Would you like to inspect the original subtitles? These are the user uploaded subtitles that are being translated:
1
00:00:00,000 --> 00:00:03,000
--> align:middle line:84%
In this section we gonna look at IP subnetting.
2
00:00:03,000 --> 00:00:06,000
--> align:middle line:84%
I�d like to teach you multiple methods for subnetting.
3
00:00:06,000 --> 00:00:10,000
--> align:middle line:84%
Subnetting is one of those subject that a lot of Network Engineers struggle with.
4
00:00:10,000 --> 00:00:13,000
--> align:middle line:84%
But in this section I hope I'll be able to simplify for you
5
00:00:13,000 --> 00:00:15,000
--> align:middle line:84%
and give you various methods.
6
00:00:15,000 --> 00:00:18,000
--> align:middle line:84%
So that any subnetting questions that you might encounter
7
00:00:18,000 --> 00:00:21,000
--> align:middle line:84%
can be completed quickly and easily.
8
00:00:21,000 --> 00:00:23,000
--> align:middle line:84%
There are 2 main parts in this section:
9
00:00:23,000 --> 00:00:27,000
--> align:middle line:84%
in the 1st part I'm gonna show you 2 methods to quickly and easily answer
10
10
00:00:27,000 --> 00:00:32,000
--> align:middle line:84%
questions like, what subnet is this host on or what is a last host in this subnet?
11
11
00:00:32,000 --> 00:00:35,000
--> align:middle line:84%
So I�m gonna teach you the binary method and then
12
12
00:00:35,000 --> 00:00:38,000
--> align:middle line:84%
I�m gonna teach you what I call the quick method.
13
13
00:00:38,000 --> 00:00:41,000
--> align:middle line:84%
I suggest that you look at both and then choose one
14
14
00:00:41,000 --> 00:00:43,000
--> align:middle line:84%
and get good at that specific method.
15
15
00:00:43,000 --> 00:00:46,000
--> align:middle line:84%
I think binary is very important to understand.
16
16
00:00:46,000 --> 00:00:50,000
--> align:middle line:84%
But as the quick method name indicates it�s very quick and easy
17
17
00:00:50,000 --> 00:00:53,000
--> align:middle line:84%
to answer subnetting questions using the quick method.
18
18
00:00:53,000 --> 00:00:57,000
--> align:middle line:84%
In the 2nd part I�d like to show you how to create multiple subnets
19
19
00:00:57,000 --> 00:01:01,000
--> align:middle line:84%
when given a specific network or subnet, so you will be able to answer questions
20
20
00:01:01,000 --> 00:01:07,000
--> align:middle line:84%
like subnet this network into at least 10 subnets or subnet this network
21
21
00:01:07,000 --> 00:01:09,000
--> align:middle line:84%
into subnets each having 10 hosts on them.
22
22
00:01:09,000 --> 00:01:15,000
--> align:middle line:84%
So in part 1 you'll be given an IP address and then you gonna need to learn
23
23
00:01:15,000 --> 00:01:20,000
--> align:middle line:84%
how to work out the subnet address, 1st host address, last host address
24
24
00:01:20,000 --> 00:01:25,000
--> align:middle line:84%
and the broadcast address and we'll gonna use 2 methods: the binary method
25
25
00:01:25,000 --> 00:01:28,000
--> align:middle line:84%
and the quick method, and as I mentioned I suggest that you choose 1 of the
26
26
00:01:28,000 --> 00:01:32,000
--> align:middle line:84%
2 methods, look at both, decide which one you prefer
27
27
00:01:32,000 --> 00:01:35,000
--> align:middle line:84%
and then learn and get good at that specific method.
28
28
00:01:35,000 --> 00:01:37,000
--> align:middle line:84%
So here's the typical example:
29
29
00:01:37,000 --> 00:01:43,000
--> align:middle line:84%
You�ve been given a host with an address of 192.168.10.18 for the /24 mask
30
30
00:01:43,000 --> 00:01:46,000
--> align:middle line:84%
and you�ll need to be able to answer the following questions:
31
31
00:01:46,000 --> 00:01:52,000
--> align:middle line:84%
what IP address with router 1 be configured with, this router being router 1
32
32
00:01:52,000 --> 00:01:57,000
--> align:middle line:84%
if it is to use the first IP address in the same subnet as PC 1?
33
33
00:01:57,000 --> 00:01:59,000
--> align:middle line:84%
Which is the PC here? The second question could be:
34
34
00:01:59,000 --> 00:02:02,000
--> align:middle line:84%
What broadcast address is used by PC 1?
35
35
00:02:02,000 --> 00:02:06,000
--> align:middle line:84%
or thirdly what IP address would router 1 be configured with if it is to use
36
36
00:02:06,000 --> 00:02:10,000
--> align:middle line:84%
the last IP address in the same subnet as PC1?
37
37
00:02:10,000 --> 00:02:13,000
--> align:middle line:84%
So the first question was, what IP address will the router use if it is
38
38
00:02:13,000 --> 00:02:15,000
--> align:middle line:84%
configured with the first IP address?
39
39
00:02:15,000 --> 00:02:19,000
--> align:middle line:84%
And the third one is: what IP address would it be configured with if it were
40
40
00:02:19,000 --> 00:02:22,000
--> align:middle line:84%
configured with the last IP address in the subnet?
41
41
00:02:22,000 --> 00:02:25,000
--> align:middle line:84%
And fourthly, a question could be what subnet is PC 1 part of?
42
42
00:02:25,000 --> 00:02:29,000
--> align:middle line:84%
so you should within 30 to 60 seconds be able to answer those question
43
43
00:02:29,000 --> 00:02:31,000
--> align:middle line:84%
and I�m gonna show you how to do that now.
44
44
00:02:31,000 --> 00:02:35,000
--> align:middle line:84%
So the first method is to use binary to work out the answers.
45
45
00:02:35,000 --> 00:02:39,000
--> align:middle line:84%
I personally prefer binary because it's used all over the place in networking.
46
46
00:02:39,000 --> 00:02:42,000
--> align:middle line:84%
And I want you get your head around how binary works.
47
47
00:02:42,000 --> 00:02:44,000
--> align:middle line:84%
It will save you a lot of time in the long run.
48
48
00:02:44,000 --> 00:02:46,000
--> align:middle line:84%
Please refer to the binary section of this course
49
49
00:02:46,000 --> 00:02:49,000
--> align:middle line:84%
if you are not sure about the basics of binary.
50
50
00:02:49,000 --> 00:02:52,000
--> align:middle line:84%
we are going to assume knowledge of binary now
51
51
00:02:52,000 --> 00:02:55,000
--> align:middle line:84%
we're just gonna use binary to work out the answer to those questions.
52
52
00:02:55,000 --> 00:02:58,000
--> align:middle line:84%
In the binary method, you need to remember 4 rules
53
53
00:02:58,000 --> 00:03:01,000
--> align:middle line:84%
to work out the network with subnet address
54
54
00:03:01,000 --> 00:03:05,000
--> align:middle line:84%
you fill the host portion of the address with binary 0's
55
55
00:03:05,000 --> 00:03:08,000
--> align:middle line:84%
So you work out which part is network and which part is host
56
56
00:03:08,000 --> 00:03:11,000
--> align:middle line:84%
and you fill the host portion with binary 0's.
57
57
00:03:11,000 --> 00:03:14,000
--> align:middle line:84%
To work out the broadcast address,
58
58
00:03:14,000 --> 00:03:17,000
--> align:middle line:84%
you fill the host portion of the address with binary 1's.
59
59
00:03:17,000 --> 00:03:22,000
--> align:middle line:84%
To work out the first host in the subnet, you fill the host portion of the
60
60
00:03:22,000 --> 00:03:26,000
--> align:middle line:84%
address with binary 0's except for the very last bit which is set to a binary 1.
61
61
00:03:26,000 --> 00:03:31,000
--> align:middle line:84%
And to work out the last host in the subnet, you fill the host portion of the
62
62
00:03:31,000 --> 00:03:35,000
--> align:middle line:84%
address with binary 1's except for the last bit which is set to binary 0.
63
63
00:03:35,000 --> 00:03:38,000
--> align:middle line:84%
If you remember those 4 rules, you will be able to work out the answer to any
64
64
00:03:38,000 --> 00:03:41,000
--> align:middle line:84%
of those questions very quickly.
65
65
00:03:41,000 --> 00:03:42,000
--> align:middle line:84%
So let�s start the basic example.
66
66
00:03:42,000 --> 00:03:52,000
--> align:middle line:84%
We have a PC with an IP address 192.168.1.18/24 or it could be written as
67
67
00:03:52,000 --> 00:03:58,000
--> align:middle line:84%
192.168.1.18 with the mask of 255.255.255.0
68
68
00:03:58,000 --> 00:04:03,000
--> align:middle line:84%
As you learnt in the IP addressing section, these 2 are equivalent subnet mask.
