Please HELP. Questions in subnetting

taktsoi

These are questions from MS 291 self paced practice exams. I got them wrong. but....i keep scratching my head....i still can't figure their correct answers....helppppppppppp.....


Question 1)
Your ISP has allocated your organization a Class C IP address range. As your head office has 48 computers, you have split the Class C range using the subnet mask 255.255.255.192. Your organization has several branch offices, each of which has 10 computers. You want to allocate only one address range per branch office. Given that the Class C range has already been divided, if you subnet the remaining addresses from that range in the most efficient manner possible, how many branch offices can you supply with public IP addresses?

A) 3 branch offices
6 branch offices
C) 2 branch offices
D) 12 branch offices
E) 24 branch offices


Question 2)
IP addresses at the local university are allocated in the following manner:

Faculty of Arts: 10.10.10.0/24
Faculty of Engineering: 10.10.12.0/23
Faculty of Science: 10.10.14.0/23
Faculty of Medicine: 10.10.16.0/24
Faculty of Economics: 10.10.17.0/24
Faculty of Law: 10.10.18.0/24

The university administration is responsible for the university’s root DNS server. Each faculty has its own DNS server. Each faculty’s DNS server will directly resolve reverse lookup queries for hosts with IP addresses within the range that it has been allocated. How many zone delegations must be made from the 10.10.in-addr.arpa reverse lookup zone hosted on the university’s root DNS server?

A) 8 delegations
16 delegations
C) 0 delegations
D) 1 delegation
E) 6 delegations

Please explain in details.

Thank you so much for your help.

royal

Question 1:

You have Class C which ends in 192. 192 is 11000000. Bit value from left to right is 128, 64, 32, 16, 8, 4, 2, and 1. Since the 2 left-most bits are turned on, that is 128 + 64 = 192. Since you have 2 bits that are turned on, you end up with 2^2 = 4 networks (bits turned on) with 2^6-2 = 62 clients (bits turned off). Since you need to split your Class C into multiple additional subnets, this is known as Virtual Length Subnetting.

So lets be as efficient as possible. Head office has 48 computers and branch offices have 10 machines. For the head office, the minimum amount of bits we need is:
2^2-2 = 2 - Not enough
2^3-2 = 6 - Not enough
2^4-2 = 12 - Enough

Since we are allocated /26 - 255.255.255.192 = 11111111.11111111.11111111.11000000 (26 consistent 1s), we can use /26, /27, and /28. We cannot use /29 because that'd only need 2 bits left (2^3-2=6) which does not meet the 10 client requirement.

So we can use the following subnets:
/26 = 192 = 11111111.11111111.11111111.11000000 = 62 clients - Main office
/27 = 224 = 11111111.11111111.11111111.11100000 = 30 clients at each BO - 2^2 = 8 Branch offices
/28 = 340 = 11111111.11111111.11111111.11110000 = 12 clients at each BO - 2^4 = 16 Branch offices

24 branch offices

royal

Question 2:

I'm pretty sure 0 delegations because any of the addresses created would fall in the 10.10 address space.

royal

Also, what were the answers given by Mspress?

taktsoi

Question 1 Answer:
Answer: 12 Branch offices
Using the 255.255.255.192 subnet mask at the head office gives you the following options:
Three branch offices using 255.255.255.192, each branch office having 62 available addresses.
Six branch offices using 255.255.255.224, each branch office having 30 available addresses.
Twelve branch offices using 255.255.255.240, each branch office having 14 available addresses.
You cannot use subnet mask 255.255.248 as this will only allow 6 available addresses and each branch office requires 10.

Question 2 Answer:
Answer: 8 delegations
A class-C IP address range is the smallest delegation that can be made for a reverse lookup zone. The faculties of Arts, Medicine, Economics, and Law will require one delegation each. The faculties of Engineering and Science will require two delegations each. Reverse lookup zone can be delegated only in class-A, class-B or class-C lots.

I m trying to figure out why question 1 needs to change to 240. the question says it has the 192 lot. so what i thought was that we will need the 192 to allocate the IP addresses....oh well....i don understand.......whatever.... on question 2 answer is 8. the answer says /24 will make 1 reverse lookup zone, this is understandable. however, why does /23 give you 2 reverse lookup zone thus giving the answer 8 delegations........

APA

/23 subnet mask will give you 510 hosts....... isn't that why you would get 2 reverse lookup zones??? Divides into two subnets pretty much...???

rhelt100

icroyal wrote:

Question 1:


So we can use the following subnets:
/26 = 192 = 11111111.11111111.11111111.11000000 = 62 clients - Main office
/27 = 224 = 11111111.11111111.11111111.11100000 = 30 clients at each BO - 2^2 = 8 Branch offices
/28 = 340 = 11111111.11111111.11111111.11110000 = 12 clients at each BO - 2^4 = 16 Branch offices

24 branch offices

I don't get why you added the two together to get 24. Here's what I'm seeing, I'm new to this so feel free to correct me:

If we make the remaining space:

/26 = 3 additional subnets with 62 clients each.
/27 = 6 additonal subnets with 30 clients each.
/28 = 12 additional subnets with 14 clients each.

