CIDR No unread replies.No replies. This week we’re still working on network addr
ID: 669301 • Letter: C
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CIDR
No unread replies.No replies.
This week we’re still working on network addressing. This week, you need to calculate the addressing that you’d need for setting up a training session at a convention center.
It will involve 250 participants and 10 instructors. Each of the individuals (participants and instructors) will have their own desktop computer and will connect to two servers that they alone manage. So that is three addresses per person, one for their computer and one each for two servers. There will be 5 networked printers, a few extra addresses for routers and switches (no more than 10) and approximately 300 addresses needed for the individual’s personal wireless devices.
The convention center has instructed you to use parts of the Class C private address space of 192.168.0.0/16. They have also requested that you minimize the amount of addresses that you use since another convention is taking place in the auditorium which will also need addressing space from the same /16 network (NOT part of this assignment).
The Convention Center network team will provide you with one Router that will isolate the South Wing of the center specifically to your needs and will also proxy Internet to you. It is your job to determine the addressing that is needed to accomplish your needs, then provide the Convention Center network team the network IP and the network’s CIDR block you request.
Once you have designed the block of address ranges you need, other items to be addressed too. The instructors have requested the following subnets within the network address block:
1. The server network (all servers),
a. The server network will have all statically assigned IPs to the devices.
2. The participants and instructors network,
a. The Participants/Instructor network will have addressing assigned through a DHCP server that resides on the same network along with a DNS and 3 networked printers.
3. The wireless network for the personal devices.
a. The wireless network will have its own DCHP server with web caching software and the other two networked printers.
For this topic, draw your designed topology; include all routers, DHCP, DNS servers and network printers. Include IP addresses on the Routers, switches, DHCP, DNS and network printers. Include the network scope on each subnet and the Supernet address of the entire network.
Explanation / Answer
Classless Inter-Domain routing (CIDR) essentially removes the idea of Class from IP addresses and allows administrations to allocate and route any valid subnet from any convenient base IP irrespective of its Class. The idea being that if you want a group of 128 IP addresses, whether you take them from a Class C or from a Class B range is not important. You simply want 128 IP addresses. Equally, if you want 512 IPv4 addresses, whether they are taken from a Class B range or are two Class C ranges is again unimportant. The table below shows two 32 address subnets, one from a nominal Class B range the other from a nominal Class C range - spot the difference!
Class
Network
Netmask
B
172.28.227.192
255.255.255.224
C
192.168.15.64
255.255.255.224
In short the key factors in a CIDR world become the Network (base) IP address and the Netmask.
IP Prefixes - the slash format (17.16/16)
To simplify writing IP addresses used in routing or filtering systems it has now become common practice to define a list of IP addresses using a slash (more properly IP Prefix) notation. The IP address to the LEFT of the SLASH (/) is the Network (base) IP address and the number (1 to 32) to the RIGHT of the SLASH (/) is the number of contiguous 1 bits in the netmask. Thus, 192.168.2.0/24 will have an associated net/subnetmask containing 24 contiguous bits or 255.255.255.0 in dotted decimal format. The following table illustrates this notation:
Line No.
IP Prefix (Slash) Form
Network Base IP
Netmask
No. of IPs
1
192.168.32/19
192.168.32.0
255.255.224.0
8192
2
172.28.127.64/27
172.28.127.64
255.255.255.224
32
3
172.28.127/24
172.28.127.0
255.255.255.0
256
4
172.28.127.130/25
172.28.127.128
255.255.255.128
128
5
10.0/16
10.0.0.0
255.255.0.0
65535
6
10.0.2.2/32
10.0.2.2
255.255.255.255
1
7
10.2.3/8
10.0.0.0
255.0.0.0
lots! (16,777,216)
8
10/8
10.0.0.0
255.0.0.0
the same lots!
3.6 IPv4 Address Calculator
The tool has two parts that are invoked separately. The first part is essentially a simple IPv4 calculator and is invoked using the IPv4 Info button. The second part provides an IPv4 bitwise AND feature. This part is invoked with the Bitwise AND button.
