Advanced Routing: Algorithms, IP Addressing, and Network Configuration

This assignment solution addresses advanced routing concepts, including the Dijkstra's shortest path algorithm, distance-vector algorithm, and the count-to-infinity problem. It covers IP addressing, subnetting, and IPv6, providing calculations for network addresses and subnet masks. The solution also details IP routing and configuration, including the configuration of a core router with EIGRP, VLSM implementation, and DHCP server setup. Furthermore, it outlines network planning, EIGRP and IGRP configurations in branch offices, internet connectivity setup using static routes, and documentation of configurations. The document also discusses implementation issues, limitations, and suggestions for improvement, such as enhancing network security and implementing wireless routers. The assignment concludes with a bibliography of relevant resources.

Running head: ADVANCED ROUTING
Advanced Routing
Name of the Student
Name of the University
Author’s Note

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ADVANCED ROUTING
Answer to Question 1: Routing protocols
a. Working of Dijkstra’s Shortest path algorithm
Step N’ D(t),p(t) D(u),p(u) D(v),p(v) D(w),p(w) D(y),p(y) D(z),p(z)
0 X ∞ ∞ 3, x 6, x 5, x 7, x
1 Xv 6, v 6, x 3, x 6, x 5, x 7, x
2 Xvu 6, v 6, x 3, x 6, x 5, x 7, x
3 Xvuw 6, v 6, x 3, x 6, x 5, x 7, x
4 Xvuwy 6, v 6, x 3, x 6, x 5, x 7, x
5 Xvuwyt 6, v 6, x 3, x 6, x 5, x 7, x
6 Xvuwytz 6, v 6, x 3, x 6, x 5, x 7, x

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ADVANCED ROUTING
b. Computation of distance tables after the initialization step and after each iteration of the
distance-vector algorithm.
Table: Node x
Cost to
x y z
x 0 3 4
From y ∞ ∞ ∞
z ∞ ∞ ∞
Cost to
x y z
x 0 3 4
From y 3 0 6
z 4 6 0
Table: Node y
Cost to
x y z
x ∞ ∞ ∞
From y 3 0 6
z ∞ ∞ ∞

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ADVANCED ROUTING
Cost to
x y z
x 0 3 4
From y 3 0 6
z 4 6 0
Table: Node z
Cost to
x y z
x ∞ ∞ ∞
From y ∞ ∞ ∞
z 4 6 0
Cost to
x y z
x 0 3 4
From y 3 0 6
z 4 6 0
c. The count to infinity problem would not occur in case of decreasing the cost of the link
because no loops could be created by decreasing the cost of the link. If two nodes are connected
that don’t have a link it would decrease the weight of the link form infinity to a finite weight.

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ADVANCED ROUTING
Answer to Question 2: IP Addressing
a. The calculation of the network address for the IP address 23.1.17/24.
Subnet 1: It requires 63 interfaces + 2 (broadcast + network) = thus 128 number of
addresses
Subnet 2: It requires 95 interfaces + 2 (broadcast + network) = thus 128 number of
addresses
Subnet 3: It requires 16 interfaces + 2 (broadcast + network) = thus 32 number of
addresses
Subnet 1: 223.1.17.128/ 25  223.1.16.129 - 223.1.16.254
Subnet 2: 223.1.17.0 /25  223.1.16.1 - 223.1.16.126
Subnet 3: 223.1.18.0/ 27  223.1.17.1 - 223.1.17.30
b. IP address within the range
128.119.40.128 – 128.119.40.191
The prefixes for the four subnets identified are given below:
i. 128.119.40.64/28
ii. 128.119.40.80/28
iii. 128.119.40.96/28, and
iv. 128.119.40.112/28
c. Short note on Ipv6 – the IPv6 address is used as a replacement of the Ipv4 address for fixing
the problems of limited number of IP address and addition if mire improvement such as auto

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configuration of the routing and the network. The length of the Ipv6 address are 128 bits and acts
as an identifier for the interfaces of the network device. A multiple IPV6 address can be assigned
to a single interface such as unicast address, anycast address and multicast address.
Answer to Question 3: IP Routing and Configuration
1. Configuration of the Core Router with EIGRP
SanJones 1

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SanJones 2
SanJones 3

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2. Allocation of one Class B IP address
3. Use of VLSM on all serial Interface
Number
of host
addresse
s
required
Number
of host
address
required
Subnet
network
address
Subnet mask Subnet
/ Value
Max
numbe
r of
hosts
possibl
e
Futur
e Use
Y/N
VLAN
Name
1 SanJose 1 192.168.5.225 255.255.255.252 /30 2 Y SanJose 1
1 192.168.5.234 255.255.255.252 /30 2 Y SanJose 1
60 192.168.5.129 255.255.255.192 /26 62 Y SanJose 1
1 192.168.0.193 255.255.255.192 /30 2 Y SanJose 1
1 SanJose 2 192.168.5.230 255.255.255.252 /30 2 N SanJose 2
1 192.168.5.233 255.255.255.252 /30 2 N SanJose 2
30 192.168.5.193 255.255.255.224 /27 30 N SanJose 2
2 SanJose 3 192.168.5.226 255.255.255.252 /30 2 N SanJose 3
2 192.168.5.229 255.255.255.252 /30 2 N SanJose 3
126 192.168.5.1 255.255.255.128 /25 126 N SanJose 3
2 Singapor
e
192.168.5.237 255.255.255.252 /30 2 N Singapor
e

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254 192.168.20.1 255.255.255.0 /24 254 Y Singapor
e
2 Auckland 192.168.20.2 255.255.255.0 /24 254 Y Auckland
30 Westasm
an
192.168.5.130 255.255.255.224 /27 30 Y Westasm
an
2 Vista 192.168.5.194 255.255.255.252 /30 2 N Vista
254 192.168.10.1 255.255.255.0 /24 254 Y Vista
254 Eastasma
n
192.168.10.2 255.255.255.0 /24 254 Y Eastasma
n
Internet 172.16.0.1 255.255.255.240 /28 14 Y Internet

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4. EIGRP configuration in Singapore and Auckland and IGRP configuration in Vista and
Eastasman branch offices

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5. Network Plan
6. Configuration of DHCP server in Vista and Eastasman

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ADVANCED ROUTING
7. Configuration of Internet Connectivity using a Static route
8. Documentation of the configurations
The main problem was faced during configuration of the interface with the IP address and
maintaining 99.9 percent availability of the IP address in the network.
9. Implementation Issues and limitations
For the implementation of the network the scalability and performance of the network
should be evaluated and the data packets should transverse to all the needs of the network.
During the testing of the network the data transfer of data may be sometimes unsuccessful due to
designing the network in the simulation software.
10. Two suggestions for the improvement of the configuration and design
The two suggestion for the improvement of the design are given below: