 255.255.192.0

A. how many valid subnets?192 is 2 MB and 6+8 (14) HB2 ^ MB-2 = 2^2-2 = 2

B. how many valid hosts/subnet?2 ^ HB - 2 = 2^14-2 = (2^6 * 256) – 2 = (64 * 256) – 2 = 16,382

C. first valid subnet and incremental value256-192 = 64... until 192 (which is invalid as it is the SM)

i.e. 64 + 64 = 128 + 64 = 192valid subnets are 64 and 128


1st is 64.0, therefore 128.0-1 = 127.2552nd is 128.0, therefore 192.0-1 = 191.255


invalid 0.01st 64.0 64.1 127.254 127.2552nd 128.0 128.1 191.254 191.255

invalid 192.0

exampleSubnet 150. 4. 0.0 /19SM 255.255.224.0

A. how many valid subnets?224 is 3 MB and 13 HB2 ^ MB-2 = 2^3-2 = 6

B. how many valid hosts/subnet?2 ^ HB-2 = 2^13-2 = (2^5 * 256) – 2 = (32 * 256) – 2 = 8190


i.e. 32 + 32 = 64 + 32 = 96 + 32 = 128 + 32 = 160 + 32 = 192 + 32 = 224valid subnets are 32, 64, 96, 128, 160, and 192


1st is 32.0, therefore 64.0-1 = 63.2552nd is 64.0, therefore 96.0-1 = 95.255


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invalid 0.01st 32.0 32.1 63.254 63.2552nd 64.0 64.1 95.254 95.2553rd 96.0 96.1 127.254 127.2554th 128.0 128.1 159.254 159.2555th 160.0 160.1 191.254 191.2556th 192.0 192.1 223.254 223.255

invalid 224.0



B. how many valid hosts/subnet?2^HB-2 = 2^12 -2 = 2^4*256 – 2 = (16 * 256) – 2 = 4094

C. first valid subnet and incremental value256-240= 16... until 240 (which is invalid as it is the SM)

i.e. 16 + 16 = 32 + 16 = 48 + 16 = 64 + 16 = 80 + 16 = 96 + 16 = 112 + 16 = 128 + 16 = 144 + 16 = 160 + 16 = 176 + 16 = 192 + 16 = 208 + 16 = 224 + 16 = 240valid subnets are 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224


1st is 16.0, therefore 32.0-1 = 31.2552nd is 32.0, therefore 48.0-1 = 47.255... 224.0, therefore 240.0-1 = 239.255


invalid 0.01st 16.0 16.1 31.254 31.2552nd 32.0 32.1 47.254 47.2553rd 48.0 48.1 63.254 63.255... ... ... ... ...14th 224.0 224.1 239.254 239.255

invalid 240.0



B. how many valid hosts/subnet?2^HB-2 = 2^9-2 = 2^1*256 = 510


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i.e. 2, 4, 6, 8, ... 252


1st is 2.0, therefore 4.0 -1 = 3.2552nd is 4.0, therefore 6.0 -1 = 5.255... 252.0, therefore 254.0 -1 = 253.255



invalid 254.0


A. how many valid subnets?255 is 8 masked-bits and 8 host-bits2^MB-2 = 2^8-2 = 254


C. first valid subnet and incremental value256-255= 1... until 254 (which is invalid as it is the SM)

i.e. 1,2,3,4,5, ... 254


1st is 1.0, therefore 2.0-1 = 1.2552nd is 2.0, therefore 3.0-1 = 2.255... 254.0, therefore 255.0-1 = 254.255



invalid 255.0

a host of 172.16.10.50 would be in subnet 10.16.10.0 of range 10.16.10.1-254 with a broadcast of 176.16.10.255

example

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Subnet 172.16.0.0SM 255.255.255.128 /25A. how many valid subnets?9 MB and 7 HB2^MB-2 = 2^9-2=2^1*256 -2 = 510


C. first valid subnet and incremental valueBecause we are subnetting more than 8 bits, we use the range 0-255 and divide within that.

i.e. 1,2,3,4...255

and each subnet is actually 2 subnets (i.e. each 4th octet value defines two ranges)i.e. 0-127 and 128-255


invalid 0.01st 0.128 0.129 0.254 0.2552nd 1.0 1.1 1.126 1.1273rd 1.128 1.129 1.254 1.2554th 2.0 2.1 2.126 2.1275th 2.128 2.129 2.254 2.2556th 3.0 3.1 3.126 3.1277th 3.128 3.129 3.254 3.2558th 4.0 4.1 4.126 4.1279th 4.128 4.129 4.254 4.255... ... ... ... ...508 254.0 254.1 254.126 254.127509 254.128 254.129 254.254 254.255510 255.0 255.1 255.126 255.127

