Faculty of Engineering and the Build Environment:
DEECE
Subject: Data Networks 2
Course code: DNT260S
Second Year: Semester 1
TITLE: ASSIGNMENT 1
Lecturer: Dr. Vipin VIPIN
Due date:31/05/2023
Student Name: Fidele Vlad MIAMBAZILA
Student no: 219393583
E-mail: 219393583@mycput.ac.za
Question 1 (Solution)
Let’s find the shortest distance between node A and node F on the network above.
Student No: 219393583
 Cost between nodes B and D: 3
 Cost between nodes D and E: 5
 Cost between nodes E and F: 8
 Cost between nodes D and F: 3
Starting Node
A
AC
ACB
ACBD
ACBDF
ACBDFE
B
5,A
C
2,A
D
inf
6,C
E
inf
10,C
F
inf
inf
inf
9,D
Using Dijkstra’s algorithm, we have:
ACBDFE  ACDF
Thus the shortest distance between nodes A and F is:
ACDF = 2+4+3 = 9
Question 2(Solution)
Student No: 219393583
Since the last digit of my student number is odd, I will make 29 subnets.
IP address: 197.1.1.0
The subnet mask for a class C IP address is: 255.255.255.0
Knowing that 29 subnets are required,
2n ≥ 29 thus we will borrow 5 bits in the last octet of our subnet mask and the
remaining 3 bits will be used for our hosts.
Each subnet will have a maximum of: 23 – 2 = 6 hosts
Considering all the information given above, the 4th octet of our subnet mask is then.
128
1
64
1
32
1
16
1
4th Octet
8
1
The subnet mask is therefore: 255.255.255.248
IP address: 197.1.1.0/29
Subnet
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Addresses
197.1.1.0 – 197.1.1.7
197.1.1.8 – 197.1.1.15
197.1.1.16 – 197.1.1.23
197.1.1.24 – 197.1.1.31
197.1.1.32 – 197.1.1.39
197.1.1.40 – 197.1.1.47
197.1.1.48 – 197.1.1.55
197.1.1.56 – 197.1.1.63
197.1.1.64 – 197.1.1.71
197.1.1.72 – 197.1.1.79
197.1.1.80 – 197.1.1.87
197.1.1.88 – 197.1.1.95
197.1.1.96 – 197.1.1.103
197.1.1.104 – 197.1.1.111
197.1.1.112 – 197.1.1.119
197.1.1.120 – 197.1.1.127
4
0
2
0
1
0
17
18
19
20
21
22
23
24
25
26
27
28
29
197.1.1.128 – 197.1.1.135
197.1.1.136 – 197.1.1.143
197.1.1.144 – 197.1.1.151
197.1.1.152 – 197.1.1.159
197.1.1.160 – 197.1.1.167
197.1.1.168 – 197.1.1.175
197.1.1.176 – 197.1.1.183
197.1.1.184 – 197.1.1.191
197.1.1.192 – 197.1.1.199
197.1.1.200 – 197.1.1.207
197.1.1.208 – 197.1.1.215
197.1.1.216 – 197.1.1.223
197.1.1.224 – 197.1.1.231
The IP address of Host 2 on subnet 6 is: 197.1.1.42
SIMULATION ON GNS3 (2 ROUTERS)
Question 3(Solution)
Class B IP address: 183.1.0.0
Subnet mask: 255.255.0.0
A
2
B
1
C
9
D
3
E
9
F
3
G
5
H
8
I
3
Based on my student number we have the following information.
Subnet 1 minimum hosts: 9AB = 921
Subnet 2 minimum hosts: 8BC = 819
Subnet 3 minimum hosts: 3DE = 339
Subnet 4 minimum hosts: 2AD = 223
921
819
339
223
32768
1
1
1
1
16384
1
1
1
1
8192
1
1
1
1
4096
1
1
1
1
Subnet 1
183.1.0.0/22 255.255.252.0
Possible hosts: 210 − 2 = 1022
Network address: 183.1.0.0
Hosts
183.1.0.1 – 255
183.1.1.0 – 255
183.1.2.0 – 255
183.1.3.0 – 254
Broadcast address: 183.1.3.255
Subnet 2
183.1.4.0/22 255.255.252.0
Possible hosts: 1024 – 2 =1022
2048
1
1
1
1
1024
1
1
1
1
512
0
0
1
1
256
0
0
0
1
128
0
0
0
0
64
0
0
0
0
32
0
0
0
0
16
0
0
0
0
8
0
0
0
0
4
0
0
0
0
2
0
0
0
0
1
0
0
0
0
Network address: 183.1.4.0
Hosts
183.1.4.1 – 255
183.1.5.0 – 255
183.1.6.0 – 255
183.1.7.0 – 254
Broadcast address: 183.1.7.255
Subnet 3
183.1.8.0/23 255.255.254.0
Possible hosts: 29 – 2 = 510
Network address: 183.1.8.0
Hosts
183.1.8.1 – 255
183.1.9.0 – 254
Broadcast address: 183.1.9.255
Subnet 4
183.1.10.0/24
255.255.255.0
Possible hosts: 28 – 2 = 254
Network address: 183.1.10.0
Hosts
183.1.10.1 – 254
Broadcast address: 183.1.10.255
SIMULATION ON GNS3 (4 ROUTERS)