COIT20261 Network Routing and Switching Written Assessment-2

COIT20261 Network Routing and Switching (Term 1, 2018)
Assignment item —Written Assessment-2
First Name:_________________________ Last Name:____________________________
Student ID: __________________________
Question Number Mark
allocated
Marks
earned
Question 1: (10 marks)
1.
2.
3.
4.
5.
6.
7.
8.
Next hop address is 150.3.0.3/16
Interface is m2
Number of hops is 2
First option, Router discarded the packets. Second option, it forwards the
packet out the interface indicated by the default route entry
Interface is m0
Next hop address is 150.3.0.3/16
Destination network is 150.3.0.0/16
q.8 Routing table of router R2:
Prefix Network address Next-hop address Interface
16 150.3.0.0 Direct -
18 150.32.0.0 150.3.0.1 m0
16 Default Network
(150.3.0.0)
- m0
Show the masks in longest mask order using CIDR format
Network address Longest Mask
150.3.0.0 255.255.255.248 (/29)
Given subnet mask is 16. Here only four IP address need. So that we reduce the
mask.
1-7 1
mark
each, q.8
3 marks

COIT20261 Network Routing and Switching (Term 1, 2018)
Assignment item —Written Assessment-2
Question 2: (5 marks)
a) The maximum size of data field in each fragment is (1500 -20= 1480 where IP
Header size 20 bytes
Number of Fragment required = (5400-20) / 1480 =3.6. So the required
fragment is 4
Fragment Bytes Starting Byte Ending Byte Flag
1st Fragment 1480 0 1479 1
2nd Fragment 1480 1480 2959 1
3rd Fragment 1480 2960 4439 1
4th Fragment (5380 –
1480-1480-
1480) = 940
4440 5379 0
First three fragment of flag is 1. Fourth fragment of flag is 0 (Here 1 is pending
fragment, 0 indicates that this is the last fragment)
2.5
b) Fragment Bytes Offset
1st Fragment 1480 0
2nd Fragment 1480 1480/8 = 185
3rd Fragment 1480 2960/8=370
4th Fragment (5380 –
1480-1480-
1480) = 940
4440=555
1.5
c) First three fragment size is 1480. Last fragment size is 940.
From all four fragment is less than the initial datagram size
1

COIT20261 Network Routing and Switching (Term 1, 2018)
Assignment item —Written Assessment-2
Question 3: (10 marks)
1. Write a brief summary of the congestion controls currently available in TCP as
covered in this Unit
Quote from first link:
“TCP was designed to slow down how fast it sends traffic when it senses
congestion, which it determines by monitoring the number of packets lost in
transport”
Our Explanation:
First, TCP sends the data with a small value of window size. If no loss occurs in
segment, TCP increasing the window size to send the data
If loss occurs in the transmission, TCP connection decreases the window size. If
it finds the unused bandwidth in the network, it automatically increases
window size. This process continues simultaneously
It has few security issues. Hackers may generate the duplicate
acknowledgement (ACK) and send to TCP end point. This may result in
congestion collapse of the network
1
2. Identify and explain two problems with current congestion controls in TCP
that are pointed out in the articles
From the first articles: loss-based congestion control can result in abysmal
throughput because it overreacts, multiplicatively decreasing the sending rate
upon packet loss
Low Bottleneck Bandwidth or Queuing Delay Increase :
If overflow the bandwidth in the path, packet loss occurs. TCP decreases the
sending rate upon the packet loss. Time delay occurs when the low bottleneck
bandwidth in the path of the network
From the second articles: loss-based congestion control causes the
"bufferbloat" problem
Bufferbloat: Undesirable latency that comes from network devices (router or
switch or etc.). Network devices should manage the flow of traffic. Otherwise it
should create interruption of network connection.
2