invalid 255.128


A. how many valid subnets?10 MB and 6 HB2^MB-2 = 2^10-2=2^2*256 -2 = 1022

B. how many valid hosts/subnet?2^HB-2 = 2^6-2= 62

C. first valid subnet and incremental value256-192=64 + 64 = 128 + 64 = 192

the 3rd octet – all 0s OR 1s lead to invalid subnets

E. valid hosts ? Notice that 0.0 and 255.192 are invalid as per the rule defined above

Network First host Last host Broadcast addressinvalid 0.0

1st 0.64 0.65 0.126 0.1272nd 0.128 0.129 0.190 0.191

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3rd 0.192 0.193 0.254 0.2554th 1.0 1.1 1.62 1.635th 1.64 1.65 1.126 1.1276th 1.128 1.129 1.190 1.1917th 1.192 1.193 1.254 1.2558th 2.0 2.1 2.62 2.639th 2.64 2.65 2.126 2.12710th 2.128 2.129 2.190 2.19111th 2.192 2.193 2.254 2.255... ... ... ... ...1020 255.0 255.1 255.62 255.631021 255.64 255.65 255.126 255.1271022 255.128 255.129 255.190 255.191

invalid 255.192

exampleSubnet 172.16.0.0/27SM 255.255.255.224

A. how many valid subnets?11 MB and 5 HB2^MB-2 = 2^13-2=2^3*256 -2 = 2044


C. first valid subnet and incremental value256-224=32, 64, 96, 128, 160, 192 and 224

the 3rd octet – all 0s OR 1s lead to an invalid subnets

E. valid hosts ? Notice that 0.0 and 255.224 are invalid as per the rule defined above

Network First host Last host Broadcast addressinvalid 0.0

1st 0.32 0.33 0.62 0.632nd 0.64 0.65 0.94 0.953rd 0.96 0.97 0.126 0.1274th 0.128 0.129 0.158 0.1595th 0.160 0.161 0.190 0.1916th 0.192 0.193 0.222 0.2237th 0.224 0.225 0.254 0.2558th 1.0 1.1 1.30 1.319th 1.32 1.33 1.62 1.6310th 1.64 1.65 1.94 1.9511th 1.96 1.97 1.126 1.12712th 1.128 1.129 1.158 1.15913th 1.160 1.161 1.190 1.19114th 1.192 1.193 1.222 1.22315th 1.224 1.225 1.254 1.255... ... ... ... ...2040 255.0 255.1 255.30 255.312041 255.32 255.33 255.62 255.632042 255.64 255.65 255.94 255.952043 255.96 255.97 255.126 255.1272044 255.128 255.129 255.158 255.1592045 255.160 255.161 255.190 255.191

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2046 255.192 255.193 255.222 255.223invalid 255.224

SUBNETTING CLASS A - 24 bits to use for subnetting

Binary decimal shorthand------------------------------------------------------------------------------------1000.0000 0000.0000 0000.0000 255.128.0. 0 /9 <----- invalid1110.0000 0000.0000 0000.0000 255.192.0. 0 /101110.0000 0000.0000 0000.0000 255.224.0. 0 /111111.0000 0000.0000 0000.0000 255.240.0. 0 /121111.1000 0000.0000 0000.0000 255.248.0. 0 /131111.1100 0000.0000 0000.0000 255.252.0. 0 /141111.1110 0000.0000 0000.0000 255.254.0. 0 /151111.1111 0000.0000 0000.0000 255.255.0. 0 /161111.1111 1000.0000 0000.0000 255.255.128.0 /171111.1111 1100.0000 0000.0000 255.255.192.0 /181111.1111 1110.0000 0000.0000 255.255.224.0 /191111.1111 1111.0000 0000.0000 255.255.240.0 /201111.1111 1111.1000 0000.0000 255.255.248.0 /211111.1111 1111.1100 0000.0000 255.255.252.0 /221111.1111 1111.1110 0000.0000 255.255.254.0 /231111.1111 1111.1111 0000.0000 255.255.255.0 /241111.1111 1111.1111 1000.0000 255.255.255.128 /251111.1111 1111.1111 1100.0000 255.255.255.192 /261111.1111 1111.1111 1110.0000 255.255.255.224 /271111.1111 1111.1111 1111.0000 255.255.255.240 /281111.1111 1111.1111 1111.1000 255.255.255.248 /291111.1111 1111.1111 1111.1100 255.255.255.252 /30

exampleSubnet 10.0.0.0 /16SM 255.255.0.0

255 is 8 masked-bits and 16 host-bits

A. how many valid subnets?2^MB-2 = 2^8-2 = 254

B. how many valid hosts/subnet?2^HB-2 = 2^16 -2 = 2^8*256 – 2 = (256 * 256) – 2 = 65534

C. first valid subnet and incremental value256-255= 1,2,3…... until 254


1st is 1.0.0, therefore 2.0.0-1 = 1.255.2552nd is 2.0.0, therefore 3.0.0-1 = 2.255.255...last is 254.0.0, therefore 255.0.0-1 = 254.255.255


invalid 0.0.01st 1.0.0 1.0.1 1.255.254 1.255.2552nd 2.0.0 2.0.1 2.255.254 2.255.2553rd 3.0.0 3.0.1 3.255.254 3.255.255

