Step 1: Analyze the network
a. Referring to the Rough Design Notes, determine the minimum number of hosts that a subnet needs to support the new network design.
1) The largest subnet must be able to support 30 hosts.
2) To support that many hosts, the number of host bits required is 2 hosts
b. What is the minimum number of subnets required for the new network design? 2
c. Can this network be subnetted according to the requirements? bisa
For example: If four subnets are required and the largest subnet has to support 128 hosts, this is a problem, because a subnet in a class C network that has been partitioned four ways can support only 62 hosts.
d. Fill in the blanks to summarize the subnetting requirements of this new network design:
This network requires 2 subnets, each supporting 29 hosts. Therefore, 255.255.255.0/27 host ID bits are reserved for the subnet ID. With those values, this network supports 4 subnets, each subnet having 30 hosts.
Step 2: Calculate the custom subnet mask
Now that the number of subnet ID bits is known, the subnet mask can be calculated. A class C network has a default subnet mask of 24 bits, or 255.255.255.0. What will the custom subnet mask be?
The custom subnet mask for this network will be 255.255.255.224or /27
Step 3: Specify the host IP addresses
Now that the subnet mask is identified, the network addressing scheme can be created. The addressing scheme includes the subnet number, the subnet broadcast address, and the range of IP addresses assignable to hosts.
a. Complete the table showing all the possible subnets for the 192.168.1.0 network.
| subnet | Subnet Address | Host IP Address Range | Broadcast Address |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.0 |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.30 |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.90 |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.120 |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.150 |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.180 |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.210 |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.240 |
| 4 | 255.255.255.224 | 30 hosts | 192.168.1.270 |
b. for it to be completed. Hosts will be assigned IP addresses as follows: (fill in the table below )
| Device | Interface | IP Address | Connect to | IP Address |
Step 4: Consider other subnetting options
What if there were more than 30 hosts that needed to be supported on either the wired or wireless portion of the network. You could borrow fewer bits, which would create fewer subnets, but each one would support a greater number of hosts per subnet.
a. How many bits would be borrowed to create four subnets? 2 buah bit
b. How many bits would be left for hosts on each subnet? 5 bit
c. What is the maximum number of hosts each subnet could support? 30 host- 2=28
d. What would the subnet mask be in dotted decimal and slash number (/#) format? 255.255.255.224or /27
e. If you start with the same 192.168.1.0 network as before and subnet it into four subnets, what would the subnet numbers be?
Step 5: Reflection
a. Does subnetting help reduce the problem of IP address depletion? Explain your answer.
ya, sangat membantu mengurangi masalah depletions pada IP address, karena bisa membedakan network identifier dengan host identifier.
b. The Rough Design Notes diagram noted that the wireless subnet would have up to 30 PCs connecting. In pairs or in small groups, discuss whether or not that creates a situation in which IP addresses might get wasted. Does it matter, and why or why not?
sia sia karena proses subnetting yang menghasilkan beberapa subjaringan dengan jumlah host yang sama telah dilakukan, maka aka nada kemungkinan di dalam segmen –segmen jaringan tersebut memiliki alamat – alamat yang tidak digunakan.
c. There are alternate methods of subnetting using CIDR and VLSM. Would VLSM be a worthwhile option for subnetting this network? Discuss in small groups. vlsm bisa menjadi solusi karena teknik subnetting dengan VLSM dapat dilakukan secara rekrusif : network identifier yang sebelumnya telah disubnetkan, di subnetkan kembali . maka dengan kata lain pengambilan kembali sisa dari bit- bit host.
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