Bài giảng IP Addressing and Subnetting

Internet Protocol (IP)

A unique identifier for host, on an IP network

32-bit binary number, usually expressed as 4 “dotted decimal” values.

Each decimal value represents 8 bits, in the range of 0 to 255

 

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n an IP network32-bit binary number, usually expressed as 4 “dotted decimal” values.Each decimal value represents 8 bits, in the range of 0 to 255Example140.179.220.200Written in binary form:	 140	.179	.220	.200	10001100.10110011.11011100.11001000We see the address in the decimal formYour computer sees it in the binary formBinary Octet:An octet is made up of eight “1”s and/or “0”s, representing the following values:128	64	32	16	8	4	2	1So the value of 140 (the first octet of our example) looks like this:1	0	0	0	1	1	0	0Binary Octet:1	0	0	0	1	1	0	0128 +	0 +	0 +	0 +	8 +	4 +	0 +	0 = 140Address Classes(32 Bit Address	 232 = 4.2 billion possible addresses)There are 5 different address classes.Only 3 are in commercial use at this time.You can determine the class of the address by looking at the first 4 bits of the IP address:Class A begin with 0xxx, or 1 to 126 decimalClass B begin with 10xx, or 128 to 191 decimalClass C begin with 110x, or 192 to 223 decimalClass D begin with 1110, or 224 to 239 decimalClass E begin with 1111, or 240 to 254 decimalNetwork vs. HostEvery IP address has 2 parts:1 identifying the network it resides on1 identifying the host address on the networkThe class of the address and the subnet mask determine which part belongs to the network address and which part belongs to the host addressIP Address Breakdowns:The class of the address determines, by default, which part is for the network (N) and which part belongs to the host (H)Class A:	NNNNNNNN.HHHHHHH.HHHHHHHH.HHHHHHHHClass B:	NNNNNNNN.NNNNNNNN.HHHHHHH.HHHHHHHHClass C:	NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH140.179.220.200Our example is a Class B addressBy default, the Network part of the address is defined by the first 2 octets: 140.179.x.xBy default, the Host part of the address is defined by the last 2 octets: x.x.220.200*Note that the network part of the address is also known as the Network AddressTwo Reserved Addresses on a Subnet:In order to specify the Network Address of a given IP address, the Host portion is set to all “0”s:140.179.0.0If all the bits in the Host portion are set to “1”s, then this specifies the broadcast address that is sent to all hosts on the network:140.179.255.255SUBNETTINGThe highly dreaded.SubnettingSubnetting an IP network can be done for various reasons including:OrganizationUse of different physical mediaPreservation of address spaceSecurityControl network trafficExample Class AMillions of Addresses AvailableOver 16,000,000EfficiencyNon-subnetted networks are wastefulDivision of networks not optimalSmaller NetworkEasier to manageSmaller broadcast domainsSubnet MaskSubnet masks are applied to an IP address to identify the Network portion and the Host portion of the address.Your computer performs a bitwise logical AND operation between the address and the subnet mask in order to find the Network Address or number.Default Subnet MasksClass A - 255.0.0.0	11111111.00000000.00000000.00000000Class B - 255.255.0.0	11111111.11111111.00000000.00000000Class C - 255.255.255.0	11111111.11111111.11111111.00000000Logical Bitwise AND OperationRemember our example?140.179.240.200It’s a Class B, so the subnet mask is:255.255.0.0We need to look at this as our computer does so we can perform the bitwise AND...Logical Bitwise AND Operation140.179.220.200	Class B address255.255.0.0	Subnet MaskIn Binary:10001100.10110011.11110000.1100100011111111.11111111.00000000.0000000010001100.10110011.00000000.00000000By doing this, the computer has found that our Network Address is 140.179.0.0Another Example:Suppose we have the address of: 206.15.143.89? Class C255.255.255.0206.15.143.00.0.0.89What class is it?What is the subnet mask?What is the Network Address?What is the host portion of the address?Why Do We Care!?You can manipulate your subnet mask in order to create more network addresses. Why?If you have a Class C network, how many individual host addresses can you have?1 to 254 Remember, you can’t have all “0”s and all “1”s in the host portion of the address. So we cannot use 206.25.143.0 (all “0”s) or 206.25.143.255 (all “1”s) as a host address.Why Do We Care!?So we have 1 Class C Network (206.15.143.0)And we have 254 host address (1 to 254)But what if our LAN has 5 networks in it and each network has no more than 30 hosts on it?Do we apply for 4 more Class C licenses, so we have one for each network?We would be wasting 224 addresses on each network, a total of 1120 addresses!SubnettingSubnetting is a way of taking an existing class license and breaking it down to create more Network Addresses.This will always reduce the number of host addresses for a given network.Subnetting makes more efficient use of the address or addresses assigned to you.How Does Subnetting Work?Additional bits can be added (changed from 0 to 1) to the subnet mask to further subnet, or breakdown, a network.When the logical AND is done by the