Question: NITC building had 4 research labs each having 24 computers. All labs are located at the 1st floor. Each computer is to be connected in the network from NCR located at 2nd floor. Prepare a bill of quality (BoQ) with necessary network resources required for complete networking. The BoQ must include estimation of all network resources required.

Physical devices requires:

  1. Router
  2. Switch
  3. CAT 6 cable
  4. RS-232 cable
  5. PCs/Laptops/workstations
  6. Server
  7. Printer
  8. IP phone
  9. Wireless router (Access Point)

Specification sheet

S.

N.

Item Description

Quantity

Units

Summary specification

1.

CAT-6 UTP

cable

100

pcs

4-pair pure copper conductor PVC jacketed Foiled/UTP cable with TIA standard color 23

AWG and having tested frequency range of 250-550 MHz, Support up to 1000Base-T at

100 meters

2.

PCs/Computer

96

pcs

Varying specification

3.

Router

1

pcs

Cisco 2901

4.

Access Point

4

pcs

150Mbps Wireless N Router; 2.4- 2.4835GHz

5.

Switch

5

pcs

Cisco 2950-24

6.

Printer

1

pcs

Varying specs

7.

IP Phone

1

pcs

 

8.

RS-232 cable

1

pcs

 

Assumption:-

  1. There are 1 server in 2nd floor
  2. There are four printer n each lab and
  3. One IP Phone in the 1st floor

Solution

Let us consider we are provided with a public address 110.100.50.0/24.

We need five subnets consisting of minimum 24, 24, 24, 24 IP for research lab and 5 IP for NCR respectively.

Initially given IP should be divided into two parts for two floors such that IP address for first floor is 100.100.50.0 /25 and that for second is 100.100.50.128 /25.

For 4 research lab in 1st floor Applying VLSM technique, Required no of host bit

2x > 24 i.e. x = 5. Subnet mask = 255.255.255.224 So, subnets in first floor become

  1. 100.100.50.0 /27 – 30 hosts
  2. 100.100.50.32 /27 – 30 hosts
  3. 200.100.50.64 /27 – 30 hosts
  4. 100.100.50.96 /27 – 30 hosts

For 2nd floor (i.e. NCR):-

We need 5 IP from IP address 100.100.50.128/5

Applying VLSM technique, Required no. of host bit

2x > 5 I.e. x=3. Subnet mask = 255.255.255.248 So, subnet in second floor become

  1. 100.100.50.128/29 - 6 hosts

Network drawn in Cisco Packet Tracer 6.1 Student Version is shown below

 

2080 numerical and design question solution