Subnetting practice problems 1

[Cluebird]

Practice your subnetting skills with the following:

1. You have a router with 9 interfaces that need IP addresses. You are assigned 195.33.44.0. Construct a complete addressing plan for your router. Use FLSM only.

2. You have two hosts that are unable to connect with each other. On one host you see the ip address 192.168.100.33 255.255.255.240. On the other host you see 192.168.100.28 255.255.255.240. What is the problem?

3. Your administrator tells you to address a host with the following address: 172.16.2.255 255.255.252.0. Will this work? Why or why not?

4. What mask will give me 18 hosts with max number of networks if I start out with 192.168.255.0?

5. What class is the following mask: 255.255.255.0?
A. Class C
B. Class B
C. Class G
D. Not enough information to decide

6. Your server administrator wants to use the 172.168.200.127 255.255.255.128 address on his Exchange server. What is your response?
A. No problem
B. Won't work because that is a local broadcast
C. Won't work because that is a network address
D. Won't work because that is a directed broadcast
E. None of the above

Enjoy

 

[bobbyforhire]

Question:
You have a router with 9 interfaces that need IP addresses. You are assigned 195.33.44.0. Construct a complete addressing plan for your router. Use FLSM only.

My Answer:

- FLSM (Fixed Lenght Subnet Mask) - From what i can remmeber a FLSM is something like a class A,B,or C. so you are wanting a fixed subnet (if it's a class C standard don't change it, leave it at 255.255.255.0)

195 = Class C
255.255.255.0
Inc of 1, the inc is in the 3rd octet

1 195.33.44.0 - 195.33.44.255
2 195.33.45.0 - 195.33.45.255
3 195.33.46.0 - 195.33.46.255
4 195.33.47.0 - 195.33.47.255
-So on-

Question:
You have two hosts that are unable to connect with each other. On one host you see the ip address 192.168.100.33 255.255.255.240. On the other host you see 192.168.100.28 255.255.255.240. What is the problem?

128 64 32 16 8 4 2 1
1 1 1 1

192.168.100.0 - 15
16 - 31
32 - 47

Answer:
The users are on a diffrent network, put them in the same subnet and your issue should resolve.


Question:
Your administrator tells you to address a host with the following address: 172.16.2.255 255.255.252.0. Will this work? Why or why not?

Answer:

No, 172.16.2.255 is the Broadcast for 172.16.2.0 network.

I do remember that there was a clause in witch you could use a .0 and a .255 but i don't think your question reached the clause...i have to look over again what had to happen for .0 and .255 to be useable.



Question:
What mask will give me 18 hosts with max number of networks if I start out with 192.168.255.0?

Answer:
There is that Max Networks again...damn you!


/27 255.255.255.192


192.168.255.0 - 31 = 28 hosts
192.168.255.32 - 63
192.168.255.64 - 95
192.168.255.96
this will give you a max of 28 hosts and max networks.



Question : 255.255.255.0 = what class
Answer: Class C address


Question: Your server administrator wants to use the 172.168.200.127 255.255.255.128 address on his Exchange server. What is your response?
A. No problem
B. Won't work because that is a local broadcast
C. Won't work because that is a network address
D. Won't work because that is a directed broadcast
E. None of the above


The answer is B, 172.168.200.127 is the Local Broadcast for 172.168.200.0 Becuase you are using a /25 subnet


172.168.200.0 - 127
172.168.200.128- 255


Hopefully i'm getting better at this.. letme know teach!

 

[bobbyforhire]

Ugh, number 5 gives you 29 useable not 28...

****READ IT TWICE******

 

[maczen]

??? Number 5?

5. What class is the following mask: 255.255.255.0?
A. Class C
B. Class B
C. Class G
D. Not enough information to decide

You got that right..

Actually, I believe that you got them all right..

B Haines
CCNA R&S, ETA FOI

 

[Cluebird]

Answers and explanations:

1. You have a router with 9 interfaces that need IP addresses. You are assigned 195.33.44.0. Construct a complete addressing plan for your router. Use FLSM only.

