ITEC647 Data Communications Assignment 1 Solution: 2019

This document presents a solution to the ITEC647 Data Communications assignment, addressing key concepts in networking. The solution begins by outlining the network stack with source, next hop, and destination addresses. It then delves into IP addressing, providing answers to Boolean operations on binary data and calculating subnet masks for given slash address blocks. The document identifies valid subnet masks and their slash forms, along with the number of networks and hosts available for various address blocks. The assignment demonstrates an understanding of fundamental networking principles, including subnetting and IP addressing, which are essential for effective data communication. Finally, the document includes a bibliography of relevant academic sources.

Running head: ITEC647 DATA COMMUNICATIONS
ITEC647 Data Communications
Name of the Student
Name of the University
Author’s Note

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ITEC647 DATA COMMUNICATIONS
Answer to Question 1 – Network Stack
Source Address – 10.54.4.6
Next Hop Address – 137.111.0.1
Next Hop Address – 154.65.0.1
Destination Address – 154.65.3.45
Answer to Question 2 – IP Addressing
a. Boolean operation for the given bits
i. not 1110 1010 1010 1110
Result 111111110101000100010101
ii. 1010101011101100 or 1110101010101110
Result in binary 1110101011101110

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ITEC647 DATA COMMUNICATIONS
iii. 1010101011101100 and 1110101010101110
Result in binary 1010101010101100
b. Subnet mask for the given slash address blocks
i. /6 –
= 252.0.0.0
ii. /17 –
= 255.255.128.0
iii. /25 –
= 255.255.255.128
c. Identification of legal subnet mask and their slash forms
i. 255.255.252.0 –
Binary form - 11111111.11111111.111111 00.00000000
Notation – (8+8+6) = /22
ii. 255.240.240.0 –
Binary form – 11111111. 11110000. 11110000.00000000
Notation – invalid
The 1st Ip address in a subnet can be 0’s in the third octet and the address of
the nodes can be represented by the octets containing 0s and 1s and thus if the
fourth octet is 0 the node octet would also be 0 for denoting the subnet.
iii. 255.255.255.226 –
Binary form – 11111111. 11111111. 11111111. 11100010
Notation – invalid

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ITEC647 DATA COMMUNICATIONS
The 1st Ip address in a subnet can be 0’s in the third octet and the address of
the nodes can be represented by the octets containing 0s and 1s and thus if the
fourth octet is 0 the node octet would also be 0 for denoting the subnet.
iv. 255.255.194.0 –
Binary form - 11111111.11111111. 11000010.00000000
Notation – invalid
The 1st Ip address in a subnet can be 0’s in the third octet and the address of
the nodes can be represented by the octets containing 0s and 1s and thus if the
fourth octet is 0 the node octet would also be 0 for denoting the subnet.
d. Network and hosts are in the network for the following address blocks are given below:
i. /18 – 255.255.192.0
Network = 16384
Hosts = 16382
ii. /25 – 255.255.255.128
Network = 128
Hosts = 126

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ITEC647 DATA COMMUNICATIONS
Bibliography
Meng, M., Lam, J., Feng, J. E., & Li, X. (2016). l1-gain analysis and model reduction
problem for Boolean control networks. Information Sciences, 348, 68-83.
Vardy, A. (2016). Subnetting For Beginners: How To Easily Master IP Subnetting And
Binary Math To Pass Your CCNA-CCNA, Networking, IT Security, ITSM.
CreateSpace Independent Publishing Platform.