69
69
00:04:03,000 --> 00:04:08,000
--> align:middle line:84%
It's very similar to saying tomato or tomato or router or router.
70
70
00:04:08,000 --> 00:04:12,000
--> align:middle line:84%
It means the same thing it's just a different way of displaying the subnet mask.
71
71
00:04:12,000 --> 00:04:18,000
--> align:middle line:84%
Now hopefully you remember that 24 means 24 bits in the subnet mask
72
72
00:04:18,000 --> 00:04:22,000
--> align:middle line:84%
in other word 8 bits in the first octet, 8 bits in the second octet
73
73
00:04:22,000 --> 00:04:27,000
--> align:middle line:84%
and the 8 bits in the third octet. That means the first 3 octets are network.
74
74
00:04:27,000 --> 00:04:33,000
--> align:middle line:84%
So 192.168.1 is the network portion of the address and this last octet is the
75
75
00:04:33,000 --> 00:04:36,000
--> align:middle line:84%
host portion and that�s how I�ve written it here.
76
76
00:04:36,000 --> 00:04:40,000
--> align:middle line:84%
The first 3 octets are displayed in black representing the network portion and
77
77
00:04:40,000 --> 00:04:44,000
--> align:middle line:84%
the last octet is shown in red which represents the host portion of the address.
78
78
00:04:44,000 --> 00:04:47,000
--> align:middle line:84%
There�s no point converting the network portion into binary.
79
79
00:04:47,000 --> 00:04:54,000
--> align:middle line:84%
So I�ve just left as it is, so the subnet is equal to 192.168.1.
80
80
00:04:54,000 --> 00:04:58,000
--> align:middle line:84%
and if you remember back to our rules to get the subnet,
81
81
00:04:58,000 --> 00:05:01,000
--> align:middle line:84%
you fill the host portion of the address with binary 0's
82
82
00:05:01,000 --> 00:05:07,000
--> align:middle line:84%
So this is the single octet, so 8 binary 0's need to fill that host portion.
83
83
00:05:07,000 --> 00:05:13,000
--> align:middle line:84%
If you convert 8 binary 0's back into decimal you'll get the value 0.
84
84
00:05:13,000 --> 00:05:18,000
--> align:middle line:84%
So the subnet is 192.168.1.0 for this specific IP address.
85
85
00:05:18,000 --> 00:05:23,000
--> align:middle line:84%
To get the first host in the subnet, you fill the host portion of the address
86
86
00:05:23,000 --> 00:05:28,000
--> align:middle line:84%
with binary 0's except for the last bit which make binary 1.
87
87
00:05:28,000 --> 00:05:30,000
--> align:middle line:84%
So an octet once again is 8 bits.
88
88
00:05:30,000 --> 00:05:34,000
--> align:middle line:84%
So there are 7 binary 0's followed by single binary 1.
89
89
00:05:34,000 --> 00:05:36,000
--> align:middle line:84%
Converting that back to decimal you get 1,
90
90
00:05:36,000 --> 00:05:41,000
--> align:middle line:84%
so the first host in this subnet is 192.168.1.1
91
91
00:05:41,000 --> 00:05:46,000
--> align:middle line:84%
To work out the last host, fill the host portion of the address with binary 1's
92
92
00:05:46,000 --> 00:05:50,000
--> align:middle line:84%
except for the last bit you set to binary 0
93
93
00:05:50,000 --> 00:05:56,000
--> align:middle line:84%
That's 7 binary 1's followed by binary 0, if you convert that back to decimal
94
94
00:05:56,000 --> 00:05:58,000
--> align:middle line:84%
you'll get the value 254.
95
95
00:05:58,000 --> 00:06:03,000
--> align:middle line:84%
So the very last host in this subnet is 192.168.1.254
96
96
00:06:03,000 --> 00:06:07,000
--> align:middle line:84%
To get the broadcast address, fill the host portion of the address
97
97
00:06:07,000 --> 00:06:14,000
--> align:middle line:84%
with binary 1's, so single octet fill with binary 1's is equal to 255 in decimal
98
98
00:06:14,000 --> 00:06:19,000
--> align:middle line:84%
So there you have it. We can work out the subnet, the first host, the last host
99
99
00:06:19,000 --> 00:06:23,000
--> align:middle line:84%
and the broadcast address very quickly and easily using the binary method.
