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Authors:
James F. Kurose, Keith W. Ross
Chapter:
The Network Layer
Exercise:
Problems
Question:13 | ISBN:9780132856201 | Edition: 6

Question

Consider a router that interconnects three subnets: Subnet 1, Subnet 2, and Subnet 3. Suppose all of the interfaces in each of these three subnets are required to have the prefix 223.1.17/24. Also suppose that Subnet 1 is required to support at least 60 interfaces, Subnet 2 is to support at least 90 interfaces, and Subnet 3 is to support at least 12 interfaces. Provide three net- work addresses (of the form a.b.c.d/x) that satisfy these constraints.

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Answer

Three net- work addresses (of the form a.b.c.d/x) that satisfy given constraints as follows:

  • 223.1.17.0/25

  • 223.1.17.128/26

  • 223.1.17.192/26

 

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Discussions

Muhammad Ehsan Ullah

A Sample Subnetting Problem

 

Consider a router that interconnects three subnets: Subnet 1, Subnet 2 and Subnet 3.  Suppose all the interfaces in each of these three subnets are required to have the prefix 230.1.17/24.

 

Also suppose that Subnet 1 is required to support up to 60 interfaces, Subnet 2 is required to support up to 90 interfaces and Subnet 3 is required to support up to 12 interfaces.

 

Provide three network addresses  (of the form a.b.c.d/x) that satisfy these constraints.

 

Calculations

 

Recall:     

• The network address cannot be used for an interface (Network prefix + all zeros).

• The broadcast address cannot be used for an interface (Network prefix + all ones)

 

Then:       

 

Subnet 1 requires 60 interfaces + 2 (network + broadcast) =>

so 64 addresses will sufice

 

Subnet 2 requires 90 interfaces + 2 (network + broadcast)  =>

so 128 addresses will sufice

 

Subnet 3 requires 12 interfaces + 2 (network + broadcast) =>

so 16 addresses will sufice

 

 

One Solution that Works:

 

Sub 2 230.1.17.0/25      à 223.1.17.00000000/25    à 223.1.17.0  to 223.1.17.127           = 128

Sub 1 230.1.17.128/26  à 223.1.17.10000000/26    à 223.1.17.128 to 223.1.17.191        =   64

Sub 3 230.1.17.192/28  à 223.1.17.11000000/28    à 223.1.17.192 to 223.1.17.207         =  16

 

Post the discussion to improve the above solution.