## Why do borrow 3 bits?

Now coming to the fore memory table with Figure 1.

If you borrow the first three bits means that we will provide the required number of subnets.

In accordance with the binary system and the power of base 2^3 gives number subnets and that is 2^3=8 the total number of subnets (of course we have to consider that the two useless, zero network and last subnets, and we are left with 6 usable subnets, which us is more than enough, because we need four subnets. using the remaining 5 bits we will calculate our hosts usable.

So the rest is our 5 host bits according to the binary system from the table in Figure 1. we know that 2^5=32

So, we have 32 IP addresses for our hosts, again we also take care that the null IP address and the last IP addresses are not usable and there will be 30 IP addresses to hosts (computers). See Figure 8.

Figure 8.

If you borrow the first three bits means that we will provide the required number of subnets.

In accordance with the binary system and the power of base 2^3 gives number subnets and that is 2^3=8 the total number of subnets (of course we have to consider that the two useless, zero network and last subnets, and we are left with 6 usable subnets, which us is more than enough, because we need four subnets. using the remaining 5 bits we will calculate our hosts usable.

So the rest is our 5 host bits according to the binary system from the table in Figure 1. we know that 2^5=32

So, we have 32 IP addresses for our hosts, again we also take care that the null IP address and the last IP addresses are not usable and there will be 30 IP addresses to hosts (computers). See Figure 8.

Figure 8.

You notice on the top of Figure 8 that the network mask changed after borrowing bits from the

Why 224? well, see Figure1 and Table. Borrowed three bits and summing the

Due to this, we now have a format of writing Subnetmask /27

Well,

At 24 bits that we had, we borrowed an additional 3 bits from the last octet (Byte).

So here we get to check operation/calculation AND.

**last octet and now has a decimal notation 224**.Why 224? well, see Figure1 and Table. Borrowed three bits and summing the

**weight values of the first, second and third bits**we get the sum of 224 (128+64+32=224), which means that we do not have more**Classfull (default) Network Mask**, now we have**changed Subnet mask**. Because of, this mathematics, we have**VLSM (Variable Length Subnet Mask)**.Due to this, we now have a format of writing Subnetmask /27

**Why /27?**Well,

**count all N**(network) address bits in the Mask (8+8+8+3=27), and see how many are there? So there are 27 (24+3=27).At 24 bits that we had, we borrowed an additional 3 bits from the last octet (Byte).

So here we get to check operation/calculation AND.

Figure 9.

- We got a number subnets (2^3=8) => (Base 2 power on 3 Host borrowed bits [which are converted into network bits] =8). To avoid any confusion, we use the first subnet)

- We have received much will each subnet have a maximum number host computers (2^5=32-2=30) => (Base 2 power on 5 host bits, this does not take into consideration the first and last IP address =30)

- To avoid any misunderstanding, the first and last IP address can not be used

Try to take any IP address from the range 192.168.1.0 - 192.168.1.255 and execute an

**operation/calculation AND-ing between the IP address and Network Mask**, and you will see that it always as a result of the network subnet to which it belongs have the same IP address.

Let's go to the next page tutorial.