Easy Sub Netting Full

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EASY SUBNETTING--BY (K.HARIKRISHNA)

IP ADDRESSING and SUBNETTING:

0.0.0.0=UNIVERSAL NETWORK I.D

255.255.255.255=UNIVERSAL BROADCASI.D

TWO Types of ipaddresses IPV4 and IPV6 WE R DEALING WITH IPV4 ADDRESSING ONLY.. IPv4 addresses

Consist of 32-bits.

Broken down into four octets (eight bits each).

Use dotted decimal format: example is 172.16.122.204. Minimum value (per octet) is 0, and the maximum value is 255. 0.0.0.0 is a Network ID.

255.255.255.255 is a Broadcast IP. IPv4 Address Classes:

Class==1stoctet, 2nd, 3rd, 4thoctet Class A Network.Host.Host.Host Class B Network.Network. Host. Host Class C Network.Network.Network.Host IN 32 bits:

NNNNNNNN .HHHHHHHH . HHHHHHHH . HHHHHHHH Class A NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C Default subnet masks:

Class A 255.0.0.0 Class B 255.255.0.0 Class C 255.255.255.0

The mask is written in slash notation as follows: Class A/8

Class B/16 Class C/24

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RANGES:

Class A: 0 – 127 1.0.0.0 through 127.0.0.0 (127 is not using for subnetting but for testing loop back stack Class B: 128 - 191 (i.e) 128.0.0.0 – 191.255.255.255

Class C: 192 –223 (i.e) 192.0.0.0 through 223.255.255.255 Class D: 224—239 (i.e) 224.0.0.0 through 239.255.255.255 Class E: 240—255 (i.e) 240.0.0.0 through 255.255.255.255 Octet priority: 0 CLASS-A 10 CLASS-B 110 CLASS-C 1110 CLASS-D 11110 CLASS-E

PRIVATE AND PUBLIC IP ADDRESS

PRIVATE uses by Internet standards groups. 10.0.0.0--10.255.255.255

172.16.0.0--172.31.255.255 192.168.0.0--192.168.255.255

This ranges are not using for subnetting

This range are fixed used for internet service providers

127.0.0.1 we are not using to assing but for testing loop back stack although it was in the range of class A address (i.e)(testing NIC is working or not )

Hint : The question may ask you which of this ip address are routable though private ip address ? Ans: apart from private (public ip ranges are routable except 127.0.0.1)

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BINARY DECIMAL AND HEXADECIMAL CONVERSION:

HEXADECIMAL = ( 0 1 2 3 4 5 6 7 8 9 a b c d e f ) that is absolutely 16 digits 0-9 (base10) (a-f)+(base 15) 128 64 32 16 8 4 2 1 ADDING DECIMAL-RESULT 1 0 0 1 0 1 0 0 128+16+4 148 0 0 0 0 1 1 1 1 8+4+2+1 15 0 1 0 1 0 1 0 1 64+16+4+1 85 1 0 0 0 0 0 1 1 128+2+1 131 0 0 0 1 0 1 1 0 16+4+1 22 1 1 1 1 1 1 1 1 128+64+32+16+8+4+2+1 255

FOR HEXADECIMAL WE USE ( 8 4 2 1 | 8 4 2 1 ) IN PLACE OF 128 64 32 16 8 4 2 1

8 4 2 1 8 4 2 1 HEXA ADDING HEXADECIMAL-RESULT 1 0 0 1 0 1 0 0 8+1=9,0+4=4 0X84 0 0 0 0 1 1 1 1 0+0=0,8+4+2+1=15 0X15 0 1 0 1 0 1 0 1 4+1=5,4+1=5, 0X55 1 0 0 0 0 0 1 1 8+0=8,2+1=3 0X83 0 0 0 1 0 1 1 0 0+1=1,4+2=6 0X16 1 1 1 1 1 1 1 1 8+4+2+1=15,8+4+2+1=15, 0XFF 1 0 0 1 1 1 0 1 8+1=9,8+4+1=13 0X9D 1 1 0 0 1 1 0 0 8+4=12,8+4=12 0XCC

---ROUTE SUMMARIZATION)- (route aggregation)

 (THE MORE ROUTER TABLE CONTAINS NETWORKS – THE MORE LOAD ON CPU(ROUTER) INCREASES . FOR THIS WE USE ROUTE SUMMARIZATION TO REDUCE LOAD ON CPU.  (THE POINT OF ROUTE SUMMARY IS TO MAKE ROUTING TABLES SMALLER.

 SMALLER ROUTING TABLES MEANS FEWER NETWORK UPDATES AND ROUTER CPU LOAD.

Most Pros would say automatic is bad, it gives no control over route distribution. Both RIP and EIGRP do auto summarization at the boundaries of classful networks. Which is Bad if you have discontinuous networks. This is when router on left has 10.1.0.0/24 and 10.2.0.0 /24 and router on right has 10.3.0.0/24 and 10.4.0.0 /24. Both routers advertise 10.0.0.0/8 summarized route to the router in the middle. That router in the middle does not know who to belive and where to send packets with destination on network 10.5.0.0.

