All language subtitles for 2. Subnetting Lab 1 Answers Part 1

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These are the user uploaded subtitles that are being translated: 1 00:00:06,440 --> 00:00:15,650 In this network we've been told to subnet 1 9 2 1 6 8 2 1 0 slash 24 into 4 subnets. 2 00:00:15,670 --> 00:00:24,190 In other words we need a subnet for site one site to the serial link and this serial link. 3 00:00:24,190 --> 00:00:29,760 Now we haven't been told to use a slash of 30 mosque on the serial links. 4 00:00:29,770 --> 00:00:37,090 We'll do that in a separate video so all we need to do after this point is take 1 9 2 1 6 8 wondered 5 00:00:37,150 --> 00:00:42,460 0 slash 24 and subnet that into 4 subnets. 6 00:00:42,460 --> 00:00:49,750 Now when we look at a network address such as 1 9 2 1 6 8 1 0 we need to determine which point is the 7 00:00:49,750 --> 00:00:54,670 network and which part is the host portion. 8 00:00:54,670 --> 00:00:58,530 Currently this portion is the network portion. 9 00:00:59,560 --> 00:01:09,860 We can't change the network portion but we can manipulate and change the host portion of the address. 10 00:01:09,910 --> 00:01:15,500 So in other words what we can do is change this lost octet. 11 00:01:15,610 --> 00:01:23,320 The reason why we can't manipulate the first three octet is because that's part of the network 24. 12 00:01:23,320 --> 00:01:28,870 Essentially means that we have 24 binary ones. 13 00:01:28,870 --> 00:01:31,800 Now you don't have spaces in an octet like this. 14 00:01:31,810 --> 00:01:41,990 I'm just doing it to make it easier to read but essentially 24 means that we have 24 ones into the subnet 15 00:01:41,990 --> 00:01:55,300 mask so that's 24 binary ones which essentially equates to 255 255 255 is zero. 16 00:01:55,300 --> 00:02:02,590 It's important that you know how to convert a decimal to binary this binary value equates to decimal 17 00:02:02,740 --> 00:02:09,000 255 so the host portion of the address is once again that portion. 18 00:02:09,190 --> 00:02:16,030 This portion is the network portion we can't manipulate the network portion but to the host portion 19 00:02:16,450 --> 00:02:20,890 consisting of eight binary zeros can be manipulated. 20 00:02:20,890 --> 00:02:28,990 Now when we subnet think of submitting kind of like stealing butts we're going to steal butts from the 21 00:02:28,990 --> 00:02:30,310 host portion. 22 00:02:30,310 --> 00:02:37,560 We need to work out how many bits are required for the number of subnets that we need in this example 23 00:02:37,570 --> 00:02:39,840 we need four subnets. 24 00:02:39,840 --> 00:02:43,990 Now when it comes to subletting there are two formulas that you need to know. 25 00:02:44,070 --> 00:02:48,680 Two to the power of n and two to the power of n minus two. 26 00:02:48,780 --> 00:02:57,590 This formula is used when we asked for subnets which is what we've been asked in this question. 27 00:02:57,690 --> 00:03:02,140 This formula we need to use when we asked for a certain number of hosts. 28 00:03:02,220 --> 00:03:10,980 So if we were asked to subnet this subnet 1 9 2 1 6 8 1 0 select 24 and create as many subnets as possible 29 00:03:11,400 --> 00:03:18,210 each having 4 hosts we would use two to the power of n where n is the number of bits that are required 30 00:03:18,600 --> 00:03:20,020 minus two. 31 00:03:20,040 --> 00:03:26,490 So if we needed to get 4 hosts we would actually need two to the power of three minus two which would 32 00:03:26,490 --> 00:03:31,500 be to to power of three which is eight minus two which means that we would be able to have six hosts 33 00:03:31,500 --> 00:03:33,440 per subnet. 34 00:03:33,510 --> 00:03:39,920 We need to subtract two because of broadcast and network address but we don't need that. 35 00:03:40,080 --> 00:03:46,590 When asked for networks two to the power of two equals four which is what we need for the number of 36 00:03:46,590 --> 00:03:49,060 subnets that we've been asked to create. 37 00:03:49,080 --> 00:03:54,020 This is the number of bits that we stealing or using for the subnet portion. 38 00:03:54,060 --> 00:03:59,780 This is the number of bits that we would keep for the host portion. 39 00:03:59,810 --> 00:04:06,470 Now we're not using that formula so I can basically scratch that from our example. 40 00:04:06,560 --> 00:04:10,930 So basically remove that to we are not using that formula. 41 00:04:11,060 --> 00:04:14,780 In this example this is the formula that we need. 42 00:04:14,780 --> 00:04:21,590 And once again that's because we've been off to four subnets we haven't been asked for hosts. 43 00:04:21,860 --> 00:04:28,310 Now two to the power of one equals t that doesn't give us enough subnets due to the power of two gives 44 00:04:28,310 --> 00:04:35,540 us four subnet two to the power of three gives us eight subnets two to the power of four would give 45 00:04:35,540 --> 00:04:41,960 us sixteen subnets we don't need that many we only need four subnets. 