$4.$ Great! You worked through the VLSM process in step $2.$ Let us now explore what can happen when a new subnet needs to be allocated on an existing network. The VLSM process described in the overview section starts with the largest subnets and works its way down to the smallest subnets. This ensures efficient subnet allocation and avoids gaps of unallocated addresses. If the new subnet is larger than some existing subnets, there will be a gap of unallocated addresses. On the other hand, if the new subnet is the same size or smaller than the smallest existing subnet, then there will not be a gap of unallocated addresses.

Let us take a look at an example. Imagine that an organization is assigned the block of addresses defined by $99\mathrm{.}216\mathrm{.}0\mathrm{.}0/13.$ Further imagine that two subnets are already allocated. The first subnet needs $100$ hosts and the second needs $13$ hosts, and were

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allocated with network identifiers as follows:

First Subnet: $0110001111011000000000000$

Second Subnet: $0110001111011000000000001000$

Would$$be subnet with identifier $0000000000$ is thus subdivided as follows.

Notice that in this example, subnet with identifier $0000000000$ supports $9$ bits for the host identifier, which allows for $512$ total addresses. However, in the preceding figure, only $256$ of those $512$ addresses are displayed for illustrative purposes.

Now imagine that the organization needs a new subnet the same size as the first subnet, $100$ hosts. We know from the first subnet calculation that the subnet identifier would be $12$ bits in length. So what do we do now? Applying the process we are already familiar with, we choose the next available $12$bit subnet identifier that has not been used previously to create a network identifier of:

Third Subnet: $0110001111011000000000010$

We now have a gap of unallocated addresses, illustrated in the following figure.

Notice that subnet $3$ is allocated beginning at address $256,$ and that addresses $144$ through $255$ are not allocated. Because of the properties of binary numbers, we must start a $128$bit subnet identifier on a value is divisible by $128.$ That is$,$ we cannot start the $128$bit subnet identifier at the next available address, $144.$

In such a situation, it is tempting to get rid of all existing subnet identifiers and to allocate the subnets from scratch. While doing so would result in an efficient allocation scheme, in reality this is rarely practicable, because it would mean a change in IP address for existing client computers, routers, and devices throughout the organization, which would be significantly disruptive. So the network is

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necessarily left with some unallocated addresses.

In summary, when a new subnet must be added to an existing network, and that subnet is larger than at least one existing subnet, applying the process of creating VLSMs covered in the overview section will result in an unavoidable gap of unallocated addresses.

$5.$ Imagine that an organization is assigned the address block $45\mathrm{.}0\mathrm{.}0\mathrm{.}0/8,$ and the organization has an existing subnet with $190$ hosts, and another existing subnet with $53$ hosts. Further imagine that after the network has been in used for some time, the organization must add another subnet that requires $105$ hosts.

First, use the process established in the overview section to create the network identifiers for the two existing subnets. Second, use the same process to create the network identifier for the new subnet $($illustrated in step $4).$ Last, identify the number of addresses that are left unallocated in the gap between the second and third subnet. Make sure to show your work.