Computer Repair Student Part.5 [Subnetting] Finnish

8/14/2019 Computer Repair Student Part.5 [Subnetting] Finnish

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Subnetting


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E-2 Appendix ESubnetting

OBJECTIVESAfter completing this chapter you will

Understand the difference between major classes of IP addresses.

When given an IP address, be able to identify its IP class.

Determine the appropriate mask to use with each IP class.

Understand and be able to subnet IP addresses.

KEY TERMSbroadcast

ISP

subnet

useable host numbers

useable subnets


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SUBNETTING OVERVIEWAny company that needs IP addresses can lease them from an ISP (Internet Serv

Provider. ISPs are organizations that provide individuals and businesses access to t

Internet. If your company decides to connect to the Internet, a network administra

would contact an ISP and make arrangements for a connection to the Internet. Yo

connection to the Internet would be through the ISPs own network. You would have

provide the ISP with the number of IP addresses the company needs. ISPs have a limit

number of IP addresses available to give out to their customers. The ISP will ask ho

many computers your company presently has and how many computers are planned f

the near future.

Public IP addresses are in short supply due to the overwhelming popularity a

success of the Internet. ISPs are reluctant to lease more IP addresses than a custom

needs. The ISP may have 100 Class C IP addresses available to lease, but that does n

mean that the ISP will lease an entire Class C just because it was requested.

What if your company doesnt have enough computers to warrant a full Class C? Fexample, the Smiley Company has 30 computers and needs Internet access. T

company predicts a growth of 20 additional computers over the next two years. If t

Smiley Company contacts an ISP and asks for a full Class C block of addresses, t

request would probably be denied. Remember that a full Class C contains 256 h

addresses, and the Smiley Company needs only 50 addresses. The ISP would probab

lease a portion of the Class C to Smiley.

As another example, the WebBook Company has more than 450 computers on

network with expected growth of 50 new computers this year. The ISP has 100 Class

available for its customers and will have to dedicate at least two Class Cs for t

WebBook Company. The WebBook Company is not centralized in one building; it hseveral offices throughout the city. Each office has 3060 computers that need Intern

access. The ISP gives the company two full Class Cs to organize as the netwo

administrator sees fit. The network administrator can subdivide the Class C addresses

enable all external sites to have access to the Internet. The ISP does not care if the Cla

C is broken into smaller segments to fit a companys needs. When an IP addresses ran

is subdivided like this, it is calledsubnetting. Although the above examples are ve

simplified, this gives you a basic understanding of the process involved in acquiring a

using IP addresses.

A subnet is a method used that divides the IP address into three parts rather than tw

A normal IP address consists of two partsa network number and a host numbRemember from the Network chapter that the number of bits used for the two pa

depends on the class of IP address. Refer back to Network Figure #14 to refresh yo

memory. When subnetting is used, the IP address has three partsa network number

subnetwork number, and a host number.

Subnetting involves borrowing bits from the host portion of the IP address a

creating the third part of the IP addressthe subnetwork number, which is common

called the subnet. Take a Class C IP address of 192.168.10.4. Because it is a Class

Subnetting Overview


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address, the first three octets are the network number192.168.10. The last octet is the

host or network device located on network 192.168.10. In this example, the host number

is 4. Without subnetting, this IP address has one network number and 256 different host

numbers. If this IP address is subnetted, the network 192.168.10 can be divided into more

than just one network. It can have a varying number of subnetworks.

Subnetting has three primary functions: (1) efficient use of one or more IP addresses,(2) reduces the money spent to lease IP addresses, and (3) divides the network into easier

to manage portions. The effects of these three functions will be seen in the sections on

how to subnet and why to subnet.

HOW TO SUBNETWhen subnetting, bits are borrowed from the host portion of the IP address.

Borrowing bits from the host creates a new number called a subnet.

When subnetting, always borrow bits from the left-most host bits. Subnetting

reduces the number of hosts, but allows more networks using a single IP address.

Take the example of a standard Class C IP address, 207.193.204.0. The number

207.193.204 is the network number, and the last octet is used for host numbers. Subnet

Figure #1 shows the bit positions for the last octet (Octet 4) for a standard Class C IP

address.

In order to subnet, bits are borrowed from the host bits to create a subnet field. The

borrowed bits make up the subnet field. Subnet Figure #2 shows two bits borrowed from

the hosts field for subnetting.

.oNkrowteNCssalC tenbuS stsoH

t 1etcO t 2etcO t 3etcO t 4etcO

821 46 23 61 8 4 2 1

.702 .391 .402

Subnet Figure #2

.oNkrowteNCsasslC stsoH1tetcO ctet 2O 3tetcO 4tetcO

821 46 23 61 8 4 2 1

.702 .391 .402

bit positions

Subnet Figure #1


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Remember that bits are always borrowed from the left-most bits in the octet. Sin

two bits are being borrowed, bit positions 128 and 64 now represent subnet numbers.

Subnet Figure #2, you can see that the IP address still contains a network number a

host numbers. The new addition to the IP address is called the subnet and is formed

taking left-most bits away from the host field. The standard Class C has eight bits th

represent hosts, but now only six host bits remain because two bits were borrowed create subnets.

SUBNETWORK NUMBERSThe subnet portion of an IP address can have varying combinations of 1s and 0s. F

example, if two bits are borrowed for subnetting, the combinations of 1s and 0s are fo

different numbers00, 01, 10, and 11. Subnet Figure #3 shows this concept.

A mathematical formula can be used to determine how many subnets are form

when borrowing bits.

The number of subnets can be found by taking 2X where x is the number of bits

borrowed. For example, if two bits are borrowed, 22 = 4 or four subnetworks. If

three bits are borrowed, 23 = 8 or eight subnetworks.

Look back at Subnet Figure #3. The 00 combination in the subnet field represe

Subnetwork 0. The 01 combination in the subnetwork column designates it

Subnetwork 64. The 64 is obtained by a 1 being set in the 64 bit position. Wsubnetwork number is designated by a bit combination of10 in the subnetwork colum

The answer is 128 because there is a 1 set in the 128bit position. Subnet Figure #4 sho

how the various combinations of 1s and 0s create different subnetwork numbers.


t 1etcO te 2tcO t 3etcO t 4etcO

821 46 23 61 8 4 2 1

.702 .391 .402 0 0

.702 .391 .402 0 1

.702 .391 .402 1 0

.702 .391 .402 1 1

Subnet Figure #3

Subnetwork Numbers


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Now that subnetwork numbers are understood, lets see how this applies to networks.Subnet Figure #5 shows two networks connected by a router. The networks are subnetted.

In Subnet Figure #5, one network is 192.168.10.64 and the other network is

192.168.10.128. Even though a company purchased one Class C IP address, two

networks can be created because of subnetting.

NUMBER OF HOSTSAn IP addressing rule is that every device on the network must have a unique IP

address. How does this rule affect subnetting? Each subnet can have a varying number of

hosts. The number of hosts on each subnet depends on how many host bits have been

borrowed to subnet. The more bits borrowed for subnetting, the fewer host bits remain for

network devices. Subnet Figure #6 shows how this works in a Class C IP address with

two bits borrowed for subnets.

