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NetAcad ITN Packet Tracer 11.5.5: Subnet an IPv4 Network

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Summary

In this tutorial on Packet Tracer 11.5.5, the focus is on subnetting an IPv4 network by transforming a single network into multiple subnets based on host requirements. Starting with a network address of 192.168.0.0/24, users will learn to create effective subnetting schemes, calculating needed hosts, and mastering subnet masks. The tutorial walks through configuration steps for PCs, switches, and router interfaces, utilizing methods like Ping to test connectivity. Attendees will determine appropriate subnet masks to accommodate host and subnet needs, making the network efficient and scalable.

Highlights

Subnetting transforms one large network into manageable parts, improving organization🌍.
Choosing the right subnet mask is crucial for meeting host and subnet requirements👌.
Efficient subnetting practices enhance network scalability and address resourcefulness🔧.

Key Takeaways

Subnetting divides a network into smaller, logical subnets optimizing IP address usage.
IPv4 scarcity demands strategic subnetting for efficient address management🚀.
Subnet masks change network and host portions of an IP address for subnet division📊.

Overview

Subnetting, a necessary skill, is essential for efficiently managing network resources. By partitioning a network into smaller subnetworks, you can ensure optimal usage of IP addresses while maintaining connectivity across different nodes. The process outlined in the tutorial includes determining the necessary number of hosts and using bit borrowing to calculate an effective subnetting strategy.

Beginning with a network address of 192.168.0.0/24, the participants explore how to effectively divide a network into smaller subnets. The focus is laid on accommodating various sized LANs and accounting for future growth by leaving room for additional subnets. The tutorial explains how to calculate and apply subnet masks such as /26, which meet the needs for multiple subnets of varying sizes.

Through hands-on configuration of switches and routers, and testing connectivity with Ping commands, the tutorial conveys the importance of setting up each network component correctly. It emphasizes the critical role subnet masks play in network efficiency and the significance of maintaining an accessible and expandable network design, highlighting common pitfalls and solutions.

Chapters

00:00 - 00:30: Introduction In this chapter titled 'Introduction,' the focus is on subnetting a customer network into multiple subnets using Packet Tracer 1155. The process involves creating a subnetting scheme based on the number of host computers required in each subnet, filling in a table with addressing information, configuring the post PCs, switches, and router interfaces, and using the Ping command to test network connectivity. Part one emphasizes creating a subnetting scheme to meet the required number of subnets.
00:30 - 03:00: Subnetting Scheme Creation The chapter focuses on creating a subnetting scheme using the network address 192.168.0.0/24. The requirements include accommodating at least 50 hosts on LAN A, 40 hosts on LAN B, and allowing for at least two additional unused subnets.
03:00 - 09:00: Subnet Mask Calculation The chapter titled 'Subnet Mask Calculation' dives into the topic of subnetting with a focus on consistent subnet mask lengths across all subnets, rather than using variable length subnet masks. It starts with identifying the number of host addresses necessary for the largest subnet, specifically noting that 50 hosts are required in the largest subnet, Lan A.
09:00 - 15:00: Subnet Assignment and IP Configuration The chapter discusses the need for subnetting in a network, specifically addressing the requirement for four subnets. It includes a subnet for LAN A, LAN B, and two additional subnets for future expansion. The chapter further explains how to subnet a given network, starting from the IP address 192.168.0.0 with a /24 subnet mask, and describes what the /24 subnet mask represents in binary.
15:00 - 20:00: Configuring the Network Devices The chapter 'Configuring the Network Devices' discusses the significance of the first 24 bits in a network mask. It explains that these bits are denoted by ones in the first three octets and zeros in the fourth octet. The ones signify the network portion of the address, while the zeros denote the portion of the address designated for hosts.
20:00 - 25:00: Testing Network Connectivity The chapter 'Testing Network Connectivity' delves into the details of subnetting a network. It explains the process of borrowing bits from the host portion to create subnet bits. The narrative outlines how the number of borrowed bits determines the number of possible subnets. Furthermore, the chapter prepares the reader to calculate the number of subnets and hosts based on given subnet masks, emphasizing practical application in networking.

