Linux+ LX104 help

espi_251

Hi everyone,

I recently took my LX103 and passed it with 690, it was not an easy exam at all, but I studied my butt off and made hundreds of flashcard to help me memorize the command switches.

I've been preparing for the LX104 for a month now, and going to take it next week, but I need help in understanding one of the questions that I came across in one of the books. Any help would be greatly appreciated.

Questions below along with possible answers:
If your IP is 192.168.33.35 and your network mask is /28, what is the address of your local network?
a. 192.168.33.16
b. 192.168.33.32
c. 192.168.33.0
d. 192.168.33.64

Thank you,

clarson

ipv4 is a 32 bit number. /28 means the first 28 bits are the network and the last 4 bits are for the host address.
the first 3 octets are in the network address. so 192.168.33 is given
and it is pretty easy to convert 35 to binary. basically it is 32 plus 3. (that also gives you the last octet as 32, ie the 3 is the host address).
so there are 0 128's, 0 64's, 1 32, 0 16, 0 8, 0 4's, 1 2 and 1 1. or
0010 0011
zero the last four bits because the network address has zero for all the host bits.
0010 0000 that is the last network octet in binary or 32 decimal
so the network address is 192.168.33.32

espi_251

Clarson, you made it look too easy!

Thank you that makes perfect sense and so much easier to understand than the methods in the book.

Joedoc164

Can you explain how you broke the networks down? That seems a lot faster than my method of writing them all out. Writing a /30 out sucks when you are asked for the network range on a test and given a last octet in the 200's

si20

Sorry to sound stupid here, where did the 35 come from? The only way I saw you made 35 was 32+3 and I don't understand how you got those two digits to add together.

thomas_

The 35 comes from the last octet of the ip address that was originally given. 32 is the subnet that the ipaddress resides in. I think he was just trying to say that 35 is really easy to convert to binary since it's easy to know which binary digits represent 32.

I do a similar thing as clarson to break down the networks. One thing clarson didn't mention is that each network address is going to increase by multiples of 16 in this example. 0 will be the first subnet, 16 the second subnet, 32 the third subnet, 48 the third, etc. You know 32 is the subnet because 35 falls between 32 and 47. 32 and 47 aren't usable because they are the subnet id(network address) and broadcast address respectively.

To find the networK WHEN i'm given the CIDR notation I tend to find the highest multiple of 8 which tells me which octet I'm working in and then from there find the value of the last bit when looking at the most significant bit direction.

/25 means I'm working in the last octet. the remaining bit is 128 so from there I know that there will be two possible networks the address can fall into, either the 0 or the 128 network.

/26 I'm working in the last octet. Now the two bits that can be used in the last octet as part of the network address are the 128 and 64. There are going to be four possible networks that increment by 64: 0, 64, 128, 192.

/27 I'll be working in the last octet. The three bits that are part of the network are the 128, 64, and 32. Here each network will increase by 32.

/28 same octet, but now there are four bits that can be used as part of the network the 128, 64, 32, and 16. Each network will increment by 16.

clarson

you can do just like thomas_ pointed out. given the number of host bits, you can figure out the size of each subnet. Such as the in the original question. 4 host bits equals a subnet size of 16. divide 35 by 16. the equal 2 with a remainder of 3. 2x16 = 32 which is the network address and the remainder of 3 is the host adress.

you do the same thing with a /30. number of host bits is 2 that means the subnet size is 4. and let say the 4th octet was 235. divide 235 by 4 that equals 58 with a remainder of 3. 58 x 4 = 232 with remainder of 3. so the network address is 232. but the 3 isn't the host address because it is actually the broadcast address for that subnet. for a subnet size of 4, 3 makes the host portion all 1's which is the broadcast address.

a remainder that is one less than the subnet size is a broadcast address. such as the original question. we had a subnet size of 16. so take any multiple of 16 then add 15 and you have a broadcast address. first network, 0 x 16 + 15 = 15 is the broadcast address of the first subnet. 1 x 16 + 15 = 31 is the broadcast address of the 2nd subnet.

and if you want really get into the numbers. ((the nth subnet - 1) x subnet size) + (subnet size - 1 ) equals broadcast address of the nth subnet. so what you are doing is working backwards. dividing the number by the subnet size gives you the (nth subnet -1 ) and the remainder is the host address or if it is the subnet size - 1, it is the broadcast address.