Router192.168.10.64 192.168.10.128

Subnet Figure #5



tenbuS

.oN

821 46 23 61 8 4 2 1

.702 .391 .402 0 0 0

.702 .391 .402 0 1 46

.702 .391 .402 1 0 821

.702 .391 .402 1 1 291

Subnet Figure #4


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tenbuS

.oN

821 46 23 61 8 4 2 1

.702 .391 .402 0 0 0 0 0 0 0 0 0

.702 .391 .402 0 0 0 0 0 0 0 1

.702 .391 .402 0 0 0 0 0 0 1 0

.702 .391 .402 0 0 0 0 0 0 1 1

.702 .391 .402 0 0 0 0 0 1 0 0

.702 .391 .402 0 0

.702 .391 .402 0 0

.702 .391 .402 0 0 1 1 1 1 1 1

.702 .391 .402 0 1 0 0 0 0 0 0 46

.702 .391 .402 0 1 0 0 0 0 0 1

.702 .391 .402 0 1 0 0 0 0 1 0

.702 .391 .402 0 1

.702 .391 .402 0 1

.702 .391 .402 0 1 1 1 1 1 1 1

.702 .391 .402 1 0 0 0 0 0 0 0 821

.702 .391 .402 1 0 0 0 0 0 0 1

.702 .391 .402 1 0 0 0 0 0 1 0

.702 .391 .402 1 0

.702 .391 .402 1 0

.702 .391 .402 1 0 1 1 1 1 1 1

.702 .391 .402 1 1 0 0 0 0 0 0 291

.702 .391 .402 1 1 0 0 0 0 0 1

.702 .391 .402 1 1 0 0 0 0 1 0

.702 .391 .402 1 1

.702 .391 .402 1 1

.702 .391 .402 1 1 1 1 1 1 1 1

Subnet Figure #6

Number of Hosts


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The easiest way to determine the total number of hosts is to count the number of bits

that are left for hosts and raise the number 2 to that power. For example, in Subnet Figure

#6, there are six hosts bits remaining. Take the number 2 and raise it to the number of host

bits26 = 64. This means that there are 64 different combinations of 1s and 0s in the host

field for each subnetwork. One of the most important rules about subnetting concerns the

first and last subnet and the first and last host addresses.

When subnetting, the first and last subnetwork numbers and the first and last host

numbers within each subnet cannot be used.

As previously discussed in the Network chapter, a network has an IP address that

cannot be used by any network device. An example of a Class C network address is

192.107.10.0. A Class B network address is 152.124.0.0 and a Class A example is

11.0.0.0. When all of the host bits are 0s, that is considered the network or subnetwork

number and that number cannot be assigned to a network device as a host number. Look

back to Subnet Figure #7. The first subnetwork shown is 0. When all host bits are 0, that

is considered the subnetwork number. Some people call it the wire. Each combination of

1s and 0s after that point is a host number on that subnetwork until the subnetwork

number changes. The only exception to this is when all host bits are set to 1. When all of

the host bits are a binary 1, this designates a broadcast for that network or subnetwork.

Using the same network numbers given above as examples, the broadcast addresses

would be 192.107.10.255 for the Class C network, 152.124.255.255 for the Class B

network, and 11.255.255.255 for the Class A network. The broadcast address cannot be

assigned to a network device. The broadcast address is used to communicate with all

network devices simultaneously.

When borrowing two host bits from a Class C IP address, subnetworks 0, 64, 128,

and 192 were created. Based on the rule stated above, subnetworks 0 and 192 cannot be

used, so all that are left are subnetworks 64 and 128. These subnetworks that can be used

are known as useable subnets. The first and the last subnet are considered unuseable

because they contain all 0s and all 1s in the host bits. Subnet 0 contains all 0s in the host

bits. Subnet 192 can contain all 1s in the host bits.

On subnetwork 64, the host numbers that are possible are 64 through 127. On

subnetwork 64, the host numbers that are useable are 65 through 126. The host numbers

that can be assigned to network devices are known as useable host numbers. For

subnetwork 128, the possible host numbers are 128 through 191, but only 129 through

190 are used. This is found by using varying combinations of 1s and 0s through the hostbits. Subnet Figure #7 illustrates this concept.


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tenbuS

.oN oNtsoH

821 46 23 61 8 4 2 1

.702 .391 .402 0 1 0 0 0 0 0 0 46

.702 .391 .402 0 1 0 0 0 0 0 1 56

.702 .391 .402 0 1 0 0 0 0 1 0 66

.702 .391 .402 0 1 0 0 0 0 1 1 76

.702 .391 .402 0 1 0 0 0 1 0 0 86

.702 .391 .402 0 1 0 0 0 1 0 1 96

.702 .391 .402 0 1 0 0 0 1 1 0 07

.702 .391 .402 0 1

.702 .391 .402 0 1

.702 .391 .402 0 1 1 1 1 1 1 0 621

.702 .391 .402 1 1 1 1 1 1 1 1 sacdaorB

.702 .391 .402 1 0 0 0 0 0 0 0 821

.702 .391 .402 1 0 0 0 0 0 0 1 921

.702 .391 .402 1 0 0 0 0 0 1 0 031

.702 .391 .402 1 0 0 0 0 0 1 1 131

.702 .391 .402 1 0 0 0 0 1 0 0 231

.702 .391 .402 1 0 0 0 0 1 0 1 331

.702 .391 .402 1 0

.702 .391 .402 1 0

.702 .391 .402 1 0 1 1 1 1 1 0 091

.702 .391 .402 1 0 1 1 1 1 1 1 191

Subnet Figure #7

Number of Hosts


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One of the hardest concepts for students to grasp is that, when determining what the

decimal number is for an octet, you must use all eight octet bits in the number. Ignore the

line drawn for the subnet bits. For example, in Subnet Figure #7, for subnetwork 64, the

first useable host number is 65. Eight bits designate the number 6501000001. Since

there is a 1 in bit position 64 and a 1 in bit position 1, 64 + 1 = 65. The subnetwork

number is 64 because the subnetwork columns have a 01 combination. The 1 is set in the64 column. The host number is 67 because the decimal IP number represents all eight bits

in an octet.

Subnet Figure #8 illustrates how a subnetted network has host numbers assigned to

each network device. Notice that each device has numbers that relate to the range valid

for each subnetwork.

Notice in Subnet Figure #8 how the router received two host numbers. This is

because, inside the router, there are two Ethernet ports. Each of these ports receives a

host number just as if it were a network card inside a computer. The host number assigned

to the routers Ethernet port corresponds to a host number on the subnetwork the Ethernet port attaches to. Since the left side of the router is connected to Subnetwork 64, the

routers port host address is .65, the first available host number on the subnetwork. The

routers port does not have to receive the first host number in the subnetwork, but it is

done this way in the figure to illustrate how a host number is assigned.

When subnetting, each device still has a unique IP address. The only difference

that subnetting makes is that each subnetwork has a range of valid host numbers

for that individual subnetwork.

MASK REVIEWIn order to subnet, the subnet mask is used and is very important to understand. In the

Network chapter you learned that a Class A IP address has a standard mask of 255.0.0.0.

A Class B IP address has a standard mask of 255.255.0.0, and a Class C IP address has a

mask of 255.255.255.0. For example, consider a computer that has an IP address of

150.150.150.150. The IP address is a Class B address. If the computer uses a standard

mask, the mask entered would be 255.255.0.0.

Router192.168.10.64 192.168.10.128

.67 .68 .69 .70 .130 .131 .132 .133

.129.65

Subnet Figure #8Host Addresses on a Subnetted Network


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Keep in mind that the mask distinguishes the network number from the host numb

Using the same example used above of 150.150.150.150 and a mask of 255.255.0

yields the network number of 150.150.0.0 because of the mask used. An IP address

193.206.52.4 with a mask of 255.255.255.0 has a network address of 193.206.52.

