Foundation Networks Assignment 2 Solutions - Electrical Engineering

This document provides comprehensive solutions to Foundation Networks Assignment 2. The solutions cover a range of topics including point-to-point links, ARQ strategies, throughput and link efficiency calculations, bit error rates, the p-persistence algorithm, CSMA/CD networks, frame length and transmission calculations, the 5-4-3 rule, IEEE 802.3 and other frame standards, DQDB bus operation, IP version 4 addressing (classes, hexadecimal conversion), and network management functions such as performance and fault management. The solutions are detailed, showing step-by-step calculations and explanations for each problem. The document is designed to aid students in understanding and solving networking problems, providing a valuable resource for their studies in electrical engineering and related fields. It includes calculations and explanations to help students learn and solve similar problems.

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Foundation Networks
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Solution 1.
A point – to - point link is operating at 10 kbps uses a stop – and - wait ARQ strategy. The
link has a propagation delay of 25ms and a total processing delay of 15ms. Data is
transmitted using a frame size of 898 bytes of which 870 bytes contain information. The
acknowledgements contain 8 bytes.
Throughput = ? = Tt x B / ( Tt + 2 Tp ) = 15 x 10 kbps / ( 15 + 2 * 25 ) = 2.3 kbps
Link efficiency = ? = ( 2.3 / 10 ) * 100 = 0.23 = 23 %
The throughput is 2.3 kbps. The link efficiency is 23 %.
Solution 2.
A frame of data of length 512 bytes is transmitted over a link with a bit error rate of 10−4.
Probability that a frame will be received erroneously = ? = 512 * 8 * 10 − 4 = 0 . 4096
The required probability is 0.4096.
Solution 3.
Explanation : The P - persistence algorithm used with CSMA / CD networks attempts to
improve bus efficiency compared with 1 - persistence and non - persistence algorithms .
The p – persistence algorithm merges the various benefits of 1 - persistence and non-
persistence algorithms. The algorithm helps in the reduction of the number of collisions

occurring in comparison with the 1 - persistence CSMA. As compared to the non –
persistence CSMA, the channel utilization is good in this algorithm. In this algorithm, if a
transmission station wants to send a frame and finds that the channel is busy, then it will wait
till the transmission ends. After this happens, it transmits the frame having the probability ‘ p
’.
Solution 4.
A CSMA / CD - based LAN operates at 100 Mbps with a utilization of 40 %.
a. The number of information bits transmitted per second = ? = 0.6 * 100 = 60 Mbits
b. The frame length is 800 bytes of which 750 convey data
The number of frame transmitted within 4 second = ? = 60 Mbits * 4 / 800 * 8 = 37500
Solution 5. “ 5 – 4 - 3 Rule ”
This rule is used for dividing the network. It states that there can be maximum 5 segments
present between 2 given nodes in a network. They can be connected by 4 repeaters /
concentrators. Out of the 5 segments , only 3 segments can have user connections. The rule is
a design method for the Ethernet computer network ( IEEE way ).

Solution 6.
IEEE 802.3 frames contain a pad and length field, while IEEE 802.4 and 5 token-based
frames contain an ED ( End – of - frame Delimiter ) field only.
The standard 802 . 3 is Ethernet , 802 . 4 is Token Bus and 802 . 5 is Token Ring. 802 . 4 has
a more deterministic nature than 802 . 3. But if the token is lost, there can be uncertainty. In
802 . 3, pad field checks if the frame satisfies minimum length criteria. It helps in the
detection of any collision. The length field gives length of data field. In 802 . 4, end delimiter
denotes end of frame and gives the position of frame check sequence field. In 802 . 5, ending
delimiter helps in identifying end of the frame. It can also identify if it is last frame or not
( for multi frame case ).
Solution 7.
A frame comprising 512 bits is transmitted at 1 Gbps.
Maximum network span = ? = 512 bits / 1 Gbps = 0.512 micro second
Velocity of propagation is 200,000 km / s.
Network span = ? = 200,000 km / s * 0.512 micro second = 0.2 * 512 m = 102.4 m
Solution 8.
A DQDB bus operates at 150 Mbps.

Assumption : The size of the frame header is negligible in comparison with the frame’s
length.
The number of slots contained in one frame, if each slot contains
a. 20 bytes
Slots = 150 * 1000000 / 8 * 20 = 937500 slots
b. 512 bytes
Slots = 150 * 1000000 / 8 * 512 = 36621 slots
Solution 9. IP version 4:
a. Three main classes of IP addresses.
Figure 1
There are a total of 5 classes – A, B, C, D, E.
In Class A , the first byte can range from 0 – 127. In Class B , the first byte can range
from 128 – 191. In Class C , the first byte can range from 192 – 223. In Class D , the
first byte can range from 224 – 239. In Class E , the first byte can range from 240 –
255.

Class Bits Use
A First bit is ‘ 0 ’ Large Organizations
B First 2 bits are ‘ 1 0 ’ Mid - Size Organizations
C First 3 bits are ‘ 1 1 0 ’ Small Organizations
D First 4 bits are ‘ 1 1 1 0 ’ Multicasting
E First 5 bits are ‘ 1 1 1 1 ’ For Fibres
b. IP address is expressed in hexadecimal form as F B 5 3 D 2 0 4 .
Conversion into dotted decimal format
F B 5 3 D 2 0 4 – Hexadecimal
1 1 1 1 1 0 1 1 0 1 0 1 0 0 1 1 1 1 0 1 0 0 1 0 0 0 0 0 0 1 0 0 – Binary
251 . 83 . 210 . 4 ( Dotted decimal format )
Solution 10. Two functions of network management :
Network management has various functions. The 2 important functions are :
1) Performance Management : The different parameters for measuring the network
performance are monitored and managed. It helps in the provision of quality service
to the clients.
2) Fault Management : It helps to detect any failure happening and leads to the isolation
of the components failed. The traffic also needs to be restored by the network.