## How to Calculate IP Broadcast address?

You've probably met once with an

Firstly immediately to ascertain,

If an IP address to see even numbers in the last octet, we can immediately eliminate, or network, subnet or IP address of the host.

For example: /28 means that we borrowed 4 bits and 4 bits we have left for the host

The weighted value of the last bit of borrowed 16

Also power the number two on the Host bits get a number of 16 (2^4=16)

If we borrowed 3 bits, the last bit has a weight value 32, and the rest of our 5 host bits is also 2^5=32)

Figure 1.

**IP address**and wondered if you could use the same IP address in a network that is subnetting, and to not take an address that is**Broadcast IP address**or one that is a network or subnetwork, or you have to get an exam question "**Which IP address is broadcast**" between 5-6 IP.Firstly immediately to ascertain,

**IP broadcast address****in the last octet**is always "**odd number**"If an IP address to see even numbers in the last octet, we can immediately eliminate, or network, subnet or IP address of the host.

**LAST borrowed Bit**will in fact determines the range between networks. So it is still counted me at the last bit borrowed or at rest Hosts bits.For example: /28 means that we borrowed 4 bits and 4 bits we have left for the host

The weighted value of the last bit of borrowed 16

Also power the number two on the Host bits get a number of 16 (2^4=16)

If we borrowed 3 bits, the last bit has a weight value 32, and the rest of our 5 host bits is also 2^5=32)

Figure 1.

Here's an example of how to quickly find out if the respective IP Broadcast or not broadcast.

I'll take the IP address 34.222.22.33 /28. This IP address belongs to 'A' class. So the default network mask is 255.0.0.0

But we see that the network address from the the Subnet /28, which means that the network mask is not the default already is 255.255.255.240 and from this we see that we are from the '

34.222.22.33 /28 (the question is whether this IP broadcast?)

Now to say that the network 34.0.0.0 /28, from which subnet this network has a total of 1,048,575 Subnets.

I'll take the IP address 34.222.22.33 /28. This IP address belongs to 'A' class. So the default network mask is 255.0.0.0

But we see that the network address from the the Subnet /28, which means that the network mask is not the default already is 255.255.255.240 and from this we see that we are from the '

**Host Octet**' borrowed 20 bits for the host and we have left 4 bits.34.222.22.33 /28 (the question is whether this IP broadcast?)

Now to say that the network 34.0.0.0 /28, from which subnet this network has a total of 1,048,575 Subnets.

Ahhhh! how much is IP? How quickly calculate whether the respective IP broadcast.

/28 = mask = 255.255.255.240 (240 means that the last octet 4 bits borrowed and 4 bits for Hosts) and consequently we borrowed the entire second and third octet.

If the last figure in the mask 240 remains to 4 Hosts bits (2^4=16) [8+8+8+(4+4)]

Watch only the last octet (byte). The number 16 means the range of addresses to hostnames that range between subnets. (Be sure you have to distinguish between '

So

/28 = mask = 255.255.255.240 (240 means that the last octet 4 bits borrowed and 4 bits for Hosts) and consequently we borrowed the entire second and third octet.

If the last figure in the mask 240 remains to 4 Hosts bits (2^4=16) [8+8+8+(4+4)]

Watch only the last octet (byte). The number 16 means the range of addresses to hostnames that range between subnets. (Be sure you have to distinguish between '

**Total hosts**' and '**Usable hosts**' => (2^4)-2So

**last octet (byte)**will have ranges of Subnets with the increase of the number 16- .0

- .16

- .32

- .48

- .64

- .80

- .96

- ...

Immediately we can see that Subnet 32 and our number 33 is the first number in that Subnet, so we know that

-------------------------------

**it is not broadcast address**.-------------------------------

Let's see with an example IP addresses

34.222.22.47 /28 (the question is whether this IP broadcast?)

Immediately we see that the sequence number in the fourth octet have the subnet in its fourth octet has the number 48 and this means that the previous Subnet has broadcast the 47th

Simply a Subnet of 48-1=47

Network range (meaning that between zero and the first network range is 16 hosts, and the first and second ... etc ..)

34.222.22.47 /28 (the question is whether this IP broadcast?)

Immediately we see that the sequence number in the fourth octet have the subnet in its fourth octet has the number 48 and this means that the previous Subnet has broadcast the 47th

Simply a Subnet of 48-1=47

**Here's the proof**Network range (meaning that between zero and the first network range is 16 hosts, and the first and second ... etc ..)

