1. ASSOSA UNIVERSITY
COLLAGE OF COMPUTING AND INFORMATICS
DEPARTMENT OF INFORMATION TECHNOLOGY
COURSE TITLE: NETWORK DESIGN
BY
No Name IDNo
1. Amare Simachew……………………………………….ETR/0027/08
SUMMITED TO: KASAHUN (MSC)
SUBMISSION DATE:16/05/2011E.C
Assosa, Ethiopia
2. Network design
Page 1
Campus X is the academic institution provides teaching and learning, community service and
researches for the country. This institution has one end to end network and different subnets
which hold the next users 125 at building A, 28 at building B, 60 at building C, 2000 at
building D, 500 at building E, 700 at building F, 10 at building G, 5 at building H and 200 at
building I. The organization or institution has nine buildings which require one subnet each and
the above users accordingly. Building D is found at the center. Building C is 1km far from
building D. Building F is 500m far from building D. All the others are 200m far from the center
building. One again the institution has only one IP address network. Assume that your
organization Y won the bid in which you are network engineer who delegates as network
designers and consultant. So, that you are expected to adjust the following requirements for the
institutions.
A. Design IP addresses accordingly and assign for each building. 10 points
B. Categorized the design by hierarchical layer. 10 points
C. Sketch the skeleton of the network topology. You are expected to list and discuss the devices
and Medias. 10 points
D. Write the documentation. 10 points
First of all we must put the given host in descending order ከትልቁ ወደ ትንሹ or from the largest to
the lowest host
Subnet Number of Hosts የአስተናጋጆች ቁጥር
Subnet D 200 0.
Subnet F 700
Subnet E 500
Subnet I 200
Subnet A 125
Subnet C 60
Subnet B 28
Subnet G 10
Subnet H 5
The Ip address given=10.5.0.0/16
Know let us start from the largest one which is D=2000 host or user
IP address=10.5.0.0/16
2n
=2000
N=11,
3. Network design
Page 2
211
=2048 is total number of host
Usable host=2n
-2=2048-2=2046
Reserved =46 host
10.5.00000000.00000000/16=10.5.00000|000.00000000/21
Number of network=2n
, where n is number of borrowed bit from host bit
Number of network=2n
, n=5, 25
=32 network id
Network number Borrowed Bit Network id
NW1 10.5.00000/000.00000000/21 10.5.0.0/21
NW2 10.5.00001/000.00000000/21 10.5.8.0/21
NW3 10.5.00010/000.00000000/21 10.5.16.0/21
. . .
. . .
NW32 10.5.11111/000.00000000/21 10.5.248.0/21
Step 1: After we create the networks we assign the first network id for the building “D”
which is 10.5.0.0/21
10.5.00000/000.00000001/21=10.5.0.1/21 is fist usable ip address
10.5.00000/111.11111110/21=10.5.7.254/21 is last usable ip address
10.5.00000/111.11111111/21=10.5.7.255/21 is Broad cast ip address
Subnet D
specification
Number of bits borrowed 5
Subnet mask 255.255.248.0
Number of usable host per subnet 2046
Number of usable subnet 32
Subnetwork address 10.5.0.0/21
First IP Host address 10.5.0.1/21
Last IP Host address 10.5.7.254/21
Broadcast address 10.5.7.255/21
Step2: Assign the second network id for F which contain 700 host
10.5.8.0/21 NW id
2n
=700
Comment [w1]: 1*27
+1*26
+1*25
=1*2^4=1*2^3=1*2^2=1*
2^1=248 so, it becomes 255.255.248.0
Comment [w2]: Since the no of Borrowed bits are
5 and 255.255 shows that the default for10.5 after u
change 10 to binary and that binary to decimal, and
change this decimal to all 11111 and change to binary it
becomes 255 and 5 is the same way as 10 and 248 is
obtained by making all 5 zeros that we are borrowed to
11111/000.00000000/21 and change to decimal
