Introduction to IP addressing, Class A, Class B, Class C, Casting and it's types ,Subnets, Subnet Mask, Routing Table, Variable length subnet masking, Classless Inter Domain Routing, Rules for CIDR, Subnetting in CIDR, VLSM in CIDR,Supernetting, Supernet Mask.

Content: 

Content covered: 

IP v4 Addressing System

Responses

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in a network host id are unique

umangchaturv's picture

how can we have 2 ipv6 addresses at the same time?

 

 in the lecture (video of subnetting )  you explain about  subnet masking there you told the Subnet Mask calculation that Point was wrong please make it correct.. In Subnet Mask all the network ID and subnet bits all should  1's and all host bits should be 0's. for example 255.255.255.128 is the Subnet Mask for network 200.1.1.1_ _ _ _ _ _ _ all left bits are host bits..

Sir , why do we need private IP address if we have MAC address.?

dashish's picture

@pritam

Sir, http uses port 80... and if two app like chrome and firefox are there.. u told they would use different ports.. how is this?? if they are sending or receiving a http request they must use port 80..

i am confused sir. pls repy!!


Content covered: 

Classful Network Architecture

Responses

Why you have not subtract 2 from NIDs of B and C class?

shashankshek's picture

Because he already have eliminated the fixed bits from couting the total network ID i.e:

For Class B:

fixed bits- 2(1,0)

total network bits- 16

total manipulative bits = 16-2

So, Total Network ID's = 214

Similarly for class C, total network id's = 221.

sardendubharti's picture

becuase he mentioned that the two NID adress were eliminated only for class A network, because the first and last is used for noIP/invalidIP  and loopback adress respectively 

Range of class D is 224-239.

IP Address: 143.27.16.14  Find the following:

  1. Class
  2. Network ID
  3. Maximum number of hosts
  4. Directed broadcast address
  5. Limited broadcast address

Responses

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sumitverma's picture

nice !!

habibkhan's picture

Nice illustration sir..

Thank you so much.

shraddhagami's picture

Very much helpful fr revision...

Thank you :)

 

GREAT SIR

Class Type

Number of IP
addresses

NID+HID (bits)

Range (First octet only)

Example

Class A

231

8 + 24

1-126

124.0.25.36

Class B

230

16 + 16

128 - 191

130.3.6.9

Class C

229

24 + 8

192 - 223

210.3.6.5

Class D : Total Number of IP addresses = 228  (224 - 239)   // Reserved for multicasting
Class E : Total Number of IP addresses = 228  (240 - 255)   // Reserved for future use

Number of hosts in any Class = 2(number of bits in HID)-2 .

Which one of the following IP addresses are not used for loop back addresses?

i. 127.1.1.10
ii. 127.10.15.7
iii. 127.0.0.0
iv. 127.255.255.255

(A) i,iii
(B) ii,iii
(C) iii,iv
(D) None

  • Explanation

    Though 127.0.0.0 , 127.255.255.255 starting with 127, but these have all 0's and all 1's in the remaining part. So, they will not be useful for loopback addressing.Hence C is the correct option.

Responses

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sumitverma's picture

127 does not belong to any class.

What is loopback address, means?

sumitverma's picture

A loopback address is primarily used as a means to validate that the locally connected physical network card is working properly and the TCP/IP stack installed. Typically, a data packet sent on a loopback address, never leaves the host system and is sent back to the source application. 

127._._._ is neither a class A nor B.

Pls Explain

 

sumitverma's picture

Yes, It does not belong to any class.
A loopback address is primarily used as a means to validate that the locally connected physical network card is working properly and the TCP/IP stack installed. Typically, a data packet sent on a loopback address, never leaves the host system and is sent back to the source application. 


Content covered: 

Subnetting and Subnet Mask

Responses

Thank You, Sir.

umangchaturv's picture

If in ques, we are given x bits for subnet bits in a network then we should substract 2 from 2^x for total no. of subnets possible or not ?

shraddhagami's picture

if we have 1-bit for subnet then we can divide a network in 2 parts.So, we loose 2*2=4 ip address.

if we have 2-bit for subnet then we can divide a network in 4 parts(b'cz 2^2).So, we loose 2*4=8 ip address.

if we have n-bit for subnet then we can divide a network in 2^n parts. So, we loose 2*(2^n) ip address.

