Consider the following classless address block:
154.78.177.3/27
List the addresses from this block that would be used as:
a) the network address,
b) the direct broadcast address, and
c) the range available for hosts to use
Show the steps you took to arrive at your answers.
154.78.177.3/27 means that this ip address is having network address to be 27 bit long ( most significant bit). i.e if we consider the ip address to be 32 bit long. So, in 154.78.177.00000011 ( last 3 number is represented as binary for simplicity) upto 154.78.177.000 (8+8+8+3 = 27) is network address and remaining 5 bit (00000) is for host ip addresses.
Note-
Network address is the first ip address of the total ip's of host.
Direct broadcast address is the last ip address of total ip's of host.
So final conclusion:
1. Network address is 154.78.177.0. This is calculated by setting all 5 bits of host id to 0.
2. Direct broadcast address is 154.78.177.31. This is calculated by setting all 5 bits of host to 1.
3. Range available for host is from 1 to 30 in the last octect. I.e from 154.78.177.1 to 154.78.177.30 ( As first and last ip is reserved for network address and direct broadcast address respectively)
Related
I am trying to learn how to calculate IP addresses from CIDR block.
For example, 10.88.135.144/28 or
10.88.135.10010000/28
From what I understand, that means first 28 bits are associated with network address while the rest 4 bits are host addresses. That would result in following IP range:
10.88.135.10010000 - 10.88.135.10011111
The first IP should be 10.88.135.144 and last IP address should be 10.88.135.159
But according to cidr.xyz. The first IP should be 10.88.135.145 and the last one should be 10.88.135.158.
I really can't figure out why. Can anyone explain the reason for me? Thanks!
Generally, the first IP is the network identifier and cannot be assigned to any device.This is used by router or switch on the network.
The last one is the broadcasting IP and cannot be assigned to any device as this IP is used by router or switch on the network to broadcast information.
https://www.quora.com/In-IP-addresses-what-is-meant-by-network-ID-and-host-ID
https://supportforums.cisco.com/t5/wan-routing-and-switching/what-is-broadcast-address/td-p/2494445#messageBodySimpleDisplay_1
Can someone explain exactly how CIDR blocks work and how it translates into 0.0.0.0/32 for example? Please use laymen’s terms or perhaps even an analogy to something not network related. Can’t seems to find an explanation that clicks with me. Thanks!!
Classless Inter-Domain Routing (CIDR) blocks are for specifying a range to IP addresses in format of IPv4 or IPv6. For the sake of simplicity I will explain rest of this in format of IPv4 however it is applicable to IPv6.
General format for CIDR Blocks: x.y.z.t/p
x, y, z and t are numbers from 0 to 255. Basically, each represents an 8 bit binary number. That's why it is range is up to 255. Combination of this numbers becomes an IPv4 IP address that must be unique to be able to identify a specific instance.
In case of AWS, p is a number from 16 to 28. It represents the number of bits that are inherited from given IP address. For example: 10.0.0.0/16 represents an IP address in following format: 10.0.x.y where x and y are any number from 0 to 255. So, actually it represents a range of IP addresses, starting from 10.0.0.0 to 10.0.255.255.
However for each CIDR block, AWS prohibits 5 possible IP addresses. Those are the first 4 available addresses and the last available address. In this case:
10.0.0.0: Network address
10.0.0.1: Reserved for VPC router
10.0.0.2: DNS server
10.0.0.3: Reserved for future use
10.0.255.255: Network broadcast
See here for official doc.
Actually this is one of the main reasons why AWS permits numeric value of p up to /28. Because for p=30, there will be 4 available values however AWS needs 5 IP address to use. In my opinion for p=29, they might found it inefficient to occupy 5 addresses to provide 3 possible IP address.