THE MASK WHEN SUBNETTINGThe way that any network device knows that a subnet is being used is through t

mask. This is why the mask is sometimes known as the subnet mask. The mask numb

stays the same up to the point that bits are borrowed. Then, in the octet where bits a

borrowed, the new mask number is found by adding the bit positions togetherthat a

borrowedto create the subnet number. Look back to Subnet Figure #3. Since this i

Class C address, the normal mask is 255.255.255.0. However, since subnetting

implemented, the mask changes to 255.255.255.192. The 192 is found by adding t

value of the bit positions being borrowedbit position 128 and bit position 64 (the tw

bit positions borrowed to create the subnet number). 128 + 64 = 192. So, the new mask255.255.255.192. If three bits are borrowed in a Class C IP address, the last octet ma

would be 224 (128 + 64 + 32). If four bits are borrowed in a Class C IP address, the l

octet mask would be 240 (128 + 64 + 32 + 16). Examples of the new mask with each

class are given later in the chapter.

SOLVING IP SUBNETTING PROBLEMSWhen asked a subnetting problem, you can be presented with several pieces

information that describe the situation. Given that information, it will be up to you

figure out the remaining pieces of information to solve the problem. The types information that you must be able to identify are as follows:

noitamrofnI

sserddaPIfossalC

k numberrowteN

ksaM

skrowtenbus/skrowtenfo.onlatoT

krowtenbus/krowtenrepstsohfo.onlatoT

srebmuntenbuS

sesserddatsacdaorB

deworrobstibfo.oN

krowtenbus/krowtenrepstsohelbaesU

stenbuselbaesU

Solving IP Subnetting Problems


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Class C Problem 1

Lets do one example without subnetting to make it simple and to explain how the

chart works. Given the XYZ Companys Class C address of 201.15.6.0 and a mask of

255.255.255.0, you should be able to fill in the following information:

Now that the purpose of the chart is clear, lets try an example with subnetting.

Suppose the XYZ Company has four different networks throughout its factory, but only

one full Class C address. One option the company can do is to divide the Class C address

into four different subnetworks. The Class C IP address is 201.15.6.0. The network

administrator decides that the new subnet mask is 255.255.255.224. The following

information is what we know so far:

noitamrofnI eulaV

sserddaPIfossalC C

rebmuNkrowteN 0.6.51.102

ksaM 422.552.552.552

skrowtenbus/skrowtenfo.onlatoT ??

krowtenbus/krowtenrepstsohfo.onlatoT ??

srebmuntenbuS ??

sesserddatsacdaorB ??deworrobstibfo.oN ??

krowtenbus/krowtenrepstsohelbaesU ??

stenbuselbaesU ??

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 0.552.552.552

skrowtenbus/skrowtenfo.onlatoT 1

krowtenbus/krowtenrepstsohfo.onlatoT 652

srebmuntenbuS stenbusoneraerehtesuacebA/N

sesserddatsacdaorB 552.6.51.102

deworrobstibfo.oN stenbusoneraerehtesuacebA/N

krowtenbus/krowtenrepstsohelbaesU 452

stenbuselbaesU stenbusoneraerehtesuacebA/N


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If the mask is given, you can solve the rest of the unknowns and fill in the chart. T

mask in this problem is 255.255.255.224. Since this is a Class C network and the defa

mask is 255.255.255.0, we know that some bits are being borrowed in the last oc

because the number in the last octet has changed to 255.255.255.224. The first step is

break the last octet into binary to see how many bits are being borrowed for

subnetting. Subnet Figure #9 shows the last octet broken down into bits.

Notice in Subnet Figure #9 how the first three bits are set to 1. Since there is a 1

the 128 column, a 1 in the 64 column, and a 1 in the 32 column, 128 + 64 + 32 = 22

This is just the process of converting decimal to binary as shown in earlier chapters.

The next step is to draw a vertical line between the 1s and the 0s. In the case of t

224 mask, a vertical line is drawn between the 32 and the 16 column. See Subnet Figu

#10 to see where the vertical line is placed.

Once you have figured out where this line goes, you can answer many question

Now you know that the number of bits borrowed is three because there are three 1s

when you translate 224 into binary. Lets update the chart.


te 1tcO t 2etcO t 3etcO t 4etcO

821 46 23 61 8 4 2 1

.552 .552 .552 1 1 1 0 0 0 0 0

Subnet Figure #10



821 46 23 61 8 4 2 1

.552 .552 .552 1 1 1 0 0 0 0 0

224 broken into bits

Subnet Figure #9



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Once you have determined the total number of subnets and the total number of ho

per subnet, you can determine the number of useable subnets and useable hosts psubnet. The number of useable subnets is the number of subnets minus 2. Since there a

eight possible subnetworks, 8 2 = 6 useable networks. The number of useable hosts

the number of hosts per subnetwork minus 2. Since there are 32 possible hosts p

subnetwork, 32 2 = 30.

We have determined that there are eight subnetworks available using the subnet ma

of 255.255.255.224. Subnet Figure #11 shows the eight different subnetworks conver

into decimal values.

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 422.552.552.552



srebmuntenbuS ??

sesserddatsacdaorB ??

deworrobstibfo.oN 3


stenbuselbaesU 6

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 422.552.552.552

skrowtenbus/skrowtenfo.onlatoT 8krowtenbus/krowtenrepstsohfo.onlatoT 23

srebmuntenbuS ??


deworrobstibfo.oN 3


stenbuselbaesU ??



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Notice in Subnet Figure #11 that the subnetworks are grouped into multiples of 32.

Also notice that the first column to the left of the line (the darker gray area) has a value

of 32. The total subnetworks are as follows: 0, 32, 64, 96, 128, 160, 192, and 224. These

are all multiples of 32.

Now, we can update our list and put in the subnet numbers as shown below:

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 422.552.552.552



srebmuntenbuS

,46.6.51.102,23.6.51.102,0.6.51.102

,061.6.51.102,821.6.51.102,69.6.51.102

422.6.51.102,291.6.51.102


deworrobstibfo.oN 3


stenbuselbaesU 6



tenbuS

.oN

821 46 23 61 8 4 2 1

.102 .51 .6 0 0 0 0 0 0 0 0 0

.102 .51 .6 0 0 1 0 0 0 0 0 23

.102 .51 .6 0 1 0 0 0 0 0 0 46

.102 .51 .6 0 1 1 0 0 0 0 0 69

.102 .51 .6 1 0 0 0 0 0 0 0 821

.102 .51 .6 1 0 1 0 0 0 0 0 061

.102 .51 .6 1 1 0 0 0 0 0 0 291

.102 .51 .6 1 1 1 0 0 0 0 0 422

Subnet Figure #11


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Instead of running 1s and 0s in the subnet columns, you can determine that the

subnetworks are in groups of 32, which is the value in the column left of the

subnet line.

Now lets determine the broadcast addresses for each subnetwork. The broadc

address can be found by placing all 1s in the host bits for each subnetwork. Subnet Figu#12 shows the broadcast address calculation.

Notice in Subnet Figure #12 how each subnetwork is shown with the subnetwo

number and the broadcast address for that subnetwork. Now for the final chart update

tenbuS egnaRsserddAPI 821 46 23 61 8 4 2 1

1 )0.6.51.102( 0 0 0 0 0 0 0 0

1 )13.6.51.102( 0 0 0 1 1 1 1 1

2 )23.6.51.102( 0 0 1 0 0 0 0 0

2 )36.6.51.102( 0 0 1 1 1 1 1 1

3 )46.6.51.102( 0 1 0 0 0 0 0 0

3 )59.6.51.102( 0 1 0 1 1 1 1 1

4 )69.6.51.102( 0 1 1 0 0 0 0 0

4 )721.6.51.102( 0 1 1 1 1 1 1 1

5 )821.6.51.102( 1 0 0 0 0 0 0 0

5 )951.6.51.102( 1 0 0 1 1 1 1 1

6 )061.6.51.102( 1 0 1 0 0 0 0 0

6 )191.6.51.102( 1 0 1 1 1 1 1 1

7 )291.6.51.102( 1 1 0 0 0 0 0 0

7 )322.6.51.102( 1 1 0 1 1 1 1 1

8 )422.6.51.102( 1 1 1 0 0 0 0 0

8 )552.6.51.102( 1 1 1 1 1 1 1 1

Subnet Figure #124th Octet in Binary



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Class C Problem 2

The Hi-IQ Company has leased one Class C address200.200.200.0. The company

has ten networks with 12 computers on each network. With this information, the chart

appears as follows:

noitamrofnI eulaV

sserddaPIfossalC C


ksaM ??

skrowtenbus/skrowtenfo.onlatoT )dedeen01(

krowtenbus/krowtenrepstsohfo.onlatoT )dedeen21(

srebmuntenbuS ??

sesserddatsacdaorB ??deworrobstibfo.oN ??


stenbuselbaesU ??