Answer: This is a Class C that must be subnetted. You cannot change the numbers in the third octet, you must make more networks using the fourth octet. New mask is 255.255.255.240 (look familiar?)

Networks are in blocks of 16 addresses, 14 usable hosts. Therefore, you can choose any nine of the available 16 new networks

195.33.44.0-15
195.33.44.16-31
195.33.44.32-47
195.33.44.48-63
195.33.44.64-79
195.33.44.80-95
195.33.44.96-111
195.33.44.112-127
195.33.44.128-143
195.33.44.144-159
195.33.44.160-175
195.33.44.176-191
195.33.44.192-207
195.33.44.208-223
195.33.44.224-239
195.33.44.240-255

2. You have two hosts that are unable to connect with each other. On one host you see the ip address 192.168.100.33 255.255.255.240. On the other host you see 192.168.100.28 255.255.255.240. What is the problem?

Answer: They are on different networks. The first is on the network from 192.168.100.32-47/28 while the second is on the network from 192.168.100.16-31/28.

3. Your administrator tells you to address a host with the following address: 172.16.2.255 255.255.252.0. Will this work? Why or why not?

Answer: This will work fine. This is a host address on the network 172.16.0.0-172.16.3.255/22

4. What mask will give me 18 hosts with max number of networks if I start out with 192.168.255.0?

Answer: 255.255.255.224 or /27

5. What class is the following mask: 255.255.255.0?
A. Class C
B. Class B
C. Class G
D. Not enough information to decide

Answer: D Without the IP address information you can't tell if this mask represents a subnetted Class A, subnetted Class B, or a Class C at its starting point. The mask only tells you how to use the address space. In this case, the mask shows 24 bits identifying networks and 8 bits identifying hosts.

6. Your server administrator wants to use the 172.168.200.127 255.255.255.128 address on his Exchange server. What is your response?
A. No problem
B. Won't work because that is a local broadcast
C. Won't work because that is a network address
D. Won't work because that is a directed broadcast
E. None of the above

Answer: D This question is not just testing subnetting but also terminology. The local broadcast is always 255.255.255.255 while the directed broadcast is the last address in the subnetwork. In this case, the address is a directed broadcast. The difference between the two addresses is that routers can be configured to forward directed broadcasts using ip helper addresses while the local broadcast can never be forwarded.

Your results....
1. Wrong. Can't touch third octet
2. Correct
3. Wrong.
4. Wrong...255.255.255.224 is /27. You got the mask correct in prefix notation but wrong in dotted decimal. Your blocks are correct but since Cisco is all or nothing, you get nothing.
5. Wrong
6. Wrong

Do you understand what your errors are based on the explanations I've given?

Need more? Wax on, wax off...

 

[bobbyforhire]

Sorry i was meaning..the question with 18 hosts... it would have gotten an avail 29 not 28

 

[bobbyforhire]

Ok, So i suck..and need to take this course back to formula.

 

[bobbyforhire]

More please....

What have i learned so far.

local broadcast (255.255.255.255)
directed broadcast X.Y.Z.255

You tossed in a few tricks there..good job. I hope i'm not wasting your time but i think im getting somewhat better

 

[maczen]

By the way, not to confuse you but one of the big advantages of the /31 that was discussed in another thread is the elimination of directed broadcasts..

B Haines
CCNA R&S, ETA FOI

 

[bobbyforhire]

so if i use 255.255.255.254 that would eleminate a direct broadcast? Is that because you only have 1 host bit to use and because you only have 1 host bit that you would have to use the direct broadcast address?

 

[maczen]

This is outside of the scope of CCNA and I just learned myself thanks to this forum but if you have x.x.x.254 or /31 you essentially have two addresses. On Cisco equipment if a /31 is used then there is no network address or broadcast address. Both are treated as host addresses. Note that this can only be implemented on a PPP connection. The lack of a broadcast address eliminates the possibility of a directed broadcast which improves security by eliminating DDoS attacks like SMURF. The best part of it is that this is supported by all of your classless routing prtocols as they multicast updates so no broadcast address is necessary so EIGRP, IS IS, OSPF and BGP are all supported.

Did I get that right guys?

