I'm trying to calculate the network address but I'm not entirely sure if I've done it correctly. Let's say the IP is: 130.243.14.29 and the Sub-mask is: 255.255.254.0
After calculating I'm getting: 130.243.14.0, is this correct or is the last number wrong? Is the broadcasting address the same thing as network address?
IP is: 130.243.14.29 and the Sub-mask is: 255.255.254.0
U have 510 IPs are useable
Network adress is 130.243.14.0
First ip address is 130.243.14.1
Last ip address is 130.243.15.254
broadcasting address is 130.243.15.255
Related
[I did my research]
I read the IPv4 and Subnetting chapters in the CCNA prep material. Watched several youtube videos and read through several forums. However, I still have a small issue because different sources use certain terms interchangeably however I want to be certain and confirm my understanding to avoid confusion.
[The Question]
Q1
Is the network address the same as the node address the same as the network id ?
Q2
Is the host address the same as the host id?
for instance:
in this ip 184.19.39.34 is 184.19 the network address == the node address == the network id?
is 39.34 the host address == the host id?
Thanks alot guys !
Q1: Basically, yes. You should note though that there are different kinds of network addresses such as IP addresses, subnet addresses/prefixes, or MAC addresses. The exact meaning of each term depends on context.
Q2: If the IP address/mask is 184.19.39.34/16 then 184.19.0.0/16 is the subnet address. 39.34 is the host part of the IP address while 184.19 is the network part. The IP address is split into these parts as indicated by the CIDR value or the network mask (255.255.0.0 for /16).
Note that this split is only relevant for routing and on the last hop, so you don't need to know the mask value of a far IP address.
Answer Q1 :
yes, But on as desired of the Mac or virtual IP or a value for the IP
Answer Q2 :
Gluttony complement each other
You can find a detailed answer here :
[https://www.geeksforgeeks.org/ip-addressing-introduction-and-classful-addressing/][1]
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
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)
Your router has the IP address of 216.83.11.65/27. You wish to connect
a new system on the network. Which of the following addresses would
you assign to the new system.
A: 216.83.11.45
B: 216.83.11.87
C: 216.83.11.95
D: 216.83.11.96
The answer is B, but I'm not sure why. Can I get a step by step breakdown?
Your host network can chose from all the bits in the IP address, that are not used to define the network.1
So when looking at the Network Configuration you have, your network has the following range (you can use a subnet calculator to get to these values):
216.83.11.65 - 216.83.11.94
This rules out A and D, because you will need a router to get there (since the IP addresses cannot be directly reached within the network).
Your network internal broadcast address will be 216.83.11.95. This rules out answer C, because a host cannot have the broadcast address of your network.
So, only answer B stands.
1 I know that this might not be understandable at all. Cannot do it better. In this case please refer to Wikipedia: Classless Inter-Domain Routing
I have 3 IP addresses.
source IP : IP1
target IP : IP2, IP3
How can I determine what target IP is the nearest one from source IP by C-code
I don't need to know exact locations of IP addresses.
Just want to know what target IP is the nearest.
Please give me some advices.
Thank you.
Your ip is in form
aaa.bbb.ccc.ddd - 4 components.
First you should process 1st component:
compare (source.aaaa - target1.aaaa) and (source.aaaa - target2.aaaa)
(by module, so diffs are always positive)
If both are equal, then proceed with next component.
If different - the one with smaller difference is the closest one.
Use traceroute?
$traceroute <IP ADDR>
Should return hop information and speed, assuming you have the IP Addresses themselves