Number of possible IP addresses can be calculated by using this formula:
NumberOfPossibleIPs = 2^(32-p) - 5
Classless Inter-Domain Routing (CIDR) block basically is a method for allocating IP addresses and IP routing. When you create a network or route table, you need to specify what range are you working in. "0.0.0.0" means that it will match to any IP address. Some IP addresses are specific, like 10.0.0.0, which will match to any IP address beginning with 10. With any IP address range, you can be more specific by using a suffix(something like /32 from your example). These allow the notation to specify number of bits to be used from Prefix(actual IP-range like 10.0.0.0). It represents the bit length of the subnet mask, as indicated above. The subnet mask is like masking when painting. You place a mask over what you DO NOT want to paint on.
For example, 10.10.0.0/16 will have 256 * 256 IP address in its range.
NOTE: Some of the IP address in a range are reserved for various purposes. According to AWS VPC documentation, following are the reserved IP addresses.
10.0.0.0: Network address.
10.0.0.1: Reserved by AWS for the VPC router.
10.0.0.2: Reserved by AWS. The IP address of the DNS server is always the base of the VPC network range plus two; however, we also reserve the base of each subnet range plus two. For VPCs with multiple CIDR blocks, the IP address of the DNS server is located in the primary CIDR. For more information, see Amazon DNS Server.
10.0.0.3: Reserved by AWS for future use.
10.0.0.255: Network broadcast address. We do not support broadcast in a VPC, therefore we reserve this address.
Hope this helps!
All of the above answers are great, but are missing something pretty important for the people who don't understand addressing.
IP addresses are literally just a string of binary, broken up into 4 "octets". Each octet is a 2^8 block; 00000000. So to a machine, an IP address looks like this (with (.) added for human-ness):
00000000(.)00000000(.)00000000(.)00000000
When we're talking about the "mask" on the IP address, it means "the bits that don't change". The /8 or /255.0.0.0 on the end of the block signifies the number of bits that are not allowed to be used by this network.
So, lets say we have a CIDR block of 10.0.0.0/8 - this can also be written in the format 10.0.0.0/255.0.0.0, and you may in fact see this for of notation in older versions of linux. You will also note that 255 is the decimal representation of the binary string 11111111 - 8 binary "ones". So what the machine sees is the following:
Net: 00001010(.)00000000(.)00000000(.)00000000
Mask: 11111111(.)00000000(.)00000000(.)00000000
The part of the mask with 0's is usable address space within the network.
So the following example addresses are valid in this network, because on the 0 parts of the masked range are changing:
00001010(.)00000001(.)00110000(.)00111000
00001010(.)00110001(.)00110100(.)00111001
When we say "cidr block" we simply mean "the human-readable shorthand way of expressing binary strings understood by a machine". In the above example, the first octet can be expressed as 10, and the latter octets 0. And the Mask can be expressed as 255 and the latter octets of 0, or; because the mask is always a sequence of 1's, then a sequence of 0's, the length of the 1's, i.e. 8
And as such, we get a cidr of 10.0.0.0/255.0.0.0, or 10.0.0.0/8
A few more examples:
-- 172.1.1.0/24
net: 10101100.00000001.00000001.00000000
mask: 11111111.11111111.11111111.00000000
^ 24 bits for the mask ^ 8 bits of usable space
-- 10.10.10.8/29
net: 00001010.00001010.00001010.00001000
mask: 11111111.11111111.11111111.11111000
^ 29 bits for the mask. ^ 3 bits of usable space
Importantly though, this is only one aspect of networking. Usually a couple of these are reserved for things. See other answers for AWS specific things. In their examples, the "first 4" ip addresses reserved for AWS will be the first 4 usable addresses, which would be
...00 - Network address
...01 - Router
...10 - DNS
...11 - Futureproofing
i already have studied about the internet IP and all those stuff, but i'm still confused about how to identify network ID from an given IP address, for example, if i have given an IP address: 192.168.250.0/24, so what will be the network ID?
can you guys help me?? and explain the method?