4. Network design
Page 3
N=10,
210
=1024 is total number of host
Usable host=2n
-2=1024-2=1022
Reserved =322 host
10.5.000010/00.00000000/22
Borrow one bit
NW=21
=2
NW1=10.5.000010/00.00000000/22=10.5.8.0/22
1*23
+0*22
+0*20
=8 so, it becomes 10.5.8.0/22
NW2=10.5.000011/00.00000000/22=10.5.12.0/22
1*23
+1*22
+0*20
=12 so, it becomes 10.5.12.0/22
We assign the first network for building “F” which is
10.5.8.0/22
10.5.000010/00.00000001/22=10.5.8.1/22 is first usable ip address
10.5.000010/11.11111110/22=10.5.11.254/22 is last usable ip address
10.5.000010/11.11111111/22=10.5.11.255/22 is broadcast ip address
Subnet F
Specification
Number of bits borrowed 6
Subnet mask 255.255.252.0
Number of usable host per subnet 1022
Number of usable subnet 64
Subnetwork address 10.5.8.0/22
First IP Host address 10.5.8.1/22
Last IP Host address 10.5.11.254/22
Broadcast address 10.5.11.255/22
Step3: Assign the NW2 for subnet “E” which contain 500 user
10.5.12.0/22
Comment [w3]: እራሱንወስደን ወደ decimal መቀየር ከዛ
10.5.8.0/22 ይመጣል
Comment [w4]: የሚመጣዉ 10/11 ንን ወደ binary
ስንቀይራቸዉ የሚመጣ ነዉ
5. Network design
Page 4
2n
=500
N=9,
29
=512 is total number of host
Usable host=2n
-2=512-2=510
Reserved =10 host
Borrow one bit
10.5.0000110/0.00000000/23
NW=21
=2
NW1=10.5.0000110/0.00000000/23=10.5.12.0/23
NW2=10.5.0000111/0.00000000/23=10.5.14.0/23
Assign the NW1 for subnet “E” which is
10.5.12.0/23 NT ID
10.5.0000110/0.00000001/23=10.5.12.1/23 is first usable ip address
10.5.0000110/1.11111110/23=10.5.13.254/23 is last usable ip address
10.5.0000110/1.11111111/23=10.5.13.255/23 is broadcast ip address
Subnet E
Specification
Number of bits borrowed 7
Subnet mask 255.255.254.0
Number of usable host per subnet 510
Number of usable subnet 128
Subnetwork address 10.5.12.0/23
First IP Host address 10.5.12.1/23
Last IP Host address 10.5.13.254/23
Broadcast address 10.5.13.255/23
Step4: Assign the NW2 for subnet “I” which contain 200 user
10.5.14.0/23
2n
=200
6. Network design
Page 5
N=8,
28
=256 is total number of host
Usable host=2n
-2=256-2=254
Reserved =54 host
Borrow one bit
10.5.00001110/.00000000/24
Number of network=21
=2
NW1=10.5.00001110/.00000000/24=10.5.14.0/24
NW2=10.5.00001111/.00000000/24=10.5.15.0/24
Assign the first network for “I” which is 10.5.14.0/24
10.5.00001110/.00000001/24=10.5.14.1/24 is first usable ip address
10.5.00001110/.11111110/24=10.5.14.254/24 is last usable ip address
10.5.00001110/.11111111/24=10.5.14.255/24 is broadcast ip address
Subnet I
Specification
Number of bits borrowed 8
Subnet mask 255.255.255.0
Number of usable host per subnet 254
Number of usable subnet 256
Subnet network address 10.5.14.0/24
First IP Host address 10.5.14.1/24
Last IP Host address 10.5.14.254/24
Broadcast address 10.5.14.255/24
Step5: Assign the NW2 for “A” which contain 125 user
10.5.15.0/24
2n
=125
N=7,
27
=128 is total number of host
7. Network design
Page 6
Usable host=2n
-2=128-2=126
Reserved =1 host
Borrow one bit
10.5.00001111.0/0000000/25
Number of network=21
=2
NW1=10.5.00001111.0/0000000/25=10.5.15.0/25
NW2=10.5.00001111.1/0000000/25=10.5.15.128/25
Assign the first network for “A” which is 10.6.15.0/25
10.5.00001111.0/0000001/25=10.5.15.1/25 is first usable ip address
10.5.00001111.0/1111110/25=10.5.15.254/25 is last usable ip address
10.5.00001111.0/1111111/25=10.5.15.255/25 is broadcast ip address
Subnet A
Specification
Number of bits borrowed 9
Subnet mask 255.255.255.128
Number of usable host per subnet 126
Number of usable subnet 512
Subnet network address 10.6.15.0/25
