1.9 Configure, verify, troubleshoot IPv4 addressing and subnetting

I admit that I skipped the troubleshooting part 1.8 because I dont know personally what are those but it seems like those are theoretical ways to torubleshoot the issue. Let me catch up with troubleshooting at the end part so that I can apply all the theories and technical aspects that I should be aware of when troubleshooting.

I personally do not know where to start discussing IPv4, maybe by defining first an IP address. Sorry, if  I kinda make you idiot that doesnt know what is IP address. Alright, so IP address is the way to present a host in a network. It seems like a physical address to locate our house in a subdivision. An IP address has already expanded from IPv4 to IPv6 address. We will deal more on IPv6 addressing later but let us identify first what is IPv4 address.

 

IPv4  address is a 32 bit, 4-octet number in a format of 192.168.1.1/24

If you wonder what is /24 it refers to the CIDR notation, CIDR notation lemme google it first, hahaha yeah!, it is the classless interdomain range. What’s that for? It is because the IP addresses has different classes that may define the number of host and networks. Here are the IP address class:

Class A 0- 126.255.255.255

Class B 128-191.255.255.255

Class C 192-223.255.255.255

Class D 224-229.255.255.255

Class E 240 – 254.255.255.255

Sir, may I ask what are their CIDR notations? Though we can just ignore writing their CIDR notations because those are classful address and CIDR  is for classless, we can say that classful address has default CIDR notation. /8 for class A, /16 for class B, /24 for class C, if you wonder where is the network for 127.x.x.x? , that is used for loopback or for self-test, class D and E are reserved or for military purposes.

**************************SUBNETTING***********************************

What are subnetting? What’s the purpose of subnetting? Subnetting is a practice of dividing the network into two or more smaller networks. The thing here is we have to save the IP addresses. Since IP addresses are limited we have to make the most of IP address.

Lemme show you how to do subnetting through examples:

1.) Given: 192.168.39.70/26

Network ID:

Broadcast ID:

Subnet Mask:

1st usable:

last usable :

# of hosts

# of subnets:

 

I cannot explain how to provide the answer for the given question but then I think it will be better if I give you first the formula;

A.# of hosts : 2^( 32-NSM) -2

NSM stands for new subnet mask/CIDR

OSM stands for old subnet mask/CIDR

The reason why there is a deduction of two because those represents network and broadcast id

B.# of subnets: 2^(NSM-OSM)

C.To determine the network ID and broadcast ID, use the finger method:

5656565

Oh yes! we have to pretend like we only have 8 fingers hahahah! But then here’s how to use the finger method:

To identify the subnet mask; the numbers from 1-32 serve as the CIDR; if the CIDR is from bottom(1-8) it means that the value of the first octet should be the green number corresponding to the CIDR number; if the CIDR is from 2nd from the bottom(9-16) it means that 255 can only be found on the first octet. Same procedure with the CIDR from (17-24) and (25-32). The next octet will be the green number corresponding to the CIDR number

For example:

/17 –> SM: 255.255.128.0

/23–> SM: 255.255.2.0

/31 –> SM: 255.255.255.254

/19 —> SM: 255.255.224.0

I hope that you  understand this technique; my college way of learning subnetting is to convert the CIDR to binary and convert it to octet and it seems like complicated

The black numbers above the green number serves as the increment; it would be easy to explain the answer on the question above while discussing on how to answer the question:

1.) Given: 192.168.39.70/26

Network ID:

First, I have to identify the subnet mask and since it is /26 then it must be 255.255.255.192 so only the 4th octet is value that we need to change when identifying the network and broadcast address. Since the value on the last octet is 70 and the /26 CIDR falls on 64 as the black numbers above the green number so the increment that we need to use is 64

Note:

For identiftying the network ID; it must be a factor of the increment but the increment must be lower than the value of the Nth octet on the given ip address

For For identiftying the broadcast ID; it must be the NEXT MULTIPLE OF  INCREMENT MINUS ONE because next multiple of increment does not include the given ip address anymore

The FIRST BIT USABLE is the first bit after the network id

The LAST BIT USABLE is the last bit before the network id

So the NETWORK ID should be 192.168.39.64

x.x.x.64 because increment is 64 and it should be less than 70

Broadcast ID: 192.168.39.127

x.x.x.127 because it must be an a factor of 64 minus one

Subnet Mask:2555.255.255.192

1st usable:192.168.39.65

x.x.x.65 because it is the 1st bit after x.x.x.64

last usable :192.168.39.126

x.x.x.126 because it is the last bit before the broadcast ID

# of hosts: Remember the formula–> # of hosts: 2^(32-NSM)-2

NSM is of course 26 so  2^(32-26)-2=62

 

# of subnets:

Remember the formula–> # of subnets: 2^(NSM-OSM)

NSM for  192.168.39.70/26 is of course 26

and the default  SM or OSM for classful addressing of  192.168.39.70 is 24

so   # of subnets: 2^(26-24)= 4

 

 

Let’s try another one:

2.) Given: 172.18.0.0 300 hosts

Find:

Network ID,Broadcast ID,Subnet Mask,1st usable,last usable ,# of subnets:

This time the number of host is given but we don’t have to worry, we will still use the same formula–># of hosts: 2^(32-NSM)-2

so 300 =  2^(32-NSM)-2  since we cannot find a perfect number for NSM; we just need to approximate 300 to the value which is multiple of 2 and more than 300.We can also omit the deduction of 2 since we are just approximating. 512 is the perfect number greater than 300 and multiple of 2. So mathematically, 512=2^9=2^(32-NSM)  and then simplfying it then 9=32-NSM; so NSM equals 23.

determining the NSM means that we can now do the same process as #1

So SM: 255.255.254.0; increment is 2.

Network ID :172.18.0.0

Broadcast ID: 172.18.1.255

Note since the next network id for the increment of 2 is 172.18.2.0, the broadcast id is 172.18.1.255

1st usable: 172.18.0.1

last usable: 172.18.1.254

I will update this part if I have time. 🙂

 

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