Would you like to inspect the original subtitles? These are the user uploaded subtitles that are being translated:
1
00:00:00,000 --> 00:00:08,000
84%
Here is another example, ABC limited have been allocated subnet 10.128.192.0/18
2
00:00:08,000 --> 00:00:11,000
84%
for several offices in the USA
3
00:00:11,000 --> 00:00:15,000
84%
Paul a network administrator once again needs to split the subnet
4
00:00:15,000 --> 00:00:19,000
84%
into smaller subnets, Paul requires 30 subnets with as many hosts
5
00:00:19,000 --> 00:00:25,000
84%
as possible on each subnet, and once again he asked you for your help.
6
00:00:25,000 --> 00:00:27,000
84%
You need to decide which formula to use.
7
00:00:27,000 --> 00:00:31,000
84%
Please note we�ve been ask for networks or subnets
8
00:00:31,000 --> 00:00:36,000
84%
so we need to use the formula 2 to the n and not the formula 2 to the n - 2.
9
00:00:36,000 --> 00:00:38,000
84%
And we need to remember to count the bits
10
10
00:00:38,000 --> 00:00:40,000
84%
from the left hand side to the right hand side.
11
11
00:00:40,000 --> 00:00:44,000
84%
So on step 2 you need to work out the number of bits required to cover the number
12
12
00:00:44,000 --> 00:00:49,000
84%
of hosts or in this case the number of networks that we�ve been asked for.
13
13
00:00:49,000 --> 00:00:57,000
84%
Paul is asked for 30 subnets so we will require 5 bits because
14
14
00:00:57,000 --> 00:01:00,000
84%
using the formula 2 to the n and substituting n with 5 will give you 32.
15
15
00:01:00,000 --> 00:01:04,000
84%
So we will actually end up having 32 subnets rather than just 30.
16
16
00:01:04,000 --> 00:01:09,000
84%
So we now know that we need to steal 5 bits from the host portion of the address
17
17
00:01:09,000 --> 00:01:11,000
84%
and allocate that to the network portion
18
18
00:01:11,000 --> 00:01:16,000
84%
because 5 binary bits are required to give us 32 networks.
19
19
00:01:16,000 --> 00:01:20,000
84%
The 3rd step is to convert the host portion of the original network into binary.
20
20
00:01:20,000 --> 00:01:32,000
84%
so the original network we were given was 10.128.192.0/18 or 10.128.192.0
21
21
00:01:32,000 --> 00:01:41,000
84%
with the mask of 255.255.192.0 now 255 tells us that the first octet is network
22
22
00:01:41,000 --> 00:01:45,000
84%
the 2nd 255 tells us that the 2nd octet is network
23
23
00:01:45,000 --> 00:01:50,000
84%
however in the third octet, the octet is not fully populated with binary 1's.
24
24
00:01:50,000 --> 00:01:55,000
84%
So in the 3rd octet there�s a split between network and host.
25
25
00:01:55,000 --> 00:02:02,000
84%
The last octet is filled with binary 0's so that entire octet is host.
26
26
00:02:02,000 --> 00:02:09,000
84%
Converting 192 into binary gives us 2 binary 1's followed by 6 binary 0's
27
27
00:02:09,000 --> 00:02:09,000
90%
0 in decimal converted to binary give us 8 binary 0's.
28
28
00:02:09,000 --> 00:02:19,000
84%
So we have converted the 3rd octet where we have both network and host bits
29
29
00:02:19,000 --> 00:02:27,000
84%
and the last octet into binary and we have drawn a line separating the network
30
30
00:02:27,000 --> 00:02:29,000
84%
and the host portion of the address.
31
31
00:02:29,000 --> 00:02:31,000
84%
How do we know that we need to draw the line here?
32
32
00:02:31,000 --> 00:02:34,000
84%
because we have 18 bits in the network mask
33
33
00:02:34,000 --> 00:02:38,000
84%
The first octet is 8 bits, the 2nd octet is 8 bits,
34
34
00:02:38,000 --> 00:02:45,000
84%
8 plus 8 is 16, plus 2 gives us 18 so this line indicates
35
35
00:02:45,000 --> 00:02:47,000
84%
the separation between network and host.
36
36
00:02:47,000 --> 00:02:54,000
84%
Now the original network once again is 10.128.192.0/18 or
37
37
00:02:54,000 --> 00:03:00,000
84%
could be written as 255.255.192.0 in dotted decimal notation.