0-network (or subnet)=34.0.0.0-(34.0.0.1-34.0.0.14)-34.0.0.15 (0=network, 15=broadcast)

1-network (or subnet)=34.0.0.16-(34.0.0.17-34.0.0.30)-34.0.0.31 (16=network, 31=broadcast)

2 network (or subnet)=34.0.0.32-(34.0.0.33-34.0.0.46)-34.0.0.47 (=32 networks, 47=broadcast)

-------------------------------

909664-network (or subnet)=34.222.22.0-(34.222.22.1-34.222.22.14)-34.222.22.15 (0=network, 15=broadcast)

909665-network (or subnet)=34.222.22.16-(34.222.22.17-34.222.22.30)-34.222.22.31 (=16 networks, 31=broadcast)

909666-network (or subnet)=34.222.22.32-(34.222.22.33-34.222.22.46)-34.222.22.47 (=32 networks, 47=broadcast)

etc ...

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### Another example: the IP address 34.222.22.95 /28

You take the first number of the network. (eg the first=34.0.0.0, look at network range) and multiply with our 1.2, 5 or 10 or 20, depends on how tall the last number in the IP address, the higher the number the multiply that with larger numbers. )

If when we are multiplying and adding the result of crossing the 255 means that we have chosen too large a number to multiply.

In this mode, first multiply the number called. Multiplier (this is the number of network range, in our case 16) with some arbitrary number and then he add a multiplier minus the number 1

If when we are multiplying and adding the result of crossing the 255 means that we have chosen too large a number to multiply.

In this mode, first multiply the number called. Multiplier (this is the number of network range, in our case 16) with some arbitrary number and then he add a multiplier minus the number 1

16*5=80 This is the subnet. 80 + 15=95 so you have broadcast (95) (number 15 is broadcast in zero network)

16*10=160 This is the subnet. 160 + 15=175, so you have broadcast (175)

16*15=240 This is the subnet. 240 + 15=255, so you have broadcast (255)

And could we like

16*5=80 This is the subnet. 80-1=79, so you have broadcast

16*10=160 This is the subnet. 160-1=159 and have broadcast

16*15=240 This is the subnet. 240-1=139 and have broadcast

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Another example, the IP address 75.32.75.35 /28

/28 .... last octet ...... 4 bits borrowed (2^4=16), the range between network 16

/28 .... last octet ...... 4 bits borrowed (2^4=16), the range between network 16

- a number 15 is broadcast because we zero the range 0-15 and the first network in the range 16-32

- 16*1=16 ......... 16+15=31 ... 31=broadcast

- 16*3=48 ......... 48+15=63 ... 63=broadcast

- 16*5=80 ......... 80+15=95 ... 95=broadcast

- 16*10=160 ......... 160+15=175 ... 175=broadcast

Another way:

The last figure in the fourth octet add the number 1 and the result divide with 16 (why the number 16, because it is the range between the network), if you get a whole number have broadcast.

Network mask /28, it means the rest is our 4 Host bits. (2^4=16)

75.32.75.15 /28

15+1=16 ......... 16:16=1 ......... this is broadcast

205.16.35.11 /28

11+1=12 ......... 12:16=0.75 ......... not broadcast

199254129111/28

111+1=112 ......... 112:16=7 ......... this is broadcast

129.130.17.143 /28

143+1=144 ......... 144:16=9 ......... this is brodcast

135.22.55.93 /28

93+1=94 ......... 94:16=5.875 ......... not broadcast

-------------------------------

205.16.35.11 /28

/28 ..... last octet ...... 4 bits borrowed (2^4=16), the range=16, total 16 hosts

Since our range of networks 16 and zero network has a range of 205.16.35.0 - 205.16.35.15 into which the address 205.16.35.11, meaning NOT broadcast

-------------------------------

199.254.129.111/28

/28 ..... last octet ..... 4 bits borrowed (2^4=16), the range=16

/28 ..... last octet ...... 4 bits borrowed (2^4=16), the range=16, total 16 hosts

Since our range of networks 16 and zero network has a range of 205.16.35.0 - 205.16.35.15 into which the address 205.16.35.11, meaning NOT broadcast

