3. IP Subnet Addressing Requirements You must use IP address space: 200.1.59.0/2

Question

3. IP Subnet Addressing Requirements You must use IP address space: 200.1.59.0/24 (256 IP addresses) for all addresses for this lab (that is, first 3 bytes of every IP must be 200.1.59), The network has 4 subnets (A-D) and 1 link. Each of the subnets must have enough IP addresses allocated to satisfy the following host requirements:

? Subnet A ? At least 10 hosts

? Subnet B ? At least 5 hosts

? Subnet C ? At least 25 hosts

? Subnet D ? At least 100 hosts

? Link 1 ? This subnet needs 2 host IP addresses, as shown in diagram.

Hint: Start by assigning addresses to the subnet with the highest number of addresses first, and then work on the subnet with the next highest number of addresses, and so on.

Explanation / Answer

Consider, 1. Subnet D which needs 100 host addresses i.e. 200.1.59. [ 0 (1bit) - 1111111 (last 7 bits) ] - Here 2^7 which is 128 host addresses. Hence IP address range will be 200.1.59.0/25 Since 1st bit is subnet and remaining will be mask. ----------------------------------------------------------------------------------------------------------------- 2. Now the remaining has to be divided into 3 subnets Consider Subnet C with 25 hosts Here 200.1.59. [ 100 (1st bit) - 11111(last 5 bits)] - Here 2^5 which is 32 host addresses. Hence IP address range will be 200.1.59.128/27 ------------------------------------------------------------------------------------------------------------------ 3. Now consider Subnet B with 10 hosts Here 200.1.59. [ 1010 (1st 4 bits) - 1111 (last 4 bits)] - i.e 2^4 which is 16 host addresses. Here IP address range will be 200.1.59.160/28 ------------------------------------------------------------------------------------------------------------------ 4. Now consider Subnet A with 5 hosts Here 200.1.59. [ 1011 (1st 4 bits) - 1111 (last 4 bits)] -i.e 2^4 which is 16 host addresses IP address range will be 200.1.59.176/28 ------------------------------------------------------------------------------------------------------------------ 5. link having 2 ip 200.1.59. [ 110000 (1st 4 bits) - 11 (last 2 bit)] - i.e 2^2 which is 4 ip addresses since if broadcast address is taken (special address). IP address range will be 200.1.59.192 / 30.

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

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