The /28 seems to satisfy the requirements here. I agree with the MSPress answer.

Everlife

Adrian_Arumugam wrote:

/23 subnet mask will give you 510 hosts....... isn't that why you would get 2 reverse lookup zones??? Divides into two subnets pretty much...???

That was my thinking exactly Adrian. However, since the subnet mask is /23, wouldn't that put all those 510 in the same logical subnet? If they're in the same subnet, why wouldn't one reverse lookup zone be enough? Can anyone explain this one, I'm lost.

EdTheLad

taktsoi wrote:

These are questions from MS 291 self paced practice exams. I got them wrong. but....i keep scratching my head....i still can't figure their correct answers....helppppppppppp.....


Question 1)
Your ISP has allocated your organization a Class C IP address range. As your head office has 48 computers, you have split the Class C range using the subnet mask 255.255.255.192. Your organization has several branch offices, each of which has 10 computers. You want to allocate only one address range per branch office. Given that the Class C range has already been divided, if you subnet the remaining addresses from that range in the most efficient manner possible, how many branch offices can you supply with public IP addresses?

A) 3 branch offices
6 branch offices
C) 2 branch offices
D) 12 branch offices
E) 24 branch offices

Isp gives you /24, You split this range into 4 networks each with 62 hosts.One network should be assigned to the head office, leaving you three to play with.
So you have 3 /26 networks, each branch requires 10 computers which means you need a block of 16 i.e. /28 for each branch.
So looking at the first /26, this can contain 4 blocks of /28, so 3 networks with /26 can support
12 blocks of /28.Answer D

Everlife

EdTheLad wrote:

Isp gives you /24, You split this range into 4 networks each with 62 hosts.One network should be assigned to the head office, leaving you three to play with.
So you have 3 /26 networks, each branch requires 10 computers which means you need a block of 16 i.e. /28 for each branch.
So looking at the first /26, this can contain 4 blocks of /28, so 3 networks with /26 can support
12 blocks of /28.Answer D

I was looking at this a little further and tell me if I'm wrong with my thinking here.

Part A - Subnet to /26

2^(32-26) - 2 = 62 Hosts
26-24 = 2
2^2 = 4 Subnets

4 Subnets - 1 reserved for your main office (62 > 48 ) = 3 Subnets for 3 remaining branches.

Part B - Subnet to /27

2^(32-27) - 2 = 30 Hosts
27-24 = 3
2^3 = 8 Subnets

8 Subnets - 2 subnets reserved for your main office (30hosts x 2 > 48 ) = 6 Subnets for remaining branches.

Part C - Subnet /28

2^(32-28 ) - 2 = 14 Hosts
28 - 24 = 4
2^4 = 16 Subnets

16 Subnets - 4 Subnets reserved for your main office (14 Hosts x 4 = 48 ) = 12 Subnets for your remaining branches.

I think the kicker for that question is you need to remember the hosts that the main office requires. Because of the higher requirement, it will demand more subnets. Pretty tricky and interesting question.

APA

taktsoi wrote:

Question 2)
IP addresses at the local university are allocated in the following manner:

Faculty of Arts: 10.10.10.0/24
Faculty of Engineering: 10.10.12.0/23
Faculty of Science: 10.10.14.0/23
Faculty of Medicine: 10.10.16.0/24
Faculty of Economics: 10.10.17.0/24
Faculty of Law: 10.10.18.0/24

Everlife wrote:

Adrian_Arumugam wrote:

/23 subnet mask will give you 510 hosts....... isn't that why you would get 2 reverse lookup zones??? Divides into two subnets pretty much...???

That was my thinking exactly Adrian. However, since the subnet mask is /23, wouldn't that put all those 510 in the same logical subnet? If they're in the same subnet, why wouldn't one reverse lookup zone be enough? Can anyone explain this one, I'm lost.

Everlife, Notice how the 10.10.13.x and 10.10.15.x subnets are missing???? That is because they are included in the engineering and science address ranges respectively.

Reverse Lookups work via in-addr.arpa zones filled with PTR(Pointer) records. The zone is listed with the subnet address in reverse followed by in-addr.arpa e.g (10.10.12.0 reverse lookup zone would be x.12.10.10.in-addr-arpa)

The PTR record identifies the last octet for the reverse lookup zone...... So if a client was 10.10.12.1 then the PTR record mapping to that client would contain the number 1.

Do you get where I'm coming from??? Only 254 hosts can be listed in the one reverse lookup zone..... Which is why both the engineering and science division will have two reverse lookup zones adding up to 510 clients each because they take up two address ranges not the normal one......

