COIT20261 Network Routing and Switching Term 3

COIT20261 Network Routing and Switching Term 3, 2019
Assignment item —Written Assessment-2
ANSWER TEMPLATE ASSIGNMENT TWO
Type your answers in the spaces provided
Marking criteria: Your answers will be marked based on technical correctness,
completeness, clarity, originality and relevance. Proper use of referencing conventions
must be followed, and marks may be deducted for failure to comply. For discussion or
research-based questions, if you decide to include a graphic (e.g., a screenshot or a
diagram) in support of your answer, the graphic must be relevant to your discussion, be
appropriately referenced, have sufficient resolution to show all its details clearly and be
of a reasonable size for normal reader viewing, with all or any text within the graphic
being legible and readable. Originality means the work is done solely by you and is
expressed in your own words. An answer is unacceptable if it is composed mostly of
quoted material from other sources and may in some cases receive no marks as a result.
First Name:_________________________ Last Name:____________________________
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Questions Mark
allocated
Mark
earned
Question 1: (10 marks) 5 each
table
Answer: Routing table of router R3:
Mask Network address Next-hop address Interface
/24 230.10.50.0 ............ m1
/24 133.0.10.0 180.0.0.4 m2
/21 210.20.40.0 180.0.0.4 m2
/18 170.16.64.0 180.20.0.30 m0
/16 130.24.0.0 180.20.0.30 m0
/8 180.0.10.0 180.0.10.10 m1
/0 0.0.0.0 180.0.10.10 m1
Routing table of router R4:
Mask Network address Next-hop address Interface
/24 133.0.10.0 ............ m1
/24 230.10.50.0 180.0.0.10 m0
/21 210.20.40.0 ............ m0
/18 170.16.64.0 180.20.0.30 m0
/16 130.24.0.0 180.20.0.30 m0
/8 180.0.10.0 180.0.10.10 m1
/0 0.0.0.0 180.0.10.10 m1
5 max
5 max

COIT20261 Network Routing and Switching Term 3, 2019
Assignment item —Written Assessment-2
Question 2: (5 marks)
a) Packet size = 5500 bytes including header, MTU = 1500 bytes, total
packets = 4
Assuming header = 20 bytes,
Size of data in Packet = (5500 – 20) = 5480 bytes
Fragment 1 size = 1500 including header or data bytes = (1500-20) =
1480, Therefore starting byte = 0, ending byte = 1479
Fragment 2 size = 1500 including header or data bytes = (1500-20) =
1480, Therefore starting byte = 1480, ending byte = 2959
Fragment 3 size = 1500 including header or data bytes = (1500-20) =
1480, Therefore starting byte = 2960, ending byte = 4439
Therefore, data bytes of fragments 4 = (5480 – 4440) bytes = 1040 bytes
Therefore total size of fragment 4 = (1040 + 20) bytes = 1060 bytes
Therefore total size of all 4 fragments = (1500 + 1500 + 1500 + 1060) =
5560 bytes
The total size of all fragments exceeds the size of the original datagram
by 60 bytes.
2.5 max
b) Fragment offset = (1500 – 20)/8 = 185,
For the last fragment = (1060-20)/8 = 130
Therefore, fragment offset of first fragment = 0
Fragment offset of second fragment = 185
Fragment offset of third fragment = (185 + 185) = 370
Fragment offset of fourth fragment = (370 + 130) = 500
1.5 max
c) The combined size of the above four fragments is greater than the
original message datagram that arrived. This is because four separate
headers are required for the four fragments instead of one in the original
datagram. Since the size of these headers are 20 bytes each, the total
size of four fragments exceeds that of the original datagram by 60 bytes.
1 max