-------------------------------

199.254.129.111/28

/28 ..... last octet ..... 4 bits borrowed (2^4=16), the range=16

- 16*5=80 ..... 80+15=95 ..... 95=Broadcast (95 smaller number than ours in the fourth octet)

- 16*6=96 ..... 96+15=111 ..... 111=Broadcast (111 is our number in the fourth octet)

- 111+1=112 ..... 112 16=7 ..... this is broadcast, (dividing we get integer and that means broadcast)

-------------------------------

129.130.17.143 /28 ..... /28 ..... last octet 4 bits borrowed (2^4=16), the range=16, total 16 hosts

129.130.17.143 /28 ..... /28 ..... last octet 4 bits borrowed (2^4=16), the range=16, total 16 hosts

- 16*10=160 ..... 160+15=175 ..... 175=broadcast

- 16*8=128 ..... 128+15=143 ..... 143=broadcast

- 143+1=144 ..... 144 16=9 ..... this is broadcast, (dividing we get integer and that means broadcast)

-------------------------------

135.22.55.93 /28 ..... /28 ..... last octet ..... 4 bits borrowed (2^4=16), the range =16, total 16 hosts

135.22.55.93 /28 ..... /28 ..... last octet ..... 4 bits borrowed (2^4=16), the range =16, total 16 hosts

- 16*5=80 ..... 80+15=95=95 ..... broadcast, our number in the fourth octet is 93, which means that it is not broadcast

or following

- 93+1=94 ..... 94 16=5.875 ..... not broadcast (dividing we get a decimal number, and it means that it is not broadcast)

********************************

### The third method of calculating Broadcast IP addresses:

As soon as we see Subnet-mask, we can immediately conclude that our network which we broadcast.

Take, for example 255.255.255.248.

We see that there is 255.255.255.(128+64+32+16+8=248) +3 hosts bits

So we have 5

2^5=32, which means that we have a network of a total of 32, left us 3 Host bits and therefore follows 2^3=8 and that means we have a total of 8 host (and we have one network and one for broadcast, so only 6 hosts is usable).

So our range between network is the

Take, for example 255.255.255.248.

We see that there is 255.255.255.(128+64+32+16+8=248) +3 hosts bits

So we have 5

**borrowed bits**for the network and we are left with 3 bits for hosts.2^5=32, which means that we have a network of a total of 32, left us 3 Host bits and therefore follows 2^3=8 and that means we have a total of 8 host (and we have one network and one for broadcast, so only 6 hosts is usable).

So our range between network is the

x.x.x.0 - x.x.x.7

x.x.x.8 - x.x.x.15

x.x.x.16 - x.x.x.23

x.x.x.24 - x.x.x.31

x.x.x.32 - x.x.x.39

x.x.x.40 - x.x.x.47

x.x.x.48 - x.x.x.55

etc ...

You notice that the range of the network increases by 8 (that they were our Hosts bits 2^3=8)

Each number in the fourth octet of IP address is in a certain range

Or exactly happens to the "

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Each number in the fourth octet of IP address is in a certain range

Or exactly happens to the "

**number of network**" or "**Broadcast**" This is not a problem if the little network, but when you have 256 or 2048**IT network**, how do you quickly get to the**IP broadcast address**?-------------------------------

Take for example the IP address 72.35.75.15 /24

This is the IP address and the default

So we left the 8 host bits 2^8=256

This means that we will have a network range of 256 IP addresses and therefore we understand that each broadcast IP addresses in the last octet have the number 255 and the third octet of us is changing, because we just borrowed bits from the third octet

This number 256 do not literally understand

The transition from one network to another address after the 255, not 256, already the integer 1, 2, 3, etc. (do not forget that the octet has 8 binary digits and eight units of 255 or 11111111=255)

This is the IP address and the default

**Class 'A'**mask is 255.0.0.0 and the Subnet mask is 255.255.255.0So we left the 8 host bits 2^8=256

This means that we will have a network range of 256 IP addresses and therefore we understand that each broadcast IP addresses in the last octet have the number 255 and the third octet of us is changing, because we just borrowed bits from the third octet

This number 256 do not literally understand

The transition from one network to another address after the 255, not 256, already the integer 1, 2, 3, etc. (do not forget that the octet has 8 binary digits and eight units of 255 or 11111111=255)