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... ... ... ... ...254th 254.0.0 254.0.1 254.255.254 254.255.255

invalid 255.0.0

exampleSubnet 10.0.0.0 /20SM 255.255.240.0

A. how many valid subnets?240 is 12 MB and 12 HB2^masked-bits-2 = 2^12-2=2^4*256 – 2 = 4094

B. how many valid hosts/subnet?2^HB-2 = 2^12-2=2^4*256 – 2 = 4094

C. first valid subnet and incremental value256-240 = 16, 32, 48, ... 240

D. broadcast address for each subnet ?


invalid 0.0.01st 0.16.0 0.16.1 0.31.254 0.31.2552nd 0.32.0 0.32.1 0.47.254 0.47.2553rd . . . .4th . . . .5th . . . .6th . . . .7th . . . .8th . . . .9th . . . .10th . . . .11th . . . .12th . . . .13th . . . .14th 0.224.0 0.224.1 0.239.254 0.239.25515th 0.240.0 0.240.1 0.255.254 0.255.25516th 1.0.0 1.0.1 1.15.254 1.15.25517th 1.16.0 1.16.1 1.31.254 1.31.255... ... ... ... ...

4094th 254.224.0 254.224.1 254.239.254 254.239.255invalid 255.240.0

exampleSM 255.255.255.192 /26Subnet 10.0.0.0

A. how many valid subnets?192 is 18 MB and 6 HB2^MB-2 = 2^18-2= 262144-2 = 262142

B. how many valid hosts/subnet?2^HB-2 = 2^6-2=64 – 2 = 62

C. first valid subnet and incremental value256-192= 64, 128, 192

D. broadcast address for each subnet ?

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E. valid hosts ?

Network First host Last host Broadcast addressinvalid 0.0. 0

1st 0.0. 64 0.0.65 0.0.126 0.0.1272nd 0.0. 128 0.0.129 0.0.190 0.0.1913rd 0.0. 192 0.0.193 0.0.254 0.0.2554th 0.1. 0 0.1.1 0.1.62 0.1.635th 0.1. 64 0.1.65 0.1.125 0.1.1276th 0.1. 128 0.1.129 0.1.190 0.1.1917th 0.1. 192 0.1.193 0.1.254 0.1.2558th 0.2. 0 0.2.1 0.2.62 0.2.639th 0.2. 64 0.2.65 0.2.126 0.2.12710th 0.2. 128 0.2.129 0.2.190 0.2.19111th 0.2. 192 0.2.193 0.2.254 0.2.255

0.3. 0 0.3.1 0.3.62 0.3.63 0.3. 192 0.3.193 0.3.254 0.3.255 0.255.192 0.255.193 0.255.254 0.255.255 1.0. 0 1.0.1 1.0.62 1.0.63 ... ... ... ... 10.0. 0 10.0.0 10.255.254 10.255.255

... ... ... ... ...

... 254.224.128 254.224.129 254.224.190 254.224.191

... ... ... ... ...

... 255.255.128 255.255.129 255.255.190 255.255.191invalid 255.255.192

Keypoints Know that 6 bits of subnetting is the most you can have with a class C address. Know that it is the subnet mask that actually determines which parts of the IP address are the Network

Number and the Host Node Only practice Class B & C in anticipation that Class A will not come up in the exam

5 WAYS THAT THE EXAM WILL TEST YOUR IP ADDRESSING KNOWLEDGE

Given an IP address + mask, what is the network/subnet number?

Given an IP address + mask, what is the network/subnet broadcast address?

Given an IP address + mask, what are the assignable IP address in that network/subnet?

Given a network number and static subnet mask, what are the valid subnet numbers?

Given a network number and static subnet mask, how many hosts per subnet, and how many subnets?

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SUBNETTING QUICK REFERENCE

x 256-x x 2^x1 128 128 1 22 192 64 2 43 224 32 3 84 240 16 4 165 248 8 5 326 252 4 6 647 254 2 7 1288 255 1 8 256

9 51210 102411 204812 4096

0 0 0 0 0816 162432 32 324048 485664 64 64 647280 808896 96 96104112 112120128 128 128 128 128136144 144152160 160 160168176 176184192 192 192 192200208 208216224 224 224232240 240248256 256 256 256 256

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