1
00:00:00,560 --> 00:00:09,170
So in this lab setup what we want to do, we want to launch www.example1.com on 172.24

2
00:00:09,170 --> 00:00:15,040
0.1, www.example2.com on 172.24.0.2.

3
00:00:15,050 --> 00:00:22,670
And example3.com on 172.24.0.3, that mean we are using three IP addresses for

4
00:00:22,670 --> 00:00:23,990
three sites.

5
00:00:24,200 --> 00:00:26,570
This is called IP based virtual hosting.

6
00:00:26,910 --> 00:00:32,240
But first of all, we have got only one address available, 172.24.0.1.

7
00:00:32,690 --> 00:00:35,540
Now we want to have another two addresses.

8
00:00:35,870 --> 00:00:37,340
So we will configure the.

9
00:00:37,340 --> 00:00:45,860
ifcfg-ens33 file and we will configure it for three IP addresses, then we will restart the

10
00:00:45,860 --> 00:00:46,310
service.

11
00:00:46,610 --> 00:00:53,900
So we will open this file with the Vi editor and what we are going to do, the entry regarding the bootprotocol

12
00:00:53,900 --> 00:00:54,950
will be the same.

13
00:00:55,340 --> 00:01:01,870
And but what we are going to do earlier this file was containing only one IP address entry and we will comment

14
00:01:01,880 --> 00:01:02,480
that line.

15
00:01:02,660 --> 00:01:08,900
And now what we are going to do that when we are going to comment this IPADDR line, NETMASK line

16
00:01:09,110 --> 00:01:16,940
and we are going to make entry regarding IPADDR1, it is going to be 172.24.0.1

17
00:01:16,940 --> 00:01:23,900
IPADDR2 it is going to be 172.24.0.2 and IPADDR3 is 172.24.0.3.

18
00:01:24,140 --> 00:01:29,840
And we are going to specify the prefix as 16 and what will happen.

19
00:01:29,840 --> 00:01:36,960
All the three IP addresses will be allocated to the network address and all will be having mask

20
00:01:37,220 --> 00:01:39,860
16 mean 255.255.0.0.

21
00:01:40,160 --> 00:01:48,470
Then we are going to restart our network service and then we will verify with which command ip a s ens.

22
00:01:48,470 --> 00:01:52,490
33 and we will see whether all the three addresses are there or not.

23
00:01:52,820 --> 00:01:58,870
So we will find that it has been allocated 172.24.0.1 already it was there.

24
00:01:59,240 --> 00:02:04,460
So another address two is there,three addresses is there because we need three addresses.

25
00:02:04,640 --> 00:02:09,590
So we are going to change that file for having another two addresses.

26
00:02:09,590 --> 00:02:10,820
One is already there.