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 422.552.552.552



srebmuntenbuS

,46.6.51.102,23.6.51.102,0.6.51.102

,061.6.51.102,821.6.51.102,69.6.51.102

422.6.51.102,291.6.51.102

sesserddatsacdaorB

,59.6.51.102,36.6.51.102,13.6.51.102

,191.6.51.102,951.6.51.102,721.6.51.102

552.6.51.102,322.6.51.102deworrobstibfo.oN 3


stenbuselbaesU 6


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To solve this problem, the first step is to determine the mask and, from there, the r

is the same as the first example. To determine what mask is needed, you must discov

how many bits to borrow. Remember to use the formula 2x = total number of subnets

= the number of useable subnets, wherex is the number of bits borrowed. If two host b

are borrowed (the minimum number for a Class C network), only two subnets are creat

(22 = 4 2 = 2). That number of subnets is not enough for the Hi-IQ Company. If thrhost bits are borrowed, six subnets are useable (23 = 8 2 = 6). Again, there are n

enough subnets. If four host bits are borrowed, 14 subnets are useable (24 = 16 2 = 1

This is the correct number of bits to borrow for the Hi-IQ Company; however, the ma

must still be determined.

When borrowing four host bits to create the subnets, the mask is found by adding t

bit values for the highest most bits. See Subnet Figure #13.

The normal Class C mask is 255.255.255.0, but we are borrowing bits from the l

octet, so we know the mask is going to be different. By adding the bit values for the b

being borrowed, the mask is found128 + 64 + 32 + 16 = 240. The mask for t

subnetted Class C address is 255.255.255.240. Updating the chart shows the following

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 042.552.552.552


krowtenbus/krowtenrepstsohfo.onlatoT )dedeen21(

srebmuntenbuS ??

sesserddatsacdaorB ??deworrobstibfo.oN 4


stenbuselbaesU 41


te 1tcO t 2etcO te 3tcO t 4etcO

821 46 23 61 8 4 2 1

.552 .552 .552 1 1 1 1 0 0 0 0

Subnet Figure #13



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Since the number of bits borrowed has been determined, it is easy to see how many

host bits are left to determine the total number of hosts per subnetwork. Look back at

Subnet Figure #13. At a quick glance, it is apparent that four bits remain for hosts. Using

the formula 2x = total number of hosts 2 = the number of useable hosts wherex is the

number of host bits remaining, if four host bits are remaining, 14 host addresses are

useable (24 = 16 2 = 14). This works well for the Hi-IQ Company since they have 12computers on each network. Updating the chart with this information provides the

following:

The only thing left to do is to figure out the subnetwork numbers and the broadcasts.

Subnet Figure #14 shows only the subnetwork numbers by placing 0s in each of the host

bits.

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 042.552.552.552


repstsohfo.onlatoTkrowtenbus/krowten

61

srebmuntenbuS ??


deworrobstibfo.oN 4

repstsohelbaesU

krowtenbus/krowten41

stenbuselbaesU 41


21/54

E-2

Notice that the first subnet has all 0s in the last octet. This is why none of the fi

subnets cannot be used as a useable subnet. Now lets get the broadcast numbers. Subn

Figure #15 shows only the broadcasts by placing 1s in each of the host bits.



tenbuS

.oN821 46 23 61 8 4 2 1

.102 .51 .6 0 0 0 0 0 0 0 0 0

.102 .51 .6 0 0 0 1 0 0 0 0 61

.102 .51 .6 0 0 1 0 0 0 0 0 23

.102 .51 .6 0 0 1 1 0 0 0 0 84

.102 .51 .6 0 1 0 0 0 0 0 0 46

.102 .51 .6 0 1 0 1 0 0 0 0 08

.102 .51 .6 0 1 1 0 0 0 0 0 69

.102 .51 .6 0 1 1 1 0 0 0 0 211

.102 .51 .6 1 0 0 0 0 0 0 0 821

.102 .51 .6 1 0 0 1 0 0 0 0 441

.102 .51 .6 1 0 1 0 0 0 0 0 061

.102 .51 .6 1 0 1 1 0 0 0 0 671

.102 .51 .6 1 1 0 0 0 0 0 0 291

.102 .51 .6 1 1 0 1 0 0 0 0 802

.102 .51 .6 1 1 1 0 0 0 0 0 422

.102 .51 .6 1 1 1 1 0 0 0 0 042

Subnet Figure #14



22/54


Notice how the last subnet has all 1s in the last octet. This is why the last subnet

cannot be used as a useable subnet. Now lets update the chart with the subnetwork

numbers and their associated broadcast addresses.



tsacdaorB

sserddA821 46 23 61 8 4 2 1

.102 .51 .6 0 0 0 0 1 1 1 1 51

.102 .51 .6 0 0 0 1 1 1 1 1 13

.102 .51 .6 0 0 1 0 1 1 1 1 74

.102 .51 .6 0 0 1 1 1 1 1 1 36

.102 .51 .6 0 1 0 0 1 1 1 1 97

.102 .51 .6 0 1 0 1 1 1 1 1 59

.102 .51 .6 0 1 1 0 1 1 1 1 111

.102 .51 .6 0 1 1 1 1 1 1 1 721

.102 .51 .6 1 0 0 0 1 1 1 1 341

.102 .51 .6 1 0 0 1 1 1 1 1 951

.102 .51 .6 1 0 1 0 1 1 1 1 571

.102 .51 .6 1 0 1 1 1 1 1 1 191

.102 .51 .6 1 1 0 0 1 1 1 1 702

.102 .51 .6 1 1 0 1 1 1 1 1 322

.102 .51 .6 1 1 1 0 1 1 1 1 932

.102 .51 .6 1 1 1 1 1 1 1 1 552

Subnet Figure #15


23/54

E-2

All important pieces of information needed to set up the network are now provide

Class C Problem 3

A network administrator for the Total Cool Company is working on a computer. T

computers IP address is 204.210.179.142 with a mask of 255.255.255.192. The netwo

administrator needs to know on which subnet the computer belongs. The informati

found by looking at the computer is the Class of IP address, the network portion of theaddress, and the mask. The following chart shows this information.

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 042.552.552.552

skrowtenbus/skrowtenfo.onlatoT 61krowtenbus/krowtenrepstsohfo.onlatoT 61

srebmuntenbuS

,61.002.002.002,0.002.002.002

,84.002.002.002,23.002.002.002

,08.002.002.002,46.002.002.002

,211.002.002.002,69.002.002.002

,441.002.002.002,821.002.002.002

,671.002.002.002,061.002.002.002

,802.002.002.002,291.002.002.002

042.002.002.002,422.002.002.002

sesserddatsacdaorB

,13.002.002.002,51.002.002.002

,36.002.002.002,74.002.002.002

,59.002.002.002,97.002.002.002

,721.002.002.002,111.002.002.002

,951.002.002.002,341.002.002.002

,191.002.002.002,571.002.002.002

,322.002.002.002,702.002.002.002

552.002.002.002,932.002.002.002

deworrobstibfo.oN 4


stenbuselbaesU 41



24/54


The first step in solving this problem is to discover how many bits are borrowed. A

normal Class C mask is 255.255.255.0, but the one on this computer is 255.255.255.192.Finding out how many bits are borrowed requires you to put 1s in Octet 4s bits until the

bit positions add up to 192. Look at Subnet Figure #16 to see how this is done.