NetAcad ITN Packet Tracer 11.5.5: Subnet an IPv4 Network Transcription

00:00 - 00:30 in Packet Tracer 1155 we are going to subnet a customer Network into multiple subnets based on the number of host computers required in each subnet we will fill in the table with all of the addressing information configure the poost PCS switches and router interfaces and then use the Ping command to test for network connectivity in part one we are to create a subnetting scheme to meet the required number of subnets and the
00:30 - 01:00 required number of host addresses we have been given a network address of 192.168.0 this is a sl24 and we are to use that to meet the following requirements the l a network is going to need at least 50 hosts the Lan B Lan is going to need at least 40 host IP addresses the customer wants at least two more subnets that are not in use for
01:00 - 01:30 future growth we will be using the same length of subnet masks for all of these subnets we're not going to be doing variable length subnet masking first we have to answer some questions how many host addresses are needed in the largest required subnet so the largest number of hosts needed are 50 hosts in Lan a and I'm just going to grab the document here the next question is is what the minimum
01:30 - 02:00 number of not host addresses but subnets so we've been asked for one subnet for Lan a one for Lan B and two more for future expansion so that comes to a total of four subnets that will be required to subnet out the network we are asked to subnet from is 192.168.0.0 sl24 what is the sl24 subnet mask in binary the sl24 means that the
02:00 - 02:30 first 24 bits are ones so the correct mask would be eight ones in the first three octets and all zeros in the fourth octet in the network mask what do the ones represent and what do the zeros represent the ones represent the portion of the address that is the network portion the zeros represent present the
02:30 - 03:00 portion that is the host portion in F to subnet a network borrow bits from the host portion and change those into subnet bits the number of borrowed defines the number of subnets it will yield in the next section we're going to be given possible subnet masks and ask to figure out how many subnets and how many hosts each would give us
03:00 - 03:30 remember that the number of host bits to the power of two defines the number of hosts per subnet if we have one bit borrowed from the host portion that's going to determine how many subnets cuz we would take two to that power 2 to the first power is two so that's how many subnets it would give us to borrow one bit from the host bits if we just took the network as a/4
03:30 - 04:00 that would be eight host bits which would be 2 to the 8 is 256 total hosts minus 2 would be 254 if we didn't do any subnetting if there's no requirement for that many hosts ip4 address space is scarce so that's why we are wanting to subnet down not only make our broadcast domains as small as possible but also to make our address space go as far as possible we don't need 254 usable host addresses we
04:00 - 04:30 have a requirement that we're going to choose how we want to subnet this starting with the sl25 where we borrow one bit what would be the dotted decimal subnet mask equivalent so we would have a sl24 plus in that last octet we're taking the very first bit here which is the most significant bit this position
04:30 - 05:00 is the 128 position so that's going to be a 128 in the fourth octet how many subnets will that give us 2 to the 1st power is two subnets and how many hosts are going to be in each of those two subnets to calculate that we look at what's remaining seven host bits so we take 2 to the 7th power which gives us 28 but we have to subtract two because
05:00 - 05:30 the first goes to the network and the last the highest goes to the broadcast address so that leaves us 126 usable hosts so the lower the number of subnets the higher the number of hosts you're going to get we are looking at 126 usable hosts a slash 26 in the next scenario 255 255.255 do then we look at this is 128
05:30 - 06:00 this position is 64 you add those together and you get 192 in the fourth octet there how many subnets would that give us borrowing two bits from the hosts is going to give us 2 to the second power is four subnets how many hosts we have six post bits left so we take 2 to the 6 is 64 subtract two to get
06:00 - 06:30 62 what about if we were to use make a sl27 as our prefix length our subnet mask that fourth octet this is the 32 position so we're adding that to the 192 that we already had that gives us 224 that's going to give us 2 to the 3 power cuz we're borrowing three bits so that's going to be eight subnet
06:30 - 07:00 and each of those eight would have 2 to the 5th power minus 2 is 30 hosts sl28 is going to use the 16 position bit so we would add that to our 224 is 240 is our mask and that will give us 2 to the 4th power is 16 subnets and we have four host bits so 2 to the 4th power we know is 16 but we
07:00 - 07:30 subtract two to get 14 if we use a slash 29 our mask now we've got the eight position bit set as a network bit so that's going to be 248 in that last doet we have five borrowed host bits to make them Network bits so 2 to the 5ifth power is 32 subnets so the higher the number of subnets the fewer hosts per subnet so we have three host bits left 2 to the 3 power is 8 - 2 is 6