Check your routing table before and after summarization, the size of routing table should be smaller when networks are summarized.

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Manual route summarization can improve routing efficiency, reduce memory consumption, and improve convergence, and reduce the length of routing tables.

Only OSPF does not support Autosummarization (must do manually). Rip 2 and EIGRP support Disabling Autosummarization but not RIP 1.

Rip1 , Rip2, and EIGRP do Autosummarization by default. Routing Protocol Route Summarization Support

Protocol Automatic Summarization at Classful Network Boundary? Capability to Turn Off Automatic Summarization? Capability to Summarize at Other Than a Classful Network Boundary? RIPv1 Yes No No

RIPv2 Yes Yes No

IGRP Yes No No

EIGRP Yes Yes Yes

OSPF No — Yes

IS-IS No — Yes

Routing table entries as:

10.128.4.0/24 /24 or 255.255.255.0 10.128.5.0/24 /24 or 255.255.255.0 10.128.7.0/24 /24 or 255.255.255.0

Convert each network into binary:

00001010.10000000.000001 00.00000000 ==== 10.128.4.0/24 or 255.255.255.0 00001010.10000000.000001 01.00000000 ==== 10.128.5.0/24 or 255.255.255.0 00001010.10000000.000001 11.00000000 ==== 10.128.7.0/24 or 255.255.255.0

( Taking most common bits in all networks and converting them into decimal )

00001010.10000000.000001xx.xxxxxxxx

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WILDCARD MASK

Here is a question that has always disturbed me from the day I came to know of these two words:

1. Subnet Mask 2. Wildcard Mask

When we configure an IP address on an interface, when we give network command in DHCP pool, when we specify the inside global range in NAT Pool we use SUBNET MASK

When we configure networks in OSPF, when we configure networks in EIGRP (not compulsory though), and in ACLs, we use WILDCARD MASK.

Why had CISCO experts decided to use two different concepts to accomplish the same aim ?

For example, 192.168.1.0 / 0.0.0.255 ( WildCard Notation) specify IP Range 192.168.1.0 to 192.168.1.255 and 192.168.1.0 / 255.255.255.0 (Subnet Mask Notation) also specify the same IP Range 192.168.1.0 to 192.168.1.255.

Then why did CISCO programmers preferred using two concepts while developing protocols Although single concept would have solved their problem..

Where am I lacking in the understanding of these concepts ?

Please note that I am not asking how to use these concepts or where to use them , Rather My question is why to use TWO concepts instead of one ??

ANSWER:

Anyway... Access Lists actually came before subnet masks. Remember way back when we lived in an evil classful world. So back in like 1985, when access-lists came about it was actually easier to code in assembler to do a NAND operation instead of an AND. Thus the wildcarding.

When we evolved into subnets (isn't everyone studying for their CCENT/CCNA exams so incredibly happy about that progress?) someone figured out not only that normal human beings weren't used to thinking "backwards" like the ACL masks, but there had to be some backwards compatibility with all the ancient IOS versions. So subnet masks being "new' took their own form. ACLs being "legacy" stayed the same.

The NAND gate were one of the faster instructions. The logic is (0,0)->1 (0,1)->1 (1,0)->1 but (1,1)->0

So if you have a address 11110000 (240) NAND 00001111 you get 11111111 For 11111111 (255) NAND 00001111 you get 11111111

So you know the first four bits are 1111 which is what you are testing and the bottom four you could not care about

And the final result is all 1's or True.

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For example: 255.255.255.255 === universal network I.D - 255.255.0.0 === your subnet mask

0 . 0 . 255. 255 === your wild-card mask

EXAMPLE:2

255.255.255.255 - 255.255.242.0

---

0. 0. 13.255

IP SUBNET-ZERO

In olden days this command ip subnet-zero was not enable in router (older) .BUT now a days it is default enabled in all the cisco routers so , don’t want to bother about this .

What this does is if (ip subnet-zero command is not enabled on router : you can’t use

FIRST AND LAST SUBNETS).

Example: 172.25.0.0/19 (FIRST) YOU CANT USE TO ASSIGN FIRST AND LAST 172.25.32.0/19 (2nd) 172.25.64.0/19 (3rd) 172.25.96.0/19 (4th) 172.25.128.0/19 (5th) 172.25.160.0/19 (6th) 172.25.192.0/19 (7th)

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Do I need to enable my router to recognize the zero subnet?

The quick answer to this question is NO. Your Cisco IOS router, by default, has the command

ip subnet-zero enabled on the router. Because of this command, the zero subnet can already be

recognized.

REFER TO the above SHOW RUNNING CONFIG You have not given (ip subnet-zero) BUT it is predefined enabled on router now a days …that means you can use first and last valid

subnets……….