Bit position 128 plus bit position 64 added together gives you 192. So, two bits are

borrowed. Updating the chart with the number of bits borrowed shows the following:



821 46 23 61 8 4 2 1

.552 .552 .552 1 1

Subnet Figure #16

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 291.552.552.552

skrowtenbus/skrowtenfo.onlatoT ??krowtenbus/krowtenrepstsohfo.onlatoT ??

srebmuntenbuS ??


deworrobstibfo.oN ??


stenbuselbaesU ??


25/54


26/54


Now, the only thing left to do is to determine the subnet numbers. Look back to

Subnet Figure #16. You can see that the line is drawn between the 32 and 64 bit positions.

A shortcut is to look at the number to the left of the line and you can tell that the

subnetwork numbers will be incremented in steps of 64, but filling in the chart with 0s in

the host bits proves this shortcut. Subnet Figure #17 shows the subnetwork numbers.

A shortcut for finding the broadcast address is to subtract 1 from the subnetwork

below the one you are working on because the broadcast address is always 1 less than the

subnetwork number. Also remember that the last subnet will always have a broadcast

address of 255. However, filling in the chart shows the full version. Subnet Figure #18

shows broadcasts for each of the subnetworks by putting 1s in the host positions.

Now, update the chart to include the subnetwork numbers and broadcast addresses:


t 1etcO t 2etcO t 3etcO t 4etcOtsacdaorB

sserddA

821 46 23 61 8 4 2 1

.402 .012 .971 0 0 1 1 1 1 1 1 36

.402 .012 .971 0 1 1 1 1 1 1 1 721

.402 .012 .971 1 0 1 1 1 1 1 1 191

.402 .012 .971 1 1 1 1 1 1 1 1 552

Subnet Figure #18


t 1etcO t 2etcO t 3etcO t 4etcO .oNtenbuS

821 46 23 61 8 4 2 1

.402 .012 .971 0 0 0 0 0 0 0 0 0

.402 .012 .971 0 1 0 0 0 0 0 0 46

.402 .012 .971 1 0 0 0 0 0 0 0 821

.402 .012 .971 1 1 0 0 0 0 0 0 291

Subnet Figure #17


27/54

E-2

Now that the chart is complete, look at the sequence of network numbers or back

Subnet Figure #17 to determine in what subnetwork IP address 204.210.179.142 fal

Subnetwork 0 is IP addresses 204.210.179.0 through 204.210.179.63. This does n

include .142, so look at the next subnetwork. Subnetwork 1 is IP address

204.210.179.64 through 204.210.179.127. Again, .142 is not in this range. Subnetwork

is IP addresses 204.210.179.128 through 204.210.179.191. This range of addresses do

include .142, so IP address 204.210.179.142 is on subnetwork 204.210.179.128. Anoth

way of solving for the subnetwork number is to and the IP address with the ma

Remember when anding that two 1s together make a 1. All other combinations of 1s a

0s make a 0. Subnet Figure #19 shows the anding of the IP address 204.210.179.1

with the mask255.255.255.192 with the result being the subnetwork number.

1

2

8

6

4

3

2

1

6 8 4 2 1

1

2

8

6

4

3

2

1

6 8 4 2 1

1

2

8

6

4

3

2

1

6 8 4 2 1

1

2

8

6

4

3

2

1

6 8 4 2

niPI

yranib1 1 0 0 1 1 0 0 1 1 0 1 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 1 1 1

niksaM

yranib1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0

nitenbuS

yranib1 1 0 0 1 1 0 0 1 1 0 1 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 0

Subnet Figure #19

noitamrofnI eulaV

sserddaPIfossalC C


ksaM 291.552.552.552



srebmuntenbuS,46.971.012.402,0.971.012.402

291.971.012.402,821.971.012.402

sesserddatsacdaorB,721.971.012.402,36.971.012.402

552.971.012.402,191.971.012.402

deworrobstibfo.oN 2


stenbuselbaesU 2



28/54


29/54

E-2

The normal subnet mask used with Class B IP addresses is 255.255.0.0. Wh

implementing subnets, the third (and possible fourth) octet number changes. The mask

found by looking at the octet(s) where host bits are borrowed and adding the bit valu

together. Subnet Figure #22 shows how the mask is obtained when borrowing three b

from a Class B address.

When three bits are borrowed, the mask has 1s set in the first three bits of the thi

octet. These bit positions are 128, 64, and 32. Add these bit values together to get 2

(128 + 64 + 32 = 224). The mask for a Class B network with three bits borrowed

255.255.224.0.

Now, for some more practice. The best way to learn Class B IP address subnetting

to practice, practice, practice.

Class B Problem 1

The PDQ, Inc. company has ten different networks located throughout the count

but has leased only one Class B address, 180.10.0.0. One option for the company is

divide the Class B address into subnetworks. The network administrator must determi

how many bits to borrow by looking at how many subnets the company needs. Since t

company has ten networks, a minimum of four bits must be borrowed. 24 = 16 and 16

= 14 useable subnet numbers. (If three bits were borrowed, there would not be enou

subnets because 23 = 8 and 8 2 = 6.) The network administrator has heard that a merg

is imminent, so, playing it safe, she decides to borrow five bits for subnetting. Borrowi

five bits allows for 30 subnetworks because 25 = 32 and 32 2 = 30. Subnet Figure #

shows how the mask is determined.

BssalC

.oNkrowteNstenbuS stsoH

tetcO

1

tetcO

2t 3etcO t 4etcO

1

28

6

4

3

2

1

6 8 4 2 1

1

28

6

4

3

2

1

6 8 4 2 1

552 552 1 1 1

Mask

Subnet Figure #22



30/54


1s are placed in all subnetwork bit positions and those bit values are added together:128 + 64 + 32 + 16 + 8 = 248. The mask used in the PDQ, Inc. network is 255.255.248.0.

The chart below shows the information gathered thus far:

noitamrofnI eulaV

sserddaPIfossalC B


ksaM 0.842.552.552

skrowtenbusfo.onlatoT 23

krowtenbus/krowtenrepstsohfo.onlatoT ??srebmuntenbuS ??


deworrobstibfo.oN 5


stenbuselbaesU 03

BssalC


tetcO

1

tetcO

2t 3etcO t 4etcO

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

552 552 1 1 1 1 1

Mask

Subnet Figure #23


31/54

E-3

The number of hosts can be determined by looking at how many host bits rema

Look back to Subnet Figure #13 and see that the number of remaining host bits is 11.

= 2,048 total number of hosts for each subnet. Subtract 2 to obtain the number of useab

hosts on each subnet: 2,048 2 = 2,046 useable hosts on each subnet. The updated ch

shows the following:

Now for the real workdetermining subnet numbers and broadcast address

Subnet Figure #24 shows the 1 and 0 patterns for determining subnetwork numbers. D

to the lack of space, not all subnetworks are shown, but enough subnets are shown

illustrate the pattern.

noitamrofnI eulaV

sserddaPIfossalC B


ksaM 0.842.552.552


krowtenbusrepstsohfo.onlatoT 840,2

srebmuntenbuS ??


deworrobstibfo.oN 5krowtenbusrepstsohelbaesU 640,2

stenbuselbaesU 03



32/54


Notice how the subnetwork numbers are in increments of eight. When doing Class A

and B subnetting, there is normally not enough time to write every combination of 1s and

0s. When first learning subnetting, you should definitely write out a few, but once yousee the pattern emerge, you should do the first couple of subnets and the last subnet.