07:30 - 08:00 hosts per subnet that'd be pretty small and finally the sl30 the subnet mask again we are taking the four position bit so we're adding four to the previous mask that gives us 252 2 to the 6th power gives us 64 subnets but each one has only only two
08:00 - 08:30 usable host addresses considering your answers which subnet masks meet the required number of minimum host addresses so we have one land that needs 50 and one land that needs 40 which one of these would give us at least 40 hosts that would be anything above the slash 26 so the sl25 and the slash 26
08:30 - 09:00 both would provide enough host and how many of those masks meets the minimum number of subnets well we only need four so pretty much all of them except for the sl25 are going to meet that requirement of at least four subnets so now that we figured all that out what subnet mask should we use which one meets both the required minimum
09:00 - 09:30 number of hosts and the minimum number of subnets required at least 40 hosts and at least four subnets so the best one that would meet that requirement would be sl26 right it'll give us four subnets and 62 hosts per subnet so the SL 26 is
09:30 - 10:00 the subnet mask we're going to need to use here and the mask for that would be 25. 255.255 192 in the fourth octet now that we've determined which subnet mask meets all of our Network requirements we're going to break it down into the subnets for each lay but instead of that sl24 it's our chosen subnet mask the sl26 the prefix is going to be the same
10:00 - 10:30 because we're using the sl26 for all of the subnets so that will go in the prefix column and the subnet mask 255.255.255.192 that is what the subnet mask is for a sl26 we know that the first Subnet is going to be 192.168.0.0 and I'm going to add some notes in here to make it easier to address
10:30 - 11:00 the interfaces and things later I want to put down what are my usable host addresses the first do0 is reserved for the network so my first usable host address is going to be one my last usable host address will be the one before the broadcast address so one nice thing to know is it's something called the magic number that you add to the network address add to get the next
11:00 - 11:30 Network address that's going to be the number of hosts possible in the sl26 there are 64 possible hosts two to the 6 power before we subtract those two reserved addresses that 64 is what you can call your magic number that you can use when you're adding to get your next subnet so the do 64 then would be the next subnet address 0 Plus 64 which
11:30 - 12:00 means that your broadcast address is going to be 63 for this subnet and the last usable host address is going to be dot 62 and making notes like this is really helpful so for my next subnet I know that's going to be 192.168.0 64 network and on that Network the first usable host will be 65 again we're going to be adding 64 to
12:00 - 12:30 get the next Network address the broadcast address of this network is going to be1 127 and the last usable host will be 1226 so our third subnet 192.168.0 do 64 + 64 is 128 and our fourth subnet is going to be 128 + 64 192
12:30 - 13:00 usable hosts will be1 129 through 190 with a broadcast address of 191 and in that last subnet our usable hosts will start with 193 because 192 is the network address and 192 + 64 would then give us 256 so the broadcast address 255 and our last usable host address
13:00 - 13:30 would be 254 so hopefully this is starting to make sense now that we have that mapped out we're going to assign the first Subnet to Lan a so I'm just going to add you don't have to do this I just find it's really helpful the second subnet is going to go to landan B and the other two subnets are going to be for future use so our first host address is going to be for the customer router interface that's connected to the land a
13:30 - 14:00 switch and we can fill in our IP addressing table so our first host address for Lan a 192.168.0.1 is going to go on the interface on the customer router that's facing Lan a so this interface right here which is gig 0000 and we know our subnet mask by now we're not going to be configuring a default gateway for the custom router
14:00 - 14:30 the second host address is going to go on the Lan a switch and that's going to be2 Lan a switch VLAN one same mask and the default gateway is going to be the customer router IP address that's in that subnet this is also going to be the default gateway for PCA so I'm going to go ahead and fill that in that is supposed to be the subnet MK
14:30 - 15:00 they're all going to be using that same subnet mask the IP address for PCA is going to be the last possible host address in that range which I can go back in my notes I can just grab that it's going to be the 62 and now that takes us to for Lan B we're going to use the first host address for the router interface the second host address for the switch and
15:00 - 15:30 the last host address for PCB okay so in that second subnet the network that begins with 64 the first host address is 65 and that goes on the router on the interface facing N Land and that will be the default gateway as well for both the switch and the PC the second usable ad um IP address then would be6 6 which is what goes on this switch