Another example: It will reject the command

Router(config)#no ip subnet-zero Router(config)#int Fa0/1

Router(config-if)#ip add 172.16.0.1 255.255.255.0 Bad mask /24 for address 172.16.0.1

Router(config-if)#ip add 172.16.1.1 255.255.255.0 Router(config-if)#exit

Router(config)#ip subnet-zero Router(config)#int Fa0/1

Router(config-if)#ip add 172.16.0.1 255.255.255.0  see now there is no error msg !!!!!!

Router(config-if)#

You can see that I cannot use the very first subnet or subnet 0, but I can use anything after that. I believe IOS v12.0+ has it by default. This use to be used to avoid confusion I believe (before my time). If the command ip subnet-zero is configured you can use the very first subnet or subnet zero. If not, you must use something other than the first subnet.

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EASY SUBNETTING--BY (K.HARIKRISHNA)

27 26 25 24 23 22 21 20

SUBNET BITS AND

VARIATION: 128 64 32 16 8 4 2 1 (+B[8]left bits) (+A[16]LEFT)

SUBNET MASK: 128 192 224 240 248 252 254 255

POWER OF 2 VARIATION OR INCREMENT OR BLOCKSIZE TABLE

8 16 32 64 128 21 TO 27 refer upper chart 1 6 16 28 256 2 4 29 512 3 2 32 32 210 1024 4 0 211 2048 4 8 48 212 4096 5 6 213 8192 6 4 64 64 64 214 16384 7 2 215 32768 8 0 80 216 65536 8 8 217 131072 9 6 96 96 218 262144 10 4 219 524288 11 2 112 220 1048576 12 0 221 2097152 12 8 128 128 128 128 222 4194304 13 6 223 8388608 14 4 144 224 16777216 15 2 16 0 160 160 16 8 17 6 176 18 4 19 2 192 192 192 20 0 20 8 208 21 6 22 4 224 224 23 2 24 0 240 24 8

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QUESTIONS TO ASK BEFORE SUBNETTING :

1.which class ? eg 1.0.0.0 class A

2.Default subnet mask or in decimal eg: /8 or 255.0.0.0 2.1 Host formula : (2h – 2 > req)

2.2 Network formula : ( 2n > req ) 3.Total bits have in this class

4.Network bits taken (based on req)

5.Left host bits (based on req ) if asked as host then, 4.1 Network bits left

5.1 Host bits taken

6.Total.No.of.Networks we get , 7.Total no.of.Host per network

8.Assigned binary form (DECIMAL-HEXDECIMAL etc.) 9.variation or increment or block size

10.variation refer’s in which octet (Step 9 – 10 is a heart of subnetting so becareful) 11.To find last network (256-variation) if it is 2nd method then ( ) 12. Last ans: ( ) Broadcast address: ( )

Benefits of subnetting

Reduced network traffic

One network will not access the data of other network without the use of router. Thus we can reduce the amount of data remain in one network. Less data less overhead, collision, or broadcast storm.

Optimized network performance

This is a result of reduced network traffic.

Simplified management

It's easier to identify and isolate network problems in a group of Smaller connected networks than within one gigantic network. Facilitated spanning of large geographical distances Because WAN links are significantly slower and more expensive than LAN links, a single large network that spans long distances can create problems in every area earlier listed. Connecting multiple smaller networks makes the system more efficient.

Powers of 2

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(FOR 2 POWER TABLE REFER easy subnetting chart)

Subnet mask

A subnet mask is a 32-bit value that allows the receiver of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address. Every IP address is composed of a network component and a host component. The subnet mask has a single purpose: to identify which part of an IP address is the network component and which part is the host component. Subnet mask value 0 represent host ID while subnet mask value 1 to 255 represents Network ID in ip address.

Classless Inter-Domain Routing (CIDR)

This slash notation is sometimes called CIDR (Classless Inter-Domain Routing) notation. It’s basically the method that ISPs (Internet service providers) use to allocate a number of Addresses to a company, a home—a customer. The slash notation is simply the number of 1s in a row in the subnet mask. The real reason to use CIDR notation is simply that it is easier to say and especially to type.

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Address Class and Default Mask

Subnetting happens when we extend the subnet mask past the default boundary for the address we are working with. So it's obvious that we first need to be sure of what the default mask is supposed to be for any given address. When faced with a subnetting question, the first thing to do is decide what class the address belongs to. And later decide what the default subnet mask is. One of the rules that Cisco devices follow is that a subnet mask must be a contiguous string of 1s followed by a contiguous string of 0s. There are no exceptions to this rule: A valid mask is always a string of 1s, followed by 0s to fill up the rest of the 32 bits. (There is no such rule in the real world, but we will stick to the Cisco rules here—it's a Cisco exam, after all.) Therefore, the only possible valid values in any given octet of a subnet mask are 0, 128, 192, 224, 240, 248, 252, 254, and 255. Any other value is invalid.

Block Size

The process of subnetting creates several smaller classless subnets out of one larger classful . The spacing between these subnets, or how many IP addresses apart they are, is called the Block Size.