Updating the chart with the subnetwork numbers yields the following:

BssalC


tetcO1

tetcO2

t 3etcO t 4etcO tenbuS.oN

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

081 01 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0

081 01 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0.8

081 01 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0.61

081 01 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0.42

081 01 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0.23081 01 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0.04

081 01 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0.84

081 01 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0.65

081 01 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.46

081 01 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0.27

081 01 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0.08

081 01 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0.88

081 01 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0.69

081 01 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0.401081 01 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0.211

081 01 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0.021

081 01 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.821

081 01

081 01

081 01 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0.842

Subnet Figure #24


33/54

E-3

All that is left to f ind in the chart is the broadcast address for each subnetwork. T

is found by placing 1s in all of the host bits after the subnetworks are found. This, towill show an emerging pattern as Subnet Figure #25 illustrates:

noitamrofnI eulaV

sserddaPIfossalC B


ksaM 0.842.552.552

skrowtenbusfo.onlatoT 23krowtenbusrepstsohfo.onlatoT 840,2

srebmuntenbuS

,0.61.01.081,0.8.01.081,0.0.01.081

,0.04.01.081,0.23.01.081,0.42.01.081

,0,46.01.081,0.65.01.081,0.84.01.081

,0.88.01.081,0.08.01.081,0.27.01.081

,0.211.01.081,0.401.01.081,0.69.01.081

0.631.01.081,0.821.01.081,0.021.01.081

0.061.01.081,0.251.01.081,0.441.01.081

0.481.01.081,0.671.01.081,0.861.01.081 0.802.01.081,0.002.01.081,0.291.01.081

0.232.01.081,0.422.01.081,0.612.01.081

0.842.01.081,0.042.01.081


deworrobstibfo.oN 5

krowtenbusrepstsohelbaesU 640,2

stenbuselbaesU 03



34/54


Do not forget to look at the entire octet when determining the decimal value for

the octet.

BssalC

.oNkrowteNtenbuS stsoH

tetcO1

tetcO2

t 3etcO t 4etcO tsacdaorB

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

081 01 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 552.7

081 01 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 552.51

081 01 0 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 552.32

081 01 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 552.13081 01 0 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 552.93

081 01 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 552.74

081 01 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 552.55

081 01 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 552.36

081 01 0 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 552.17

081 01 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 552.97

081 01 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 552.78

081 01 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 552.59

081 01 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 552.301

081 01 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 552.111

081 01 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 552.911

081 01 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 552.721

081 01 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 552.531

081 01 1 1 1 1 1 1 1 1 1 1 1

081 01 1 1 1 1 1 1 1 1 1 1 1

081 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 552.552

Subnet Figure #25


35/54

E-3

The pattern that emerges is that the third octet increments by eight each time and t

fourth octet is always 255. Also notice that the broadcast number is one less than t

subnetwork number that follows. The completed chart lists below:

noitamrofnI eulaV

sserddaPIfossalC BrebmuNkrowteN 0.0.01.081

ksaM 0.842.552.552



srebmuntenbuS

,0.61.01.081,0.8.01.081,0.0.01.081

,0.04.01.081,0.23.01.081,0.42.01.081

,0,46.01.081,0.65.01.081,0.84.01.081

,0.88.01.081,0.08.01.081,0.27.01.081

,0.211.01.081,0.401.01.081,0.69.01.0810.631.01.081,0.821.01.081,0.021.01.081

0.061.01.081,0.251.01.081,0.441.01.081

0.481.01.081,0.671.01.081,0.861.01.081

0.802.01.081,0.002.01.081,0.291.01.081

0.232.01.081,0.422.01.081,0.612.01.081

0.842.01.081,0.042.01.081

sesserddatsacdaorB

,552.51.01.081,552.7.01.081

,552.93.01.081,552.13.01.081

,552.55.01.081,552.74.01.081,552.17.01.081,552.36.01.081

,552.78.01.081,552.97.01.081

,552.301.01.081,552.59.01.081

,552.911.01.081,552.111.01.081

,552.531.01.081,552.721.01.081

,552.151.01.081,552.341.01.081

,552.571.01.081,552.761.01.081

,552.191.01.081,552.381.01.081

,552.702.01.081,552.991.01.081

,552.322.01.081,552.512.01.081

,552.932.01.081,552.132.01.081

552.552.01.081

deworrobstibfo.oN 5


stenbuselbaesU 03



36/54


Class B Problem 2

One of the hardest concepts for students to grasp is when borrowed host bits are in

more than one octet. In the next scenario, the Top Hats Co. has 2,000 locations

throughout the world. Each location has a network with approximately 20 computers.

The Top Hats Co. has leased the Class B IP address of 189.208.0.0. The first step in

solving this problem is determining how many bits to borrow. Subnet Table #1 helps with

this decision:

Looking at Subnet Table #1, you can see that borrowing 11 bits allows for the 2,000

Top Hats Co.s locations throughout the world. By borrowing 11 bits, there are also

enough remaining host bits to accommodate the computers at each site. The followingchart summarizes the information gathered so far:

noitamrofnI eulaV

sserddaPIfossalC B


ksaM ??

skrowtenbusfo.onlatoT 840,2

krowtenbusrepstsohfo.onlatoT 23

srebmuntenbuS ??


deworrobstibfo.oN 11

krowtenbusrepstsohelbaesU 03

stenbuselbaesU 640,2

deworroB

stiB

tsoHfo.oN

stiB

latoT

stenbuS

elbaesU

stenbuSstsoHlatoT

elbaesU

stsoH

5 11 23 03 8402 6402

6 01 43 23 4201 2201

7 9 821 621 215 015

8 8 652 452 652 452

9 7 215 015 821 621

01 6 4201 2201 46 26

11 5 8402 6402 23 03

21 4 6904 4904 61 41

31 3 2918 0918 8 6

Subnet Table #1


37/54

E-3

To determine what mask is needed throughout the Top Hats Co.s network, place 1s

the subnetwork field and add the bit value positions together for each octet. Subn

Figure #26 shows this concept:

The first two octets are the standard 255.255 numbers. The third octet is filled w

1s, so the third octet mask is 255(128 + 64 + 32 + 16 + 8 + 4 + 2 +1 = 255). The fou

octet has three bits set, so the mask is 224(128 + 64 + 32 = 224). The final mask

this problem is 255.255.255.244, and the information can be inserted into the chart:

Now, the subnetworks must be determined. Subnet Figure #27 shows the breakdow

of the subnets with 1s and 0s.

Keep in mind when putting 1s and 0s in multiple subnetwork octets that you

must treat them as one big group.

noitamrofnI eulaV

sserddaPIfossalC BrebmuNkrowteN 0.0.802.981

ksaM 422.552.552.552


krowtenbusrepstsohfo.onlatoT 23

srebmuntenbuS ??





BssalC


tetcO

1

tetcO

2t 3etcO t 4etcO

1

2

8

46 23 61 8 4 2 1

1

2

8

46 23 61 8 4 2

552 552 1 1 1 1 1 1 1 1 1 1 1

Mask

Subnet Figure #26



38/54


Updating the chart could take several pages with these subnetworks, but entering in afew of them shows the following:

BssalC


tetcO1

tetcO2

t 3etcO t 4etcO stenbuS

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2

981 802 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0

981 802 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 23.0

981 802 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 46.0

981 802 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 69.0

981 802 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 821.0981 802 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 061.0

981 802 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 291.0

981 802 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 422.0

981 802 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0.1

981 802 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 23.1

981 802 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 46.1

981 802 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 69.1

981 802 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 821.1

981 802 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 061.1981 802 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 291.1

981 802 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 422.1

981 802 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0.2

981 802 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 23.2

981 802

981 802

981 802 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 422.552

Subnet Figure #27


39/54

E-3

The last bit of information left to f ind is the broadcast numbers. Simply put 1s in t

host bits for each subnetwork. Subnet Figure #28 shows this process.

noitamrofnI eulaV

sserddaPIfossalC B


ksaM 422.552.552.552

skrowtenbusfo.onlatoT 840,2krowtenbusrepstsohfo.onlatoT 23

srebmuntenbuS

,46.0.802.981,23.0.802.981,0.0.802.981

,821.0.802.981,69.0.802.981

,291.0.802.981,061.0.802.981

,0.1.802.981,422.0.802.981

,422.1.802.981,23.1.802.981

,46.2.802.981,23.2.802.981,0.2.802.981

...,821.2.802.981,69.2.802.981

422.552.802.981sesserddatsacdaorB ??