IPv4 Calculator: Enter an IPv4 address with its /Prefix (or slash) notation in the IPv4/Prefix: box and click IPv4 Info. The calculator will populate No. of IPs (covered by the /Prefix or slash value), IPv4 Netmask in dotted decimal format (a dotted hex format is also shown enclosed in []) represented by the /Prefix, IPv4 Range lists the start and end IPv4 addresses covered by the /prefix (a dotted hex format is also shown enclosed in []). Finally, for those of you who need it (ignore it if not) the reverse map zone name of the IPv4 address is calculated based on the /Prefix value (IPv4 addresses below this name will be defined in the reverse zone file) and is shown in FQDN format in Reverse Zone.
Alternatively, enter a single IPv4 address (that lies anywhere in the desired range) without the /prefix in IPv4/Prefix and the number of desired IP addresses (if it is not a power of 2 it will be rounded up to the nearest power of 2) in No. of IPs then click IPv4 Info. The calculator will populate IPv4/Prefix (with the calculated /Prefix to cover the required or calculated number of IP addresses), No. of IPs will be updated if needed (due to any power of 2 rounding required) and IPv4 Netmask, IP Range and Reverse Zone will be populated normally.
The Clear button zaps ALL entries in IPv4 Calculator only.
Notes:
1. Validation and other errors are shown in the IPv4 Netmask box for no very good reason.
2. When entering a /prefix IPv4 address you only need to enter as many dotted elements as are required by the /prefix (more is not a problem), if you enter less than the required number the calculator silently pads with zeros to the right.
Bitwise AND: This calculator Bitwise ANDs any two IPv4 addresses. The bit patterns of the two IPv4 addresses and shown together with the bit pattern of the result together with its dotted decimal (and a dotted hex) value.
Enter the two IPv4 address (without any /Prefix notation) in the boxes labeled IPv4 1: and IPv4 2: (order is not important). If you omitIPv4 1: the value in IPv4/Prefix: will be used. If you omit IPv4 2: the calculated value in IPv4 Netmask: will be used. Click theBitwise AND button to activate.
The calculator places the binary value (bit pattern) of IPv4 1: in Binary 1: and that of IPv4 2: in Binary 2:. Binary AND: contains the binary value of the result and IPv4 Result: shows the dotted decimal form (and a dotted hex form enclosed in []).
Clear will zap all entries in the Bitwise AND part only.
Sample Exercise 1
Now that you have an understanding of subnetting, put this knowledge to use. In this example, you are given two address / mask combinations, written with the prefix/length notation, which have been assigned to two devices. Your task is to determine if these devices are on the same subnet or different subnets. You can do this by using the address and mask of each device to determine to which subnet each address belongs.
Determining the Subnet for DeviceA:
Looking at the address bits that have a corresponding mask bit set to one, and setting all the other address bits to zero (this is equivalent to performing a logical "AND" between the mask and address), shows you to which subnet this address belongs. In this case, DeviceA belongs to subnet 172.16.16.0.
Determining the Subnet for DeviceB:
From these determinations, DeviceA and DeviceB have addresses that are part of the same subnet.
Sample Exercise 2
Given the Class C network of 204.15.5.0/24, subnet the network in order to create the network in Figure 3 with the host requirements shown.
Figure 3
Looking at the network shown in Figure 3, you can see that you are required to create five subnets. The largest subnet must support 28 host addresses. Is this possible with a Class C network? and if so, then how?
You can start by looking at the subnet requirement. In order to create the five needed subnets you would need to use three bits from the Class C host bits. Two bits would only allow you four subnets (22).
Since you need three subnet bits, that leaves you with five bits for the host portion of the address. How many hosts does this support? 25 = 32 (30 usable). This meets the requirement.
Therefore you have determined that it is possible to create this network with a Class C network. An example of how you might assign the subnetworks is:
Class
Network
Netmask
B
172.28.227.192
255.255.255.224
C
192.168.15.64
255.255.255.224
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