Network ID and Broadcast ID

The first address in a network number is called the network address, or wire number. This address is used to uniquely identify one segment or broadcast domain from all the other segments in the network.

The Broadcast ID

The last address in the network number is called the directed broadcast address and is used to represent all hosts on this network segment. it is the common address of all hosts on that Network ID. This should not be confused with a full IP broadcast to the address of

255.255.255.255, which hits every IP host that can hear it; the Broadcast ID hits only hosts on a common subnet. A directed broadcast is similar to a local broadcast.

The main difference is that routers will not propagate local broadcasts between segments, but they will, by default, propagate directed broadcasts.

Host Addresses

Any address between the network address and the directed broadcast address is called a host address for the segment. You assign these middle addresses to host devices on the segment, such as PCs, servers, routers, and switches.

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Method of Subnetting

There is several method of subnetting. Different author different approach to calculate the subnets. You should choose the method you can understand and perform subnetting easily. Whatever approach you choose need conversion of decimal to binary. Cram up this chart

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

To convert a decimal number into binary, you must turn on the bits (make them a 1) that would add up to that number, as follows:

187 = 10111011 = 128+32+16+8+2+1 224 = 11100000 = 128+64+32

To convert a binary number into decimal, you must add the bits that have been turned on (the 1s), as follows:

10101010 = 128+32+8+2 = 170 11110000 = 128+64+32+16 = 240

The IP address 138.101.114.250 is represented in binary as 10001010.01100101.01110010.11111010

The subnet mask of 255.255.255.224 is represented in binary as 11111111.11111111.11111111.11100000

Practical approach of subnetting

When faced with a subnetting question, the first thing to do is decide what class the address belongs to. for examples:

192.168.1.1

The first octet is between 192 and 223 so it is a Class C address

Default mask for Class C: is 255.255.255.0

In exam default subnet mask is not subnetted. Now write down the given ip address as shown here. Write down the default side of IP as it is and reset of part where actual subnetting will perform in binary

192.168. 1 .00000001 255.255.255.00000000 (default maks)

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Step 1:- calculate the CIDR value

CIDR are the on bit in subnet mask. As you can see in our example we have on bit only in default side.

255.255.255.00000000

So our CIDR value is 24 + 0 = 24

Step 2:- calculate the Subnet mask

To calculate the subnet mask use the binary to decimal chart given above. Add the decimal place value of on network bit.

<==H bit 255.255.255.00000000 N bit==>

In our example we are using on default mask so our subnet mask will be 255.255.255.0

Step 3:- calculate the Total Host(now a days this formula not usable) To calculate the total host count the H bit and use this formula

Total host = 2H

<==H bit 255.255.255.00000000 Total host = 28_{= 256}

Step 4:- calculate the Valid Host(this is valid formula)

Subtract 2 from Total host Every network or subnet has two reserved addresses that cannot be assigned to a host. These addresses are called the Network ID and the Broadcast ID, respectively. They are the first and last IPs in any network or subnet. We lose those two IP addresses from the group of values that could be assigned to hosts.

Total host - 2 256 -2 = 254

Step 5:- calculate the Network

To calculate the Network count the N bit and use this formula

Network = 20

255.255.255.00000000 N bit==>

Network = 20 = 1

Step 6:- Find out the block Size

Finding block size is very easy just subtract the subnet mask from 256

256 – Subnet mask (only the last octal, don’t include the default subnet mask)

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Step 7:- Write down the subnet chart Network 1

CIDR Value /24 IP Sunetmask

Net ID 192.168.1.0 255.255.255.0

First Valid Host 192.168.1.1 255.255.255.0 Last Valid Host 192.168.1.254 255.255.255.0 Broadcast ID 192.168.1.255 255.255.255.0

Subnetting of CIDR /25

Now do the subnetting of CIDR /25 using same method

Step 1:- calculate the CIDR value CIDR = sum of all on bit in subnet mask

255.255.255.10000000

So our CIDR value is 24 + 1 = 25

Step 2:- calculate the Subnet mask Add the decimal place value of on network bit.

<==H bit 255.255.255.10000000 N bit==>

In our example we have one on bit and as you can see in decimal chart the place value of 1000000 is 128 so our subnet mask will be 255.255.255.128

Step 3:- calculate the Total Host

Total host = 2H

<==H bit 255.255.255.10000000 Total host = 27 = 128

Step 4:- calculate the Valid Host Subtract 2 from Total host

Total host - 2 128-2 = 126

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Step 5:- calculate the Network

To calculate the Network count the N bit and use this formula

Network = 21

255.255.255.10000000 N bit==>

Network = 21 = 2

Step 6:- Find out the block Size

256 – Subnet mask (only the last octal, don’t include the default subnet mask)

256 - 128 = 128

With help of block size you can easy find out the network ID and broadcast ID of all possible networks as we have 8 bits in one octal those can give maximum of 28 = 256 decimal number

We start from 0 so it will end up on 255 (Do not get confuse because we are counting from 0 not from 1 so the last digit will be 255 not 256. It will 256 only when you count from 1 ). All

subnetting will perform between these two numbers.

