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Routing and Switching Assignment #2: Instructor: Ali Ahmad Siddiqui

The document contains two questions regarding IP address distribution and subnetting. Question 1 asks to distribute a range of IP addresses equally among 7 companies and identify which company owns a given IP. Question 2 asks to distribute IP addresses in blocks to 6 departments and identify which department a given IP belongs to, along with leftover addresses. The answers provide the detailed steps and calculations to divide the IP ranges into subnets and assign them to the corresponding companies/departments based on the requirements.

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Shahzad Ali
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151 views4 pages

Routing and Switching Assignment #2: Instructor: Ali Ahmad Siddiqui

The document contains two questions regarding IP address distribution and subnetting. Question 1 asks to distribute a range of IP addresses equally among 7 companies and identify which company owns a given IP. Question 2 asks to distribute IP addresses in blocks to 6 departments and identify which department a given IP belongs to, along with leftover addresses. The answers provide the detailed steps and calculations to divide the IP ranges into subnets and assign them to the corresponding companies/departments based on the requirements.

Uploaded by

Shahzad Ali
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Routing and Switching

Assignment #2
Instructor: Ali Ahmad Siddiqui

Student Name: _________________


Registration Number: _________________
Program: _________________

Question 1:
Distribute the range of IP address equally among 7 companies. One of the IP address from
the range is 100.2.30.40/11. Which company owns the given IP address? Write down all
subnets with their network and broadcast addresses

ANSWER:
100.2.30.40/11 has range from 100.2.30.41 - 100.2.30.255 with total 214 addresses
Now if I would like to divide the above subnet into 2 with 106 addresses each, then It would be

The 1st address is network address and last is broadcast address in any subnet. so including them
would be total IP addresses but those are not usable IP addresses, so now it will be 108-2 = 106

100.2.30.40/12 - network address with range 100.2.30.41 - 100.2.30.147

100.2.30.148/12 - broadcast address and 2nd one being 100.2.30.149/12 with range 100.2.30.149
- 100.2.30.254

Now divide into 12 - 100.2.30.40/13 with 16 addresses each

100.2.30.40/13 range 100.2.30.41 - 100.2.30.57

100.2.30.59/13 range 100.2.30.60 - 100.2.30.76

100.2.30.78/13 range 100.2.30.79 - 100.2.30.95

100.2.30.97/13 range 100.2.30.98 - 100.2.30.114

100.2.30.116/13 range 100.2.30.117 - 100.2.30.133

100.2.30.135/13 range 100.2.30.136 - 100.2.30.152

100.2.30.154/13 range 100.2.30.155 - 100.2.30.171

100.2.30.173/13 range 100.2.30.174 - 100.2.30.190


100.2.30.192/13 range 100.2.30.193 - 100.2.30.209

100.2.30.211/13 range 100.2.30.213 - 100.2.30.229

100.2.30.231/13 range 100.2.30.232 - 100.2.30.248

100.2.30.250/13 range 100.2.30.251 - 100.2.30.254

Like wise you can divide one large subnet into multiple small subnets. You should know the rule
of octet and subnetting.

Subnets can be divided into groups(2,4,8,16….etc) and each group can have
2,4,8,16,32,64,128,255,1204….etc

Question 2:
Distribute the block of IP addresses in to 6 departments. First two gets 400 IP addresses
each. Third department has 150 employees and 180 IP addresses are equally divided
among last three departments. One of the IP address from the block is 10.1.2.3 /8. Which
department the given IP address belongs to and how many IP addresses will be left for
future use

ANSWER:

Department Subnet Host Address Host Address Broadcast Required Subnet


NO Address Start Range End Range Address Address Capacity
Dept 1 10.0.0.0 10.0.0.1 10.0.1.254 10.0.1.254 400 512
Dept 2 10.0.2.0 10.0.2.1 10.0.3.254 10.0.3.254 400 512
Dept 3 Student Name:
10.0.4.0 _______________________
10.0.4.1 10.0.5.254 10.0.5.254 150 512
Dept 4 10.0.6.0 10.0.6.1 10.0.7.254 10.0.7.254 60 512
Dept 5 10.0.8.0 10.0.8.1 10.0.9.254 10.0.9.254 60 512
Dept 6 10.0.10.0 10.0.10.1 10.0.11.254 .0.11.254 60 512

Registration Number: _______________________


Network Address Block: 10.1.2.3/8

Program: _______________________
Subnet Mask: 255.255.254.0

No. of Hosts/Subnet: 512

Number of Subnets: 32768

IP Address left for future use in IP block: 16777216 - 3072 = 16774144

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