First IP Host address 10.5.15.1/25
Last IP Host address 10.5.15.254/25
Broadcast address 10.5.15.255/25
Step6: Assign the NW2 for “C” which contain 28 user
10.5.15.128/25
2n
=60
N=6,
26
=64 is total number of host
Usable host=2n
-2=64-2=62
Reserved =2 host
8. Network design
Page 7
Borrow one bit
10.5.00001111.10/000000/26
Number of Network=21
=2
NW1=10.5.00001111.10/000000/26=10.5.15.128/26
NW2=10.5.00001111.11/000000/26=10.5.15.192/26
Assign the first network for “C” which is 10.5.15.128/26
10.5.00001111.10/000001/26=10.5.15.129/26 is first usable ip address
10.5.00001111.10/111110/26=10.5.15.190/26 is last usable ip address
10.5.00001111.10/111111/26=10.5.15.191/26 is broadcast ip address
Subnet C
Specification
Number of bits borrowed 10
Subnet mask 255.255.255.192
Number of usable host per subnet 62
Number of usable subnet 1024
Subnet network address 10.5.15.128/26
First IP Host address 10.5.15.128/26
Last IP Host address 10.5.15.190/26
Broadcast address 10.5.15.191/26
Step7: Assign the NW2 for building “B” which contain 28 user
10.5.15.192/26
2n
=28
N=5,
25
=32 is total number of host
Usable host=2n
-2=32-2=30
Reserved =2 host
Borrow one bit
10.5.00001111.110/00000/27
9. Network design
Page 8
Number of network create=21
=2
NW1=10.5.00001111.110/00000/27=10.5.15.192/27
NW2=10.5.00001111.111/00000/27=10.5.15.224/27
Assign the first network for “B” which is 10.5.15.192/27
10.5.00001111.110/00001/27=10.5.15.193/27 is first usable ip address
10.5.00001111.110/11110/27=10.5.15.222/27 is last usable ip address
10.5.00001111.110/11111/27=10.5.15.223/27 is broadcast ip address
Subnet B
Specification
Number of bits borrowed 11
Subnet mask 255.255.255.224
Number of usable host per subnet 30
Number of usable subnet 2048
Subnet network address 10.5.15.192/27
First IP Host address 10.5.15.193/27
Last IP Host address 10.5.15.222/27
Broadcast address 10.5.15.223/27
Step8: Assign the NW2 for building “G” which contain 10 user
10.5.15.224/27
2n
=10
N=4,
25
=16 is total number of host
Usable host=2n
-2=16-2=14
Reserved =4 host
Borrow one bit
10.5.00001111.1110/0000/28
Number of network= 21
=2
NW1= 10.5.00001111.1110/0000/28=10.5.15.224/28
NW2= 10.5.00001111.1111/0000/28=10.5.15.240/28
10. Network design
Page 9
Assign the first network for “G” which is 10.5.15.224/28
10.5.00001111.1110/0001/28=10.5.15.225/28 is first usable ip address
10.5.00001111.1110/1110/28=10.5.15.238/28 is last usable ip address
10.5.00001111.1110/1111/28=10.5.15.239/28 is broadcast ip address
Subnet G
Specification
Number of bits borrowed 12
Subnet mask 255.255.255.240
Number of usable host per subnet 14
Number of usable subnet 4096
Subnet network address 10.5.15.224/28
First IP Host address 10.5.15.225/28
Last IP Host address 10.5.15.238/28
Broadcast address 10.5.15.239/28
Step9: Assign the NW2 for building “H” which contain 5 user because the reserved host is not
sufficient
10.5.15.240/28
2n
=5
N=3,
23
=8 is total number of host
Usable host=2n
-2=8-2=6
Reserved =1 host
Borrow one bit
10.5.00001111.11110/000/29
Number of network=21
=2
NW1=10.5.00001111.11110/000/29=10.5.15.240/29
NW2=10.5.00001111.11111/000/29=10.5.15.248/29
Assign the first network for “H” which is 10.5.15.240/29
10.5.00001111.11110/001/29=10.5.15.241/29 is first ip address
11. Network design
Page 10
10.5.00001111.11110/110/29=10.5.15.246/29 is last ip address
10.5.00001111.11110/111/29=10.5.15.247/29 is broadcast ip address
Subnet H
Specification
Number of bits borrowed 13
Subnet mask 255.255.255.248
Number of usable host per subnet 6