Create a table of x Columns where x is the number of your network

First ip of first network will always be 0 and last ip of last network will be 255 fill its in chart Now you have network ID of first network and broadcast ID of last network.

Now add block size in the first ip of first network to get the network ID of second network and so on till we get the network id of last network

First network ID 0

Second Network ID 0 +128 = 128 Fill this in Chart.

As you can see from 128 next network is started so the last IP of first network will be 127 fill it in chart. With this method you can fill the last ip of all networks.

Now you have first ip ( network ID ) of all networks and the last ip (Broadcast ID) of all networks. At this point you can easily fill the valid ip in each network. As valid hosts are all ip address those fall between network ip and host ip.

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Step 7:- Write down the subnet chart

CIDR /25 Network 1 Network 2 Net ID 192.168.1.0 192.168.1.128 First Valid Host 192.168.1.1 192.168.1.129 Last Valid Host 192.168.1.126 192.168.1.254 Broadcast ID 192.168.1.127 192.168.1.255

Binary ANDing

Binary ANDing is the process of performing multiplication to two binary numbers. In the decimal numbering system, ANDing is addition: 2 and 3 equals 5. In decimal, there are an countless number of answers when ANDing two numbers together. However, in the binary numbering system, the AND function give up only two possible outcomes, based on four different combinations. These answers, can be displayed as a truth table:

0 and 0 = 0 1 and 0 = 0 0 and 1 = 0 1 and 1 = 1

You use ANDing most often when comparing an IP address to its subnet mask. The end result of ANDing these two numbers together is to give up the network number of that address.

Example Question

What is the network number of the IP address 192.168.100.115 if it has a subnet mask of 255.255.255.240?

Answer

Step 1 Convert both the IP address and the subnet mask to binary:

192.168.100.115 = 11000000.10101000.01100100.01110011 255.255.255.240 = 11111111.11111111.11111111.11110000

Step 2 Perform the AND operation to each pair of bits—1 bit from the address ANDed to the corresponding bit in the subnet mask. Refer to the truth table for the possible outcomes: 192.168.100.115 = 11000000.10101000.01100100.01110011

255.255.255.240 = 11111111.11111111.11111111.11110000 ANDed result = 11000000.10101000.01100100.01110000

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Step 3 Convert the answer back into decimal:

11000000.10101000.01100100.01110000 = 192.168.100.112

The IP address 192.168.100.115 belongs to the 192.168.100.112 network when a mask of 255.255.255.240 is used.

My easy method

Conversion of decimal to binary and vice versa to get network ID is too time consuming process in exam. So I found this easy method.

Step 1:- Decide from which class this IP belongs and what's its default subnet mask

As given IP have 192 in its first octal so it’s a class C IP. And default subnet mask of class C is 255.255.255.0

Step2:- Find out the block size. ( As we describe above)

256 -240 = 16

Step3:- Write down all possible network using block size till we do not get our host partition in middle of two network

0,16,32,48,64,80,96,112,128,

As our host number is 115 which fall in the network of 112 so our network ID is

192.168.1.112,

And our host's broad cast ID is 192.168.1.127 as from 128 onward next network will start. Easy as I promise

WHY (HOST-2) to find valid host

NOTE:---- We use 30 bits only out of 32 bits for subnetting bcoz of (host – 2 ) formula

we need atleast 2 host for assigning and /31 and /32 we won’t get 2 valid host TRY IT ??? for example:

1.(/31) -- octet is 4th_{and we get block size of 2}

192.168.10.0 and what is the First Host -- Last Host and BROADCAST I.D 192.168.10.2

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VLSM: variable length subnetmask

The main purpose of vlsm is save host addresses as possible. RULES: Two rules of vlsm

 Based on requirement :

 Host: take highest to lowest host requirement  Subnet: take lowest to highest requirement

 The important thing is not to repeat network (very imp heart of vlsm)  Octet and variation is important.

Example:1 (10.0.0.0/8) – 500,6,12,2000,14,30,65 host required

This question is based on host requirement as mentioned above highest to lowest host requirements: H=2h > req Host bits: N.S.M=32-h Which octet$

VAR Subnet range Broadcast i.d and sub.mask 2000 11 (/21) 3rd ,8 10.0.0.0 10.0.7.255-- (255.255.248.0) 500 9 (/23) 3rd ,2 10.0.8.0 10.0.9.255-- (255.255.252.0) 62 6 (/26) 4th ,64 10.0.10.0 10.0.10.63--(255.255.255.192) 30 5 (/27) 4th, ,32 10.0.10.64 10.0.10.95—(255.255.255.224) 14 4 (/28) 4th ,16 10.0.10.96 10.0.10.111(255.255.255.240) 12 4 (/28) 4th ,16 10.0.10.112 10.0.10.127--(255.255.255.240) 6 3 (/29) 4th ,8 10.0.10.128 10.0.10.135--(255.255.255.248) Example:2 ( 7.0.0.0/8) -- /9,/15,/24,/24,/11,/29,/29,/24,/16,/25,/24,/16,/30,/30,/30 Or