so in the original question the address was 35. subnet size is 16. divide 35 by 16 equals 2 plus the remainder. so 2 equals the nth subnet - 1. so 32 is the 3rd /28 subnet. and the remainder of 3 is the host address because it isn't 15.

or if you wanted the 5th host of the 7th subnet of /29. /29 has a subnet size of 8. so it would be. ((7-1) x8 + 5
48 + 5 or 53. you want to know the network. it would be 7 -1 x 8 = 48 or working backwards, take 53/8 = 6 + a remainder. 6x8 = 48. that is the network address. 48/8 = 6 add one 6+1 = 7 so it is the 7th subnet.
the remainder is 5. the broadcast address would have a remainder of 8-1 or 7. 5 isn't 7 so it is the 5th host on the 7th /29 subnet.

I probably have you all confused now. work through some number and see how it works for you.

espi_251

si20 wrote: »

Sorry to sound stupid here, where did the 35 come from? The only way I saw you made 35 was 32+3 and I don't understand how you got those two digits to add together.

Si20, I know you are planning to take the LX104 in the near future. The last post Clarson posted definitely confused me, but then again I'm new to IT certs.

I would recommend you watch the YouTube video, link below, and use his slide rule to answer some of the practice questions. It has been somewhat helpful to me, but I need to practice a lot more to feel comfortable with subnetting.

https://www.youtube.com/watch?v=GSX1GlaznKM

After you've watched the video and managed to stay awake somehow, try the question below and see if you can answer it using the slide rule.

Your boss wants you to create a subnet scheme that gives your company eight
networks with at least 10 hosts per network. Which subnet mask for a Class C
leased network address meets those objectives?
a. 255.255.255.192
b. 255.255.255.224
c. 255.255.255.240
d. 255.255.255.248

Best of luck with your LX104.

Joedoc164

Clarson, I don't know how I didn't see it that way before. I'll leave the formula for the times I'm not taking a timed test, there is always enough to remember as it is. Thanks.

si20

Thanks epsi I really appreciate it. Networking like this is definitely my weakness because i've never really had to or needed to do it in the workplace. I will definitely check the video and make notes as I go along. It could be useful for the LX0-104!

Andy from Sactown

If your IP is 192.168.33.35 and your network mask is /28, what is the address of your local network?

a. 192.168.33.16
b. 192.168.33.32
c. 192.168.33.0
d. 192.168.33.64

/28 CIDR = 255.255.255.240 = 11111111.11111111.11111111.11110000

So you're using the 4 high order bits from the last octet as part of the network address, leaving the 4 low order bits for host addresses. 2^4 = 16 addresses per IP block (sub-network) (16-2= 14 hosts per block after subtracting the network and broadcast addresses) and also 2^4 = 16 IP blocks.

So you're looking at IP blocks (sub-networks) like this:

======BLOCK 1========
192.168.33.0 network
192.168.33.1-14 hosts
192.168.33.15 broadcast
======BLOCK 2========
192.168.33.16 network
192.168.33.17-30 hosts
192.168.33.31 broadcast
======BLOCK 3========
192.168.33.32 network
192.168.33.33-46 hosts
192.168.33.47 broadcast
...
======BLOCK 16=======
192.168.33.240 network
192.168.33.241-254 hosts
192.168.33.255 broadcast

You can clearly see that the 192.168.33.35 host address resides in block 3, with a network address (answer) of: 192.168.33.32



Now that's the explanation. If I got this on a test, I'd look at the /28 and say to myself, "okay so 4 host bits.. that's 16.. so networks at 0, 16, 32, 48.. I'm looking for .. 35. Okay, which is 32-- gotcha, B."

But these are low-hanging fruit, sub-netting the last octet and asking for low (numerically) host addresses. If they asked for 192.168.33.221/28, I'd probably do as clarson suggested and visualize the last octet as:

net | host
11110000 mask
11011101 last octet of address
========== Logical AND ==
11010000 last part of network address = 208 decimal, for a subnetted network address of 192.168.33.208

Which is easier than counting 16, 32, 48, 64, 80, 96, 108, 126, 144, 160, 176, 194, 208, but you can still do it, especially if you're super smart and start at 16x10 = 160... but this quickly becomes unreasonable as the number of hosts increases. It's just the method I use mostly out of habit for small stuff. Visualizing the netmask / address interactions in binary is really the way to go.