46 00:04:42,230 --> 00:04:50,300 So all we need to do is use two bits which means we can steal two bits from the host portion for our 47 00:04:50,300 --> 00:04:51,470 subnet portion. 48 00:04:52,220 --> 00:05:03,560 So the host portion is now only six bits in length the subnet portion is now two bits in length. 49 00:05:03,560 --> 00:05:04,790 So write it like that. 50 00:05:04,850 --> 00:05:11,570 Once again there's obviously no spaces here in binary but that hopefully just makes it easier to read. 51 00:05:11,690 --> 00:05:20,740 So we're going to steal two bits from the host portion and allocate to that to the subnet portion. 52 00:05:20,770 --> 00:05:23,880 So how many bits are part of the network and subnet. 53 00:05:24,100 --> 00:05:28,240 We've got our 8 bits from our original example. 54 00:05:28,240 --> 00:05:33,470 An additional eight but from our original example plus eight bits. 55 00:05:33,580 --> 00:05:38,040 So that's sixteen plus one plus one. 56 00:05:38,080 --> 00:05:48,010 So in other words this is now a slash twenty six subnet it's no longer slash 24 because we've stolen 57 00:05:48,040 --> 00:05:52,950 bits from the host portion for the network portion. 58 00:05:53,170 --> 00:05:58,570 So the first network that we have is 1 9 2 1 6 8 1 dot 0. 59 00:05:58,750 --> 00:06:03,530 Look at to these 8 2 binary bits to make it easier to read. 60 00:06:03,640 --> 00:06:11,230 Notice we've got eight binary bits I've just split them up with spaces here to make it easier to demonstrate 61 00:06:11,710 --> 00:06:14,150 which party's subnet in which parties host. 62 00:06:14,170 --> 00:06:21,670 But notice there are eight binary but say eight binary bits equates to zero in decimal. 63 00:06:21,670 --> 00:06:26,450 So the network is 1 9 2 1 6 8 wondered 0 slash 26. 64 00:06:26,500 --> 00:06:28,680 That's our first network. 65 00:06:28,810 --> 00:06:31,270 Now what's the second network. 66 00:06:31,270 --> 00:06:34,830 The second network is 1 and 2 1 6 8 1. 67 00:06:34,990 --> 00:06:41,350 Dot and what we do now is we just go through the different binary options. 68 00:06:41,350 --> 00:06:44,440 This is 0 0 this is 0 1. 69 00:06:44,500 --> 00:06:49,030 This one would be 1 0 and this would be 1 1. 70 00:06:49,030 --> 00:06:52,930 So those are the different binary options that we have. 71 00:06:52,930 --> 00:07:01,200 So if we'll look at this network that second binary but to set to a one what does this equal to this 72 00:07:01,210 --> 00:07:04,920 equals to sixty four in a decimal. 73 00:07:04,960 --> 00:07:08,350 So the next network is 64. 74 00:07:08,350 --> 00:07:10,180 What is is this equal. 75 00:07:10,300 --> 00:07:12,710 It looks like this. 76 00:07:12,940 --> 00:07:15,650 Which equals 128. 77 00:07:15,650 --> 00:07:24,340 Now as soon as you've worked out this second subnet you can simply do addition by that number. 78 00:07:24,340 --> 00:07:32,620 So zero plus 64 64 plus 64 is 128 plus 64 is 192. 79 00:07:32,830 --> 00:07:39,760 But if you look at the binary it's 128 plus 64 which equals 192. 80 00:07:39,760 --> 00:07:46,460 So there are are four subnets that we've been asked to work out in this example. 81 00:07:46,540 --> 00:07:51,370 So we've got four subnets the first one is for side one the second one is for the link between right 82 00:07:51,370 --> 00:07:58,060 of one an Internet router third one is for side 2 fourth one is for the link between right at 2 and 83 00:07:58,090 --> 00:08:05,480 the Internet router so the first subnet would be 1 0 2 1 6 8 wondered slash 26. 84 00:08:05,800 --> 00:08:08,540 And I'll zoom in here to make it clearer. 85 00:08:08,920 --> 00:08:11,190 So that's our first subnet. 86 00:08:11,350 --> 00:08:21,810 Per our calculations second subnet is 1 9 2 1 6 8 1 64 slash 26. 87 00:08:21,840 --> 00:08:27,870 Third one is 1 9 2 1 6 8 1 128 slash 26. 88 00:08:27,870 --> 00:08:38,320 And the last one is 1 9 2 1 6 8 1 192 flash 26 so those subnets are these subnets that we worked out 89 00:08:39,540 --> 00:08:46,700 so we need to now configure the network with this information in the first step we need to work out 90 00:08:46,700 --> 00:08:53,760 to the subnets but now we need to configure the route as per the instructions so as an example the lost 91 00:08:53,790 --> 00:08:58,950 IP in the subnet should be configured on rather one second lost IP address should be configured on the 92 00:08:58,950 --> 00:09:07,390 switch third lost IP address should be configured on the DHB server per these instructions we also need 93 00:09:07,390 --> 00:09:15,150 to configure a DHB pool on the DHB server so I'm going to start with the subject first and get that 94 00:09:15,160 --> 00:09:18,040 working and then I'll move to the other subnets. 10068

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