Number of usable subnet 8192
Subnet network address 10.5.15.240/29
First IP Host address 10.5.15.241/29
Last IP Host address 10.5.15.246/29
Broadcast address 10.5.15.247/29
B. Categorized the design by hierarchical layer.
The design of the network can be categorized in to three hierarchical layers and the ip address
that we subnet in the above are subnated in hierarchical format.
The core layer provides an optimized and reliable transport structure by forwarding traffic at
very high speeds.
The distribution layer shows several multilayer switches and link connections to the core
layer. The distribution layer shows several multilayer switches and link connections to the
core layer.
Access layer shows PCs, access switches, VPN gateways, printers, steelworker, home office,
and wireless router. Also shown in this layer are redundant links to the distribution layer.
C. Sketch the skeleton of the network topology. You are expected to list and discuss the
devices and Medias.
When we design the network for one organization we use different media and devices. From
these we used:
Network device:-
Layer 3 router 4300 series (core layer):- It is a device that logically segments networks at the
IP subnet level. A router is responsible for routing IP packets between different IP
networks. Routers do this using an IP routing table.
12. Network design
Page 11
Layer 3 Switch 3560 distribution switch: is a device used to distribute the network in to
different Vlan and we create those Vlan in this device
Access Switch 2960:- The Access layer, located within a campus building, aggregates end users
from different workgroups and provides uplinks to the Building Distribution layer.
In addition to network device, there are hard ware requirement to design network for campus and
for the other organization .these are:
Firewall: is a device which is used as guard to filter incoming and outgoing data in the network.
Fiber patch panel: They make it easy to terminate fiber optic cables and provide access to the
cable's individual fibers for cross connection.
Media converter: - is hardware device which is used to convert the fiber cable which is come
from ethio telecom to LAN cable.
Patch panel: which is used to connect and manage incoming and outgoing LAN cables.
Rack or cabinet: a device which is used to hold switch and patch panel.
Media:
Fiber optic cable: used for carry the internet from the distribution switch to each buildings
Fiber Patch cord: is cable that carries the connection to the switch and connect to the SFP
module.
LAN Cat 6E cable: commonly referred to as Cat 6, is a standardized twisted pair cable
for Ethernet , which is used to transfer high bandwidth of the data.
Consol cable: is a type of cable used for configuring the network device.
13. Network design
Page 12
1.1. Introduction
Campus X is the academic institution provides teaching and learning, community service and
researches for the country and contains 3628 users. It has nine building, those are building A, B,
C, D, E, F, G, H, I each of building contain their own number of users or host from the total
users. For those building we design the network to deliver good network access for the user. In
order to design network we assign the ip address for each building and use network device such
us layer 3 routers, layer 3 switch, and access switch and media also needed such us fiber optic
cable, unshielded twisted pair cable. And also ip address 10.5.0.0/16 for the campus x we design
this ip address in to different subnet.