100
100
00:06:23,000 --> 00:06:29,000
--> align:middle line:84%
So if you were ask to configure the 1st IP address in the same subnet as PC 1
101
101
00:06:29,000 --> 00:06:33,000
--> align:middle line:84%
on the routers interface you would know that you need to configure the IP address
102
102
00:06:33,000 --> 00:06:43,000
--> align:middle line:84%
192.168.1.1/24 or 192.168.1.1 255.255.255.0
103
103
00:06:43,000 --> 00:06:48,000
--> align:middle line:84%
You could be expected to type the commands unto the router to configure it with
104
104
00:06:48,000 --> 00:06:52,000
--> align:middle line:84%
the relevant IP address, later in this course you'll learn the commands to do that
105
105
00:06:52,000 --> 00:06:57,000
--> align:middle line:84%
But for now, realize that you might need to work out first, last, broadcast and
106
106
00:06:57,000 --> 00:07:00,000
--> align:middle line:84%
subnet addresses When giving a specific IP address.
107
107
00:07:00,000 --> 00:07:03,000
--> align:middle line:84%
This is our first example using the binary method
108
108
00:07:03,000 --> 00:07:05,000
--> align:middle line:84%
I�d like to show you more complicated example now.
109
109
00:07:05,000 --> 00:07:12,000
--> align:middle line:84%
So in this example you've been given an IP address 172.16.35.123/20
110
110
00:07:12,000 --> 00:07:15,000
--> align:middle line:84%
or in other words 255.255.240.0
111
111
00:07:15,000 --> 00:07:18,000
--> align:middle line:84%
This is a little bit more complicated because the mask
112
112
00:07:18,000 --> 00:07:21,000
--> align:middle line:84%
does not fall on the octet boundary,
113
113
00:07:21,000 --> 00:07:26,000
--> align:middle line:84%
in other words, it's not a simple /8 or /16 or /24 subnet.
114
114
00:07:26,000 --> 00:07:29,000
--> align:middle line:84%
The network and host portion lies somewhere in the third octet.
115
115
00:07:29,000 --> 00:07:35,000
--> align:middle line:84%
So just to reiterate, /20 means that 20 bits of the 32 bit IP address
116
116
00:07:35,000 --> 00:07:39,000
--> align:middle line:84%
which are used for networks or subnet portion and the remaining 12 bits are used
117
117
00:07:39,000 --> 00:07:41,000
--> align:middle line:84%
as the host portion of the address.
118
118
00:07:41,000 --> 00:07:45,000
--> align:middle line:84%
So somewhere in the third octet, the network host portion changes.
119
119
00:07:45,000 --> 00:07:52,000
--> align:middle line:84%
So once again our IP address is 172.16.35.123/20
120
120
00:07:52,000 --> 00:07:57,000
--> align:middle line:84%
Now the first octet is 8 bits in size , the second octet is 8 bits in size.
121
121
00:07:57,000 --> 00:07:59,000
--> align:middle line:84%
So adding those together you get 16 bits.
122
122
00:07:59,000 --> 00:08:05,000
--> align:middle line:84%
The third octet is 8 bits in size, so adding those 3 together gives you 24 bits.
123
123
00:08:05,000 --> 00:08:10,000
--> align:middle line:84%
So /20 in the subnet mask means that the split between subnet
124
124
00:08:10,000 --> 00:08:13,000
--> align:middle line:84%
and host occurs somewhere in the third octet.
125
125
00:08:13,000 --> 00:08:17,000
--> align:middle line:84%
So once again the first octet is 8 bits, second octet is 8 bits thus
126
126
00:08:17,000 --> 00:08:20,000
--> align:middle line:84%
20 bits puts the split somewhere in the third octet.
127
127
00:08:20,000 --> 00:08:24,000
--> align:middle line:84%
So what you need to do is convert the third and fourth octet
128
128
00:08:24,000 --> 00:08:27,000
--> align:middle line:84%
into binary because they both have host bits.
129
129
00:08:27,000 --> 00:08:30,000
--> align:middle line:84%
There�s no need to convert the first 2 octets into binary
130
130
00:08:30,000 --> 00:08:33,000
--> align:middle line:84%
because no host bits are found in the first 2 octets.
