Computer Dynamics is a microcomputer software development company that has a 300

Question

Computer Dynamics is a microcomputer software development company that has a 300-computer network. The company is located in three adjacent five-story buildings in an office park, with about 100 computers in each building. The LANs in each building are similar, but one building has the data center on the second floor. There are no other office locations. The current network is poorly designed for its current needs and must be completely replaced.

1. Develop a logical design for this enterprise campus that considers the seven network architecture components. There are no other campuses, so you can omit WAN access.

2. Document the assumptions you made in approaching this design.

3. Explain why you have designed the network in this way.

Explanation / Answer

ANSWER:

For this question, we need to assign the IP addresses to company's computer network.

Things to keep in mind.

1. We require picking the set of personal IP address range because business workers need to converse with each other.

2. So if we want to converse with in LAN system only then we use private IP address range

Private IPV4 Address range is as follows

(1.1) 10.0.0.0 – 10.255.255.255

(1.2) 172.16.0.0 – 172.31.255.255

(1.3) 192.168.0.0 --192.168.255.255

2. Here we use the concept of subletting to divide the network among building and for efficient utilisation of IP address.

Niumber of buildings = No. of Subnets = 3

Number of computers on each building = No of hosts required = 100

2^n - 2 = Number of subnest, where n is number of subnet bits

2^n - 2 = 3 ,

So n = 3 is best suited value

So we need 3 subnet bits

No of bits required for host

2^2 - 2 = 100

So n = 7 is best suited value

So total bits necessary for organization is 7 + 3 = 10 and smallest amount loss of IP address we use range 192.168.0.0

So Network Id of whole Network is 192.168.252.0 and it will be assign to data center.

So Network ID of first subnet is 192.168.252.128

First IP address of first structure is 192.168.252.129

Last address of First building is 192.168.252.228

Remaining 25 addresses will be unassigned which can be used in outlook in first building.
So Network ID of 2nd subnet is 192.168.253.0
First IP address of first building is 192.168.253.1
Last address of First building is 192.168.253.100
Remaining 25 addresses will be unassigned which can be used in future in 2nd building.
So Network ID of first subnet is 192.168.253.128
First IP address of first building is 192.168.253.129
Last address of First building is 192.168.253.228
Remaining 25 addresses will be unassigned which can be used in future in initial building.
This is how we can manage the network

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