And total no. of subnets possible =2^n

hope u understand.

yes we need to subtract for no of hosts for each subnet to get no of hosts

but in case of no of subnwts we have 2^x only

For the given subnet mask 255.255.255.224 , Find the number of subnets in each class.


Responses

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jagmt's picture

So we can't have only 1 bit in subnet(according to RFC950 as specified above) because #Subnets = 2^1 - 1 = 0?
Also in above question what is the correct answer it is still confusing!
I also looked at this it is still ambiguous.
http://gateoverflow.in/1265/gate2007-67-isro2016-72?show=1265#q1265
However according to wikipedia-answer should be 64 and 1022
https://en.wikipedia.org/wiki/Subnetwork#Subnet_and_host_counts

nikhilkumawat's picture

why we remove start or last ip address??

 

sardendubharti's picture

becaus efirst ip adress is used as dircet broadcast adress. such that if any host from other network wants to send packet to all the host present in any network

and second used for knowing the adress of own network 

According to explainatioin its ans should be for subnets : 2^6

Sir,What About VLSM?

For one of class B network, all odd bit positions are selected for subnetmask bits. What is the possible subnetmask?

(A) 255.255.170.170
(B) 255.255.85.85
(C) 255.255.255.240
(D) None

  • Explanation

    Given that, it is class B netwrk. Means first two octents are for NID. Have to keep all 1's in NID part.
    All odd bit positions are selected for subnetmask bits. These bits are from HID part.

    11111111.11111111.10101010.10101010 = 255.255.170.170
     1 octect  2 octect   3 octect   4 octect

Responses

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singhdl935's picture

I think 255.255.170.170 is not a practical mask (discontinuous mask).

every other host will fall in alternate network.this is only a theoretical question.

For the given subnet mask 11111111.11111111.10101010.10101010, can we say that NID+SID=17 and HID=15? Since 18th bit is 0, it should be a part of HID and hence all subsequent bits are part of HID. Is it correct?

No, according to this question, because this is discontinuous mask.

dashish's picture

@pritam , @sumitverma

Here odd bits are to be taken.. Now, i think in this soln, the numbering has been started from 0.

if i start it from 1, then the ans i got was 255.255.85.85 which is also in the option..

is there any standard rule/convention of nubering bits from 0 only??

255.255.58.0 is the subnet mask for a particular network. Which of the following pairs of IP addresses could belong to this network?

(A) 100.25.135.10 and 100.25.167.10
(B) 192.212.3.5 and 192.212.4.6
(C) 140.34.23.65 and 140.35.222.123
(D)177.90.204.44 and 177.90.136.89

  • Explanation

    If we perform bitwise ANDing between the subnet mask and ip address, network id can be obtained.
    For option (A): 100.25.135.10 and 100.25.167.10
    We have subnet mask as
    255.255.58.0   =  11111111. 11111111.   00111010. 00000000
    100.25.135.10 =  01100100.00011001. 10000111.  00001010 
                              -------------------------------------------------------
                                0110100.00011001.00000010.00000000   
                            = 100.25.2.0 
    255.255.58.0   =  11111111. 11111111.   00111010. 00000000
    100.25.167.10 =  01100100.00011001. 10100111.  00001010 
                              -------------------------------------------------------
                                0110100.00011001.  00100010.00000000   
                            = 100.25.36.0 
    Since both network address are different . Hence It is incorrect.

    For option (B): 192.212.3.5 and 192.212.4.6
    255.255.58.0   =  11111111. 11111111.  00111010. 00000000
    192.212.3.5     =  11000000.11010100. 00000011.00000101
                              -------------------------------------------------------
                                 11000000.11010100.00000010.00000000
                           = 192.121.2.0
    255.255.58.0   =  11111111. 11111111.   00111010. 00000000
    192.212.4.6     =  11000000.11010100. 00000100.00000110
                              ------------------------------------------------------
                                 11000000.11010100.00000000.00000000
                           = 192.121.0.0
    Since both network address are different . Hence It is incorrect.

    For option (C): 140.34.23.65 and 140.35.222.123
    255.255.58.0   =  11111111.  11111111.   00111010. 00000000
    140.34.23.65   =   10001100.00100010.00010111. 01000001
                                 -------------------------------------------------------
                                 10001100.00100010.00010010.00000000
                              =140.34.18.0
    255.255.58.0   =    11111111.  11111111.   00111010. 00000000
    140.35.222.123=   10001100.00100011.  11011110. 01111011
                                 -------------------------------------------------------
                                 10001100.00100011.00011010.00000000
                              =140.35.26.0
    Since both network address are different . Hence It is incorrect.

