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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.

Understanding IP Addressing [Part-1]
Content covered: 

IP v4 Addressing System

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Understanding IP Addressing [Part-2]
Content covered: 

Classful Network Architecture

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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
Classful IPv4 addressing:

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 .

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IP addressing

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

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.

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Understanding the basics of Subnetting and Subnet Mask
Content covered: 

Subnetting and Subnet Mask

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

Finding subnet mask

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

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

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Subnet masking

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

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.
 

5Comments
CIDR Representation

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 ?
 

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.

12Comments

  • 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
Parikshit parikshitshe 27 Dec 2016 11:23 am

where are the notes and quizzes?

Meet Shah smartmeet 27 Dec 2016 11:36 am

Where is  Quiz?

Pritam Prasun pritam 27 Dec 2016 01:25 pm

Adding soon

yusuf quadri yusufquadri 27 Dec 2016 02:59 pm

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

Sumit Verma sumitverma 27 Dec 2016 05:04 pm

@yusufquadri Thank you for pointing this out.

saurabh sharma saurabh11 27 Dec 2016 10:54 pm

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

 

Arjun arjunsinghra 28 Dec 2016 11:43 am

Great thanks sir 

Sadhana Maurya sadhana82 28 Dec 2016 12:11 pm

Thanks Sir!

Sadhana Maurya sadhana8 28 Dec 2016 07:16 pm

thanks!

Sahastranshu Kr sahastranshu 28 Dec 2016 08:14 pm

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

 

Supriya Singh supriyasingh 29 Dec 2016 12:35 am

Good

Shraddha shraddhagami 29 Dec 2016 04:45 pm

How can i go for flow control??

Sadhana Maurya sadhana8 29 Dec 2016 07:44 pm

How can i get access to flow control??

Abaa wakawaka 30 Dec 2016 09:21 am

What is the procedure to access the next videos?

partha pps121 30 Dec 2016 06:49 pm

day 2 onwards links are not activated Sir.. 

Chetna Wadhwa chetnawadhwa 2 Jan 2017 05:35 am

Other links are not opening.. 

Niharika dniharika 4 Jan 2017 08:29 pm

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

Utkarsh utkarshsaxena5 6 Jan 2017 07:23 pm

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

titas sarkar titassarkar 13 Jan 2017 04:05 pm

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.

Meet Shah smartmeet 14 Jan 2017 01:19 pm

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! 

Lokesh Jajoo lokeshjajoo1995 24 Jan 2017 10:58 pm

the next day videos have no link :( ..

Abhilash abhilashchaddar 25 Jan 2017 01:04 pm

Very helpful

Harshita Sharma harshitasharma 6 Feb 2017 07:19 pm

Thank You Sir you are doing a great and noble job.  :)

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