0 …… 75.0.0.0 …… 75.0.0.1 - 75.0.0.254 …… 75.0.0.255

1 …… 75.0.1.0 …… 75.0.1.1 - 75.0.1.254 …… 75.0.1.255

2 …… 75.0.2.0 …… 75.0.2.1 - 75.0.2.254 …… 75.0.2.255

3 …… 75.0.3.0 …… 75.0.3.1 - 75.0.3.254 …… 75.0.3.255

4 …… 75.0.4.0 …… 75.0.4.1 - 75.0.4.254 …… 75.0.4.255

5 …… 75.0.5.0 …… 75.0.5.1 - 75.0.5.254 …… 75.0.5.255

6 …… 75.0.6.0 …… 75.0.6.1 - 75.0.6.254 …… 75.0.6.255

7 …… 75.0.7.0 …… 75.0.7.1 - 75.0.7.254 …… 75.0.7.255

8 …… 75.0.8.0 …… 75.0.8.1 - 75.0.8.254 …… 75.0.8.255

-------------------------------

But do not you just watch the last octet

Some IP addresses should be seen next to last octet

One of these IP address is 135.22.55.93 /20 (/20 or 255.255.240.0)

This is a

Subnetting we borrowed 4 bits of the third octet and the remaining 4 bits plus 8 bits in the fourth octet remains for the host

Given that we would borrow nothing from the fourth octet, we will be based on the variability of the third octet (byte)

This means that the remaining 4 bits in the third octet defines the range between Subnets, 2^4=16

I will try and explain otherwise, Last borrowed Bites in fact determines the range between networks.

To get you know (if you do not know what I'm saying), you should know '

So our fourth bit from the third octet, which we borrowed, he has a weight value 16.

Here is our last octet constant 255 and second last (third) octet is changeable

Some IP addresses should be seen next to last octet

One of these IP address is 135.22.55.93 /20 (/20 or 255.255.240.0)

This is a

**Class 'B**' IP address and a default mask 255.255.0.0Subnetting we borrowed 4 bits of the third octet and the remaining 4 bits plus 8 bits in the fourth octet remains for the host

Given that we would borrow nothing from the fourth octet, we will be based on the variability of the third octet (byte)

This means that the remaining 4 bits in the third octet defines the range between Subnets, 2^4=16

I will try and explain otherwise, Last borrowed Bites in fact determines the range between networks.

To get you know (if you do not know what I'm saying), you should know '

**weighted value**' individual positional Bits in one octet (see Figure 1).So our fourth bit from the third octet, which we borrowed, he has a weight value 16.

Here is our last octet constant 255 and second last (third) octet is changeable

0 …… 135.22.0.0 …… 135.22.0.1 - 135.22.15.254 …… 135.22.15.255

1 …… 135.22.16.0 …… 135.22.16.1 - 135.22.31.254 …… 135.22.31.255

2 …… 135.22.32.0 …… 135.22.32.1 - 135.22.47.254 …… 135.22.47.255

3 …… 135.22.48.0 …… 135.22.48.1 - 135.22.63.254 …… 135.22.63.255 => here belongs to our IP addresses

4 …… 135.22.64.0 …… 135.22.64.1 - 135.22.79.254 …… 135.22.79.255

5 …… 135.22.80.0 …… 135.22.80.1 - 135.22.95.254 …… 135.22.95.255

6 …… 135.22.96.0 …… 135.22.96.1 - 135.22.111.254 …… 135.22.111.255

7 …… 135.22.112.0 …… 135.22.112.1 - 135.22.127.254 …… 135.22.127.255

8 …… 135.22.128.0 …… 135.22.128.1 - 135.22.143.254 …… 135.22.143.255

9 …… 135.22.144.0 …… 135.22.144.1 - 135.22.159.254 …… 135.22.159.255

10 …… 135.22.160.0 …… 135.22.160.1 - 135.22.175.254 …… 135.22.175.255

11 …… 135.22.176.0 …… 135.22.176.1 - 135.22.191.254 …… 135.22.191.255

12 …… 135.22.192.0 …… 135.22.192.1 - 135.22.207.254 …… 135.22.207.255

13 …… 135.22.208.0 …… 135.22.208.1 - 135.22.223.254 …… 135.22.223.255

14 …… 135.22.224.0 …… 135.22.224.1 - 135.22.239.254 …… 135.22.239.255

15 …… 135.22.240.0 …… 135.22.240.1 - 135.22.255.254 …… 135.22.255.255

I hope you have understood how you can calculate

**the broadcast IP address**