131
131
00:08:33,000 --> 00:08:44,000
--> align:middle line:84%
So 00100011 is the binary equivalent of decimal 35, 0 followed by 4 binary 1's
132
132
00:08:44,000 --> 00:08:50,000
--> align:middle line:84%
followed by 0 followed by 2 binary 1's is the binary equivalent of decimal a 123.
133
133
00:08:50,000 --> 00:08:57,000
--> align:middle line:84%
So all we�ve done is convert 35 and 123 into their binary equivalents.
134
134
00:08:57,000 --> 00:09:00,000
--> align:middle line:84%
Please see the binary section of this course if you are not
135
135
00:09:00,000 --> 00:09:03,000
--> align:middle line:84%
sure how to convert decimal into binary and back again.
136
136
00:09:03,000 --> 00:09:09,000
--> align:middle line:84%
20 bits of subnet mask put us in this position because we have 8 bits
137
137
00:09:09,000 --> 00:09:12,000
--> align:middle line:84%
plus 8 bits in the first 2 octets which give us a total
138
138
00:09:12,000 --> 00:09:19,000
--> align:middle line:84%
of 16 binary bits plus 1 is 17 plus 1 is 18 plus 1 is 19 plus 1 is 20
139
139
00:09:19,000 --> 00:09:25,000
--> align:middle line:84%
So we can draw a line at this point to split the network and host portion
140
140
00:09:25,000 --> 00:09:29,000
--> align:middle line:84%
To the left of this line, the binary bits indicate network or subnet,
141
141
00:09:29,000 --> 00:09:33,000
--> align:middle line:84%
to the right of this line the binary bits indicates host.
142
142
00:09:33,000 --> 00:09:39,000
--> align:middle line:84%
So there are split between network and host portions in this address at this line.
143
143
00:09:39,000 --> 00:09:43,000
--> align:middle line:84%
So to work out the answers you need to remember the 4 rules.
144
144
00:09:43,000 --> 00:09:48,000
--> align:middle line:84%
The first rule states that, to work out the network or subnet address,
145
145
00:09:48,000 --> 00:09:51,000
--> align:middle line:84%
you need to fill the host portion of the address with binary 0's.
146
146
00:09:51,000 --> 00:09:54,000
--> align:middle line:84%
So to the right of the line we need to fill
147
147
00:09:54,000 --> 00:09:58,000
--> align:middle line:84%
this entire portion of the address with binary 0's.
148
148
00:09:58,000 --> 00:10:02,000
--> align:middle line:84%
So once again 20 bits of subnet mask put us in this position
149
149
00:10:02,000 --> 00:10:05,000
--> align:middle line:84%
and we fill everything to the right with binary 0's.
150
150
00:10:05,000 --> 00:10:12,000
--> align:middle line:84%
So we end up having 172.16 the first 4 bits of the third octet
151
151
00:10:12,000 --> 00:10:15,000
--> align:middle line:84%
remain the same and all remaining bits are set to binary 0.
152
152
00:10:15,000 --> 00:10:21,000
--> align:middle line:84%
When you convert that back to decimal, please note firstly that these 4 bits
153
153
00:10:21,000 --> 00:10:26,000
--> align:middle line:84%
in red plus the following 4 bits in green are part of the third octet.
154
154
00:10:26,000 --> 00:10:31,000
--> align:middle line:84%
So 0010 followed by 0000 when converted from binary
155
155
00:10:31,000 --> 00:10:36,000
--> align:middle line:84%
to decimal gives us a value of 32. The fourth octet is filled with 0's.
156
156
00:10:36,000 --> 00:10:42,000
--> align:middle line:84%
So converting an octet filled with binary 0's to decimal give us a value of 0.
157
157
00:10:42,000 --> 00:10:47,000
--> align:middle line:84%
So the subnet address is 172.16.32.0
158
158
00:10:47,000 --> 00:10:53,000
--> align:middle line:84%
Now to work out the first host address, fill the host portion of the address
159
159
00:10:53,000 --> 00:10:58,000
--> align:middle line:84%
with binary 0's except for the last bit which is set to binary 1.
160
160
00:10:58,000 --> 00:11:04,000
--> align:middle line:84%
So here you go, the first host is equal to 172.16.0010
161
161
00:11:04,000 --> 00:11:09,000
--> align:middle line:84%
remember, we don�t touch this portion of the address, we only change the host
162
162
00:11:09,000 --> 00:11:13,000
--> align:middle line:84%
portion of the address, so we fill the host portion with binary 0's
163
163
00:11:13,000 --> 00:11:16,000
--> align:middle line:84%
except for the last bit which we set to binary 1.