Hope this clarifies things for you

If anyone notices anything wrong in my post please feel free to correct me as I can assure you I am not a expert (Well not yet anyways! )

APA

I just noticed the 10.10.11.x subnet is missing as well.......... I'm thinking thats just a typo or something to throw you off....?

Everlife

A.P.A wrote:

Do you get where I'm coming from??? Only 254 hosts can be listed in the one reverse lookup zone..... Which is why both the engineering and science division will have two reverse lookup zones adding up to 510 clients each because they take up two address ranges not the normal one......

Awesome thanks APA!

EdTheLad

Everlife wrote:

EdTheLad wrote:

I was looking at this a little further and tell me if I'm wrong with my thinking here.

Part A - Subnet to /26

2^(32-26) - 2 = 62 Hosts
26-24 = 2
2^2 = 4 Subnets

4 Subnets - 1 reserved for your main office (62 > 48 ) = 3 Subnets for 3 remaining branches.

Part B - Subnet to /27

2^(32-27) - 2 = 30 Hosts
27-24 = 3
2^3 = 8 Subnets

8 Subnets - 2 subnets reserved for your main office (30hosts x 2 > 48 ) = 6 Subnets for remaining branches.

Part C - Subnet /28

2^(32-28 ) - 2 = 14 Hosts
28 - 24 = 4
2^4 = 16 Subnets

16 Subnets - 4 Subnets reserved for your main office (14 Hosts x 4 = 48 ) = 12 Subnets for your remaining branches.

I think the kicker for that question is you need to remember the hosts that the main office requires. Because of the higher requirement, it will demand more subnets. Pretty tricky and interesting question.

I dont think you have got a grip on subnetting from your reply, Part B i don't know where you got that from!
You are given a class C address space which means a range from 0->255 in the last octet.
You first subnet with a /26 mask giving you
0->63,64->127,128->191,192->255
The head office uses 48 Pcs, you must assign a full /26 block and lose a few addresses.
Head office uses 0->63 range

Each branch office needs 10pcs therefore you need a mask /28 which will use a block of 16 addresses.Assigning this block will give
64->79
80->95
96->111
112->127
128->143
144->159
160->175
176->191
192->207
208->223
224->239
240->255

So as you can see you get 12 branches and your whole address space has been used.I assume
another ip address is used for the links between the branch offices,otherwise you would have to remove 3 branch offices.So for the ptp links between branch offices you would have a /30 block.
You could remove the first branch block 64->79 and split it into 4 /30 blocks
64->67
68->71
72->75
76->79
If you did this with 3 of the original branch allocations you would have 12 ptp links and 9 branches possible, since 9 isn't a possible answer i think they have bothered to bring in this complexity.

Everlife

Subnetting has definitely been giving me some problems, so the help is appreciated. Let me run through what I think you're trying to say (I tend to need to see it worked out in detail.)

Say they are using a 192.168.0.0 address scheme.

The head office would be 192.168.0.0 /26 (192.168.0.0 - 192.168.0.63)

This leaves you with 3 /26 address blocks left, which you will further subnet into 12 /28 address blocks giving you the following:

Branch 1 -> 192.168.0.0 /28 (192.168.0.64-192.168.0.79)
Branch 2 -> 192.168.0.0 /28 (192.168.0.80-192.168.0.95)
Branch 3 -> 192.168.0.0 /28 (192.168.0.96-192.168.0.111)
Branch 4 -> 192.168.0.0 /28 (192.168.0.112-192.168.0.127)
Branch 5 -> 192.168.0.0 /28 (192.168.0.128-192.168.0.143)
Branch 6 -> 192.168.0.0 /28 (192.168.0.144-192.168.0.159)
Branch 7 -> 192.168.0.0 /28 (192.168.0.160-192.168.0.175)
Branch 8 -> 192.168.0.0 /28 (192.168.0.176-192.168.0.191)
Branch 9 -> 192.168.0.0 /28 (192.168.0.192-192.168.0.207)
Branch 10-> 192.168.0.0 /28 (192.168.0.208-192.168.0.223)
Branch 11-> 192.168.0.0 /28 (192.168.0.224-192.168.0.239)
Branch 12-> 192.168.0.0 /28 (192.168.0.240-192.168.0.255)

Satisfies 48 hosts for the head office and the 10 hosts per branch office.

Am I grasping what you are saying Ed?

EdTheLad

Everlife wrote:

Am I grasping what you are saying Ed?

Yes, perfect!

Everlife

Awesome. I've read through the Microsoft chapters on it, but really need to get off my butt and take a read through the Syngress. Looks like I have some work for the weekend.

Thanks for you help!

APA

Everlife wrote:

A.P.A wrote:

Do you get where I'm coming from??? Only 254 hosts can be listed in the one reverse lookup zone..... Which is why both the engineering and science division will have two reverse lookup zones adding up to 510 clients each because they take up two address ranges not the normal one......

Awesome thanks APA!

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