WS 14 (Subnet 2)
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Pennsylvania State University *
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Dec 6, 2023
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IST 220: Networking and Telecommunications
Work Sheet 14
(Subnet 2 and CIDR)
Instructions:
- Use your class notes and text book to complete this work sheet.
- You can discuss questions with other students and instructor.
- Submit your answers (Word Document) to Canvas using the drop box before the deadline
Student Name:
Open Ended:
1.
If an organization wants to divide a network into 5 subnets with the subnet mask of 24 bits, how
many bits should be reserved for host ID
?
24 (total bits) - 3 (subnet bits) = 21 bits for host IDs.
2.
Refer to previous question, how many maximum numbers of subnets an organization can create.
2^ (3-1) = 4
3.
Refer to previous question (Question # 1), what should be the subnet mask for each new subnet?
24 + 3 = 27 bits
4.
What is the purpose of subnet mask?
The subnet mask is essential in IP addressing because it facilitates network organization, efficient use
of IP address space, and data routing between multiple subnets.
5.
What is the difference between subnet mask and network prefix?
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Mask of the subnet
A subnet mask is a 32-bit dotted-decimal value (for example, 255.255.255.0).
It is most encountered in the context of IPv4 addressing.
The subnet mask is used to determine the network and host components of an IP address
.
Network Prefix:
A network prefix is frequently coupled with Classless Inter-Domain Routing (CIDR) notation in IPv4
or used directly in IPv6 addressing.
In CIDR notation, the network prefix is stated as an IP address followed by a forward slash and the
prefix length (e.g., 192.168.1.0/24).
Problem Solving – Show all steps (do not just write
the answer)
6.
An organization would like to divide its network so that there are five separate subnets for each department.
They will need 23 hosts in each subnet. The organization main IP address is
192.168.162.
0/24. Complete
each of the following:
NOTE:
If you create more than five subnets, list the extra ones too.
24 (original bits) + 5 = 29 bits.
2^(32 - subnet mask) = 2^3 = 8
Subnet
Network address
Host addresses
Broadcast address
Subnet mask: 255.255.255. (
or 24 in CIDR
)
First subnet
192.168.162.
192.168.162.
- 192.168.162.
192.168.162.
Second subnet
192.168.162.
192.168.162.
- 192.168.162.
192.168.162.
Third subnet
192.168.162.
192.168.162.
- 192.168.162.
192.168.162.
Fourth subnet
192.168.162.
192.168.162.
- 192.168.162.
192.168.162.
Fifth subnet
192.168.162.
192.168.162.
- 192.168.162.
192.168.162.
Sixth subnet ?
192.168.162.40/29
192.168.162.41
- 192.168.162.46
: 192.168.162.47
?
?
?
7.
An organization would like to divide its network (219.7.9.0/24) so that there are 32 separate subnets. They
will need 6 hosts in each subnet. Complete the following table:
NOTE:
Because there are so many subnets, don't write them all out (unless you just want to). If you
can do the first ten and know what the last one is, you get the idea.
Subnet
Network address
Host addresses
Broadcast address
Subnet mask: 255.255.255 (
or 24 in CIDR
)
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First subnet
219.7.9./28___
219.7.9. 1___- 219.7.9. _14__
219.7.9. 15___
Second subnet
219.7.9. /28
219.7.9. _17__
- 219.7.9.30___
219.7.9. 31___
Third subnet
219.7.9.
/28
___
___
219.7.9. 33__ -
219.7.9.46 ___
219.7.9.47 ___
Fourth subnet
219.7.9. /28___
219.7.9.49 ___
-
219.7.9.62 ___
219.7.9.63 ___
Fifth subnet
219.7.9. /28___
219.7.9.65 ___
-
219.7.9.78 ___
219.7.9.79 ___
Sixth subnet
219.7.9./28
219.7.9. 81
-
219.7.9.94
219.7.9.95
Seventh subnet
219.7.9./28
219.7.9.97
-
219.7.9.110
219.7.9. 111
Eighth subnet
219.7.9./28
219.7.9.113 -
219.7.9. 126
219.7.9. 127
Ninth subnet
219.7.9./28
219.7.9. 129-
219.7.9. 142
219.7.9. 143
Tenth subnet
219.7.9./28
219.7.9. 145-
219.7.9.158
219.7.9. 159
.
.
Thirty-second
subnet
219.7.9./28
219.7.9. 481-
219.7.9. 494
219.7.9.
495
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