Consider the exact same question, just prose with a smaller netmask / larger host addressing space:

If your IP is 10.94.161.234 and your network mask is /11, what is the address of your local network?

[No A,B,C & D] We're going to find out exactly!

/11 CIDR = 255.224.0.0 netmask = 11111111.11100000.00000.00000

Now doing the exact same thing we did before still applies:

---

net

---

|

---

host

---

11111111.11100000.00000000.00000000 - Netmask (255.254.0.0)
00001010.01011110.10100001.11101010 - IP Address (10.94.161.234)
========================== Logical AND ==
00001010.01000000.00000000.00000000 Network Address = 10.64.0.0

Which is a hell (!) of a lot easier than trying to count in intervals of 2^21 host bits = 2,097,152 IP blocks distributed across 3 byte sized octets (not even going to touch on how one would begin to do that).

Here's a follow-up question to keep you in the binary IP representation mindset.. what is the broadcast address from the previous example (10.64.0.0/11 network)? More importantly.. how would you find it, super fast?!

si20

I'll be honest - this seems to be way over my head. I kinda understand, but if I got this on an exam, i'd definitely get it wrong. From the practice tests i've sat from the Robb Tracy book, the networking questions *seemed* to be much easier than this. Some of them being too easy e.g what runs on port 21? FTP. Obviously there are questions much harder than that, but you get my drift.

Andy - I bow down to your knowledge. I know when i've been defeated and this is it. Networking is definitely not my strong subject and I know it.

Andy from Sactown

si20 wrote: »

I'll be honest - this seems to be way over my head. I kinda understand, but if I got this on an exam, i'd definitely get it wrong. From the practice tests i've sat from the Robb Tracy book, the networking questions *seemed* to be much easier than this. Some of them being too easy e.g what runs on port 21? FTP. Obviously there are questions much harder than that, but you get my drift.

Andy - I bow down to your knowledge. I know when i've been defeated and this is it. Networking is definitely not my strong subject and I know it.

Boooo! Ports questions shouldn't be that generic and easy. FTP runs on tcp/20 for data channel and tcp/21 for initial control channel.

si20, if you actually take the time to break down subnetting, I think you'll find it's not way over your head [capabilities] and is really rewarding to master something that "seems" overly technical, but isn't really that bad.

espi_251

Hey guys,

I passed the lx104 today!Whew!! It was not an easy exam at all, much harder than 103 for me. I had more than 15 fill in the blanks if not more. Some were fairly easy but others were extremely difficult.

Si20, couple of tips for you-- make sure you have a good understand of the X server, and detailed understanding of configuration files listed in the exam objectives.

Also, you must know how to subnet and feel comfortable finding number of hosts.

I wish you all the best.

si20

espi_251 wrote: »

Hey guys,

I passed the lx104 today!Whew!! It was not an easy exam at all, much harder than 103 for me. I had more than 15 fill in the blanks if not more. Some were fairly easy but others were extremely difficult.

Si20, couple of tips for you-- make sure you have a good understand of the X server, and detailed understanding of configuration files listed in the exam objectives.

Also, you must know how to subnet and feel comfortable finding number of hosts.

I wish you all the best.

Congrats espi !! Wow. it sounds like an absolutely horrific exam. I've yet to hear someone say it was easy. I'll go through the chapter on X server again. I'm actually shocked it has much on the X server because it's only worth approx 8% of the exam isn't it? 15+ fill in the blanks!?! Wow! That is shocking. I only had 4-5 of them on the LX0-103. The LX0-104 sounds like a beast. Really well done matey!

espi_251

I was more shocked when I saw that many fill in the blanks. It felt like every other question was a fill in the blank.

That was my understanding as well and that's why I didn't spend too much time studying for it. But yeah it was really surprising to see that many question on one chapter or topic.

You'll obviously have different exam so keep doing what you are doing and you'll be fine.

si20

Thanks! I've just spent the last 1.5 hours working on subnetting, so if I get asked "how many valid hosts are in this subnet" I should be able to answer with confidence! The thing i'm worried about is the sheer scale of questions. X Server alone is a big subject, but Security, Networking, System Essentials etc. The Linux+ is one hard cert. Congrats once again! I'll let you know how it goes on Tuesday..