NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting
AIM: Assigning IP addresses to a network using VLSM and FLSM
Reference Book: Chapter no 9 , CCNA Complete Guide 2nd Edition by Yap Chin Hoong
1. Familiar your-self with IP basics; types of Class address; range of each class.
2. Design a topology for 3 network segments connected in a star topology through router ,
a. 4 host (use switch)
b. 2 host and (use Switch)
c. 10 computer host + one DHCP Server (use Hub)
Consider Class C Address 192.168.10.0/24
Following Particle will demonstrate the difference between FLSM and VLSM Techniques.
3. You need to create an IP plan for given network. You are given an address space of
192.168.1.0/24, and need to subnet it. This involves finding the mask, subnets, and ranges. You
need four subnets, and that the four subnets need the following numbers of hosts:
 105 hosts
 50 hosts
 20 hosts
 10 hosts
Consider yourself as Network Administrator you need to create an IP plan for this network by
following methods
•
•
•
Unstructured FLSM
FLSM
VLSM
4. We’re given the address space 172.16.55.0/24 (a Class “C” sized piece of a Class “B” network),
There are two routers, connected by a pair of point– to-point WAN links (for fault-tolerance). In
addition to the P2P links, R1 has four directly-attached LAN subnets, as follows:
 50 hosts
 20 hosts
 10 hosts
 5 hosts
R2 also has four directly-attached LAN subnets, as follows:
 40 hosts
 25 hosts
 12 hosts
 4 hosts
5. A HQ site is connected to 20 remote sales offices in a hub and spoke topology (using point-topoint Frame Relay sub interfaces). We’re given the address space 10.100.200.0/22. We need to
subnet to support a HQ LAN subnet of 100 hosts, 20 sales office LANs of 25 hosts each, and the
20 WAN links connecting HQ to the remote sites.
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NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting
6. Given the address space 172.30.20.0/24, we need a HQ LAN subnet with 50 hosts, ten sales
office LANs with five hosts each, and leased lines (running PPP) connecting HQ to the sales
offices.
7. A Class C network192.168.100.0/24 is assigned to you.
8.
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NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting
9.
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NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting
Solution 1) Familiar yourself with the basic concept of IP addressing and Subnetting
Sol 2)
--------------------------------------------------------------------------------------------------------------------------Sol 3)
Let’s see if we can do it with standard subnet masking (where the mask is the same for all subnets).
We start out the usual way, verifying that the total number of hosts (185) will fit within our “/24”
address space (which will support up to 254 hosts). Next, we observe that the number of hosts on the
largest subnet is 105, which requires seven host bits (and allows for up to 126 hosts per subnet). Since
there are four subnets, we need two subnet bits (which allows up to four subnets).
Oh-oh … big problem! Seven plus two is nine, but we have only eight bits to work with (the first 24
are fixed by the address space). Put another way, 32 minus 9 is 23, which means a “/23” mask.
Unfortunately, this is to the left of the “/24” specified by the address space, so we can’t do it. Since
standard subnet masking won’t work, let’s try VLSM.
In VLSM, we vary the mask to suit the size of the subnet. In other words, since not all subnets require
the same number of hosts, why use the same mask? Put yet another way, if we don’t waste address
space on one subnet, we can use that address space on another. Let’s give it a try…
As with standard subnet masking, we verify that the total number of hosts (185) will fit into the
available address space (“/24”), which it will. Next, we handle the largest subnet (105 hosts), which
requires seven host bits, or a “/25” mask (32 – 7 = 25, which would actually support up to 126 hosts).
Thus, the first subnet is 192.168.1.0/25:

105 hosts => 192.168.1.0/25
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NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting
With a “/25 mask (seven host bits), the subnets increment by 128, so the next subnet begins at
192.168.1.128. The next largest subnet contains 50 hosts. Since 50 hosts require six host bits, the
mask for this subnet is a “/26” (which would support up to 62 hosts). Thus, the second subnet is
192.168.1.128/26:


105 hosts => 192.168.1.0/25
50 hosts => 192.168.1.128/26
With a “/26” mask (six host bits), the subnets increment by 64, so the next subnet begins at
192.168.1.192 (128 + 64 = 192). The next largest subnet contains 20 hosts. Since 20 hosts require five
host bits, the mask for this subnet is a “/27” (supporting up to 30 hosts). Thus, the third subnet is
192.168.1.192/27:



105 hosts => 192.168.1.0/25
50 hosts => 192.168.1.128/26
20 hosts => 192.168.1.192/27
With a “/27” mask (five host bits), the subnets increment by 32, so the next subnet begins at
192.168.1.224 (192 + 32 = 224). The next largest subnet contains 10 hosts. Since 10 hosts require four
host bits, the mask for this subnet is a “/28” (allowing up to 14 hosts). Thus, the fourth subnet is
192.168.1.224/28:




