Hello, I was wondering if you could help me answer the 1DS question, in chapter

Question

Hello,

I was wondering if you could help me answer the 1DS question, in chapter 6 of the Top-Down Network Design (3rd Edition). Question below. I'm not sure how to address the network...

Here is a brief synopsis of the scenario. EMC’s network will support about 200 employees. The network will include a data center and a new state-of-the-art manufacturing facility. Users in the campus network will access the servers in the data center from their PCs. For online sales, EMC plans to have a DMZ that connects a web server, a DNS server, and an e-mail server. EMC also plans to open a branch sales office in a city that is about 500 miles from EMC’s headquarters. EMC also intends to allow various retail outlets (not their own stores – the outlets sell EMC and other company’s products) to sell their products. For the purposes of this exercise, you should only consider one such retail establishment (the first one they have arranged to sell their product through), not all of the outlets that will sell their product.

Design and document an IP addressing scheme to meet EMC’s needs.

Specify which IP address blocks will be assigned to different modules of your network’s design.

Document whether you will use public or private addressing for each network module.

Document whether you will use manual or dynamic addressing for each module. Specify where (if anywhere) route summarization will occur.

You may not be able to create addressing for VPNs, but indicate (in a statement) where you would place VPNs.

You must address each of the above points in your design in order to be eligible for full credit. You must be able to make a valid argument for why your design will work for EMC.

Explanation / Answer

Considering total 200 people in the company and 1 IP assigned to each user.

Considering all listed requirements

Total number of host is x.
Find the subnet size, that is largest power of 2 that allows all of your hosts and required addresses for your subnet.

Subnet A: 2 hosts

Subnet B: 6 hosts

Subnet C: 46 hosts

Subnet D: 124 hosts

Considering our Initial IP: 192.168.0.0/24

IP: 192.168.0.0

Subnet Mask: 255.255.255.0

/24 means 24 1's so the subnet mask is 1111 1111.1111 1111.1111 1111.0000 0000 -- which is 255.255.255.0

So let's look into on octet-4 only. For now, it looks like this: .0000 0000 -- and our first subnet needs 2+2 hosts (you need to add 2 to cover for the 2 that you will lose later: the Network Address and the Broadcast Address), so to start with the farthest bit on the right.

2^1 >= 4? No -- so I leave that bit set to zero. Now I move left 1 bit and

Is 2^2 >= 4? Yes -- so I set that bit to zero and put 1's in the remaining bits to the left.

So now octet-4 looks like: 1111 1100 -- we started at the far right and moved 2 times to the left or 2^2, which is 4. And 4 is definitely greater than or equal to 4.

And so on...

VPN

Place your VPN outside your DMZ with limited access to your internal LAN, Data center. You may need to place firewalls in parallel for the same.

Now find the farthest '1' to the right, and figure out its bit value. So octet-4 looks like this: 1111 1100 -- and the farthest one is in the 6th spot (going left-to-right) which is 128, 64, 32, 16, 8, the 4's spot. So my networks will be broken into groups of 4.

So, I have Subnet-A: (using Subnet Mask: /30)
* 172.20.0.0 -- Network IP <-- Unusable
- 192.168.0.1 -- 1st usable IP
- 192.168.0.2 -- Last usable IP
- 192.168.0.3 -- Broadcast IP <--- Not in use

- 192.168.0.4 -- Start your next Subnet

One subnet down, let's do the next one. We simply reset octet-4 back to all zeroes and start again.

So let's conside on octet-4 only. For now, it looks like this: .0000 0000 -- and our second subnet needs 6+2 hosts (you need to add 2 to cover for the 2 that you lose later: the Network Address and the Broadcast Address), so start with the farthest bit on the right and I:
2^1 >= 8? No -- so I leave that bit set to zero. Now I move left 1 bit and
2^2 >= 8? No -- so I leave that bit set to zero. Now I move left 1 bit and
2^3 >= 8? Yes -- so I set that bit to zero and put 1's in the remaining bits to the left. So now octet-4 looks like: 1111 1000 -- we started at the far right and moved 3 times to the left or 2^3, which is 8. And 8 is definitely greater than or equal to 8.

New subnet mask for Subnet-B looks like:
- 1111 1111.1111 1111.1111 1111.1111 1000 = 255.255.255.248 or /29 (we added 5 more 1's to the initial subnet mask)

find the farthest '1' to the right, and figure out its bit value. So octet-4 looks like this: 1111 1000 ---- ---- and the farthest one is the 5th spot (going left to right) which is 128, 64, 32, 16, the 8's spot. So the networks will be broken into groups of 8.

So, I have Subnet-B: (Using Subnet Mask: /29 -- also, remember where we left off after Subnet-A... but we need to start at the next multiple of 8...)
* 192.168.0.8 -- Network IP <-- Unusable
- 192.168.0.9 -- 1st usable IP
- 192.168.0.10 -- 2nd usable IP
- 192.168.0.11 -- 3rd usable IP
- 192.168.0.12 -- 4th usable IP
- 192.168.0.13 -- 5th usable IP
- 192.168.0.14 -- Last usable IP
- 192.168.0.15 -- Broadcast IP <-- No Use

- 192.168.0.16 -- (Probable) Start of next Subnet/Network

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