B Haines
CCNA R&S, ETA FOI

 

[Cluebird]

I'm asking you questions at the level of difficulty you might expect on an exam. If you're learning by making mistakes you're doing great! Mistakes are how we learn. Don't get discouraged, though. You're doing fine.

A few more to try:

1. A network administrator wants to subnet his address space using FLSM techniques. He has 45 hosts on each of five networks. He is using 194.44.55.0. What subnet mask would you suggest?
A. 255.255.255.224
B. 255.255.255.192
C. Can't be done
D. 255.255.255.128

2. Your Windows server administrator wants to implement NAT for a network using the following scope of addresses: 202.92.233.46-202.92.233.61 255.255.255.240 Will this work? Why or why not?

3. You administer a small branch office with 15 hosts. Your central administrator sends you an e-mail with your addresses which are 234.55.33.0 255.255.255.224. Will you have any issues using this address space? Why or why not.

4. You receive a Cisco router from your local reseller. It is a 2811 with a WIC-2T installed. Based on the total number of hardware interfaces, what subnet mask will you need to use if you have the 192.225.33.0 network?

5. Given the address 208.23.24.249 255.255.255.192. Is this a network address, a host address, or a directed broadcast?

6. You look at the output from ipconfig on an old Windows PC. You see the address 172.55.33.251 255.255.255.254. What is the network address and the directed broadcast for this host?

RTFQ. Carefully. Thar' be dragoons.

 

[burtsbees]

I like the dragoon with #1 and #5...lol
Good one...

Burt

 

[Cluebird]

On #6 change the address to 172.55.33.251 255.255.254.0. You won't see /31 on CCENT.

 

[bobbyforhire]

don't tappy tappy tappy on me I have to spend a few hours with the fiance. I will get back on later tonight and see how bad i've gotten after i let my noggen cool down!


ugh i can't help it now.. Ok ill do one.

A network administrator wants to subnet his address space using FLSM techniques. He has 45 hosts on each of five networks. He is using 194.44.55.0. What subnet mask would you suggest?
A. 255.255.255.224
B. 255.255.255.192
C. Can't be done
D. 255.255.255.128


Ok just looking at this has got me sceptical. I keep making the small things hard. He has 45 hosts on each FIVE networks that means this network guy had to of done something like this.

Now when you say he has 45 hosts are you saying that thats all he planned for? i mean did he really span out for more?

How can i answer this issue without my personal "why in the **** did he do that"? I keeps poping in my head and just tossing me around!@#

I would say D because 45 is not an incroment. thus i can't give you a subnet for an incroment that doesn't exist.

 

[bobbyforhire]

Ugh, i mean C

 

[maczen]

Plus there would be no network/broadcast addy's...

Hey guys.. I am looking and looking but the #5 Dragoon is eluding me!

B Haines
CCNA R&S, ETA FOI

 

[maczen]

C is right but the tricky part there is FLSM not VLSM.. So it has to be C...

B Haines
CCNA R&S, ETA FOI

 

[bobbyforhire]

Ok this one keeps going over my head. Ok if this guy has 5 networks and 45 hosts. i would recomend 255.255.255.224. this would give him his 5 networks and it would give him his more than needed 45 hosts.

When cisco asks me questions like this do they want the best answer that applys? because if he only wanted 45 hosts and only 5 networks i couldn't do it. But if i had to i would use /27.

 

[burtsbees]

Well, bobby, the answer indeed is C, but not for the reason you think---he has 45 hosts, meaning that a .224, or /27 is not enough, because that only gives 30 usable IPs. So the next one is /26, or .192, which will give 62 hosts per subnet, which is plenty. But how many networks does that give?
net 1=194.44.55.0, hosts 194.44.55.1-194.44.55.62, broadcast 194.44.55.63
net 2=.64, hosts=.65-.126, broad .127
net 3=.128, hosts .129-190, broad=.191
net 4=.192, hosts=.193-.254, braod=.255

I'll let you do network 5...bit of a trick question---that's why cluebird said to RTFQ---Cisco will throw curves at you like that...
Mac---I meant the dragoon in #6...sorry

Burt