38
38
00:03:00,000 --> 00:03:05,000
84%
So once again the network portion is 10.128 the network host portion is 192
39
39
00:03:05,000 --> 00:03:11,000
84%
and the host portion is 0. We are going to take 5 bits from the host portion
40
40
00:03:11,000 --> 00:03:13,000
84%
and allocate that to the subnet, so the network portion is 10.129
41
41
00:03:13,000 --> 00:03:19,000
84%
and then on the 3rd octet is the first 2 bits are the network
42
42
00:03:19,000 --> 00:03:23,000
84%
and we count 5 bits from the left hand side to the right hand side
43
43
00:03:23,000 --> 00:03:32,000
84%
so 12345 and we draw a line here indicating that this 5 bits are subnet
44
44
00:03:32,000 --> 00:03:35,000
84%
and all bits to the right of the second line are host.
45
45
00:03:35,000 --> 00:03:39,000
84%
So we have now stolen 5 bits from the host portion and allocated
46
46
00:03:39,000 --> 00:03:42,000
84%
that to the subnet portion of the address.
47
47
00:03:42,000 --> 00:03:45,000
84%
So we need to work out what the new subnet mask is.
48
48
00:03:45,000 --> 00:03:49,000
84%
It�s equal to the number of bits in the network and subnet portion of the address.
49
49
00:03:49,000 --> 00:03:52,000
84%
So it�s equal to this portion of the address
50
50
00:03:52,000 --> 00:03:55,000
84%
plus the extra 5 bits allocated to the subnet portion.
51
51
00:03:55,000 --> 00:03:58,000
84%
Just to remind you once again 1 octet is 8 bits.
52
52
00:03:58,000 --> 00:04:03,000
84%
so the first octet is 8 bits, the 2nd octet is 8 bits
53
53
00:04:03,000 --> 00:04:05,000
84%
so that gives you a total of 16 bits.
54
54
00:04:05,000 --> 00:04:10,000
84%
We�ve got 2 bits in the 3rd octet which are part of the network plus
55
55
00:04:10,000 --> 00:04:15,000
84%
5 additional bits which have been allocated to subnet, so that gives us 7 bits.
56
56
00:04:15,000 --> 00:04:18,000
84%
so the total number of bits in the network subnet portion is equal
57
57
00:04:18,000 --> 00:04:23,000
84%
to 8 plus 8 plus 2 plus 5 which equals 23 bits
58
58
00:04:23,000 --> 00:04:28,000
84%
you could also work this backward once again, there are 32 bits in an IPv4 address
59
59
00:04:28,000 --> 00:04:32,000
84%
and notice in the host portion there are 8 bits in the last octet
60
60
00:04:32,000 --> 00:04:37,000
84%
allocated to host plus 1 bit in the 3rd octet
61
61
00:04:37,000 --> 00:04:42,000
84%
so 1 plus 8 equals 9, 32 less 9 gives you 23.
62
62
00:04:42,000 --> 00:04:46,000
84%
Either method is fine, the result is the same 23 bits have now been
63
63
00:04:46,000 --> 00:04:52,000
84%
allocated to network and subnet where's before only 18 bits were allocated.
64
64
00:04:52,000 --> 00:04:55,000
84%
So now it's possible to work out to new subnet.
65
65
00:04:55,000 --> 00:04:57,000
84%
Once again, to work out the subnet go through the various
66
66
00:04:57,000 --> 00:05:01,000
84%
binary combinations for the subnet portion of the address.
67
67
00:05:01,000 --> 00:05:04,000
84%
So this portion in green mark the subnet
68
68
00:05:04,000 --> 00:05:09,000
84%
so the first network or subnet is equal to 10.128.
69
69
00:05:09,000 --> 00:05:16,000
84%
a 2 binary bits which part of the original network plus 5 additional
70
70
00:05:16,000 --> 00:05:19,000
84%
binary bits which would now allocated to subnet.
71
71
00:05:19,000 --> 00:05:24,000
84%
So once again the subnet mask is /23 which can be written
72
72
00:05:24,000 --> 00:05:30,000
84%
in dotted decimal notation as 255.255.254.0
73
73
00:05:30,000 --> 00:05:34,000
84%
To work out the first subnet, fill the subnet portion of the address
74
74
00:05:34,000 --> 00:05:39,000
84%
with 0's and populate the host portion of the address with 0s.
75
75
00:05:39,000 --> 00:05:45,000
84%
please note this 2 binary 1's. the 5 green binary 0's that are part of the subnet
76
76
00:05:45,000 --> 00:05:49,000
84%
and the 1 red binary 0 that�s part of the host portion
77
77
00:05:49,000 --> 00:05:51,000
84%
all form part of the same subnet.
78
78
00:05:51,000 --> 00:05:58,000
84%
So 11 followed by 6 binary 0's equals 192 in decimal.
79
79
00:05:58,000 --> 00:06:05,000
84%
To work out the 2nd network or subnet, we go through binary combination.