Thank you in advance
You need to convert the IP address and mask to binary, then logically AND them, then convert back to decimal:
192.168.250.0 -----> 11000000.10101000.11111010.00000000
24 bit mask -------> 11111111.11111111.11111111.00000000
=================================== <--- Logical AND
Network address ---> 11000000.10101000.11111010.00000000 ---> 192.168.250.0
This answer on Network Engineering describes how to do all your IP operations.
It s simple, your IP address is 192.168.250.0/24 and you should know that IP address is in a fact 32 bits value of zeros and ones. So it s easy for human to say listen this first 24 bits are network ID (network part) and remaining 8 bits are host ID (host part), but machines need parameter that will use to determine what bits are in network and what bits are ih host part. That s why we have subnet mask, that map IP address in a way that binary ones "1" map bits that belongs to network part and binary zeros "0" map bits that belongs to host part.
So your IP address in binary form is
11000000.10101000.11111010.00000000 when we add subnet mask then you have
11111111.11111111.11111111.00000000 or 255.255.255.0 in decimal form
-------------------------------------
11000000.10101000.11111010. bits maped by binary ones (network part)
.00000000 bits maped by binary zeros (host part)
If you count number of ones in subnetmask you will see that there are 24 binary ones, and that s why we use other notations for presenting subnet mask, and that is by something we call network prefix or "/24".
When host want to send packets he first must determine network address, but how he do that? Every host have IP address and subnet mask let say
192.168.250.6 and subnetmask /24 == 255.255.255.0 or in binary
11000000.10101000.11111010.00000110
11111111.11111111.11111111.00000000
------------------------------------- network hosts use logical AND operation
11000000.10101000.11111010.00000000
(notice that network address is address that have binary zeros in host part of IP address. Logical AND use two values (0 or 1), results of this logical operation is "1" only when both values are "1" (or you can easily rememmber it has to be "1" AND "1" to get "1" as result :) ).
After this host examine destination IP let s say 192.168.10.6, and host use again same operation and if result show that network part is identical with his IP than thay communicate directly or if they have different networkID hosts use gateway since destination is at another network.
I figured out the subnet mask for both subnets 1 and 2. My problem is I can't grasp how the subnet turns to 172.20.11.254 and 172.20.13.254 respectively? I assume this is VSLM, but not certain. I'm just learning this. I got 172.20.8.0 and 172.20.6.0 as my subnet and I know that is wrong now. Thanks for any help you can provide.
To determine which subnet mask will work for the 172.20.0.0 network, first look at the number of hosts required for each subnet:
Subnet1 (connected to FastEthernet0/0) has 672 hosts. To support 672 hosts, a subnet mask of /22 is required (10 host bits in the 2n-2 formula will afford 1022 host addresses in the subnet).
Subnet2 (connected to FastEthernet0/1) has 258 hosts. To support 258 hosts, a subnet mask of /23 is required (9 host bits in the 2n-2 formula will afford 510 host addresses in the subnet).
With a network address of 172.20.0.0 and the masks needed to fit the requirements, you need to configure the following IP address and subnet masks:
For the FastEthernet0/0 connection:
172.20.8.0/22 is the third possible subnet. (172.20.0.0/22 is the first possible subnet and 172.20.4.0/22 is the second possible subnet.)
172.20.11.254 is the last possible IP address in the subnet.
255.255.252.0 is the decimal version of a 22-bit mask.
For the FastEthernet0/1 connection:
172.20.12.0/23 is the next available subnet that does not overlap.
172.20.13.254 is the last possible IP address in the subnet.
255.255.254.0 is the decimal version of a 23-bit mask.