40/54


Of course updating the list with the broadcasts is quite lengthy too, but some have

been inserted into the chart to illustrate the point.

BssalC


tetcO1

tetcO2

t 3etcO t 4etcO stenbuS

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2

981 802 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 13.0

981 802 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 36.0

981 802 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 59.0

981 802 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 721.0

981 802

981 802

981 802 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 322.1

981 802 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 552.1

981 802 0 0 0 0 0 0 1 0 0 0 0 1 1 1 1 13.2

981 802 0 0 0 0 0 0 1 0 0 0 1 1 1 1 1 36.2

981 802

981 802 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 552.552

Subnet Figure #28


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E-4

Of course, each of the broadcast addresses is in groups of 32 just like t

subnetworks are. Do not forget to treat each octet as a group of eight bits wh

determining subnetwork numbers and broadcast addresses!

noitamrofnI eulaV

sserddaPIfossalC B


ksaM 422.552.552.552

skrowtenbusfo.onlatoT 840,2krowtenbusrepstsohfo.onlatoT 23

srebmuntenbuS

,46.0.802.981,23.0.802.981,0.0.802.981

,821.0.802.981,69.0.802.981

,291.0.802.981,061.0.802.981

,0.1.802.981,422.0.802.981

,422.1.802.981,23.1.802.981

,46.2.802.981,23.2.802.981,0.2.802.981

...,821.2.802.981,69.2.802.981

422.552.802.981

sesserddatsacdaorB

59.0.802.981,36.0.802.981,13.0.802.981

,322.1.802.981,721.0.802.981

,13.2.802.981,552.1.802.981

,59.2.802.981,36.2.802.981

,36.452.802.981721.2.802.981

,721.452.802.981,59.452.802.981

,191.452.802.981,951.452.802.981

,36.552.802.981,552.452.802.981

,721.552.802.981,59.552.802.981

,191.552.802.981,951.552.802.981

552.552.802.981






42/54


Class B Problem 3

The network administrator is working on a computer with the IP address of

157.208.190.144. The mask shows as 255.255.255.192. On what subnet is the computer

attached? The information known so far is summarized in the following table:

The first step is to determine how many host bits have been borrowed for subnetting.

Subnet Figure #29 shows the mask (the borrowed bits).

As seen in Subnet Figure #29, there are ten borrowed host bits. Knowing this

information, the total number of subnets, useable subnets, total number of hosts, and

useable hosts can be determined by using the formula 2x (wherex is either the number of

bits borrowed or the remaining bits) and then entered into the chart as shown below.

BssalC.oNkrowteN

stenbuS stsoH

tetcO

1

tetcO

2t 3etcO t 4etcO

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

552 552 1 1 1 1 1 1 1 1 1 1

Mask

Subnet Figure #29

noitamrofnI eulaV

sserddaPIfossalC B


ksaM 291.552.552.552

skrowtenbusfo.onlatoT ??

krowtenbus/krowtenrepstsohfo.onlatoT ??

srebmuntenbuS ??


deworrobstibfo.oN ??


stenbuselbaesU ??


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E-4

The next thing is to determine the subnetwork numbers. Subnet Figure #30 show

partial illustration of subnetwork numbers for this problem.

noitamrofnI eulaV

sserddaPIfossalC B


ksaM 291.552.552.552

skrowtenbusfo.onlatoT 420,1repstsohfo.onlatoT

krowtenbus/krowten46

srebmuntenbuS ??







44/54


45/54

E-4

However, to determine the broadcast address for each subnetwork, the same method

used as shown beforeput 1s in all of the host addresses and determine the decim

value for the octet.

A shortcut for solving a problem that gives an IP address and a mask and asks for

the subnetwork number is to put the IP address and mask in binary and and thetwo numbers together. Then convert the result to dotted decimal notation.

Remember when anding, the only way to get a 1 is by anding two 1s together.

Subnet Figure #31 shows this process.

Subnet in dotted decimal notation: 157.208.190.128

CLASS A SUBNETTINGClass A subnetting is handled the same as Class Bs and Cs with the exception of h

many host bits can be borrowed for subnetting. With Class A IP addresses, the first oc

(eight bits) represents the network number and the last three octets (24 bits) represe

host bits. Subnet Figure #32 shows this concept.

When subnetting Class A IP addresses, bits are borrowed from the left-most host b

and can extend across octets 2, 3, and 4 because these are the Class A host bits. Subn

Figure #33 shows a Class A IP address with eleven bits borrowed from the first a

second octets.

krowteNAssalC

.oNstsoH


Subnet Figure #32

1

2

8

6

4

3

2

1

6 8 4 2 1

1

2

8

6

4

3

2

1

6 8 4 2 1

1

2

8

6

4

3

2

1

6 8 4 2 1

1

2

8

6

4

3

2

1

6 8 4 2

niPIyranib

1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 0 1 0 1 1 1 1 1 0 1 0 0 1 0 0 0

niksaM

yranib1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0

nitenbuS

yranib1 0 0 1 1 1 0 1 1 1 0 1 0 0 1 0 1 0 1 1 1 1 1 0 1 0 0 0 0 0 0

Subnet Figure #31



46/54

E-46 Appendix CSubnetting

With Class A subnets, the same formula, 2x = total number of subnets (wherex is the

number of bits borrowed) is still used. The useable subnets is found by subtracting 2 from

the result, just like it was done with Class B and Class C subnets. In Subnet Figure #33,11 bits are borrowed from octets 2 and 3. 211 = 2,048 total subnets and subtracting 2

yields the useable subnets2,048 2 = 2,046.

The same formula is also used for determining total number of hosts. In Subnet

Figure #33, 13 host bits remain. 213 = 8,192 total host addresses. Subtracting 2 yields the

useable host addresses8,192 2 = 8,190. Keep consistent in how you solve IP

subnetting problems and no exam can trip you up.

The normal subnet mask used with Class A IP addresses is 255.0.0.0. When

implementing subnets, the second, third, and fourth octets can be used and therefore the

mask changes for these octets. Subnet Figure #34 shows how the mask is obtained when

borrowing 11 bits from a Class A address.

AssalC

krowteN

.oN

stsoH


12

8

64

32

16

8 4 2 1 12

8

64

32

16

8 4 2 1 12

8

64

32

16

8 4 2 1

552 1 1 1 1 1 1 1 1 1 1 1

Mask

Subnet Figure #34

AssalC

krowteN

.oN

stsoH


1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

Subnet Figure #33


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E-4

When borrowing 11 bits, the mask has 1s set in the second octet and the first thr

bits of the third octet. Octet 2 is all 1s, so the mask for octet 2 is 255. Octet 3 has 1s set

the first three bit positions. Add these bit values together to get 224. (128 + 64 + 32

224). The mask for a Class A network with 11 bits borrowed is 255.255.224.0.

The best way to learn Class A addresses (as it has been for the other classes) is

practice. They are done the exact same way as the other addresses except there are mohost bits from which to borrow.