    For option (D) :177.90.204.44 and 177.90.136.89
    255.255.58.0   =    11111111.  11111111.   00111010. 00000000
    177.90.204.44 =    10110001. 01011010. 11001100. 00101100
                                   ------------------------------------------------------
                               =  10110001.01011010. 00001000.00000000
                               =  177.90.8.0
    255.255.58.0   =    11111111.  11111111.   00111010. 00000000
    177.90.136.89=    10110001. 01011010. 10001000. 01011001
                                   ------------------------------------------------------
                               =  10110001.01011010. 00001000.00000000
                               =  177.90.8.0
    Since both network address are same . Hence It is the correct answer.
     

Responses

saurabh2612's picture

we don't need to check the option A and B because opt.  A> is class A and option B> is class C, and the subnet mask given is of class B.

saurabh2612's picture

a

i didnt understood what is being done??

can some1 tell when will todays content upload

Consider the following block of ip addresses:
150.10.20.32
150.10.20.33
150.10.20.34
-----------------
-----------------
150.10.20.47
Check if it can be represented into CIDR representation or not ?
 

  • Explanation

    Step1: IP addresses should be contiguous.
    Here IP addresses are contiguous.
    Step2: Check total size of the block. It should be power of 2.
    Here total size is 16, i.e, 16 ip addresses are present in the block, which is 24.
    Step3: First ip address in the block should be evenly divisible by size of block.
    Here first ip address is 150.10.20.32 = 150.10.20.(00100000)and block size is 16 = 24 .
    We can check if the number of zeros present at the end of the ip address are greater or equal to log2(Size of block), then this address will be evenly divisible by the size of the block.
    So it is divisible here.

    Hence this block can be represented into CIDR representation.
    Here block size is 24, so 4 bits can be used for HID and remaining 28 bits can be used for NID.
    So the CIDR representation will look like, 150.10.20.32/28.

Responses

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habibkhan's picture

Nice illustration ..

akhileshgang's picture

thanks good concept

 

lovelyverma's picture

How total size of the block is 16 here? plz explain

sumitverma's picture

Starting from 32 to 47 (last octet), total 16.

lovelyverma's picture

ok thx sir

  • This quiz contains 5 questions on the topic IP Addressing
  • Lean well before you attempt the quiz
  • You can attempt the quiz unlimited number of times.

Difficulty Level:  basic
No. of Questions:  5

Awesome, you did good to complete todays course.

To continue your preparation while we publish tomorrow’s topic,

please subscribe to our youtube channel.

 

Responses

where are the notes and quizzes?

smartmeet's picture

Where is  Quiz?

pritam's picture

Adding soon

in cidr example the last ip address in the series is wrong 

sumitverma's picture

@yusufquadri Thank you for pointing this out.

saurabh11's picture

Thanks Sir,It is very helpful for me and my friends.

 

arjunsinghra's picture

Great thanks sir 

Thanks Sir!

thanks!

sahastranshu's picture

when i can get access to flow control..??

 

Good

shraddhagami's picture

How can i go for flow control??

How can i get access to flow control??

What is the procedure to access the next videos?

day 2 onwards links are not activated Sir.. 

chetnawadhwa's picture

Other links are not opening.. 

In QUIZ asking about Direct broadcast address and limited broadcast address why option A is wrong??

If the subnet mask is 255.255.254.0, then how many hosts can be configured in this network ? how to find

255.255.254.0 -> Convert it to Binary 

11111111.11111111.11111110. 00000000   -> the 0's will be the host ids . Since there are 9 0's you will have (2^9) -2 hosts in this network. -2 (1 Network ID and the last one is for Broadcast)

so 510 Hosts can be configured in this network.

smartmeet's picture

Actually this question is incorrect as 255.255.254.0 is Class-C addres so first 3 octate i.e. Network ID is fixed as 1 in subnet mask and then further we can make upto 2^8 subnets by selecting bits from HID(which are of 8 bits, here.) 

Thus, minimum subnet mask starts with 255.255.255.0 for Class-C address, if we need 2 subnets so we have to select 1-bit and SM will be 255.255.255.192.

Coreect me if I'm wrong! 

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