80
80
00:06:05,000 --> 00:06:09,000
84%
The next binary combination is 4 binary 0's followed by binary 1
81
81
00:06:09,000 --> 00:06:15,000
84%
taking the whole octet into account that equals 194 in decimal.
82
82
00:06:15,000 --> 00:06:19,000
84%
Please note the host portion is always set to binary 0's.
83
83
00:06:19,000 --> 00:06:22,000
84%
So the last octet is once again 0.
84
84
00:06:22,000 --> 00:06:27,000
84%
So the second network or subnet is 10.128.194.0
85
85
00:06:27,000 --> 00:06:31,000
84%
Now you probably already guess what the 3rd one is gonna be
86
86
00:06:31,000 --> 00:06:33,000
84%
because we're going up in multiples of 2.
87
87
00:06:33,000 --> 00:06:38,000
84%
But if we go to the whole process again getting the next binary value
88
88
00:06:38,000 --> 00:06:42,000
84%
would be 3 binary 0's followed by binary 1 followed by binary 0.
89
89
00:06:42,000 --> 00:06:47,000
84%
And converting that whole octet back into decimal will give us 196.
90
90
00:06:47,000 --> 00:06:51,000
84%
So we know that we're going in multiples of 2,
91
91
00:06:51,000 --> 00:07:00,000
84%
so the first 1 is 192, then 194, then 196, then 198, then 200, 202, 204, etc.
92
92
00:07:00,000 --> 00:07:03,000
84%
all the way up to the last subnet.
93
93
00:07:03,000 --> 00:07:09,000
84%
To work out the last subnet, fill the subnet portion of the address with binary 1's
94
94
00:07:09,000 --> 00:07:14,000
84%
so we end up having 10.128 followed by 7 binary 1's,
95
95
00:07:14,000 --> 00:07:17,000
84%
followed by binary 0 in the 3rd octet.
96
96
00:07:17,000 --> 00:07:23,000
84%
7 binary 1's followed by binary 0 in an octet is equal to 254.
97
97
00:07:23,000 --> 00:07:27,000
84%
The last octet is once again equal to 0.
98
98
00:07:27,000 --> 00:07:33,000
84%
So the last subnet is 10.128.254.0 with the /23 mask
99
99
00:07:33,000 --> 00:07:42,000
84%
or it can be written as 10.128.254.0 with the mask of 255.255.254.0
100
100
00:07:42,000 --> 00:07:46,000
84%
I hope that�s helped you learn how to subnet based on a requirement
101
101
00:07:46,000 --> 00:07:50,000
84%
first specific number of hosts or specific number of networks.
102
102
00:07:50,000 --> 00:07:52,000
84%
So what have we covered?
103
103
00:07:52,000 --> 00:07:56,000
84%
We look at the reason for subnetting, subnetting is very important for this course
104
104
00:07:56,000 --> 00:07:59,000
84%
and it�s important that you have a good understanding of subnetting.
105
105
00:07:59,000 --> 00:08:03,000
84%
so we spent time looking at the binary method and the quick method
106
106
00:08:03,000 --> 00:08:06,000
84%
for determining the subnet address, broadcast address
107
107
00:08:06,000 --> 00:08:11,000
84%
first host address and last host address for a given IP address.
108
108
00:08:11,000 --> 00:08:14,000
84%
I also showed you how to create multiple subnets
109
109
00:08:14,000 --> 00:08:19,000
84%
based on specific host or network requirements.
11654
Can't find what you're looking for?
Get subtitles in any language from opensubtitles.com, and translate them here.
Afrikaans
Albanian
Amharic
Arabic
Armenian
Azerbaijani
Basque
Belarusian
Bengali
Bosnian
Bulgarian
Catalan
Cebuano
Chichewa
Chinese (Simplified)
Chinese (Traditional)
Corsican
Croatian
Czech
Danish
Dutch
English
Esperanto
Estonian
Finnish
French
Frisian
Galician
Georgian
German
Greek
Gujarati
Haitian Creole
Hawaiian
Hungarian
Icelandic
Indonesian
Irish
Italian
Japanese
Kazakh
Khmer
Korean
Kurdish (Kurmanji)
Kyrgyz
Lao
Latvian
Lithuanian
Luxembourgish
Macedonian
Malagasy
Malay
Maltese
Maori
Mongolian
Myanmar (Burmese)
Nepali
Norwegian
Pashto
Polish
Romanian
Russian
Samoan
Serbian
Sesotho
Shona
Sinhala
Slovak
Slovenian
Somali
Spanish
Swahili
Swedish
Tajik
Thai
Turkish
Ukrainian
Urdu
Uzbek
Vietnamese