Use the following commands to configure the SFO interfaces:
SFO>enable
SFO#configure terminal
SFO(config)#interface FastEthernet0/0
SFO(config-if)#ip address 172.20.11.254 255.255.252.0
SFO(config-if)#no shutdown
SFO(config-if)#interface FastEthernet0/1
SFO(config-if)#ip address 172.20.13.254 255.255.254.0
SFO(config-if)#no shutdown
SFO(config-if)#exit
SFO(config)#exit
SFO#copy run start
I detect some desperation, so let's see if I can convey and understandable explanation. :-)
172.20.0.0 seems to be the address space destined for you to use in this exercise. That is a class B network (255.255.0.0, or /16 netmask), but since we're going to subnet it variably, you can safely forget that. For example, you could subnet all of it it in small, class C subnets (all with a mask of 255.255.255.0, or /24), and if you did you would use 172.20.0.0/24 for one network, 172.20.1.0/24 for another, 172.20.2.0/24 for another, and so on. But if you did that, each subnet would be able to hold no more than 254 hosts (that is because you leave the last octet - 8 bits - for the host portion, and you have to reserve two - the first and last - for the subnet address and the broadcast address: 2^8-2=254).
But 254 hosts is not enough for your needs, since you have requirements for 672 and 258.
If you use a smaller sized mask (meaning larger sized network -> more hosts) like a /23 (255.255.254.0) you now have 9 bits for the host portion, therefore you can acommodate 2^9-2=510 hosts, big enough for 258, but not for 672. So for the latter you will need a /22 network (255.255.252.0), which will leave 10 bits for the host portion thus allowing 2^10-2=1022.
With each bit you reduce in the netmask, you double your network size. So if a /24 goes from 172.20.0.0 to 172.20.0.255 (the single '0' class C network), a /23 goes from 172.20.0.0 to 172.20.1.255 (two class C networks, '0' and '1'). And a /22 goes from 172.20.0.0 to 172.20.3.255 (four class C networks). In each case the first address is considered the network address and is not assigned to any device, and the last one is the broadcast address, and is not assigned either.
So, back to your example, they choose to assign the 3rd /22 network (1st being from 172.20.0.0 to 172.20.3.255, 2nd being from 172.20.4.0 to 172.20.7.255, and 3rd being from 172.20.8.0 to 172.20.11.255) to that particular subnet. So 172.20.8.0/22 it is. And they choose to assign the 7th /23 subnet possible (1st is '0' and '1' class C's, 2nd is '2' and '3' class C's, and so on) to the other subnet. So 172.20.12.0/23 it is for it. Remember that they cannot overlap!
Now, as to why they chose the .254 addresses for the router interfaces, that is just a convention. Router interfaces are usually configured to use either the first usable (.1) IP address or the last usable (.254) IP address in their subnets, at least on the LAN side. Note that your subnets' broadcast addresses are 172.20.11.255 for the /22 and 172.20.13.255 for the /23. In both cases they picked for the router interfaces the address which is one below them, i.e. the last usable address. But it could have been any one in the corresponding range.
Did that help?
I am developing a web application.
I want to deploy it in a LAN.
I would like to know what will be the maximum number of systems possible in a LAN?
The maximum number of nodes on a LAN depends on the media type, network protocol, and (at least for the IPv4 protocol) the network address class.
For example, a Class C IPv4 network (mask 255.255.255.0) on ethernet could have up to 254 nodes. (0 and 255 are reserved).
A class B IPv4 network such as 192.168.x.x (mask 255.255.0.0) might have up to 16381 nodes. (0 and 16383 are reserved). But in practice the number of ethernet nodes within the same collision domain is likely to be much less, depending on the number of hubs and their layout.
In a LAN like the one you say (192.168.1.56) is the maximum number of systems 253 if I'm not wrong. 255 is the highest number in an IP address and the last number and first number are reserved for broadcast address and network address.
Or it would be that the subnet mask is other than 255.255.255.0 then it can be even less than 253.
And about the handling of assignment of IP addresses that depends on how the network and systems are configured. Most of the networks are configured so that every PC gets his IP address from a DHCP server (router most of the times). If it goes through DHCP then one PC can have now IP address 192.168.1.56 and when it reboots it can have address 192.168.1.105. Also a PC can have a static IP address so that it does not change.
What do you mean with "maximum number of systems"? Amount of IP-adresses? That depends on the IP class and subclass.