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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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Related Questions
An ISP is granted a block of addresses starting with 130.100.0.0/16. The ISP needs to distribute
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1
2
4
3.
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QUESTION 3:
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Network Subnet
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A
B
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D
E
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4. Network D has 190 hosts
5. Network E has 15 hosts
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FGH
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.
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%3D
%3D
2001:540:FACE:20:40/32
2001:540:FACE:20:40/64
2001:540:FACE::40:20/32
2001:540:FACE::40:20/64
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45-
An organization currently requires 40 hosts for its sales department.
Given the IP Address 194.12.1.0 /24, design an optimal address scheme usingTradition Subnetting method, which will enable the connection of the sales department’s 40 hosts in one subnet and uses the remaining subnets for other departments’ hosts in the future.
What will be the network address of 2nd subnet?
Select one:
A.
194.12.11.64
B.
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C.
194.12.1.64
D.
94.12.1.95
E.
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44- An organization currently requires 40 hosts for its sales department.
Given the IP Address 194.12.1.0 /24, design an optimal address scheme usingTradition Subnetting method, which will enable the connection of the sales department’s 40 hosts in one subnet and uses the remaining subnets for other departments’ hosts in the future.
What will be the total number of usable (assignable) addresses per subnet?
Select one:
A.
62
B.
65
C.
63
D.
64
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. Consider a network with prefix 182.168.56.128/26. Give an example of one IP address (of form xxx.xxx.xxx.xxx) that can be assigned to this network. Suppose an ISP owns the block of addresses of the form 170.168.56.64/26. Suppose it wants to create four subnets from this block, with each block having the same number of IP addresses. What are the prefixes (of form a.b.c.d/x) of the four subnets?
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You are given with an IP address of 112.200.67.132/19. You are asked to make 6 subnets of size 60, 200, 25, 64, 4 and 12. Now, based on your calculation answer the followings:
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An ISP is granted a block of addresses starting with 150.80.0.0/14 . The ISP wants to distribute these blocks to customers as follows:A- the first group has 50 customers, each needs 1024 IP addresses. B- the second group has 4 customers, each needs 512 IP addressesC- the third group has 64 customers, each needs 16 IP addressesSo, Design all the subnets? And find the amount of available IPs after all allocations.
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• Given the IP address 165.165.8.0/21. We want to divide this networks into 7
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• Network 3: 225 hosts
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Answer the following questions:
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2. The first IP address in Network 2 is:
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49-
An organization currently requires 40 hosts for its sales department.
Given the IP Address 194.12.1.0 /24, design an optimal address scheme usingTradition Subnetting method, which will enable the connection of the sales department’s 40 hosts in one subnet and uses the remaining subnets for other departments’ hosts in the future.
What will be the last host address of 4th subnet?
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A.
194.12.1.254
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194.12.1.192
C.
194.12.1.253
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194.12.1.255
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Assignment Module 11
For the following Ips, write the network name, give 2 Ips in the same network,
find the number of the possible Ips
Net name
1" ip
2nd IP
no. of valid IPs
• 11.45.3.5 →
• 180.90.27.89-→
• 190.35.177.98 →
• 192.76.19.1
• 200.30.50.90>
• 100.100.100.100>
107.39.50.20→
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48-
An organization currently requires 40 hosts for its sales department.
Given the IP Address 194.12.1.0 /24, design an optimal address scheme usingTradition Subnetting method, which will enable the connection of the sales department’s 40 hosts in one subnet and uses the remaining subnets for other departments’ hosts in the future.
What will be the range of usable (assignable) addresses in 4th subnet?
Select one:
A.
194.12.1.192 to 194.12.1.255
B.
194.12.1.193 to 194.12.1.254
C.
194.12.1.193 to 194.12.1.253
D.
194.12.1.192 to 194.12.1.254
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