105 hosts => 192.168.1.0/25
50 hosts => 192.168.1.128/26
20 hosts => 192.168.1.192/27
10 hosts => 192.168.1.224/28
With a “/28” mask (four host bits), the subnets increment by 16, so the next subnet begins at
192.168.1.240 (224 + 16 = 240), but since we’ve taken care of all four of the required subnets, we
don’t need that address space. Using VLSM, not only did we solve the problem, but we even have
some address space left over!
--------------------------------------------------------------------------------------------------------------------------Solution 4)
Let’s see if we can solve this problem using standard subnet masking. The largest subnet must support
50 hosts (and thus needs six host bits), and there are a total of ten subnets (which requires four subnet
bits). Since six plus four is ten, and we only have eight bits to work with, we can’t do it with standard
subnet masking. Looking at it another way, since a “/22” (32 minus 10) is to the left of “/24”, it
won’t work.
Okay then, let’s try VLSM. As before, we start out with the largest subnet, and work our way down.
A subnet with 50 hosts requires a “/26” mask:

50 hosts => 172.16.55.0/26
With a “/26” mask (six host bits), the subnets increment by 64, so the next subnet begins at
172.16.55.64. The next largest subnet contains 40 hosts. Since 40 hosts require six host bits, the mask
for this subnet is also a “/26”. Thus, the second subnet is 172.16.55.64/26:


50 hosts => 172.16.55.0/26
40 hosts => 172.16.55.64/26
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NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting
With a “/26” mask (six host bits), the subnets increment by 64, so the next subnet begins at
172.16.55.128 (64 + 64 = 128). The next largest subnet contains 25 hosts, and therefore requires five
host bits, or a “/27”. Thus, the next subnet is 172.16.55.128/27:



50 hosts => 172.16.55.0/26
40 hosts => 172.16.55.64/26
25 hosts => 172.16.55.128/27
With a “/27” mask (five host bits), the subnets increment by 32, so the next subnet begins at
172.16.55.160 (128 + 32 = 160). The next largest subnet contains 20 hosts, and therefore requires five
host bits, or another “/27”. Thus, the next subnet is 172.16.55.160/27:




50 hosts => 172.16.55.0/26
40 hosts => 172.16.55.64/26
25 hosts => 172.16.55.128/27
20 hosts => 172.16.55.160/27
With a “/27” mask (five host bits), the subnets increment by 32, so the next subnet begins at
172.16.55.192 (160 + 32 = 192). The next largest subnet contains 12 hosts, and therefore requires four
host bits, or a “/28”. Thus, the next subnet is 172.16.55.192/28:





50 hosts => 172.16.55.0/26
40 hosts => 172.16.55.64/26
25 hosts => 172.16.55.128/27
20 hosts => 172.16.55.160/27
12 hosts => 172.16.55.192/28
With a “/28” mask (four host bits), the subnets increment by 16, so the next subnet begins at
172.16.55.208 (192 + 16 = 208). The next largest subnet contains 10 hosts, and therefore also requires
four host bits, or another “/28”. Thus, the next subnet is 172.16.55.208/28:






50 hosts => 172.16.55.0/26
40 hosts => 172.16.55.64/26
25 hosts => 172.16.55.128/27
20 hosts => 172.16.55.160/27
12 hosts => 172.16.55.192/28
10 hosts => 172.16.55.208/28
With a “/28” mask (four host bits), the subnets increment by 16, so the next subnet begins at
172.16.55.224 (208 + 16 = 224). The next largest subnet contains five hosts, and therefore requires
three host bits, or a “/29”. Thus, the next subnet is 172.16.55.224/29:







50 hosts => 172.16.55.0/26
40 hosts => 172.16.55.64/26
25 hosts => 172.16.55.128/27
20 hosts => 172.16.55.160/27
12 hosts => 172.16.55.192/28
10 hosts => 172.16.55.208/28
5 hosts => 172.16.55.224/29
With a “/29” mask (three host bits), the subnets increment by eight, so the next subnet begins at
172.16.55.232 (224 + 8 = 232). The next largest subnet contains four hosts, and therefore requires
three host bits, or another “/29”. Thus, the next subnet is 172.16.55.232/29:
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NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting








50 hosts => 172.16.55.0/26
40 hosts => 172.16.55.64/26
25 hosts => 172.16.55.128/27
20 hosts => 172.16.55.160/27
12 hosts => 172.16.55.192/28
10 hosts => 172.16.55.208/28
5 hosts => 172.16.55.224/29
4 hosts => 172.16.55.232/29
That takes care of the eight LAN subnets, now let’s do the two WAN links. Taking up where we left
off, with a “/29” mask (three host bits), the subnets increment by eight, so the next subnet begins at
172.16.55.240 (232 + 8 = 240). The WAN links need two hosts each, and therefore we can use a “/30”
for each, and with a “/30” mask, the subnets increment by four. Therefore, the two WAN subnets
could be:


2 hosts => 172.16.55.240/30
2 hosts => 172.16.55.244/30
The next available subnet would be 172.16.55.248, but since we’ve covered all ten required subnets,
we don’t need it at present. Note that we could also have used “/31” masks on the WAN links (per
RFC 3021), in which case (incrementing by two, instead of four) they could be:


2 hosts => 172.16.55.240/31
2 hosts => 172.16.55.242/31
In this case, the address space starting at 172.16.55.244 is unused (with a “/31” mask, the subnets
increment by two).
--------------------------------------------------------------------------------------------------------------------------Solution 5)
First, we find the total number of hosts, which is 100 + (20 x 25) + (20 x 2) = 640. Since we have a
“/22” address space, we have ten bits to work with. Ten bits is off the chart, but if we extend the
powers of two chart two more columns to the left, we see that two to the tenth is 1,024 (allowing for
1,022 hosts), so we’re good to proceed. Also, using the methods we developed in “Subnet Masking
Shortcuts”, with a “/22” mask (255.255.252.0) the subnets increment by four in the third octet, so the
next subnet up would be 10.100.204.0/22. In other words, we have the range of addresses from
10.100.200.0 to 10.100.203.255 available. Anything outside of that range is outside of our allowable
address space.
We’ll start with the largest subnet, the HQ LAN, which requires 100 hosts (seven host bits), and
therefore a “/25” mask:

100 hosts => 10.100.200.0/25
With a “/25” mask (seven host bits), the subnets increment by 128, so the next subnet begins at
10.100.200.128. The next largest subnet, a sales LAN, contains 25 hosts, and therefore requires five
host bits, or a “/27”. Thus, the next subnet is 10.100.200.128/27. With a “/27” mask (five host bits),
the subnets increment by 32, so the next subnet begins at 10.100.200.160. In fact, all of the sales
LANs require 25 hosts, and therefore “/27” masks. Thus, the twenty sales LANs could be:

10.100.200.128/27
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NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting







10.100.200.160/27
10.100.200.192/27
10.100.200.224/27
10.100.201.0/27
(thirteen more subnets incrementing by 32 go here)
10.100.202.192/27
10.100.202.224/27
That takes care of the 21 LAN subnets (HQ and the 20 remote sites), so now let’s do the WAN links.
Taking up where we left off at 10.100.200.224/27, with five host bits the subnets increment by 32, so
the next available subnet begins at 10.100.203.0. The WAN links need two hosts each, so we can use
a “/30” for each (in which case they increment by four). Therefore, the twenty WAN subnets
could be:

10.100.203.0 (4, 8 … 72, 76) /30
The next available subnet would be 10.100.203.80, but since we’ve covered all required subnets, we
don’t need it at present. We could also have used “/31” masks on the WAN links, in which case they
could be:

10.100.203.0 (2, 4 … 36, 38) /31
In this case, the address space starting at 10.100.203.40 remains unused.
-------------------------------------------------------------------------------------------------------------------Solution 6)
In this case, the largest subnet is the HQ LAN (50 hosts), which requires six host bits, or a
“/26” mask:

50 hosts => 172.30.20.0/26
With a “/26” mask (six host bits), the subnets increment by 64, so the next subnet begins at
172.30.20.64. The next largest subnet is a sales office LAN, which contains five hosts, and there are
ten of them, all requiring three host bits, or “/29” masks. Thus, the ten sales office LAN subnets
could be:

172.30.20.64 (72, 80, 88 … 128, 136) /29
The next subnet starts at 172.30.20.144, and we still need to allocate address space for the ten WAN
links. Because these are P2P links, we can use either “/30” or “/31” subnets. Let’s use “/30”, which
increment by four:

172.30.20.144 (148, 152, 156 … 176, 180) /30
That takes care of that, with the addresses 184 and above currently unused. Now comes the
“flexibility” part. Since we have quite a bit of unused address space, we could get fancy and allocate
addresses below 100 for the HQ LAN, those in the range 100 to 199 for sales office LANs, and 200
and above for the WAN links, thus:

HQ LAN => 172.30.20.0/26 (still allows for 62 hosts)
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NETWORK LABORATORY
Lab 4: IP Addressing and Subnetting


Sales LANs => 172.30.20.104 (112, 120 … 168, 176) /29
WANs => 172.30.20.200/30 (204, 208 … 236, 240) /30
With the scheme just outlined, the addresses in the ranges from 172.30.20.64 to 103, and
172.30.20.244 to 255 are unused.
---------------------------------------------------------------------------------------------------------Solution 7) Try your self
00000000 = .0/26
Cannot use because it has been subnetted
00000000 = .0/30
Network E
00000100 = .4/30
Network F
00001000 = .8/30
Network G
00001100 = .12/30
Network H
00010000 = .16/28
Future growth
01000000 = .64/26
Network A
10000000 = .128/26 Cannot use because it has been subnetted
10000000 = .128/27 Network B
10100000 = 160/27
Cannot use because it has been subnetted
10100000
160/28
Network C
10110000
176/28
Network D
11000000 = .192/26 Future growth
Solution 8)
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