May ask as : network req 200-350-50 etc

This question is based on network or subnet requiment as mentioned above lowest to high req: Rearranging as : /9, /11, /15, /16, /16, /24, /24, /24, /25, /29, /29, /30, /30, /30, /30

Subnet mask Oct $ var Subnet range Broadcast i.d Sub.mask Hostbits= Sm-32 No.of host No.of networks /9 2nd , 128 7.0.0.0 255.128.0.0 23 8388606| 2 /11 2nd , 32 7.128.0.0 255.224.0.0 21 2097150| 8 /15 2nd , 2 7.160.0.0 255.254.0.0 17 131070 |128 /16 2nd , 1 7.162.0.0 255.255.0.0 16 65534 |256 /16 2nd ,1 7.163.0.0 255.255.0.0 16 65534 |256 /24 3rd , 1 7.164.0.0 255.255.255.0 8 254 |65536 /24 3rd , 1 7.164.1.0 255.255.255.0 8 254 |65536 /24 3rd , 1 7.164.2.0 255.255.255.0 8 254 |65536 /24 3rd , 1 7.164.3.0 255.255.255.0 8 254 |65536 /25 4th , 128 7.164.4.0 255.255.255.128 7 126 |131072 …….. 7.164.4.128 |

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Flsm : Fixed length subnet mask questions to solve:

Fixed length subnet mask will not save host addresses there are (fixed length) Questions:  10.0.0.0 req 7 networks  11.0.0.0 req 12 networks  172.16.0.0 req 2 networks  172.39.0.0 req 6 networks  7.0.0.0 req 1000 networks  39.0.0.0 req 200 networks  172.30.0.0 req 500 networks  172.172.0.0 req 30 networks  192.168.17.0 req 30 networks  192.192.192.0 req 7 networks  10.0.0.0 req 32 hosts  7.0.0.0 req 1020 hosts  172.19.0.0 req 800 hosts  172.172.0.0 req 250 hosts  192.168.10.0 req 2 hosts  192.168.168.0 req 30 hosts

 192.168.1.0 req 4 subnets each subnet needs atleast 10 host  200.100.20.0 req 9 subnets each subnet needs atleast 10 host  130.100.0.0 req 30 subnets subnet needs atleast 1000 host  192.168.100.0/30 1st

:method of finding last subnet  192.168.10.0/28 1st

:method of finding last subnet  99.99.99.99/23 2nd

: method of finding last subnet  37.38.39.40/16 2nd

: method of finding last subnet

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SUBNETTING EXERCISES II

Question – 1: 200.0.8.66 /26 – Give the answer for each of the items below:

Subnet mask in dotted decimal: Subnet Mask in Binary:

Number of Subnets: Number of hosts: Subnet address: First usable host: Last usable host: Broadcast address:

Question – 2: 192.168.1.39 /28– Give the answer for each of the items below:

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Question – 7: 10.1.5.10 /18– Give the answer for each of the items below: Subnet mask in dotted decimal:

Subnet Mask in Binary: Number of Subnets: Number of hosts: Subnet address: First usable host: Last usable host: Broadcast address:

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(24)

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Number of Subnets: Number of hosts: Subnet address: First usable host:

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Last usable host: Broadcast address:

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Number of Subnets: Number of hosts:

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Subnet address: First usable host: Last usable host: Broadcast address:

ANSWER FOR SUBNETTING EXERCISES II Question-1:200.0.8.66 /26

Subnet Mask in Binary: 11111111. 11111111. 11111111. 11000000

Number of Subnets: 2 Number of hosts: 62 Subnet address: 200.0.8.64 First usable host: 200.0.8.65 Last usable host: 200.0.8.126 Broadcast address: 200.0.8.127 Question – 2: 192.168.1.39 /28 Subnet Mask in Binary:

11111111.11111111.11111111.11110000

Subnet Mask in Dotted Decimal: 255.255.255.240 Number of Subnets: 14

Number of hosts: 14

Subnet address: 192.168.1.32 First usable host: 192.168.1.33 Last usable host: 192.168.1.46 Broadcast address: 192.168.1.47 Question – 3: 192.168.1.54 /30 Subnet Mask in Binary:

11111111.11111111.11111111.11111100

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11111111.11111111.100000000.00000000 Subnet Mask in Dotted Decimal: 255.255.128.0 Number of Subnets: 510

Number of hosts: 32,766 Subnet address: 10.18.0.0 First usable host: 10.18.0.1 Last usable host: 10.18.127.254 Broadcast address: 10.18.127.255 Question – 5: 10.18.1.97 /17 Subnet Mask in Binary:

Number of hosts: 510 Subnet address: 172.16.8.0 First usable host: 172.16.8.1 Last usable host: 172.16.9.254 Broadcast address: 172.16.9.255 Question – 7: 10.1.5.10 /18 Subnet Mask in Binary:

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11111111.11111111.11111111.11110000

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11111111.11111111.11111111.11000000

Subnet address: 172.22.112.0 First usable host: 172.22.112.1 Last usable host: 172.22.112.62 Broadcast address: 172.22.112.63 Question – 13: 10.8.122.39 /27

Subnet Mask in Binary: 11111111.1111111.1111111.11100000 Subnet Mask in Dotted Decimal: 255.255.255.224

Number of Subnets: 524286 Number of hosts: 30

Subnet address: 10.8.122.32 First usable host: 10.8.122.33 Last usable host: 10.8.122.62 Broadcast address: 10.8.122.63 Question – 14: 169.254.55.77 /25

Subnet Mask in Binary: 1111111.11111111.11111111.10000000 Subnet Mask in Dotted Decimal: 255.255.255.128

Number of Subnets: 510 Number of hosts: 126

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11111111.11111111.11111111.11100000

11111111.11111111.11111111.11111100

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11111111.11111111.11111111.11100000

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11111111.11111111.11111111.11111000

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11111111.11111111.11111111.11111000

Number of hosts: 4094 Subnet address: 129.7.160.0 First usable host: 129.7.160.1 Last usable host: 129.7.175.254

SUBNETTING QUESTIONS III:

Question: What is the first valid host on the subnetwork that the node 192.168.171.13/26 belongs to?

Answer: 192.168.171.1

Question: Which subnet does host 172.24.49.31/25 belong to?

Answer: 172.24.49.0

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Answer: 172.29.89.255

Question: What valid host range is the IP address 172.21.20.84/26 a part of?

Answer: 172.21.20.65 through to 172.21.20.126

Question: Which subnet does host 172.19.0.149 255.255.254.0 belong to?

Answer: 172.19.0.0

Question: How many subnets and hosts per subnet can you get from the network 192.168.111.0 255.255.255.248?

Answer: 32 subnets and 6 hosts

Question: How many subnets and hosts per subnet can you get from the network 172.22.0.0/26?

Answer: 172.16.208.1

Answer: 10.250.144.0

Question: What is the last valid host on the subnetwork 172.26.255.224 255.255.255.240?

Answer: 172.26.255.238

Answer: 172.19.128.1

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Question: What is the broadcast address of the network 172.23.168.0/23? Answer: 172.23.169.255

Answer: 172.26.178.158

Question: What is the broadcast address of the network 172.26.208.0/20?

Answer: 172.26.223.255

Question: What is the last valid host on the subnetwork 172.24.48.0/20?

Answer: 172.24.63.254

Answer: 192.168.119.128

Question: What is the first valid host on the subnetwork that the node 172.18.201.61 255.255.254.0 belongs to?

Answer: 172.18.200.1

Question: What is the broadcast address of the network 172.20.54.0 255.255.254.0?

Answer: 172.20.55.255

Answer: 192.168.10.160

Answer: 192.168.128.1

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192.168.56.20 255.255.255.192 belongs to? Answer: 192.168.56.1

Answer: 192.168.178.145

Question: You are designing a subnet mask for the 172.24.0.0 network. You want 70 subnets with up to 400 hosts on each subnet. What subnet mask should you use?

Answer: 255.255.254.0

Answer: 172.24.160.1

Answer: 192.168.95.32

Answer: 192.168.141.64

Answer: 255.255.255.192

Answer: 172.19.77.128

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Answer: 10.165.31.255

Answer: 172.19.19.254

Answer: 172.29.162.0

Answer: 172.17.238.1

Answer: 172.31.164.0

Answer: 172.27.11.254

Answer: 10.241.207.254

Answer: 172.23.135.255

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Answer: 172.18.118.1

Answer: 172.18.5.255

Answer: 172.16.221.255

Answer: 10.63.79.254

Answer: 172.16.93.255

Answer: 192.168.234.95

Answer: 172.26.9.254

Answer: 172.23.42.1

Answer: 192.168.181.31

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Answer: 10.180.224.0

Answer: 172.17.75.0

Answer: 172.29.252.64

Answer: 172.19.146.129

Answer: 172.27.55.193

Answer: 255.255.248.0

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Answer: 172.25.184.1

Answer: 172.29.26.0

Answer: 172.24.16.0

Question: What valid host range is the IP address 172.21.111.213 255.255.255.224 a part of? Answer: 172.21.111.193 through to 172.21.111.222

Answer: 192.168.141.78

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Answer: 172.22.1.255

Answer: 10.119.0.0

Answer: 172.19.117.254

Answer: 192.168.199.255

Answer: 192.168.104.169

Answer: 172.26.180.255

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Answer: 192.168.70.103

SUBNETTING QUESTIONS IV:

1 – Assume that 4 bits have been borrowed. Identify the subnet addresses (choose 3) a 192.168.14.8 b 192.168.14.16 c 192.168.14.24 d 192.168.14.32 e 192.168.14.148 f 192.168.14.208