computer, the result will give it a new Network (or Subnet) Address.Remember, an address of all “0”s or all “1”s cannot be used in the last octet (or host portion). All “0”s signify the Network Address and all “1”s signify the broadcast addressSo How Does This Work?We ask our ISP for a Class C license.They give us the Class C bank of 206.15.143.0This gives us 1 Network (206.15.143.0) with the potential for 254 host addresses (206.15.143.1 to 206.15.143.254).But we have a LAN made up of 5 Networks with the largest one serving 25 hosts.So we need to Subnet our 1 IP address...So How Does This Work?To calculate the number of subnets (networks) and/or hosts, we need to do some math:Use the formula 2n-2 where the n can represent either how many subnets (networks) needed OR how many hosts per subnet needed.Magic FormulaSo How Does This Work?We know we need at least 5 subnets. So 23-2 will give us 6 subnet addresses (Network Addresses).We know we need at least 25 hosts per network. 25-2 will give us 30 hosts per subnet (network).This will work, because we can steal the first 3 bits from the host’s portion of the address to give to the network portion and still have 5 (8-3) left for the host portion:Break it down:Let’s go back to what portion is what:We have a Class C address:NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHHWith a Subnet mask of:11111111.11111111.11111111.00000000We need to steal 3 bits from the host portion to give it to the Network portion:NNNNNNNN.NNNNNNNN.NNNNNNNN.NNNHHHHHBreak it down:NNNNNNNN.NNNNNNNN.NNNNNNNN.NNNHHHHHThis will change our subnet mask to the following:11111111.11111111.11111111.11100000Above is how the computer will see our new subnet mask, but we need to express it in decimal form as well:	255.255.255.224	128+64+32=224What address is what?Which of our 254 addresses will be a Subnet (or Network) address and which will be our host addresses?Because we are using the first 3 bits for our subnet mask, we can configure them into eight different ways (binary form):	What address is what?Which of our 254 addresses will be a Subnet (or Network) address and which will be our host addresses?Because we are using the first 3 bits for our subnet mask, we can configure them into eight different ways (binary form):	000	001	010	011	100	101	110	111What address is what?We cannot use all “0”s or all “1”s	000	001	010	011	100	101	110	111We are left with 6 useable network numbers.Network (Subnet) AddressesRemember our values:	128	64	32	16	8	4	2	1	EqualsNow our 3 bit configurations:	0	0	1	H	H	H	H	H	32	0	1	0	H	H	H	H	H	64	0	1	1	H	H	H	H	H	96	1	0	0	H	H	H	H	H	128	1	0	1	H	H	H	H	H	160	1	1	0	H	H	H	H H	192Network (Subnet) Addresses	0	0	1	h	h	h	h	h	32	0	1	0	h	h	h	h	h	64	0	1	1	h	h	h	h	h	96	1	0	0	h	h	h	h	h	128	1	0	1	h	h	h	h	h	160	1	1	0	h	h	h	h	h	192Each of these numbers becomes the Network Address of their subnet...Network (Subnet) Addresses206.15.143.32206.15.143.64206.15.143.96206.15.143.128206.15.143.160206.15.143.192host AddressesThe device assigned the first address will receive the first number AFTER the network address shown before.206.15.143.33 or 32+10	0	1	0	0	0	0	1And the last address in the Network will look like this: 206.15.143.620	0	1	1	1	1	1	0*Remember, we cannot use all “1”s, that is the broadcast address (206.15.143.63)Host AddressesThe next network will start at 206.15.143.64The first IP address on this subnet network will receive:206.15.143.650	1	0	0	0	0	0	1And the last address in the Network will receive: 206.15.143.940	1	0	1	1	1	1	0*Remember, the broadcast address (206.15.143.95)Can you figure out the rest?Network:	Host Range206.15.143.32	206.15.143.32 to 206.15.143.62206.15.143.64	206.15.143.65 to 206.15.143.94206.15.143.96	206.15.143.97 to 206.15.143.126206.15.143.128	206.15.143.129 to 206.15.143.158206.15.143.160	206.15.143.161 to 206.15.143.190206.15.143.192	206.15.143.193 to 206.15.143.222How the computer finds the Network Address:200.15.143.89 An address on the subnet225.225.225.224 The new subnet maskWhen the computer does the Logical Bitwise AND Operation it will come up with the following Network Address (or Subnet Address):11001000.00001111.10001111.01011001= 200.15.143.8911111111.11111111.11111111.11100000 = 255.255.255.22411001000.00001111.10001111.01000000 = 200.15.143.64This address falls on our 2nd Subnet (Network)ReviewWe have one class C license.We need to subnet that into 12 possible networks.Each network needs a maximum of 10 hosts.How many bits do we need to take?24-2=144 bits need to be taken from the host portion and given to the network portion.ReviewWill that leave enough bits for the host portion? We need a maximum of 10 on each network24-2=14If we take 4 away, that leaves us with 4. That is enough for our individual networks of 10 hosts each.ReviewOur new subnet mask will look like this:11111111.11111111.11111111.11110000255.255.255.240 	128+64+32+16= 240Our subnet, or network addresses will be:206.15.143.16	 206.15.143.32	 206.15.143.48206.15.143.64	 206.15.143.80	 206.15.143.96206.15.143.112	 206.15.143.128	 206.15.143.144206.15.143.160	 206.15.143.176	 206.15.143.192206.15.143.208	 206.15.143.224

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