164
164
00:11:16,000 --> 00:11:21,000
--> align:middle line:84%
Once again the third octet consist of 4 binary bits which are part of the network
165
165
00:11:21,000 --> 00:11:25,000
--> align:middle line:84%
or subnet and 4 binary bits which a part of the host portion.
166
166
00:11:25,000 --> 00:11:30,000
--> align:middle line:84%
Those 4 red binary bits together with those 4 green binary bits
167
167
00:11:30,000 --> 00:11:33,000
--> align:middle line:84%
give us a decimal value of 32.
168
168
00:11:33,000 --> 00:11:39,000
--> align:middle line:84%
The last octet is 7 binary 0's followed by single binary 1
169
169
00:11:39,000 --> 00:11:45,000
--> align:middle line:84%
which is equal to decimal 1, so the first host address is 172.16.32.1
170
170
00:11:45,000 --> 00:11:49,000
--> align:middle line:84%
To work out the last host address, fill the host portion of the address
171
171
00:11:49,000 --> 00:11:54,000
--> align:middle line:84%
with binary 1's except for the last bit which is set to binary 0.
172
172
00:11:54,000 --> 00:12:02,000
--> align:middle line:84%
So the last host is equal to 172.16.0010 followed 4 binary 1's, in the third octet
173
173
00:12:02,000 --> 00:12:07,000
--> align:middle line:84%
followed by 7 binary 1's in the fourth octet and then lastly a binary 0.
174
174
00:12:07,000 --> 00:12:13,000
--> align:middle line:84%
So the third octet consist of this 4 red bits and this 4 green bits.
175
175
00:12:13,000 --> 00:12:18,000
--> align:middle line:84%
So if you convert 0010 followed by 4 1's into decimal
176
176
00:12:18,000 --> 00:12:22,000
--> align:middle line:84%
you'll get the decimal equivalent which is 47, 7 binary 1's
177
177
00:12:22,000 --> 00:12:30,000
--> align:middle line:84%
followed by binary 0, gives you the equivalent decimal value of 254.
178
178
00:12:30,000 --> 00:12:37,000
--> align:middle line:84%
So the last host in this subnet is 172.16.47.254
179
179
00:12:37,000 --> 00:12:41,000
--> align:middle line:84%
To work out the broadcast address, fill the host portion
180
180
00:12:41,000 --> 00:12:44,000
--> align:middle line:84%
of the address with binary 1's.
181
181
00:12:44,000 --> 00:12:50,000
--> align:middle line:84%
So the broadcast address is equal to 172.16.0010
182
182
00:12:50,000 --> 00:12:53,000
--> align:middle line:84%
Once again, the network and subnet portion are left as is.
183
183
00:12:53,000 --> 00:12:56,000
--> align:middle line:84%
And then you fill the host portion with binary 1's.
184
184
00:12:56,000 --> 00:13:02,000
--> align:middle line:84%
So the third octet is equal to 0010 and 4 binary 1's
185
185
00:13:02,000 --> 00:13:06,000
--> align:middle line:84%
which once again gives you the decimal equivalent of 47.
186
186
00:13:06,000 --> 00:13:12,000
--> align:middle line:84%
8 binary 1's in octet gives you the decimal value 255.
187
187
00:13:12,000 --> 00:13:18,000
--> align:middle line:84%
So the broadcast address would be 172.16.47.255
188
188
00:13:18,000 --> 00:13:28,000
--> align:middle line:84%
So in summary, an address of 172.16.35.123 with the subnet mask of /20 resides
189
189
00:13:28,000 --> 00:13:37,000
--> align:middle line:84%
190
172:16:32,000 --> 172:16:47,255
on subnet and has a broadcast address of
190
191
00:13:37,000 --> 00:13:42,000
--> align:middle line:84%
the first host in the same subnet is 172.16.32.1
191
192
00:13:42,000 --> 00:13:47,000
--> align:middle line:84%
and the last host in the same subnet is 172.16.47.254
192
193
00:13:47,000 --> 00:13:52,000
--> align:middle line:84%
This was a slightly more complicated example, but I�m hoping that by now
193
194
00:13:52,000 --> 00:13:55,000
--> align:middle line:84%
you have learned the process to work out subnet first host, last host and
194
195
00:13:55,000 --> 00:14:00,000
--> align:middle line:84%
broadcast address when given an IP address and its subnet mask.
25471
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
