Foundation Networks Assignment 2 Solution - Networking Fundamentals

This document presents a comprehensive solution to Foundation Networks Assignment 2, covering a range of networking concepts. The solution begins with calculations related to data transmission rates, throughput, and stop-and-wait ARQ strategies, including the determination of throughput for a point-to-point link. It then explores topics such as bit error rates, persistence algorithms (p-persistence, 1-persistence, and non-persistence), and the 5-4-3 rule in Ethernet networks. The solution also analyzes the advantages and disadvantages of IEEE standards like 802.3, 802.4, and 802.5. Further, it delves into calculations of signal span, slot durations, and the number of slots containing frames. The assignment also addresses IP address classes (A, B, and C), hexadecimal to decimal conversion, and the conversion of decimal to binary. Finally, it outlines the two key functions of network management: performance management and fault management, providing a well-rounded understanding of the subject matter.

Running head: FOUNDATION NETWORKS
Assignment 2
Foundation Networks
Name of the Student
Name of the University
Author’s Note

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Answer to Question 1:
Rate of data transmission = 15 kbps
Frame size, n = 896 bytes
Information bytes per frame, k = 868 bytes
Delay in propagation, td = 30 ms
Acknowledgement size, ta = 8 bytes
Round trip delay in process, tp = 10 ms
tf = 896 * 8/8000
= 0.896
ta = 8/8 /8000
= 0.008
tp = 15 * 8/8000
= 0.015
td = 30 * 8 /8000
= 0.03
tf + ta + tp + 2 td = 0.896 +0.008+ 0.015 + 0.06
= 0.979
Throughput k = 868 *8/0.979 = 6944/0.979 = 7.092 kbps

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a = td / tf = 0.03 /0.896
=0.03
1/(1+2a) = 1/(1+0.06)
= 94.33%
Answer to Question 2:
BER, E = 0.001.
Probability of a bit being error free = 1 - 0.001 = 0.999.
If the frame length, n, is 256 bytes then the probability of the frame being error free is (0.999)256
= 0.774 (0.774042818860)
The probability of a frame being in error is given by: P = 1 – 0.774 = 0.226
Answer to Question 3:
p-persistence If the station wanting transmission needs to
wait for a random time and it is picked
between 1-0. Then the availability of the
station is needed to be checked again, if it is
found busy then a random number is needed
to be picked again until an appropriate
number is found and an available station is
found available
1-persistence The algorithm used for the station that needs

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to wait for the transmission until the channel
is available and transmit data over it [2]. It is
riskiest since there are multiple messages
waiting and can cause collision and wait for
the station becoming available and transmit
the data suddenly.
Non-persistence It is a combination of 1persistent and non-
persistent for reducing the collision numbers
that can occur in the algorithms. A time
management slot is used in the station for the
transmission of the messages. A random
number is needed to be picked and it
transmits if the number < probability.
Answer to Question 4:
a.
1 Mbps = 1,000,000 bps
Convert Mbps to bps
50 x 1,000,000 = 50,000,000 bps
With a utilization of 60%
50,000,000 x 0.6 = 30,000,000 bps

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Thus, number of information bits transmitted per second is 30,000,000
b.
From a. 30,000,000 bps convert to Mbps
30,000,000 / 1,000,000 = 30 Mbps
S = 30 Mbps
Number of frame = Average number of bit per frame/(S × transmission time)
Number of frame = (800*8)/(30*5)
= 6400/150
Thus, number of frame transmitted within 5 second = 42.67
Answer to Question 5:
5-4-3 is used for demonstrating the pattern of usage of Ethernet network and specifying
the number of repeaters and segments of the network. The rule can help in avoidance of conflicts
and specifies that number of segments of the network i.e. 5 (max) and it is needed to be
connected with 4 repeaters and the 3 segments of the network can be connected with end users
[4].
Answer to Question 6:
IEEE Advantage Disadvantage
802.3 It uses a large platform and
specialist are needed for its
installation.
There are different
components used for collision
avoidance

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It uses basic protocols
New stations can be added
without shutting down the
whole network.
The used protocols cannot
work for providing real-time
update
There is limitation of cable
length
On increases loads the
chances of collision can
increase.
802.4 Good and reliable cable is
used
There are no limitations for
frame
It is sued for handling high
loads on the network
It is a complicated protocol
Lots of delay is faced for low
network loads.
802.5 Point to point connection
All transmission media can
be used
Collapse of cable can be
avoided with the use of token
ring for the wire center [1]
Suited for the large and shot
frame
It can handle high load
Delay is noticed for low loads
on the network.

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efficiently
Answer to Question 7:
tf = 512 × 1 ns
= 0.512 μs
Span = v × tf
= 2×108 × 0.512/2 μs
=51.20 m.
Answer to Question 8:
Duration of 1 bit = 1/(150*106)
= 66.67 ns
a. 40 bytes
Number of slots = 125 μs/(40*8*66.67*10-9)
= 13.02
Thus, number of slots containing 1 frame is 13.02
b. 4096 bits
Number of slots = 125 μs/(4096*8*66.67*10-9)
= 0.572

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Answer to Question 9:
a. The following are the three classes of IP addresses i.e. A, B and C and the number of
network, number of hosts and the address range are tabulated below:
Class Network Number Hosts Address Range
A 27-2 224-2 1.0.0.1 –
126.255.255.254
B 214 216-2 128.0.0.1 –
191.255.255.254
C 221 28-2 192.0.0.1 –
223.255.255.254
b. AF81D32C = 10 15 8 1 13 3 2 12
Conversion of hexadecimal to decimal
10 * 167 = 2684354560
15 * 167 = 4,026,531,840
8 * 167 = 2,147,483,648
1 * 167 = 268,435,456
13 * 167 = 3,489,660,928
3 * 167 = 805,306,368
2 * 167 = 536,870,912
12* 167 = 3,221,225,472
Dotted decimal format = 171.798.691.84
Conversion of decimal to binary
(171)10 = (10101011)2

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(798)10 = (1100011110)2
(691)10 = (1010110011)2
(84)10 = (1010100)2
Answer to Question 10:
The following are the two functions of network management:
Performance management – it is used for management and monitoring the different
parameters and performance measures of a network [3]. It is an essential function for enabling
service provider to improve the quality of service guaranteed to clients and ensuring that the
client comply with requirements.
Fault management – it is used for finding failures and isolation of the components that
have failed. The traffic is needed to be restored that gets disrupted for any failure and is
considered as a separate function.

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References
[1]C. Kao, Network data envelopment analysis. .
[2]R. Lemaire, "The Functions of Network Executives: A Goal-directed Network Management
Framework", Academy of Management Proceedings, vol. 2017, no. 1, p. 16172, 2017. Available:
10.5465/ambpp.2017.16172abstract.
[3]E. management, "Explain different functions of network management", Ques10.com, 2020.
[Online]. Available: https://www.ques10.com/p/5516/explain-different-functions-of-network-
managemen-2/. [Accessed: 05- Feb- 2020].
[4]R. Stair and G. Reynolds, Fundamentals of information systems. .