CS206 Assignment 2 Solution and Discussion
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Assignment No. 02
Semester: Fall 2019
CS206: Introduction to Network Design & Analysis Total Marks: 20Due Date: 28 Nov 2019
Instructions:
Please read the following instructions carefully before submitting assignment:
You need to use MS word document to prepare and submit the assignment on VU-LMS.
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Assignment is copied (partial or full) from any source (websites, forums, students, etc).Objective:
To enhance the learning about IP allocation.
Assignment
Scenario:
An Internet Service Provider (ISP) is providing its services to 3500 customers in a residential area. If every customer is provided with a unique static IP address to avail Internet services then answer the following questions:
Question No. 1:
a) If IP address range can only be purchased as a block of 2n then how many minimum number of IP addresses should be purchased by the ISP for providing services to its customers?
b) If the ISP extends its services to another small residential area then how many maximum number of users can be accommodated using the remaining available static IP addresses from the first area?
Question No. 2:
In continuation of scenario given above, consider that maximum number of active (online) customers at a time are not more than 950 even after combining both the residential areas.a) Considering the maximum number of active users, static IP addressing is not suitable for the ISP. Which alternative approach (or protocol) can be used to minimize the number of IP address requirement of the ISP to continue its service for customers?
b) How many minimum number of IP addresses should be purchased by the ISP for providing services to its customers considering the maximum active users combining both the areas?
Best of Luck!
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Solution 1:
a) As total number of customers are 3500 in the area, so the ISP will need 3500 hundred IP addresses for assigning unique IP address to the customers. But IP addresses can only be purchased in a range according to multiple of 2 i.e., number of IP addresses should be according to 2n. So,
Minimum number of static addresses required by the ISP = 2n (for minimum ‘n’ such that 2n >=3500)
= 212 = 4096
So, minimum number of static IP addresses purchased by the ISP = 4096b) Number of remaining IP address = 4096 – 3500 = 596
A total of 596 IP addresses are available to be used in the extended area for providing services. So, maximum number of users in the extended area are 596 that can be provided service using the same static IP addresses.Solution 2:
a) As active number of users are much lesser, so we don’t need to assign static IP address to each customer. Dynamic address allocation technique can be used with a relatively small number of available addresses to be assigned to active users only. As the hosts join and leave, the DHCP (Dynamic Host Configuration Protocol) server allocates an arbitrary address from its current pool of available addresses. Each time a host leaves, its address is returned to the pool. DHCP is ideally suited to these situations, as there are many users coming and going, and addresses are needed for only a limited amount of time
b) As total number of active customers are no more than 950 at a time. The ISP does not need 4,096 IP addresses. Only a block of 1024 IP addresses are sufficient along with the DHCP protocol to provide seamless services to its customers.
Minimum number of IP addresses required by the ISP = 2n (for minimum ‘n’ such that 2n >=950)
= 210 = 1024 > 950
So, many minimum number of IP addresses purchased by the ISP = 1024