Class A Problem 1

The Super Duper Company has 5,000 locations worldwide. In each location, the

are more than 1000 computers. The Super Duper Company has leased one Class A

address, 19.0.0.0. One option for the company is to divide the class A address in

subnetworks. The first task for the network administrator is to determine how many b

to borrow for subnetting. Subnet Table #2 summarizes some borrowed bits w

corresponding number of subnets.

Looking at Subnet Table #2, one can see that, to assign subnetwork numbers to 5,0

locations, the Super Duper Company must borrow 13 host bits. This also allows for 2,0

host addresses per location. Subnet Figure #35 shows how the subnet mask is determin

using the 13 host bits for subnetworks.

deworroB

stiB

tsoHfo.oN

stiB

latoT

stenbuS

elbaesU

stenbuSstsoHlatoT

elbaesU

stsoH

5 91 23 03 882425 682425

6 81 46 26 441262 241262

7 71 821 621 270131 070131

8 61 652 452 63556 43556

9 51 215 015 86723 6672301 41 4201 2201 48361 28361

11 31 8402 6402 2918 0918

21 21 6904 4904 6904 4904

31 11 2918 0918 8402 6402

41 01 48361 28361 4201 2201

51 9 86723 66723 215 015

61 8 63556 43556 652 452

71 7 270131 070131 821 621

Subnet Table #2



48/54


With 13 bits borrowed, the mask has the standard 1s in the first octet (255), 1s in the

second octet (255), and 1s in five bits of the third octet (248128 + 64 + 32 + 16 + 8).

So, the mask for the Super Duper Company is 255.255.248.0. The chart below shows the

information determined so far.

Determining subnetwork numbers is done exactly the same way as when Class B and

Class C addresses are subnetted. Subnet Figure #36 shows a partial view of the Class A

subnetworks. Octet 1 is not divided into bit positions because it is always 19. Octet 4is not subdivided because it always contains 0s for the subnetwork number. For

presentation, only Octets 2 and 3 are shown where the subnetting occurs; Octet 4 is not

shown.

noitamrofnI eulaV

sserddaPIfossalC A


ksaM 0.842.552.552


krowtenbusrepstsohfo.onlatoT 840,2srebmuntenbuS ??





AssalC

krowteN.oN

stsoH


1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

552 1 1 1 1 1 1 1 1 1 1 1 1 1

Mask

Subnet Figure #35


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E-4

As you can see in Subnet Figure #36, there are 13 borrowed host bits. Subnetwo

numbers increment in groups of eight (19.0.0.0, 19.0.8.0, 19.0.16.0, and so on up

19.255.248.0). A few of the subnetwork numbers are filled into the chart so you can s

the trend. To determine the broadcast address for each subnetwork, the same method us

with Class C and Class B addresses is usedput all 1s in the host address and determ

the decimal value (or take the shortcut and subtract 1 from the next subnetwork numbe

The final chart is as follows:

AssalC

krowteN.oN

stsoH

t 1etcO t 2etcO t 3etcO stenbuS

1

2

8

6

4

3

2

1

68 4 2 1

1

2

8

6

4

3

2

1

68 4 2 1

91 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0.0.91

91 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0.8.0.91

91 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0.61.0.91

91 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0.42.0.91

91 0 0 0

91 0 0 0

91 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0.842.0.91

91 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0.0.1.91

91 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0.8.1.91

91 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0.61.1.91

91 0 0 0

91 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0.842.552.91

Mask

Subnet Figure #36



50/54


Writing all of the subnet numbers and broadcast addresses would take up page after

page of this text, so enough numbers are inserted for you to get the idea of the pattern.

Once you do a couple of numbers and see the patterns, you can determine all of the

subnet numbers.

noitamrofnI eulaV

sserddaPIfossalC A


ksaM 0.842.552.552



srebmuntenbuS

0.23.0.91,0.42.0.91,0.61.0.91,0.8.0.91

,0.8.1.91,0.0.1.91,0.842.0.91hguorht

0.0.2.91,842.1.91hguorht0.61.1.91

0.842.552.91hguorht

sesserddatsacdaorB

,552.13.0.91,552.32.0.91,552.51.0.91

,552.552.0.91hguorht.cte552.93.0.91

.cte552.13.1.91,552.51.1.91,552.7.1.91

552.7.2.91,552.552.1.91hguorht552.552.552.91hguorht





51/54

E-5

Name _____________________________

SUBNETTING REVIEW QUESTIONS

1. A company has received a Class C IP address for their four networks. How ma

bits need to be borrowed?

2. A company uses a Class C mask of 255.255.255.224. What is the maximu

number of hosts per subnetwork?

3. How many bits are borrowed in a Class C address if the mask is 255.255.255.24

4. Given the IP address 199.14.180.4, what class IP address is this?

5. Given the IP address of 201.60.250.91 and a mask of 255.255.255.248, what is t

subnetwork number?

6. Given the IP address 210.199.184.66 and the fact that a company borrows th

bits to subnet, what mask is used?

7. What is the standard subnet mask for a Class C address?

8. What is the maximum number of bits that can be borrowed when using a Class

address?

9. What is the minimum number of bits that can be borrowed when using a Class

address?

10. Given the IP address 204.16.8.0 and a mask of 255.255.255.240. What is the fi

useable subnetwork number?

11. Given the IP address 197.56.2.141 and a mask of 255.255.255.192, what is t

broadcast address for this subnetwork?

Subnetting Review Questions


52/54


12. Given the broadcast address of 202.202.159.159 and a mask of 255.255.255.248,

what is the subnetwork number?

13. A company has a policy of only 25 hosts per subnet. They have 20 networks. How

many Class C addresses does the company need?

14. How many bits are set in a standard Class C mask?

15. What is the maximum number of hosts on a Class C network?

16. Given the mask of 255.255.255.224 and an IP address of 200.200.200.200, on what

subnetwork is the device?

17. Given an IP address of 193.15.10.105 and a mask of 255.255.255.252, what is thesubnetwork number?

18. Given the mask of 255.255.255.248 and the fact that a Class C address is being

used, how many hosts are on each subnet?

19. Given the IP address of 206.19.1.186 and a mask of 255.255.255.192, what are the

two unuseable subnets?

20. Given the IP address of 199.199.144.43 and a mask of 255.255.255.224, what isthe last useable subnetwork number?

21. Given the IP address 130.14.207.39 and a mask of 255.255.240.0, how many total

subnets are available?

22. Given the IP address 130.14.207.39 and a mask of 255.255.240.0, what is the

subnetwork number associated with this IP address?

23. Given the IP address 188.188.188.188 and a mask of 255.255.255.128, what is the

subnetwork number associated with this IP address?

24. Given the IP address 191.10.59.63 and with six bits borrowed, what is the mask?


53/54

E-5

25. A company has a Class B IP address. What is the maximum number of bits that c

be borrowed and still have 100 hosts per subnetwork?

26. A company is leasing a Class A IP address and has 3000 networks. How many b

do they need to borrow?

27. Given the IP address 15.200.166.41 and a mask of 255.252.0.0, what is

subnetwork number?

28. What is the mask when 15 bits are borrowed and a Class A network address

being used?

29. Given the IP address 14.168.29.180 and a mask of 255.255.192.0, how many b

are borrowed for subnetting?

30. Given the IP address 14.168.29.180 and a mask of 255.255.192.0, what is t

broadcast for this subnetwork?

31. How many bits are set with a Class A subnet mask of 255.255.240.0?

32. Given the IP address 120.150.150.150 and a mask of 255.255.240.0, what is

subnetwork number and broadcast address?

Subnetting Review Questions


54/54

Documents

Computer Repair Student Part.5 [Subnetting] Finnish