4. IP Addresses and Subnets (10 pt. 5 bonus pt) (show all the details/work) a. W

Question

4. IP Addresses and Subnets (10 pt. 5 bonus pt) (show all the details/work) a. What is the 32-bit binary equivalent of the IP address 223.1.3.28 pt) b. Consider a subnet with prefix 128.119.40.128/26. Give an example of one IP address (of form xxx.xxx.xxx.xxx or a.b.c.d) that can be assigned to this network (must show all the details/work include all assumptions). (1 pt) c. 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 network addresses (of the form a.b.c.d/x) that satisfy these constraints. (8 pt) (show all the details/work - include all assumptions) d. Suppose an ISP owns the block of addresses of the form 130.64.128.0/21. The ISP wants to divide its address block into four equal-sized contiguous address blocks and give one of these address blocks out to each of up to four organizations (A, B, C, and D) that are supported by this ISP (i) Provide four network addresses (of the form a.b.c.d to w.x.y.z) for organization A, B, C, and D (ii) Provide four network addresses (of the form a.b.c.d/x) for organization A, B, C, and D. show all the details/work - include all assumptions) 6 bonus points-2.5 each)

Explanation / Answer

Answers :-

a.) 11011111.00000001.00000011.00011011

b.)Broadcast address of given subnet = 128.119.40.10111111/26 = 128/119/40.191/26 Any IP address in the range of 128.119.40.129 to 128.119.40.190 (between and excluding network and broadcast addresses) can be assigned. b. It is required to create four subnets from the block of addresses 128.119.40.64/26. To do so, we need to borrow 2 bits for the network part from the host part because borrowing 2 bits gives a combination of 2^2= 4 subnets. After doing so, we now have 28 bits for the network part and 4 bits for the host part. a. This means that each of the four subnets will have a prefix of a.b.c.d/28, and will be capable of supporting up to 2^4 - 2= 14 host addresses.

c.)

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.