2 – Assuming a subnet mask of 255.255.224.0, which of the following would be a valid host address? (choose 3) a 124.78.103.0 b 125.67.32.0 c 125.78.160.0 d 126.78.48.0 e 176.55.96.0 f 186.211.100.0

3 – Which of the following are private IP addresses? ( choose 3) a 172.168.33.1 b 10.35.66.70 c 192.168.99.5 d 172.18.88.90 e 192.169.77.89 f 127.33.55.16

4 – The router reads each bit to determine the class of an address. Which of the following binary numbers would the router identify as a public class A address? (choose 3)

a 00001010.10101100.11001100.00000111 b 00011111.11110011.11111111.00111011 c 01011101.11100001.11001100.11011011 d 10000000.11111000.11000111.11110011 e 00010111.11011011.11000001.11001100

5 – What is the maximum number of subnets that can be assigned to networks when using the address 172.16.0.0 with a subnet mask of 255.255.240. ?

a 16 b 32 c 30

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d 14

e this is an invalid subnet mask for the Network

6 – Which network mask should you place on a class C address to accommodate a user requirement of two sub networks with a maximum of 35 hosts on each network?

A 255.255.255.192 B 255.255.255.224 C 255.255.255.240 D 255.255.255.248

7 – How many valid host IP addresses are available on the following network/subnetwork? 198.197.196.16/30 a 2 b 30 c 254 d 16,382 e 65,534

8- Given an IP address of 172.16.2.160 and a subnet mask of 255.255.255.192, to which subnet does the host belong?

A 172.16.2.32 B 172.16.2.64 C 172.16.2.96 D 172.16.2.128 E 172.16.2.192

9 – Given the following IP address from the class B address range: 172.35.21.12

Your network plan requires no more than 126 hosts on a subnet that includes this address. When you configure the IP address in Cisco IOS software, which value should you use as the subnet mask? A 255.255.0.0

B 255.255.128.0 C 255.255.255.128 D 255.255.255.252

10 - You are given an ip adress 132.15.136.2/18 what subnet is the host ip on ? A 132.15.136.0

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B 132.15.128.0 C 132.15.192.0 D 132.15.64.0

11- In order to have 5 subnets and 17 hosts on each subnet, how many bits of subnetting will you use on the class B address 162.13.0.0/16 ?

A 255.255.128.0 B 255.255.224.0 C 255.255.240.0 D 255.255.248.0 SUBNETTING QUESTIONS:V

EXERCISE

Do all of the following. Please DO NOT USE calculators. All fit into 8 bits.

1. Given the IP address below, determine the network portion and host portion. a. 182.44.55.77/24 b. 10.11.12.15/8 c. 192.168.24.222/26 d. 163.22.115.32/22 e. 136.123.79.84/30 f. 111.222.15.29/20 g. 126.15.46.77/18

2. Given the IP address below, determine the class of the network and the default subnet mask. a. 182.44.55.77 b. 10.11.12.15 c. 192.168.24.222 d. 163.22.115.32 e. 136.123.79.84 f. 111.222.15.29 g. 126.15.46.77

3. Given the IP address below, determine the network address, the broadcast address and the range of usable addresses. a. 182.44.55.77/24

b. 10.11.12.15/8 c. 192.168.24.222/26 d. 163.22.115.32/22 e. 136.123.79.84/30

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f. 111.222.15.29/20 g. 126.15.46.77/18

Do all of the following. Please DO NOT USE calculators. All fit into 8 bits. 1. Convert the following decimal numbers to binary.

a. 25 b. 75 c. 48 d. 122 e. 222 f. 127 g. 63 h. 31 i. 15

2. Convert the following hexadecimal numbers to binary. a. 2 b. A4 c. 55 d. E3 e. F7 f. 8B g. 6C h. D9 i. CD

3. Convert the following binary numbers to decimal. a. 11001010 b. 1110000 c. 00011001 d. 11011101 e. 11110001 f. 11110000 g. 10101010 h. 10101

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i. 11011

4. Convert the following binary numbers to hexadecimal. a. 11001010 b. 1110000 c. 00011001 d. 11011101 e. 11110001 f. 11110000 g. 10101010 h. 10101 i. 11011

5. Convert these IP addresses into binary. a. 10.11.12.15 b. 192.168.24.222 c. 163.22.115.32 d. 236.123.79.84 e. 111.222.15.29 f. 126.15.46.77

SUBNETTING QUESTIONS:VI—VII -- REAL QUESTIONS:

Answer: 255.255.240.0

Question 1

Which IP address can be assigned to an Internet interface? A. 10.180.48.224 B. 9.255.255.10 C. 192.168.20.223 D. 172.16.200.18 Answer: B Question 2

What will happen if a private IP address is assigned to a public interface connected to an ISP?

A. Addresses in a private range will be not routed on the Internet backbone. B. Only the ISP router will have the capability to access the public network.