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ECE 363
Assignment 3
Spring 2023
1. Consider IP address X: 193.101.50.44/26 (a) Which of the IP addresses below is on the same network as X:
IP address
On the same network as X (Yes or No)?
193.101.50.1
0
Yes
193.100.50.1
1
No
193.101.50.6
5
No
To find out if any of the IP addresses belong to the same network as X, we
need to look at their network prefixes. The /26 prefix means that the subnet
mask is 255.255.255.192 when written in the usual format. By performing a
logical AND operation between the subnet mask and the IP addresses, we
can isolate their network parts. Then, we can compare this network part with
the network part for X to see if any of them share the same network.
193.101.50.44 (X): 11000001 01100101 00110010 00101100
&&
255.255.255.192: 11111111 11111111 11111111 11000000
______________________________________________
=
11000001 01100101 00110010 00000000
193.101.50.10: 11000001 01100101 00110010 00001010
&&
255.255.255.192: 11111111 11111111 11111111 11000000
______________________________________________
=
11000001 01100101 00110010 00000000
193.100.50.11: 11000001 01100100 00110010 00001011
&&
255.255.255.192: 11111111 11111111 11111111 11000000
______________________________________________
=
11000001 01100100 00110010 00000000
193.101.50.65: 11000001 01100101 00110010 01000001
&&
255.255.255.192: 11111111 11111111 11111111 11000000
______________________________________________
=
11000001 01100101 00110010 01000000
Comparing the network prefixes, we can see that 193.101.50.10 is the only
IP on the same network as X.
(b) With subnet mask 255.255.255.192, what is the maximum
number of hosts on the subnet?
A subnet mask that corresponds to /26 indicates that there are 6 bits
reserved for host addresses. This means there are 64 possible addresses for
hosts. However, two addresses—the one with all zeros (the network address)
and the one with all ones (the broadcast address)—cannot be assigned to
ECE 363
Assignment 3
Spring 2023
individual hosts. Therefore, the maximum number of hosts that can be
assigned is 64 minus 2, which equals 62.
2. An IP packet has the following information in its header arrives at
a WLAN:
...
length
ID
fragflag
offset
...
5000
x
0
0
Since the maximum transmission unit (MTU) of the WLAN is 2308
bytes, the packet will be fragmented into how many packets? What
will be the length, ID, fragflag and offset values in their IP headers?
Looking at the packet provided, we can see that the total size of the IP
packet header and data is 5000 bytes. The IP header typically takes up 20
bytes for IPv4. For each fragment, except the last one, there will also be an
additional 20 bytes for the IP header. The rest of the space is used for the
actual data payload.
Data payload size per fragment = MTU - IP header size = 2308 - 20 =
2288 bytes.
Number of fragments required = Total length / Data payload size per
fragment
= 5000 / 2288 = 2.19
Because fragments cannot be divided into smaller parts, we require three
fragments to encompass the entire packet. The initial two fragments will
contain 2308 bytes each (comprising the IP header and data payload), while
the final fragment will consist of the remaining 424 bytes, comprising only
the data payload.
The ID field assists the receiving host in recognizing which packet a newly
received fragment belongs to. Consequently, all fragments belonging to the
same packet will share the same ID value. The fragflag is necessary to
determine when all fragments of a packet have been received. Hence, all
fragments except the final one have this bit set to 1.
The offset value indicates where a particular fragment starts within the current packet. These values increase in chunks of 8 bytes. So, the offset for the initial fragment is 0. For the second fragment, it's calculated as 2288 divided by 8, which equals 286. Likewise, for the third fragment, it's calculated as 4576 divided by 8, resulting in 572. Here's how the three data segments look visually:
...
length
ID
fragflag
offset
...
2308
x
1
0
...
length
ID
fragflag
offset
...
2308
x
1
286
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58-
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Given the IP Address 192.168.15.0 /24 and above described network requirements:
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A.
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53-
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A.
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B.
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Question 3
3.1 An end device was given the IP address of 198.168.4.35/28. Consider this address and indicate:
3.1.1 The network ID/address to which the host belongs.
3.1.2 The network broadcast address to which the host belongs.
3.1.3 The total number of hosts available in the network.
3.1.4 The total number of networks.
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3.2.1 How is that possible?
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Assume that a large number of consecutive IP addresses are available starting at 198.16.32.0 and suppose that two organizations, A, B, C, and D, request 1000, 3000, 2000, and 4500 addresses, respectively, in that order. For each of these, give the first possible IP address within the assigned subnet, the last possible IP address within the assigned subnet, and the subnet representation in slash notation (w.x.y.z/s).
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Subnet mask: 255.255.0.0
Computer “B”
IP address: 172.16.200.200
Subnet mask: /16
What should be done to allow these computers to send and receive data to and from each other?
A.) Nothing needs to be changed.
B.) Delete the subnet mask from each computer.
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D.) Remove the hub between the two computers and replace it with a switch.
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46-
An organization currently requires 40 hosts for its sales department.
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Select one:
A.
194.12.11.95
B.
194.12.1.128
C.
94.12.1.191
D.
194.121.1.192
E.
194.12.1.191
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c
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d
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q no 5:
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Router1 to Router2 (R2)
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CIDR Mask
Link
0.0.0.0/0
L4
128.4.6.0/24
L3
128.4.8.0/24
L5
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VoIP always uses TCP to ensure that voice data is transmitted reliably.
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47-
An organization currently requires 40 hosts for its sales department.
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What will be the first host address in 3rd subnet?
Select one:
A.
194.12.1.190
B.
194.121.1.192
C.
194.12.11.191
D.
94.12.1.128
E.
194.12.1.129
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3.2.1 How is that possible?
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c. Network 3 needs to support 2 host addresses.
d. Network 4 needs to support 520 host addresses.
e. Network 5 needs to support 230 host addresses.
f. Network 6 needs to support 62 host addresses.
g. Network 7 needs to support 2 host addresses.
h. Network 8 needs to support 2 host addresses.
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