15:30 - 16:00 virtual land interface and the last usable host address is the 126 and that's going to go on PCB and that takes care of the addressing table now we can start configuring so grab the customer router go to the command line interface enable for privileged exact
16:00 - 16:30 configuration mode con t for short first set the enable secret password to class 123 enable secret because we're going to encrypt that and then the actual password is class one two 3 with a Capital C make sure it looks just like that so your grading works next set the console login password does Cisco 123
16:30 - 17:00 and that dot there that's the period at the end of the sentence the password won't have a dot in it so we go line console 0 password uppercase C Isco one two 3 remember the command that always goes hand inand with password is login for password checking next host name you can do the host name command from line
17:00 - 17:30 configuration mode and customer router uppercase C uppercase R is what the grading will check for the prompt has changed to reflect the new host name now we're going to put our IP addresses on our interfaces looking back at our table g0000 so if I say do show IP interface brief we're going to have to also administrate enable them with the no
17:30 - 18:00 shout command because they are disabled interface G 0/0 IP address 19216801 followed by the subnet mask which is the sl26 1992 there and of course no shut to enable see a message that it's coming up interface g01 IP address is going to be
18:00 - 18:30 0.65 with the same network mask and no shut fix my document here and get that mask filled in there we go show IP interface brief now shows up and addressed then save the running configuration copy run for
18:30 - 19:00 short and that looks like all we need to do on the customer router now we're going to go over to the switches because they need IP addresses too so that they can be managed l a switch command line interface enable conf T if I do do show IP interface brief VLAN 1 is currently un addressed and administratively
19:00 - 19:30 disabled but not for long l a switch has the address sorry about that that one is supposed to be two which is the second usable IP address in that subnet select an interface VLAN one IP address is just paste that in and my mask and remember to do the no shut we also have to tell this switch
19:30 - 20:00 who is its default gateway so that it can communicate or be communicated with with other networks and that is going to be IP default gateway and that will be the one interface on the router there and at this point I like to see make sure I can ping the default gateway so like if I show my ARP table right now I've got an entry here for the switch's IP address as well as its Mac address and if I ping the default gateway make sure I can everything's
20:00 - 20:30 working and I can reach the router that first ping failed because it was doing the ARP request because it didn't know the MAC address so now if you check the ARP table you see there's the default gateway router interface address and its corresponding Mac address so the next time that it goes to Ping it won't have to ARP because it already has that information in the table the other switch now will need to be at address which is the Lan B
20:30 - 21:00 switch enable con T interface VLAN one IP address for VLAN one on the Lan B switch is going to be 66 the second usable address in that. 64 Network
21:00 - 21:30 and no shut and it's default gateway IP default gateway 192.168.0 65 and I just want to make sure it can ping its default gateway yes now the PCS have to be inter configured with their information so PC a and Lan a that one
21:30 - 22:00 got the last usable host address in the subnet so we go to desktop IP configuration 1921 168 0.62 well this is going to be a static address the subnet mask will autop populate but it's going to be a classful mask we've got to change that cuz we're subnetting that's going to be92 there and make sure to put in the default gateway information which is going to be
22:00 - 22:30 0.1 the customer router is the default gateway and over on [Music] PCB 126 again fix that default mask there and its default gateway is g01 on customer router which is
22:30 - 23:00 65 okay and finally use the Ping command to test network connectivity PCA should be able to communicate with its default gateway PCB should be able to Ping its Gateway and the two then should be able to Ping each other get our addresses here what we can see
23:00 - 23:30 them so PCA is going to want to ping one as its default gateway and that is successful PCB is going to want to be able to Ping its default gateway and if you forget you can do a IP
23:30 - 24:00 config and that'll tell you who the default gateway is and that is successful so if everything is working end to end then these two PCS will be able to Ping each other so PCB ping PCA
24:00 - 24:30 and that is also successful so everything is looking good these are all going to be [Music] yeses if they're not yeses go back and check all of your IP and subnet mask configurations default gateways also you should see 100% completion if not hit the check results button and look at the assessment items
24:30 - 25:00 and look for red X's once you're done with everything make sure you save it thank you