Design Practice Question
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:
- Router
- Switch
- CAT 6 cable
- RS-232 cable
- PCs/Laptops/workstations
- Server
- Printer
- IP phone
- 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:-
- There are 1 server in 2nd floor
- There are four printer n each lab and
- 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
- 100.100.50.0 /27 – 30 hosts
- 100.100.50.32 /27 – 30 hosts
- 200.100.50.64 /27 – 30 hosts
- 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
- 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