1.2. Logical design
1.2.3. Naming
As we know naming is way of giving appropriate name for the device and for the other. The
naming must be easy and understandable for the designer to troubleshoot the problem easily. In
the campus x we assign appropriate name for the device and buildings.
14. Network design
Page 13
As in every organization, campus x has its own naming convention for all the network devices.
The naming convention in campus x has follow :{ Device}-{Building}-{Floor}-Number}.
Device is type of the device; it can be Switch, core switch, and distribution switch
(SW|CSW|DSW). Building is to describe where the device will be
located (A|B|C|D|E|F|G|H|I). Floor is describing the floor of Building. Number is describes how
many devices with the same attribute.
We assigned the name for building and devices are listed below:-
The names of networking device which are assigned in each building are:
SW1-A-F1….Access switch one located at floor one building A
SW2-A-F2….Access switch two located at floor two at building A
SW3-A-F3…Access switch three located at floor three at building A
SW1-B-F1….Access switch one located at floor one at building B
SW2-B-F2…. Access switch two located at floor two at building B
SW3-B-F3…. Access switch three located at floor three at building B
SW1-C-F1…. Access switch one located at floor one at building C
SW2-C-F2… Access switch two located at floor two at building C
SW3-C-F3… Access switch three located at floor three at building C
DSW1-D-F1….Distributed switch one located at floor one at building D
CSW1-D-F1…Core switch one located at floor one at building D
SW1-D-F1…. Access switch one located at floor one at building D
SW2-D-F2… Access switch two located at floor two at building D
SW3-D-F3… Access switch three located at floor three at building D
SW1-E-F1…. Access switch one located at floor one at building E
SW2-E-F2….. Access switch two located at floor two at building E
SW3-E-F3…. Access switch three located at floor three at building E
SW1-F-F1…. Access switch one located at floor one at building F
15. Network design
Page 14
SW2-F-F2…. Access switch two located at floor two at building F
SW3-F-F3…. Access switch three located at floor three at building F
SW1-G-F1…. Access switch one located at floor one at building G
SW2-G-F2…. Access switch two located at floor two at building G
SW3-G-F3…. Access switch three located at floor three at building G
SW1-H-F1….. Access switch one located at floor one at building H
SW2-H-F2….. Access switch two located at floor two at building H
SW3-H-F3….. Access switch three located at floor three at building H
SW1-I-F1….. Access switch one located at floor one at building I
SW2-I-F2…. .Access switch two located at floor two at building I
SW3-I-F3…. .Access switch three located at floor three at building I
1.2.4. Addressing
Addressing is way of assigning unique ip address for the device. The ip address in campus x is
divided based on building as follow:
No Name of building Network ID /ip address assigned
1 Building A 10.5.15.0/25
2 Building B 10.5.15.193/27
3 Building C 10.5.15.126/26
4 Building D 10.5.0.0/21
5 Building E 10.5.12.0/23
6 Building F 10.5.8.0/22
7 Building G 10.5.15.224/28
8 Building H 10.5.15.240/29
9 Building I 10.5.14.0/24
1.3. Physical design
Layer 3 core switch 4300 series and Layer 3 Switch 3560 distribution switch in data center
connected by using multimode fiber optic cable. And from distribution switch to each of
16. Network design
Page 15
buildings are connected by using single mode fiber optic cable and fiber patch panel is placed in
all building connected with access switch by using patch cord. And also access switch or layer 2
2960 switches in building are connected with patch panel each other by cat6e network cable and
also with end device (computers) by using cat 6e network cable which is used for carry
bandwidth than other network cable. In the building the internet is extended from the first to
second and to the third floor of building by using cat6e network cable.
Conclusion
Campus x is enhancing Local Area Network to accommodate their needs. Using the top-down
network design process to provides teaching and learning, community service and researches for
the country. The building block components hierarchical structure networks are the core layer,
the distribution layer and the access layer. Core layer is designed using Layer 3 switch,
Distribution layer at each building is design with using Layer 3 switch and Access layer is design
with using layer 2 switches.